def test_meijerint_indefinite_abs():
# issue sympy/sympy#4311
assert meijerint_indefinite(x*abs(9 - x**2), x) is not nan
# issue sympy/sympy#7165
assert meijerint_indefinite(abs(y - x**2), y) is not nan
# issue sympy/sympy#8733
assert meijerint_indefinite(abs(x + 1), x) is not nan
def test_meijerint_indefinite_abs():
# issue sympy/sympy#4311
assert meijerint_indefinite(x * abs(9 - x**2), x) is not nan
# issue sympy/sympy#7165
assert meijerint_indefinite(abs(y - x**2), y) is not nan
# issue sympy/sympy#8733
assert meijerint_indefinite(abs(x + 1), x) is not nan
def t(fac, arg):
g = meijerg([a], [b], [c], [d], arg)*fac
subs = {a: randcplx()/10, b: randcplx()/10 + I,
c: randcplx(), d: randcplx()}
integral = meijerint_indefinite(g, x)
assert integral is not None
assert verify_numerically(g.subs(subs), integral.diff(x).subs(subs), x)
def t(fac, arg):
g = meijerg([a], [b], [c], [d], arg)*fac
subs = {a: randcplx()/10, b: randcplx()/10 + I,
c: randcplx(), d: randcplx()}
integral = meijerint_indefinite(g, x)
assert integral is not None
assert verify_numerically(g.subs(subs), integral.diff(x).subs(subs), x)
def test_sympyissue_8368():
assert meijerint_indefinite(
cosh(x) * exp(-x * t),
x) == ((-t - 1) * exp(x) +
(-t + 1) * exp(-x)) * exp(-t * x) / 2 / (t**2 - 1)
def test_sympyissue_6860():
assert meijerint_indefinite(x**x**x, x) is None
def test_meijerint():
s, t, mu = symbols('s t mu', extended_real=True)
assert integrate(
meijerg([], [], [0], [], s * t) *
meijerg([], [], [mu / 2], [-mu / 2], t**2 / 4),
(t, 0, oo)).is_Piecewise
s = symbols('s', positive=True)
assert integrate(x**s*meijerg([[], []], [[0], []], x), (x, 0, oo)) == \
gamma(s + 1)
assert integrate(x**s * meijerg([[], []], [[0], []], x), (x, 0, oo),
meijerg=True) == gamma(s + 1)
assert isinstance(
integrate(x**s * meijerg([[], []], [[0], []], x), (x, 0, oo),
meijerg=False), Integral)
assert meijerint_indefinite(exp(x), x) == exp(x)
# TODO what simplifications should be done automatically?
# This tests "extra case" for antecedents_1.
a, b = symbols('a b', positive=True)
assert simplify(meijerint_definite(x**a, x, 0, b)[0]) == \
b**(a + 1)/(a + 1)
# This tests various conditions and expansions:
meijerint_definite((x + 1)**3 * exp(-x), x, 0, oo) == (16, True)
# Again, how about simplifications?
sigma, mu = symbols('sigma mu', positive=True)
i, c = meijerint_definite(exp(-((x - mu) / (2 * sigma))**2), x, 0, oo)
assert simplify(i) == sqrt(pi) * sigma * (erf(mu / (2 * sigma)) + 1)
assert c
i, _ = meijerint_definite(exp(-mu * x) * exp(sigma * x), x, 0, oo)
# TODO it would be nice to test the condition
assert simplify(i) == 1 / (mu - sigma)
# Test substitutions to change limits
assert meijerint_definite(exp(x), x, -oo, 2) == (exp(2), True)
# Note: causes a NaN in _check_antecedents
assert expand(meijerint_definite(exp(x), x, 0, I)[0]) == exp(I) - 1
assert expand(meijerint_definite(exp(-x), x, 0, x)[0]) == \
1 - exp(-exp(I*arg(x))*abs(x))
# Test -oo to oo
assert meijerint_definite(exp(-x**2), x, -oo, oo) == (sqrt(pi), True)
assert meijerint_definite(exp(-abs(x)), x, -oo, oo) == (2, True)
assert meijerint_definite(exp(-(2*x - 3)**2), x, -oo, oo) == \
(sqrt(pi)/2, True)
assert meijerint_definite(exp(-abs(2 * x - 3)), x, -oo, oo) == (1, True)
assert meijerint_definite(
exp(-((x - mu) / sigma)**2 / 2) / sqrt(2 * pi * sigma**2), x, -oo,
oo) == (1, True)
# Test one of the extra conditions for 2 g-functinos
assert meijerint_definite(exp(-x) * sin(x), x, 0,
oo) == (Rational(1, 2), True)
# Test a bug
def res(n):
return (1 / (1 + x**2)).diff(x, n).subs({x: 1}) * (-1)**n
for n in range(6):
assert integrate(exp(-x)*sin(x)*x**n, (x, 0, oo), meijerg=True) == \
res(n)
# This used to test trigexpand... now it is done by linear substitution
assert simplify(integrate(exp(-x) * sin(x + a), (x, 0, oo),
meijerg=True)) == sqrt(2) * sin(a + pi / 4) / 2
# Test the condition 14 from prudnikov.
# (This is besselj*besselj in disguise, to stop the product from being
# recognised in the tables.)
a, b, s = symbols('a b s')
assert meijerint_definite(meijerg([], [], [a/2], [-a/2], x/4)
* meijerg([], [], [b/2], [-b/2], x/4)*x**(s - 1), x, 0, oo) == \
(4*2**(2*s - 2)*gamma(-2*s + 1)*gamma(a/2 + b/2 + s)
/ (gamma(-a/2 + b/2 - s + 1)*gamma(a/2 - b/2 - s + 1)
* gamma(a/2 + b/2 - s + 1)),
And(0 < -2*re(4*s) + 8, 0 < re(a/2 + b/2 + s), re(2*s) < 1))
# test a bug
assert integrate(sin(x**a)*sin(x**b), (x, 0, oo), meijerg=True) == \
Integral(sin(x**a)*sin(x**b), (x, 0, oo))
# test better hyperexpand
assert integrate(exp(-x**2)*log(x), (x, 0, oo), meijerg=True) == \
(sqrt(pi)*polygamma(0, Rational(1, 2))/4).expand()
# Test hyperexpand bug.
n = symbols('n', integer=True)
assert simplify(integrate(exp(-x)*x**n, x, meijerg=True)) == \
lowergamma(n + 1, x)
# Test a bug with argument 1/x
alpha = symbols('alpha', positive=True)
assert meijerint_definite((2 - x)**alpha*sin(alpha/x), x, 0, 2) == \
(sqrt(pi)*alpha*gamma(alpha + 1)*meijerg(((), (alpha/2 + Rational(1, 2),
alpha/2 + 1)), ((0, 0, Rational(1, 2)), (-Rational(1, 2),)), alpha**2/16)/4, True)
# test a bug related to 3016
a, s = symbols('a s', positive=True)
assert simplify(integrate(x**s*exp(-a*x**2), (x, -oo, oo))) == \
a**(-s/2 - Rational(1, 2))*((-1)**s + 1)*gamma(s/2 + Rational(1, 2))/2
def test_sympyissue_8368():
assert meijerint_indefinite(cosh(x)*exp(-x*t), x) == (
(-t - 1)*exp(x) + (-t + 1)*exp(-x))*exp(-t*x)/2/(t**2 - 1)
def test_sympyissue_6860():
assert meijerint_indefinite(x**x**x, x) is None
def test_meijerint():
s, t, mu = symbols('s t mu', extended_real=True)
assert integrate(meijerg([], [], [0], [], s*t)
* meijerg([], [], [mu/2], [-mu/2], t**2/4),
(t, 0, oo)).is_Piecewise
s = symbols('s', positive=True)
assert integrate(x**s*meijerg([[], []], [[0], []], x), (x, 0, oo)) == \
gamma(s + 1)
assert integrate(x**s*meijerg([[], []], [[0], []], x), (x, 0, oo),
meijerg=True) == gamma(s + 1)
assert isinstance(integrate(x**s*meijerg([[], []], [[0], []], x),
(x, 0, oo), meijerg=False),
Integral)
assert meijerint_indefinite(exp(x), x) == exp(x)
# TODO what simplifications should be done automatically?
# This tests "extra case" for antecedents_1.
a, b = symbols('a b', positive=True)
assert simplify(meijerint_definite(x**a, x, 0, b)[0]) == \
b**(a + 1)/(a + 1)
# This tests various conditions and expansions:
meijerint_definite((x + 1)**3*exp(-x), x, 0, oo) == (16, True)
# Again, how about simplifications?
sigma, mu = symbols('sigma mu', positive=True)
i, c = meijerint_definite(exp(-((x - mu)/(2*sigma))**2), x, 0, oo)
assert simplify(i) == sqrt(pi)*sigma*(erf(mu/(2*sigma)) + 1)
assert c
i, _ = meijerint_definite(exp(-mu*x)*exp(sigma*x), x, 0, oo)
# TODO it would be nice to test the condition
assert simplify(i) == 1/(mu - sigma)
# Test substitutions to change limits
assert meijerint_definite(exp(x), x, -oo, 2) == (exp(2), True)
# Note: causes a NaN in _check_antecedents
assert expand(meijerint_definite(exp(x), x, 0, I)[0]) == exp(I) - 1
assert expand(meijerint_definite(exp(-x), x, 0, x)[0]) == \
1 - exp(-exp(I*arg(x))*abs(x))
# Test -oo to oo
assert meijerint_definite(exp(-x**2), x, -oo, oo) == (sqrt(pi), True)
assert meijerint_definite(exp(-abs(x)), x, -oo, oo) == (2, True)
assert meijerint_definite(exp(-(2*x - 3)**2), x, -oo, oo) == \
(sqrt(pi)/2, True)
assert meijerint_definite(exp(-abs(2*x - 3)), x, -oo, oo) == (1, True)
assert meijerint_definite(exp(-((x - mu)/sigma)**2/2)/sqrt(2*pi*sigma**2),
x, -oo, oo) == (1, True)
# Test one of the extra conditions for 2 g-functinos
assert meijerint_definite(exp(-x)*sin(x), x, 0, oo) == (Rational(1, 2), True)
# Test a bug
def res(n):
return (1/(1 + x**2)).diff(x, n).subs({x: 1})*(-1)**n
for n in range(6):
assert integrate(exp(-x)*sin(x)*x**n, (x, 0, oo), meijerg=True) == \
res(n)
# This used to test trigexpand... now it is done by linear substitution
assert simplify(integrate(exp(-x)*sin(x + a), (x, 0, oo), meijerg=True)
) == sqrt(2)*sin(a + pi/4)/2
# Test the condition 14 from prudnikov.
# (This is besselj*besselj in disguise, to stop the product from being
# recognised in the tables.)
a, b, s = symbols('a b s')
assert meijerint_definite(meijerg([], [], [a/2], [-a/2], x/4)
* meijerg([], [], [b/2], [-b/2], x/4)*x**(s - 1), x, 0, oo) == \
(4*2**(2*s - 2)*gamma(-2*s + 1)*gamma(a/2 + b/2 + s)
/ (gamma(-a/2 + b/2 - s + 1)*gamma(a/2 - b/2 - s + 1)
* gamma(a/2 + b/2 - s + 1)),
And(0 < -2*re(4*s) + 8, 0 < re(a/2 + b/2 + s), re(2*s) < 1))
# test a bug
assert integrate(sin(x**a)*sin(x**b), (x, 0, oo), meijerg=True) == \
Integral(sin(x**a)*sin(x**b), (x, 0, oo))
# test better hyperexpand
assert integrate(exp(-x**2)*log(x), (x, 0, oo), meijerg=True) == \
(sqrt(pi)*polygamma(0, Rational(1, 2))/4).expand()
# Test hyperexpand bug.
n = symbols('n', integer=True)
assert simplify(integrate(exp(-x)*x**n, x, meijerg=True)) == \
lowergamma(n + 1, x)
# Test a bug with argument 1/x
alpha = symbols('alpha', positive=True)
assert meijerint_definite((2 - x)**alpha*sin(alpha/x), x, 0, 2) == \
(sqrt(pi)*alpha*gamma(alpha + 1)*meijerg(((), (alpha/2 + Rational(1, 2),
alpha/2 + 1)), ((0, 0, Rational(1, 2)), (-Rational(1, 2),)), alpha**2/16)/4, True)
# test a bug related to 3016
a, s = symbols('a s', positive=True)
assert simplify(integrate(x**s*exp(-a*x**2), (x, -oo, oo))) == \
a**(-s/2 - Rational(1, 2))*((-1)**s + 1)*gamma(s/2 + Rational(1, 2))/2
def _eval_integral(self,
f,
x,
meijerg=None,
risch=None,
conds='piecewise'):
"""
Calculate the anti-derivative to the function f(x).
The following algorithms are applied (roughly in this order):
1. Simple heuristics (based on pattern matching and integral table):
- most frequently used functions (e.g. polynomials, products of trig functions)
2. Integration of rational functions:
- A complete algorithm for integrating rational functions is
implemented (the Lazard-Rioboo-Trager algorithm). The algorithm
also uses the partial fraction decomposition algorithm
implemented in apart() as a preprocessor to make this process
faster. Note that the integral of a rational function is always
elementary, but in general, it may include a RootSum.
3. Full Risch algorithm:
- The Risch algorithm is a complete decision
procedure for integrating elementary functions, which means that
given any elementary function, it will either compute an
elementary antiderivative, or else prove that none exists.
Currently, part of transcendental case is implemented, meaning
elementary integrals containing exponentials, logarithms, and
(soon!) trigonometric functions can be computed. The algebraic
case, e.g., functions containing roots, is much more difficult
and is not implemented yet.
- If the routine fails (because the integrand is not elementary, or
because a case is not implemented yet), it continues on to the
next algorithms below. If the routine proves that the integrals
is nonelementary, it still moves on to the algorithms below,
because we might be able to find a closed-form solution in terms
of special functions. If risch=True, however, it will stop here.
4. The Meijer G-Function algorithm:
- This algorithm works by first rewriting the integrand in terms of
very general Meijer G-Function (meijerg in Diofant), integrating
it, and then rewriting the result back, if possible. This
algorithm is particularly powerful for definite integrals (which
is actually part of a different method of Integral), since it can
compute closed-form solutions of definite integrals even when no
closed-form indefinite integral exists. But it also is capable
of computing many indefinite integrals as well.
- Another advantage of this method is that it can use some results
about the Meijer G-Function to give a result in terms of a
Piecewise expression, which allows to express conditionally
convergent integrals.
- Setting meijerg=True will cause integrate() to use only this
method.
5. The Heuristic Risch algorithm:
- This is a heuristic version of the Risch algorithm, meaning that
it is not deterministic. This is tried as a last resort because
it can be very slow. It is still used because not enough of the
full Risch algorithm is implemented, so that there are still some
integrals that can only be computed using this method. The goal
is to implement enough of the Risch and Meijer G-function methods
so that this can be deleted.
"""
from diofant.integrals.deltafunctions import deltaintegrate
from diofant.integrals.heurisch import heurisch, heurisch_wrapper
from diofant.integrals.rationaltools import ratint
from diofant.integrals.risch import risch_integrate
if risch:
try:
return risch_integrate(f, x, conds=conds)
except NotImplementedError:
return
# if it is a poly(x) then let the polynomial integrate itself (fast)
#
# It is important to make this check first, otherwise the other code
# will return a diofant expression instead of a Polynomial.
#
# see Polynomial for details.
if isinstance(f, Poly) and not meijerg:
return f.integrate(x)
# Piecewise antiderivatives need to call special integrate.
if f.func is Piecewise:
return f._eval_integral(x)
# let's cut it short if `f` does not depend on `x`
if not f.has(x):
return f * x
# try to convert to poly(x) and then integrate if successful (fast)
poly = f.as_poly(x)
if poly is not None and not meijerg:
return poly.integrate().as_expr()
if risch is not False:
try:
result, i = risch_integrate(f,
x,
separate_integral=True,
conds=conds)
except NotImplementedError:
pass
else:
if i:
# There was a nonelementary integral. Try integrating it.
return result + i.doit(risch=False)
else:
return result
# since Integral(f=g1+g2+...) == Integral(g1) + Integral(g2) + ...
# we are going to handle Add terms separately,
# if `f` is not Add -- we only have one term
# Note that in general, this is a bad idea, because Integral(g1) +
# Integral(g2) might not be computable, even if Integral(g1 + g2) is.
# For example, Integral(x**x + x**x*log(x)). But many heuristics only
# work term-wise. So we compute this step last, after trying
# risch_integrate. We also try risch_integrate again in this loop,
# because maybe the integral is a sum of an elementary part and a
# nonelementary part (like erf(x) + exp(x)). risch_integrate() is
# quite fast, so this is acceptable.
parts = []
args = Add.make_args(f)
for g in args:
coeff, g = g.as_independent(x)
# g(x) = const
if g is S.One and not meijerg:
parts.append(coeff * x)
continue
# g(x) = expr + O(x**n)
order_term = g.getO()
if order_term is not None:
h = self._eval_integral(g.removeO(), x)
if h is not None:
parts.append(coeff *
(h + self.func(order_term, *self.limits)))
continue
# NOTE: if there is O(x**n) and we fail to integrate then there is
# no point in trying other methods because they will fail anyway.
return
# c
# g(x) = (a*x+b)
if g.is_Pow and not g.exp.has(x) and not meijerg:
a = Wild('a', exclude=[x])
b = Wild('b', exclude=[x])
M = g.base.match(a * x + b)
if M is not None:
if g.exp == -1:
h = log(g.base)
elif conds != 'piecewise':
h = g.base**(g.exp + 1) / (g.exp + 1)
else:
h1 = log(g.base)
h2 = g.base**(g.exp + 1) / (g.exp + 1)
h = Piecewise((h1, Eq(g.exp, -1)), (h2, True))
parts.append(coeff * h / M[a])
continue
# poly(x)
# g(x) = -------
# poly(x)
if g.is_rational_function(x) and not meijerg:
parts.append(coeff * ratint(g, x))
continue
if not meijerg:
# g(x) = Mul(trig)
h = trigintegrate(g, x, conds=conds)
if h is not None:
parts.append(coeff * h)
continue
# g(x) has at least a DiracDelta term
h = deltaintegrate(g, x)
if h is not None:
parts.append(coeff * h)
continue
# Try risch again.
if risch is not False:
try:
h, i = risch_integrate(g,
x,
separate_integral=True,
conds=conds)
except NotImplementedError:
h = None
else:
if i:
h = h + i.doit(risch=False)
parts.append(coeff * h)
continue
# fall back to heurisch
try:
if conds == 'piecewise':
h = heurisch_wrapper(g, x, hints=[])
else:
h = heurisch(g, x, hints=[])
except PolynomialError:
# XXX: this exception means there is a bug in the
# implementation of heuristic Risch integration
# algorithm.
h = None
else:
h = None
if meijerg is not False and h is None:
# rewrite using G functions
try:
h = meijerint_indefinite(g, x)
except NotImplementedError:
from diofant.integrals.meijerint import _debug
_debug('NotImplementedError from meijerint_definite')
res = None
if h is not None:
parts.append(coeff * h)
continue
# if we failed maybe it was because we had
# a product that could have been expanded,
# so let's try an expansion of the whole
# thing before giving up; we don't try this
# at the outset because there are things
# that cannot be solved unless they are
# NOT expanded e.g., x**x*(1+log(x)). There
# should probably be a checker somewhere in this
# routine to look for such cases and try to do
# collection on the expressions if they are already
# in an expanded form
if not h and len(args) == 1:
f = f.expand(mul=True, deep=False)
if f.is_Add:
# Note: risch will be identical on the expanded
# expression, but maybe it will be able to pick out parts,
# like x*(exp(x) + erf(x)).
return self._eval_integral(f,
x,
meijerg=meijerg,
risch=risch,
conds=conds)
if h is not None:
parts.append(coeff * h)
else:
return
return Add(*parts)
