3.3.95 \(\int \frac {1}{x^2 (a+b x^2)^2 (c+d x^2)} \, dx\)

Optimal. Leaf size=144 \[ -\frac {b^{3/2} (3 b c-5 a d) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2} (b c-a d)^2}-\frac {3 b c-2 a d}{2 a^2 c x (b c-a d)}-\frac {d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{3/2} (b c-a d)^2}+\frac {b}{2 a x \left (a+b x^2\right ) (b c-a d)} \]

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Rubi [A]  time = 0.21, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {472, 583, 522, 205} \begin {gather*} -\frac {b^{3/2} (3 b c-5 a d) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2} (b c-a d)^2}-\frac {3 b c-2 a d}{2 a^2 c x (b c-a d)}-\frac {d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{3/2} (b c-a d)^2}+\frac {b}{2 a x \left (a+b x^2\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x^2)^2*(c + d*x^2)),x]

[Out]

-(3*b*c - 2*a*d)/(2*a^2*c*(b*c - a*d)*x) + b/(2*a*(b*c - a*d)*x*(a + b*x^2)) - (b^(3/2)*(3*b*c - 5*a*d)*ArcTan
[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(5/2)*(b*c - a*d)^2) - (d^(5/2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(c^(3/2)*(b*c - a*d)^
2)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 472

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*(e*x
)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*e*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d)*(
p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n*(
p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p
, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {1}{x^2 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx &=\frac {b}{2 a (b c-a d) x \left (a+b x^2\right )}-\frac {\int \frac {-3 b c+2 a d-3 b d x^2}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{2 a (b c-a d)}\\ &=-\frac {3 b c-2 a d}{2 a^2 c (b c-a d) x}+\frac {b}{2 a (b c-a d) x \left (a+b x^2\right )}+\frac {\int \frac {-3 b^2 c^2+2 a b c d+2 a^2 d^2-b d (3 b c-2 a d) x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{2 a^2 c (b c-a d)}\\ &=-\frac {3 b c-2 a d}{2 a^2 c (b c-a d) x}+\frac {b}{2 a (b c-a d) x \left (a+b x^2\right )}-\frac {d^3 \int \frac {1}{c+d x^2} \, dx}{c (b c-a d)^2}-\frac {\left (b^2 (3 b c-5 a d)\right ) \int \frac {1}{a+b x^2} \, dx}{2 a^2 (b c-a d)^2}\\ &=-\frac {3 b c-2 a d}{2 a^2 c (b c-a d) x}+\frac {b}{2 a (b c-a d) x \left (a+b x^2\right )}-\frac {b^{3/2} (3 b c-5 a d) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2} (b c-a d)^2}-\frac {d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{3/2} (b c-a d)^2}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 123, normalized size = 0.85 \begin {gather*} \frac {b^{3/2} (5 a d-3 b c) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2} (a d-b c)^2}+\frac {b^2 x}{2 a^2 \left (a+b x^2\right ) (a d-b c)}-\frac {1}{a^2 c x}-\frac {d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{3/2} (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*x^2)^2*(c + d*x^2)),x]

[Out]

-(1/(a^2*c*x)) + (b^2*x)/(2*a^2*(-(b*c) + a*d)*(a + b*x^2)) + (b^(3/2)*(-3*b*c + 5*a*d)*ArcTan[(Sqrt[b]*x)/Sqr
t[a]])/(2*a^(5/2)*(-(b*c) + a*d)^2) - (d^(5/2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(c^(3/2)*(b*c - a*d)^2)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^2 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/(x^2*(a + b*x^2)^2*(c + d*x^2)),x]

[Out]

IntegrateAlgebraic[1/(x^2*(a + b*x^2)^2*(c + d*x^2)), x]

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fricas [A]  time = 1.63, size = 1003, normalized size = 6.97 \begin {gather*} \left [-\frac {4 \, a b^{2} c^{2} - 8 \, a^{2} b c d + 4 \, a^{3} d^{2} + 2 \, {\left (3 \, b^{3} c^{2} - 5 \, a b^{2} c d + 2 \, a^{2} b d^{2}\right )} x^{2} + {\left ({\left (3 \, b^{3} c^{2} - 5 \, a b^{2} c d\right )} x^{3} + {\left (3 \, a b^{2} c^{2} - 5 \, a^{2} b c d\right )} x\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) - 2 \, {\left (a^{2} b d^{2} x^{3} + a^{3} d^{2} x\right )} \sqrt {-\frac {d}{c}} \log \left (\frac {d x^{2} - 2 \, c x \sqrt {-\frac {d}{c}} - c}{d x^{2} + c}\right )}{4 \, {\left ({\left (a^{2} b^{3} c^{3} - 2 \, a^{3} b^{2} c^{2} d + a^{4} b c d^{2}\right )} x^{3} + {\left (a^{3} b^{2} c^{3} - 2 \, a^{4} b c^{2} d + a^{5} c d^{2}\right )} x\right )}}, -\frac {4 \, a b^{2} c^{2} - 8 \, a^{2} b c d + 4 \, a^{3} d^{2} + 2 \, {\left (3 \, b^{3} c^{2} - 5 \, a b^{2} c d + 2 \, a^{2} b d^{2}\right )} x^{2} + 4 \, {\left (a^{2} b d^{2} x^{3} + a^{3} d^{2} x\right )} \sqrt {\frac {d}{c}} \arctan \left (x \sqrt {\frac {d}{c}}\right ) + {\left ({\left (3 \, b^{3} c^{2} - 5 \, a b^{2} c d\right )} x^{3} + {\left (3 \, a b^{2} c^{2} - 5 \, a^{2} b c d\right )} x\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right )}{4 \, {\left ({\left (a^{2} b^{3} c^{3} - 2 \, a^{3} b^{2} c^{2} d + a^{4} b c d^{2}\right )} x^{3} + {\left (a^{3} b^{2} c^{3} - 2 \, a^{4} b c^{2} d + a^{5} c d^{2}\right )} x\right )}}, -\frac {2 \, a b^{2} c^{2} - 4 \, a^{2} b c d + 2 \, a^{3} d^{2} + {\left (3 \, b^{3} c^{2} - 5 \, a b^{2} c d + 2 \, a^{2} b d^{2}\right )} x^{2} + {\left ({\left (3 \, b^{3} c^{2} - 5 \, a b^{2} c d\right )} x^{3} + {\left (3 \, a b^{2} c^{2} - 5 \, a^{2} b c d\right )} x\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) - {\left (a^{2} b d^{2} x^{3} + a^{3} d^{2} x\right )} \sqrt {-\frac {d}{c}} \log \left (\frac {d x^{2} - 2 \, c x \sqrt {-\frac {d}{c}} - c}{d x^{2} + c}\right )}{2 \, {\left ({\left (a^{2} b^{3} c^{3} - 2 \, a^{3} b^{2} c^{2} d + a^{4} b c d^{2}\right )} x^{3} + {\left (a^{3} b^{2} c^{3} - 2 \, a^{4} b c^{2} d + a^{5} c d^{2}\right )} x\right )}}, -\frac {2 \, a b^{2} c^{2} - 4 \, a^{2} b c d + 2 \, a^{3} d^{2} + {\left (3 \, b^{3} c^{2} - 5 \, a b^{2} c d + 2 \, a^{2} b d^{2}\right )} x^{2} + {\left ({\left (3 \, b^{3} c^{2} - 5 \, a b^{2} c d\right )} x^{3} + {\left (3 \, a b^{2} c^{2} - 5 \, a^{2} b c d\right )} x\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + 2 \, {\left (a^{2} b d^{2} x^{3} + a^{3} d^{2} x\right )} \sqrt {\frac {d}{c}} \arctan \left (x \sqrt {\frac {d}{c}}\right )}{2 \, {\left ({\left (a^{2} b^{3} c^{3} - 2 \, a^{3} b^{2} c^{2} d + a^{4} b c d^{2}\right )} x^{3} + {\left (a^{3} b^{2} c^{3} - 2 \, a^{4} b c^{2} d + a^{5} c d^{2}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x^2+a)^2/(d*x^2+c),x, algorithm="fricas")

[Out]

[-1/4*(4*a*b^2*c^2 - 8*a^2*b*c*d + 4*a^3*d^2 + 2*(3*b^3*c^2 - 5*a*b^2*c*d + 2*a^2*b*d^2)*x^2 + ((3*b^3*c^2 - 5
*a*b^2*c*d)*x^3 + (3*a*b^2*c^2 - 5*a^2*b*c*d)*x)*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) -
2*(a^2*b*d^2*x^3 + a^3*d^2*x)*sqrt(-d/c)*log((d*x^2 - 2*c*x*sqrt(-d/c) - c)/(d*x^2 + c)))/((a^2*b^3*c^3 - 2*a^
3*b^2*c^2*d + a^4*b*c*d^2)*x^3 + (a^3*b^2*c^3 - 2*a^4*b*c^2*d + a^5*c*d^2)*x), -1/4*(4*a*b^2*c^2 - 8*a^2*b*c*d
+ 4*a^3*d^2 + 2*(3*b^3*c^2 - 5*a*b^2*c*d + 2*a^2*b*d^2)*x^2 + 4*(a^2*b*d^2*x^3 + a^3*d^2*x)*sqrt(d/c)*arctan(
x*sqrt(d/c)) + ((3*b^3*c^2 - 5*a*b^2*c*d)*x^3 + (3*a*b^2*c^2 - 5*a^2*b*c*d)*x)*sqrt(-b/a)*log((b*x^2 + 2*a*x*s
qrt(-b/a) - a)/(b*x^2 + a)))/((a^2*b^3*c^3 - 2*a^3*b^2*c^2*d + a^4*b*c*d^2)*x^3 + (a^3*b^2*c^3 - 2*a^4*b*c^2*d
+ a^5*c*d^2)*x), -1/2*(2*a*b^2*c^2 - 4*a^2*b*c*d + 2*a^3*d^2 + (3*b^3*c^2 - 5*a*b^2*c*d + 2*a^2*b*d^2)*x^2 +
((3*b^3*c^2 - 5*a*b^2*c*d)*x^3 + (3*a*b^2*c^2 - 5*a^2*b*c*d)*x)*sqrt(b/a)*arctan(x*sqrt(b/a)) - (a^2*b*d^2*x^3
+ a^3*d^2*x)*sqrt(-d/c)*log((d*x^2 - 2*c*x*sqrt(-d/c) - c)/(d*x^2 + c)))/((a^2*b^3*c^3 - 2*a^3*b^2*c^2*d + a^
4*b*c*d^2)*x^3 + (a^3*b^2*c^3 - 2*a^4*b*c^2*d + a^5*c*d^2)*x), -1/2*(2*a*b^2*c^2 - 4*a^2*b*c*d + 2*a^3*d^2 + (
3*b^3*c^2 - 5*a*b^2*c*d + 2*a^2*b*d^2)*x^2 + ((3*b^3*c^2 - 5*a*b^2*c*d)*x^3 + (3*a*b^2*c^2 - 5*a^2*b*c*d)*x)*s
qrt(b/a)*arctan(x*sqrt(b/a)) + 2*(a^2*b*d^2*x^3 + a^3*d^2*x)*sqrt(d/c)*arctan(x*sqrt(d/c)))/((a^2*b^3*c^3 - 2*
a^3*b^2*c^2*d + a^4*b*c*d^2)*x^3 + (a^3*b^2*c^3 - 2*a^4*b*c^2*d + a^5*c*d^2)*x)]

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giac [A]  time = 0.37, size = 164, normalized size = 1.14 \begin {gather*} -\frac {d^{3} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt {c d}} - \frac {{\left (3 \, b^{3} c - 5 \, a b^{2} d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} \sqrt {a b}} - \frac {3 \, b^{2} c x^{2} - 2 \, a b d x^{2} + 2 \, a b c - 2 \, a^{2} d}{2 \, {\left (a^{2} b c^{2} - a^{3} c d\right )} {\left (b x^{3} + a x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x^2+a)^2/(d*x^2+c),x, algorithm="giac")

[Out]

-d^3*arctan(d*x/sqrt(c*d))/((b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*sqrt(c*d)) - 1/2*(3*b^3*c - 5*a*b^2*d)*arctan(
b*x/sqrt(a*b))/((a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2)*sqrt(a*b)) - 1/2*(3*b^2*c*x^2 - 2*a*b*d*x^2 + 2*a*b*c -
2*a^2*d)/((a^2*b*c^2 - a^3*c*d)*(b*x^3 + a*x))

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maple [A]  time = 0.02, size = 169, normalized size = 1.17 \begin {gather*} \frac {b^{2} d x}{2 \left (a d -b c \right )^{2} \left (b \,x^{2}+a \right ) a}+\frac {5 b^{2} d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \left (a d -b c \right )^{2} \sqrt {a b}\, a}-\frac {b^{3} c x}{2 \left (a d -b c \right )^{2} \left (b \,x^{2}+a \right ) a^{2}}-\frac {3 b^{3} c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \left (a d -b c \right )^{2} \sqrt {a b}\, a^{2}}-\frac {d^{3} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{\left (a d -b c \right )^{2} \sqrt {c d}\, c}-\frac {1}{a^{2} c x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(b*x^2+a)^2/(d*x^2+c),x)

[Out]

1/2*b^2/a/(a*d-b*c)^2*x/(b*x^2+a)*d-1/2*b^3/a^2/(a*d-b*c)^2*x/(b*x^2+a)*c+5/2*b^2/a/(a*d-b*c)^2/(a*b)^(1/2)*ar
ctan(1/(a*b)^(1/2)*b*x)*d-3/2*b^3/a^2/(a*d-b*c)^2/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)*c-1/c*d^3/(a*d-b*c)^2/
(c*d)^(1/2)*arctan(1/(c*d)^(1/2)*d*x)-1/a^2/c/x

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maxima [A]  time = 2.26, size = 178, normalized size = 1.24 \begin {gather*} -\frac {d^{3} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt {c d}} - \frac {{\left (3 \, b^{3} c - 5 \, a b^{2} d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} \sqrt {a b}} - \frac {2 \, a b c - 2 \, a^{2} d + {\left (3 \, b^{2} c - 2 \, a b d\right )} x^{2}}{2 \, {\left ({\left (a^{2} b^{2} c^{2} - a^{3} b c d\right )} x^{3} + {\left (a^{3} b c^{2} - a^{4} c d\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x^2+a)^2/(d*x^2+c),x, algorithm="maxima")

[Out]

-d^3*arctan(d*x/sqrt(c*d))/((b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*sqrt(c*d)) - 1/2*(3*b^3*c - 5*a*b^2*d)*arctan(
b*x/sqrt(a*b))/((a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2)*sqrt(a*b)) - 1/2*(2*a*b*c - 2*a^2*d + (3*b^2*c - 2*a*b*d
)*x^2)/((a^2*b^2*c^2 - a^3*b*c*d)*x^3 + (a^3*b*c^2 - a^4*c*d)*x)

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mupad [B]  time = 1.00, size = 2400, normalized size = 16.67

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b*x^2)^2*(c + d*x^2)),x)

[Out]

(atan((a^5*d*x*(-c^3*d^5)^(3/2)*4i + b^5*c^8*d*x*(-c^3*d^5)^(1/2)*9i + a^2*b^3*c^6*d^3*x*(-c^3*d^5)^(1/2)*25i
- a*b^4*c^7*d^2*x*(-c^3*d^5)^(1/2)*30i)/(4*a^5*c^5*d^8 - 9*b^5*c^10*d^3 + 30*a*b^4*c^9*d^4 - 25*a^2*b^3*c^8*d^
5))*(-c^3*d^5)^(1/2)*1i)/(b^2*c^5 + a^2*c^3*d^2 - 2*a*b*c^4*d) - (1/(a*c) - (x^2*(3*b^2*c - 2*a*b*d))/(2*a^2*c
*(a*d - b*c)))/(a*x + b*x^3) - (atan((((x*(144*a^6*b^10*c^10*d^3 - 912*a^7*b^9*c^9*d^4 + 2272*a^8*b^8*c^8*d^5
- 2784*a^9*b^7*c^7*d^6 + 1744*a^10*b^6*c^6*d^7 - 592*a^11*b^5*c^5*d^8 + 192*a^12*b^4*c^4*d^9 - 64*a^13*b^3*c^3
*d^10) + ((5*a*d - 3*b*c)*(-a^5*b^3)^(1/2)*(1280*a^9*b^9*c^11*d^3 - 192*a^8*b^10*c^12*d^2 - 3520*a^10*b^8*c^10
*d^4 + 4992*a^11*b^7*c^9*d^5 - 3520*a^12*b^6*c^8*d^6 + 512*a^13*b^5*c^7*d^7 + 960*a^14*b^4*c^6*d^8 - 640*a^15*
b^3*c^5*d^9 + 128*a^16*b^2*c^4*d^10 + (x*(5*a*d - 3*b*c)*(-a^5*b^3)^(1/2)*(256*a^10*b^10*c^13*d^2 - 1536*a^11*
b^9*c^12*d^3 + 3584*a^12*b^8*c^11*d^4 - 3584*a^13*b^7*c^10*d^5 + 3584*a^15*b^5*c^8*d^7 - 3584*a^16*b^4*c^7*d^8
+ 1536*a^17*b^3*c^6*d^9 - 256*a^18*b^2*c^5*d^10))/(4*(a^7*d^2 + a^5*b^2*c^2 - 2*a^6*b*c*d))))/(4*(a^7*d^2 + a
^5*b^2*c^2 - 2*a^6*b*c*d)))*(5*a*d - 3*b*c)*(-a^5*b^3)^(1/2)*1i)/(4*(a^7*d^2 + a^5*b^2*c^2 - 2*a^6*b*c*d)) + (
(x*(144*a^6*b^10*c^10*d^3 - 912*a^7*b^9*c^9*d^4 + 2272*a^8*b^8*c^8*d^5 - 2784*a^9*b^7*c^7*d^6 + 1744*a^10*b^6*
c^6*d^7 - 592*a^11*b^5*c^5*d^8 + 192*a^12*b^4*c^4*d^9 - 64*a^13*b^3*c^3*d^10) + ((5*a*d - 3*b*c)*(-a^5*b^3)^(1
/2)*(192*a^8*b^10*c^12*d^2 - 1280*a^9*b^9*c^11*d^3 + 3520*a^10*b^8*c^10*d^4 - 4992*a^11*b^7*c^9*d^5 + 3520*a^1
2*b^6*c^8*d^6 - 512*a^13*b^5*c^7*d^7 - 960*a^14*b^4*c^6*d^8 + 640*a^15*b^3*c^5*d^9 - 128*a^16*b^2*c^4*d^10 + (
x*(5*a*d - 3*b*c)*(-a^5*b^3)^(1/2)*(256*a^10*b^10*c^13*d^2 - 1536*a^11*b^9*c^12*d^3 + 3584*a^12*b^8*c^11*d^4 -
3584*a^13*b^7*c^10*d^5 + 3584*a^15*b^5*c^8*d^7 - 3584*a^16*b^4*c^7*d^8 + 1536*a^17*b^3*c^6*d^9 - 256*a^18*b^2
*c^5*d^10))/(4*(a^7*d^2 + a^5*b^2*c^2 - 2*a^6*b*c*d))))/(4*(a^7*d^2 + a^5*b^2*c^2 - 2*a^6*b*c*d)))*(5*a*d - 3*
b*c)*(-a^5*b^3)^(1/2)*1i)/(4*(a^7*d^2 + a^5*b^2*c^2 - 2*a^6*b*c*d)))/(((x*(144*a^6*b^10*c^10*d^3 - 912*a^7*b^9
*c^9*d^4 + 2272*a^8*b^8*c^8*d^5 - 2784*a^9*b^7*c^7*d^6 + 1744*a^10*b^6*c^6*d^7 - 592*a^11*b^5*c^5*d^8 + 192*a^
12*b^4*c^4*d^9 - 64*a^13*b^3*c^3*d^10) + ((5*a*d - 3*b*c)*(-a^5*b^3)^(1/2)*(192*a^8*b^10*c^12*d^2 - 1280*a^9*b
^9*c^11*d^3 + 3520*a^10*b^8*c^10*d^4 - 4992*a^11*b^7*c^9*d^5 + 3520*a^12*b^6*c^8*d^6 - 512*a^13*b^5*c^7*d^7 -
960*a^14*b^4*c^6*d^8 + 640*a^15*b^3*c^5*d^9 - 128*a^16*b^2*c^4*d^10 + (x*(5*a*d - 3*b*c)*(-a^5*b^3)^(1/2)*(256
*a^10*b^10*c^13*d^2 - 1536*a^11*b^9*c^12*d^3 + 3584*a^12*b^8*c^11*d^4 - 3584*a^13*b^7*c^10*d^5 + 3584*a^15*b^5
*c^8*d^7 - 3584*a^16*b^4*c^7*d^8 + 1536*a^17*b^3*c^6*d^9 - 256*a^18*b^2*c^5*d^10))/(4*(a^7*d^2 + a^5*b^2*c^2 -
2*a^6*b*c*d))))/(4*(a^7*d^2 + a^5*b^2*c^2 - 2*a^6*b*c*d)))*(5*a*d - 3*b*c)*(-a^5*b^3)^(1/2))/(4*(a^7*d^2 + a^
5*b^2*c^2 - 2*a^6*b*c*d)) - ((x*(144*a^6*b^10*c^10*d^3 - 912*a^7*b^9*c^9*d^4 + 2272*a^8*b^8*c^8*d^5 - 2784*a^9
*b^7*c^7*d^6 + 1744*a^10*b^6*c^6*d^7 - 592*a^11*b^5*c^5*d^8 + 192*a^12*b^4*c^4*d^9 - 64*a^13*b^3*c^3*d^10) + (
(5*a*d - 3*b*c)*(-a^5*b^3)^(1/2)*(1280*a^9*b^9*c^11*d^3 - 192*a^8*b^10*c^12*d^2 - 3520*a^10*b^8*c^10*d^4 + 499
2*a^11*b^7*c^9*d^5 - 3520*a^12*b^6*c^8*d^6 + 512*a^13*b^5*c^7*d^7 + 960*a^14*b^4*c^6*d^8 - 640*a^15*b^3*c^5*d^
9 + 128*a^16*b^2*c^4*d^10 + (x*(5*a*d - 3*b*c)*(-a^5*b^3)^(1/2)*(256*a^10*b^10*c^13*d^2 - 1536*a^11*b^9*c^12*d
^3 + 3584*a^12*b^8*c^11*d^4 - 3584*a^13*b^7*c^10*d^5 + 3584*a^15*b^5*c^8*d^7 - 3584*a^16*b^4*c^7*d^8 + 1536*a^
17*b^3*c^6*d^9 - 256*a^18*b^2*c^5*d^10))/(4*(a^7*d^2 + a^5*b^2*c^2 - 2*a^6*b*c*d))))/(4*(a^7*d^2 + a^5*b^2*c^2
- 2*a^6*b*c*d)))*(5*a*d - 3*b*c)*(-a^5*b^3)^(1/2))/(4*(a^7*d^2 + a^5*b^2*c^2 - 2*a^6*b*c*d)) + 144*a^6*b^8*c^
7*d^5 - 624*a^7*b^7*c^6*d^6 + 976*a^8*b^6*c^5*d^7 - 656*a^9*b^5*c^4*d^8 + 160*a^10*b^4*c^3*d^9))*(5*a*d - 3*b*
c)*(-a^5*b^3)^(1/2)*1i)/(2*(a^7*d^2 + a^5*b^2*c^2 - 2*a^6*b*c*d))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(b*x**2+a)**2/(d*x**2+c),x)

[Out]

Timed out

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