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

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

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Rubi [A]  time = 0.20, 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^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} (b c-a d)^2}+\frac {d^{3/2} (5 b c-3 a d) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 c^{5/2} (b c-a d)^2}-\frac {2 b c-3 a d}{2 a c^2 x (b c-a d)}-\frac {d}{2 c x \left (c+d x^2\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.26, size = 123, normalized size = 0.85 \begin {gather*} -\frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} (a d-b c)^2}+\frac {d^{3/2} (5 b c-3 a d) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 c^{5/2} (b c-a d)^2}+\frac {d^2 x}{2 c^2 \left (c+d x^2\right ) (b c-a d)}-\frac {1}{a c^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(a*c^2*x)) + (d^2*x)/(2*c^2*(b*c - a*d)*(c + d*x^2)) - (b^(5/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(3/2)*(-(b
*c) + a*d)^2) + (d^(3/2)*(5*b*c - 3*a*d)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(2*c^(5/2)*(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 ) \left (c+d x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.52, size = 1005, normalized size = 6.98 \begin {gather*} \left [-\frac {4 \, b^{2} c^{3} - 8 \, a b c^{2} d + 4 \, a^{2} c d^{2} + 2 \, {\left (2 \, b^{2} c^{2} d - 5 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} x^{2} - 2 \, {\left (b^{2} c^{2} d x^{3} + b^{2} c^{3} 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 ) + {\left ({\left (5 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{3} + {\left (5 \, a b c^{2} d - 3 \, a^{2} c d^{2}\right )} 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 b^{2} c^{4} d - 2 \, a^{2} b c^{3} d^{2} + a^{3} c^{2} d^{3}\right )} x^{3} + {\left (a b^{2} c^{5} - 2 \, a^{2} b c^{4} d + a^{3} c^{3} d^{2}\right )} x\right )}}, -\frac {2 \, b^{2} c^{3} - 4 \, a b c^{2} d + 2 \, a^{2} c d^{2} + {\left (2 \, b^{2} c^{2} d - 5 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} x^{2} - {\left ({\left (5 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{3} + {\left (5 \, a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x\right )} \sqrt {\frac {d}{c}} \arctan \left (x \sqrt {\frac {d}{c}}\right ) - {\left (b^{2} c^{2} d x^{3} + b^{2} c^{3} 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 ({\left (a b^{2} c^{4} d - 2 \, a^{2} b c^{3} d^{2} + a^{3} c^{2} d^{3}\right )} x^{3} + {\left (a b^{2} c^{5} - 2 \, a^{2} b c^{4} d + a^{3} c^{3} d^{2}\right )} x\right )}}, -\frac {4 \, b^{2} c^{3} - 8 \, a b c^{2} d + 4 \, a^{2} c d^{2} + 2 \, {\left (2 \, b^{2} c^{2} d - 5 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} x^{2} + 4 \, {\left (b^{2} c^{2} d x^{3} + b^{2} c^{3} x\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + {\left ({\left (5 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{3} + {\left (5 \, a b c^{2} d - 3 \, a^{2} c d^{2}\right )} 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 b^{2} c^{4} d - 2 \, a^{2} b c^{3} d^{2} + a^{3} c^{2} d^{3}\right )} x^{3} + {\left (a b^{2} c^{5} - 2 \, a^{2} b c^{4} d + a^{3} c^{3} d^{2}\right )} x\right )}}, -\frac {2 \, b^{2} c^{3} - 4 \, a b c^{2} d + 2 \, a^{2} c d^{2} + {\left (2 \, b^{2} c^{2} d - 5 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} x^{2} + 2 \, {\left (b^{2} c^{2} d x^{3} + b^{2} c^{3} x\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) - {\left ({\left (5 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{3} + {\left (5 \, a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x\right )} \sqrt {\frac {d}{c}} \arctan \left (x \sqrt {\frac {d}{c}}\right )}{2 \, {\left ({\left (a b^{2} c^{4} d - 2 \, a^{2} b c^{3} d^{2} + a^{3} c^{2} d^{3}\right )} x^{3} + {\left (a b^{2} c^{5} - 2 \, a^{2} b c^{4} d + a^{3} c^{3} 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)/(d*x^2+c)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.38, size = 164, normalized size = 1.14 \begin {gather*} -\frac {b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \sqrt {a b}} + \frac {{\left (5 \, b c d^{2} - 3 \, a d^{3}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}\right )} \sqrt {c d}} - \frac {2 \, b c d x^{2} - 3 \, a d^{2} x^{2} + 2 \, b c^{2} - 2 \, a c d}{2 \, {\left (a b c^{3} - a^{2} c^{2} d\right )} {\left (d x^{3} + c x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.30, size = 178, normalized size = 1.24 \begin {gather*} -\frac {b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \sqrt {a b}} + \frac {{\left (5 \, b c d^{2} - 3 \, a d^{3}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}\right )} \sqrt {c d}} - \frac {2 \, b c^{2} - 2 \, a c d + {\left (2 \, b c d - 3 \, a d^{2}\right )} x^{2}}{2 \, {\left ({\left (a b c^{3} d - a^{2} c^{2} d^{2}\right )} x^{3} + {\left (a b c^{4} - a^{2} c^{3} 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)/(d*x^2+c)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.83, size = 432, normalized size = 3.00 \begin {gather*} -\frac {\frac {1}{a\,c}+\frac {x^2\,\left (3\,a\,d^2-2\,b\,c\,d\right )}{2\,a\,c^2\,\left (a\,d-b\,c\right )}}{d\,x^3+c\,x}+\frac {\mathrm {atan}\left (\frac {b\,c^5\,x\,{\left (-a^3\,b^5\right )}^{3/2}\,4{}\mathrm {i}+a^8\,b\,d^5\,x\,\sqrt {-a^3\,b^5}\,9{}\mathrm {i}+a^6\,b^3\,c^2\,d^3\,x\,\sqrt {-a^3\,b^5}\,25{}\mathrm {i}-a^7\,b^2\,c\,d^4\,x\,\sqrt {-a^3\,b^5}\,30{}\mathrm {i}}{-9\,a^{10}\,b^3\,d^5+30\,a^9\,b^4\,c\,d^4-25\,a^8\,b^5\,c^2\,d^3+4\,a^5\,b^8\,c^5}\right )\,\sqrt {-a^3\,b^5}\,1{}\mathrm {i}}{a^5\,d^2-2\,a^4\,b\,c\,d+a^3\,b^2\,c^2}+\frac {\mathrm {atan}\left (\frac {a^5\,d^3\,x\,{\left (-c^5\,d^3\right )}^{3/2}\,9{}\mathrm {i}+b^5\,c^{10}\,d\,x\,\sqrt {-c^5\,d^3}\,4{}\mathrm {i}-a^4\,b\,c\,d^2\,x\,{\left (-c^5\,d^3\right )}^{3/2}\,30{}\mathrm {i}+a^3\,b^2\,c^2\,d\,x\,{\left (-c^5\,d^3\right )}^{3/2}\,25{}\mathrm {i}}{9\,a^5\,c^8\,d^7-30\,a^4\,b\,c^9\,d^6+25\,a^3\,b^2\,c^{10}\,d^5-4\,b^5\,c^{13}\,d^2}\right )\,\left (3\,a\,d-5\,b\,c\right )\,\sqrt {-c^5\,d^3}\,1{}\mathrm {i}}{2\,\left (a^2\,c^5\,d^2-2\,a\,b\,c^6\,d+b^2\,c^7\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((b*c^5*x*(-a^3*b^5)^(3/2)*4i + a^8*b*d^5*x*(-a^3*b^5)^(1/2)*9i + a^6*b^3*c^2*d^3*x*(-a^3*b^5)^(1/2)*25i
- a^7*b^2*c*d^4*x*(-a^3*b^5)^(1/2)*30i)/(4*a^5*b^8*c^5 - 9*a^10*b^3*d^5 + 30*a^9*b^4*c*d^4 - 25*a^8*b^5*c^2*d^
3))*(-a^3*b^5)^(1/2)*1i)/(a^5*d^2 + a^3*b^2*c^2 - 2*a^4*b*c*d) - (1/(a*c) + (x^2*(3*a*d^2 - 2*b*c*d))/(2*a*c^2
*(a*d - b*c)))/(c*x + d*x^3) + (atan((a^5*d^3*x*(-c^5*d^3)^(3/2)*9i + b^5*c^10*d*x*(-c^5*d^3)^(1/2)*4i - a^4*b
*c*d^2*x*(-c^5*d^3)^(3/2)*30i + a^3*b^2*c^2*d*x*(-c^5*d^3)^(3/2)*25i)/(9*a^5*c^8*d^7 - 4*b^5*c^13*d^2 - 30*a^4
*b*c^9*d^6 + 25*a^3*b^2*c^10*d^5))*(3*a*d - 5*b*c)*(-c^5*d^3)^(1/2)*1i)/(2*(b^2*c^7 + a^2*c^5*d^2 - 2*a*b*c^6*
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)/(d*x**2+c)**2,x)

[Out]

Timed out

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