3.3.59 \(\int \frac {1}{x^3 (a+b x^2) (c+d x^2)^3} \, dx\)

Optimal. Leaf size=178 \[ \frac {b^4 \log \left (a+b x^2\right )}{2 a^2 (b c-a d)^3}-\frac {d^2 \left (3 a^2 d^2-8 a b c d+6 b^2 c^2\right ) \log \left (c+d x^2\right )}{2 c^4 (b c-a d)^3}-\frac {\log (x) (3 a d+b c)}{a^2 c^4}+\frac {d^2 (3 b c-2 a d)}{2 c^3 \left (c+d x^2\right ) (b c-a d)^2}+\frac {d^2}{4 c^2 \left (c+d x^2\right )^2 (b c-a d)}-\frac {1}{2 a c^3 x^2} \]

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Rubi [A]  time = 0.21, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {446, 88} \begin {gather*} -\frac {d^2 \left (3 a^2 d^2-8 a b c d+6 b^2 c^2\right ) \log \left (c+d x^2\right )}{2 c^4 (b c-a d)^3}+\frac {b^4 \log \left (a+b x^2\right )}{2 a^2 (b c-a d)^3}-\frac {\log (x) (3 a d+b c)}{a^2 c^4}+\frac {d^2 (3 b c-2 a d)}{2 c^3 \left (c+d x^2\right ) (b c-a d)^2}+\frac {d^2}{4 c^2 \left (c+d x^2\right )^2 (b c-a d)}-\frac {1}{2 a c^3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^2)*(c + d*x^2)^3),x]

[Out]

-1/(2*a*c^3*x^2) + d^2/(4*c^2*(b*c - a*d)*(c + d*x^2)^2) + (d^2*(3*b*c - 2*a*d))/(2*c^3*(b*c - a*d)^2*(c + d*x
^2)) - ((b*c + 3*a*d)*Log[x])/(a^2*c^4) + (b^4*Log[a + b*x^2])/(2*a^2*(b*c - a*d)^3) - (d^2*(6*b^2*c^2 - 8*a*b
*c*d + 3*a^2*d^2)*Log[c + d*x^2])/(2*c^4*(b*c - a*d)^3)

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 446

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

Rubi steps

\begin {align*} \int \frac {1}{x^3 \left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^2 (a+b x) (c+d x)^3} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {1}{a c^3 x^2}+\frac {-b c-3 a d}{a^2 c^4 x}-\frac {b^5}{a^2 (-b c+a d)^3 (a+b x)}-\frac {d^3}{c^2 (b c-a d) (c+d x)^3}-\frac {d^3 (3 b c-2 a d)}{c^3 (b c-a d)^2 (c+d x)^2}-\frac {d^3 \left (6 b^2 c^2-8 a b c d+3 a^2 d^2\right )}{c^4 (b c-a d)^3 (c+d x)}\right ) \, dx,x,x^2\right )\\ &=-\frac {1}{2 a c^3 x^2}+\frac {d^2}{4 c^2 (b c-a d) \left (c+d x^2\right )^2}+\frac {d^2 (3 b c-2 a d)}{2 c^3 (b c-a d)^2 \left (c+d x^2\right )}-\frac {(b c+3 a d) \log (x)}{a^2 c^4}+\frac {b^4 \log \left (a+b x^2\right )}{2 a^2 (b c-a d)^3}-\frac {d^2 \left (6 b^2 c^2-8 a b c d+3 a^2 d^2\right ) \log \left (c+d x^2\right )}{2 c^4 (b c-a d)^3}\\ \end {align*}

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Mathematica [A]  time = 0.42, size = 171, normalized size = 0.96 \begin {gather*} \frac {1}{4} \left (-\frac {2 b^4 \log \left (a+b x^2\right )}{a^2 (a d-b c)^3}-\frac {2 d^2 \left (3 a^2 d^2-8 a b c d+6 b^2 c^2\right ) \log \left (c+d x^2\right )}{c^4 (b c-a d)^3}-\frac {4 \log (x) (3 a d+b c)}{a^2 c^4}+\frac {2 d^2 (3 b c-2 a d)}{c^3 \left (c+d x^2\right ) (b c-a d)^2}+\frac {d^2}{c^2 \left (c+d x^2\right )^2 (b c-a d)}-\frac {2}{a c^3 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*x^2)*(c + d*x^2)^3),x]

[Out]

(-2/(a*c^3*x^2) + d^2/(c^2*(b*c - a*d)*(c + d*x^2)^2) + (2*d^2*(3*b*c - 2*a*d))/(c^3*(b*c - a*d)^2*(c + d*x^2)
) - (4*(b*c + 3*a*d)*Log[x])/(a^2*c^4) - (2*b^4*Log[a + b*x^2])/(a^2*(-(b*c) + a*d)^3) - (2*d^2*(6*b^2*c^2 - 8
*a*b*c*d + 3*a^2*d^2)*Log[c + d*x^2])/(c^4*(b*c - a*d)^3))/4

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^3 \left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/(x^3*(a + b*x^2)*(c + d*x^2)^3),x]

[Out]

IntegrateAlgebraic[1/(x^3*(a + b*x^2)*(c + d*x^2)^3), x]

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fricas [B]  time = 18.49, size = 640, normalized size = 3.60 \begin {gather*} -\frac {2 \, a b^{3} c^{6} - 6 \, a^{2} b^{2} c^{5} d + 6 \, a^{3} b c^{4} d^{2} - 2 \, a^{4} c^{3} d^{3} + 2 \, {\left (a b^{3} c^{4} d^{2} - 6 \, a^{2} b^{2} c^{3} d^{3} + 8 \, a^{3} b c^{2} d^{4} - 3 \, a^{4} c d^{5}\right )} x^{4} + {\left (4 \, a b^{3} c^{5} d - 19 \, a^{2} b^{2} c^{4} d^{2} + 24 \, a^{3} b c^{3} d^{3} - 9 \, a^{4} c^{2} d^{4}\right )} x^{2} - 2 \, {\left (b^{4} c^{4} d^{2} x^{6} + 2 \, b^{4} c^{5} d x^{4} + b^{4} c^{6} x^{2}\right )} \log \left (b x^{2} + a\right ) + 2 \, {\left ({\left (6 \, a^{2} b^{2} c^{2} d^{4} - 8 \, a^{3} b c d^{5} + 3 \, a^{4} d^{6}\right )} x^{6} + 2 \, {\left (6 \, a^{2} b^{2} c^{3} d^{3} - 8 \, a^{3} b c^{2} d^{4} + 3 \, a^{4} c d^{5}\right )} x^{4} + {\left (6 \, a^{2} b^{2} c^{4} d^{2} - 8 \, a^{3} b c^{3} d^{3} + 3 \, a^{4} c^{2} d^{4}\right )} x^{2}\right )} \log \left (d x^{2} + c\right ) + 4 \, {\left ({\left (b^{4} c^{4} d^{2} - 6 \, a^{2} b^{2} c^{2} d^{4} + 8 \, a^{3} b c d^{5} - 3 \, a^{4} d^{6}\right )} x^{6} + 2 \, {\left (b^{4} c^{5} d - 6 \, a^{2} b^{2} c^{3} d^{3} + 8 \, a^{3} b c^{2} d^{4} - 3 \, a^{4} c d^{5}\right )} x^{4} + {\left (b^{4} c^{6} - 6 \, a^{2} b^{2} c^{4} d^{2} + 8 \, a^{3} b c^{3} d^{3} - 3 \, a^{4} c^{2} d^{4}\right )} x^{2}\right )} \log \relax (x)}{4 \, {\left ({\left (a^{2} b^{3} c^{7} d^{2} - 3 \, a^{3} b^{2} c^{6} d^{3} + 3 \, a^{4} b c^{5} d^{4} - a^{5} c^{4} d^{5}\right )} x^{6} + 2 \, {\left (a^{2} b^{3} c^{8} d - 3 \, a^{3} b^{2} c^{7} d^{2} + 3 \, a^{4} b c^{6} d^{3} - a^{5} c^{5} d^{4}\right )} x^{4} + {\left (a^{2} b^{3} c^{9} - 3 \, a^{3} b^{2} c^{8} d + 3 \, a^{4} b c^{7} d^{2} - a^{5} c^{6} d^{3}\right )} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^2+a)/(d*x^2+c)^3,x, algorithm="fricas")

[Out]

-1/4*(2*a*b^3*c^6 - 6*a^2*b^2*c^5*d + 6*a^3*b*c^4*d^2 - 2*a^4*c^3*d^3 + 2*(a*b^3*c^4*d^2 - 6*a^2*b^2*c^3*d^3 +
8*a^3*b*c^2*d^4 - 3*a^4*c*d^5)*x^4 + (4*a*b^3*c^5*d - 19*a^2*b^2*c^4*d^2 + 24*a^3*b*c^3*d^3 - 9*a^4*c^2*d^4)*
x^2 - 2*(b^4*c^4*d^2*x^6 + 2*b^4*c^5*d*x^4 + b^4*c^6*x^2)*log(b*x^2 + a) + 2*((6*a^2*b^2*c^2*d^4 - 8*a^3*b*c*d
^5 + 3*a^4*d^6)*x^6 + 2*(6*a^2*b^2*c^3*d^3 - 8*a^3*b*c^2*d^4 + 3*a^4*c*d^5)*x^4 + (6*a^2*b^2*c^4*d^2 - 8*a^3*b
*c^3*d^3 + 3*a^4*c^2*d^4)*x^2)*log(d*x^2 + c) + 4*((b^4*c^4*d^2 - 6*a^2*b^2*c^2*d^4 + 8*a^3*b*c*d^5 - 3*a^4*d^
6)*x^6 + 2*(b^4*c^5*d - 6*a^2*b^2*c^3*d^3 + 8*a^3*b*c^2*d^4 - 3*a^4*c*d^5)*x^4 + (b^4*c^6 - 6*a^2*b^2*c^4*d^2
+ 8*a^3*b*c^3*d^3 - 3*a^4*c^2*d^4)*x^2)*log(x))/((a^2*b^3*c^7*d^2 - 3*a^3*b^2*c^6*d^3 + 3*a^4*b*c^5*d^4 - a^5*
c^4*d^5)*x^6 + 2*(a^2*b^3*c^8*d - 3*a^3*b^2*c^7*d^2 + 3*a^4*b*c^6*d^3 - a^5*c^5*d^4)*x^4 + (a^2*b^3*c^9 - 3*a^
3*b^2*c^8*d + 3*a^4*b*c^7*d^2 - a^5*c^6*d^3)*x^2)

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giac [B]  time = 0.34, size = 357, normalized size = 2.01 \begin {gather*} \frac {b^{5} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, {\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )}} - \frac {{\left (6 \, b^{2} c^{2} d^{3} - 8 \, a b c d^{4} + 3 \, a^{2} d^{5}\right )} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, {\left (b^{3} c^{7} d - 3 \, a b^{2} c^{6} d^{2} + 3 \, a^{2} b c^{5} d^{3} - a^{3} c^{4} d^{4}\right )}} + \frac {18 \, b^{2} c^{2} d^{4} x^{4} - 24 \, a b c d^{5} x^{4} + 9 \, a^{2} d^{6} x^{4} + 42 \, b^{2} c^{3} d^{3} x^{2} - 58 \, a b c^{2} d^{4} x^{2} + 22 \, a^{2} c d^{5} x^{2} + 25 \, b^{2} c^{4} d^{2} - 36 \, a b c^{3} d^{3} + 14 \, a^{2} c^{2} d^{4}}{4 \, {\left (b^{3} c^{7} - 3 \, a b^{2} c^{6} d + 3 \, a^{2} b c^{5} d^{2} - a^{3} c^{4} d^{3}\right )} {\left (d x^{2} + c\right )}^{2}} - \frac {{\left (b c + 3 \, a d\right )} \log \left (x^{2}\right )}{2 \, a^{2} c^{4}} + \frac {b c x^{2} + 3 \, a d x^{2} - a c}{2 \, a^{2} c^{4} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^2+a)/(d*x^2+c)^3,x, algorithm="giac")

[Out]

1/2*b^5*log(abs(b*x^2 + a))/(a^2*b^4*c^3 - 3*a^3*b^3*c^2*d + 3*a^4*b^2*c*d^2 - a^5*b*d^3) - 1/2*(6*b^2*c^2*d^3
- 8*a*b*c*d^4 + 3*a^2*d^5)*log(abs(d*x^2 + c))/(b^3*c^7*d - 3*a*b^2*c^6*d^2 + 3*a^2*b*c^5*d^3 - a^3*c^4*d^4)
+ 1/4*(18*b^2*c^2*d^4*x^4 - 24*a*b*c*d^5*x^4 + 9*a^2*d^6*x^4 + 42*b^2*c^3*d^3*x^2 - 58*a*b*c^2*d^4*x^2 + 22*a^
2*c*d^5*x^2 + 25*b^2*c^4*d^2 - 36*a*b*c^3*d^3 + 14*a^2*c^2*d^4)/((b^3*c^7 - 3*a*b^2*c^6*d + 3*a^2*b*c^5*d^2 -
a^3*c^4*d^3)*(d*x^2 + c)^2) - 1/2*(b*c + 3*a*d)*log(x^2)/(a^2*c^4) + 1/2*(b*c*x^2 + 3*a*d*x^2 - a*c)/(a^2*c^4*
x^2)

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maple [A]  time = 0.02, size = 322, normalized size = 1.81 \begin {gather*} -\frac {a^{2} d^{4}}{4 \left (a d -b c \right )^{3} \left (d \,x^{2}+c \right )^{2} c^{2}}+\frac {a b \,d^{3}}{2 \left (a d -b c \right )^{3} \left (d \,x^{2}+c \right )^{2} c}-\frac {b^{2} d^{2}}{4 \left (a d -b c \right )^{3} \left (d \,x^{2}+c \right )^{2}}-\frac {a^{2} d^{4}}{\left (a d -b c \right )^{3} \left (d \,x^{2}+c \right ) c^{3}}+\frac {3 a^{2} d^{4} \ln \left (d \,x^{2}+c \right )}{2 \left (a d -b c \right )^{3} c^{4}}+\frac {5 a b \,d^{3}}{2 \left (a d -b c \right )^{3} \left (d \,x^{2}+c \right ) c^{2}}-\frac {4 a b \,d^{3} \ln \left (d \,x^{2}+c \right )}{\left (a d -b c \right )^{3} c^{3}}-\frac {b^{4} \ln \left (b \,x^{2}+a \right )}{2 \left (a d -b c \right )^{3} a^{2}}-\frac {3 b^{2} d^{2}}{2 \left (a d -b c \right )^{3} \left (d \,x^{2}+c \right ) c}+\frac {3 b^{2} d^{2} \ln \left (d \,x^{2}+c \right )}{\left (a d -b c \right )^{3} c^{2}}-\frac {3 d \ln \relax (x )}{a \,c^{4}}-\frac {b \ln \relax (x )}{a^{2} c^{3}}-\frac {1}{2 a \,c^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b*x^2+a)/(d*x^2+c)^3,x)

[Out]

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

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maxima [B]  time = 1.28, size = 364, normalized size = 2.04 \begin {gather*} \frac {b^{4} \log \left (b x^{2} + a\right )}{2 \, {\left (a^{2} b^{3} c^{3} - 3 \, a^{3} b^{2} c^{2} d + 3 \, a^{4} b c d^{2} - a^{5} d^{3}\right )}} - \frac {{\left (6 \, b^{2} c^{2} d^{2} - 8 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} \log \left (d x^{2} + c\right )}{2 \, {\left (b^{3} c^{7} - 3 \, a b^{2} c^{6} d + 3 \, a^{2} b c^{5} d^{2} - a^{3} c^{4} d^{3}\right )}} - \frac {2 \, b^{2} c^{4} - 4 \, a b c^{3} d + 2 \, a^{2} c^{2} d^{2} + 2 \, {\left (b^{2} c^{2} d^{2} - 5 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} x^{4} + {\left (4 \, b^{2} c^{3} d - 15 \, a b c^{2} d^{2} + 9 \, a^{2} c d^{3}\right )} x^{2}}{4 \, {\left ({\left (a b^{2} c^{5} d^{2} - 2 \, a^{2} b c^{4} d^{3} + a^{3} c^{3} d^{4}\right )} x^{6} + 2 \, {\left (a b^{2} c^{6} d - 2 \, a^{2} b c^{5} d^{2} + a^{3} c^{4} d^{3}\right )} x^{4} + {\left (a b^{2} c^{7} - 2 \, a^{2} b c^{6} d + a^{3} c^{5} d^{2}\right )} x^{2}\right )}} - \frac {{\left (b c + 3 \, a d\right )} \log \left (x^{2}\right )}{2 \, a^{2} c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^2+a)/(d*x^2+c)^3,x, algorithm="maxima")

[Out]

1/2*b^4*log(b*x^2 + a)/(a^2*b^3*c^3 - 3*a^3*b^2*c^2*d + 3*a^4*b*c*d^2 - a^5*d^3) - 1/2*(6*b^2*c^2*d^2 - 8*a*b*
c*d^3 + 3*a^2*d^4)*log(d*x^2 + c)/(b^3*c^7 - 3*a*b^2*c^6*d + 3*a^2*b*c^5*d^2 - a^3*c^4*d^3) - 1/4*(2*b^2*c^4 -
4*a*b*c^3*d + 2*a^2*c^2*d^2 + 2*(b^2*c^2*d^2 - 5*a*b*c*d^3 + 3*a^2*d^4)*x^4 + (4*b^2*c^3*d - 15*a*b*c^2*d^2 +
9*a^2*c*d^3)*x^2)/((a*b^2*c^5*d^2 - 2*a^2*b*c^4*d^3 + a^3*c^3*d^4)*x^6 + 2*(a*b^2*c^6*d - 2*a^2*b*c^5*d^2 + a
^3*c^4*d^3)*x^4 + (a*b^2*c^7 - 2*a^2*b*c^6*d + a^3*c^5*d^2)*x^2) - 1/2*(b*c + 3*a*d)*log(x^2)/(a^2*c^4)

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mupad [B]  time = 1.50, size = 314, normalized size = 1.76 \begin {gather*} -\frac {\frac {1}{2\,a\,c}+\frac {x^4\,\left (3\,a^2\,d^4-5\,a\,b\,c\,d^3+b^2\,c^2\,d^2\right )}{2\,a\,c^3\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}+\frac {x^2\,\left (9\,a^2\,d^3-15\,a\,b\,c\,d^2+4\,b^2\,c^2\,d\right )}{4\,a\,c^2\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}}{c^2\,x^2+2\,c\,d\,x^4+d^2\,x^6}-\frac {\ln \left (d\,x^2+c\right )\,\left (3\,a^2\,d^4-8\,a\,b\,c\,d^3+6\,b^2\,c^2\,d^2\right )}{-2\,a^3\,c^4\,d^3+6\,a^2\,b\,c^5\,d^2-6\,a\,b^2\,c^6\,d+2\,b^3\,c^7}-\frac {b^4\,\ln \left (b\,x^2+a\right )}{2\,\left (a^5\,d^3-3\,a^4\,b\,c\,d^2+3\,a^3\,b^2\,c^2\,d-a^2\,b^3\,c^3\right )}-\frac {\ln \relax (x)\,\left (3\,a\,d+b\,c\right )}{a^2\,c^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*x^2)*(c + d*x^2)^3),x)

[Out]

- (1/(2*a*c) + (x^4*(3*a^2*d^4 + b^2*c^2*d^2 - 5*a*b*c*d^3))/(2*a*c^3*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)) + (x^2*
(9*a^2*d^3 + 4*b^2*c^2*d - 15*a*b*c*d^2))/(4*a*c^2*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)))/(c^2*x^2 + d^2*x^6 + 2*c*
d*x^4) - (log(c + d*x^2)*(3*a^2*d^4 + 6*b^2*c^2*d^2 - 8*a*b*c*d^3))/(2*b^3*c^7 - 2*a^3*c^4*d^3 + 6*a^2*b*c^5*d
^2 - 6*a*b^2*c^6*d) - (b^4*log(a + b*x^2))/(2*(a^5*d^3 - a^2*b^3*c^3 + 3*a^3*b^2*c^2*d - 3*a^4*b*c*d^2)) - (lo
g(x)*(3*a*d + b*c))/(a^2*c^4)

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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**3/(b*x**2+a)/(d*x**2+c)**3,x)

[Out]

Timed out

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