3.250 \(\int \frac {1}{x^4 (a+b x^2) (c+d x^2)^2} \, dx\)

Optimal. Leaf size=189 \[ \frac {b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} (b c-a d)^2}+\frac {-5 a^2 d^2+2 a b c d+2 b^2 c^2}{2 a^2 c^3 x (b c-a d)}-\frac {d^{5/2} (7 b c-5 a d) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 c^{7/2} (b c-a d)^2}-\frac {2 b c-5 a d}{6 a c^2 x^3 (b c-a d)}-\frac {d}{2 c x^3 \left (c+d x^2\right ) (b c-a d)} \]

[Out]

1/6*(5*a*d-2*b*c)/a/c^2/(-a*d+b*c)/x^3+1/2*(-5*a^2*d^2+2*a*b*c*d+2*b^2*c^2)/a^2/c^3/(-a*d+b*c)/x-1/2*d/c/(-a*d
+b*c)/x^3/(d*x^2+c)+b^(7/2)*arctan(x*b^(1/2)/a^(1/2))/a^(5/2)/(-a*d+b*c)^2-1/2*d^(5/2)*(-5*a*d+7*b*c)*arctan(x
*d^(1/2)/c^(1/2))/c^(7/2)/(-a*d+b*c)^2

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Rubi [A]  time = 0.27, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {472, 583, 522, 205} \[ \frac {-5 a^2 d^2+2 a b c d+2 b^2 c^2}{2 a^2 c^3 x (b c-a d)}+\frac {b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} (b c-a d)^2}-\frac {d^{5/2} (7 b c-5 a d) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 c^{7/2} (b c-a d)^2}-\frac {2 b c-5 a d}{6 a c^2 x^3 (b c-a d)}-\frac {d}{2 c x^3 \left (c+d x^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^2)*(c + d*x^2)^2),x]

[Out]

-(2*b*c - 5*a*d)/(6*a*c^2*(b*c - a*d)*x^3) + (2*b^2*c^2 + 2*a*b*c*d - 5*a^2*d^2)/(2*a^2*c^3*(b*c - a*d)*x) - d
/(2*c*(b*c - a*d)*x^3*(c + d*x^2)) + (b^(7/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(5/2)*(b*c - a*d)^2) - (d^(5/2)*
(7*b*c - 5*a*d)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(2*c^(7/2)*(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^4 \left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx &=-\frac {d}{2 c (b c-a d) x^3 \left (c+d x^2\right )}+\frac {\int \frac {2 b c-5 a d-5 b d x^2}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{2 c (b c-a d)}\\ &=-\frac {2 b c-5 a d}{6 a c^2 (b c-a d) x^3}-\frac {d}{2 c (b c-a d) x^3 \left (c+d x^2\right )}-\frac {\int \frac {3 \left (2 b^2 c^2+2 a b c d-5 a^2 d^2\right )+3 b d (2 b c-5 a d) x^2}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{6 a c^2 (b c-a d)}\\ &=-\frac {2 b c-5 a d}{6 a c^2 (b c-a d) x^3}+\frac {2 b^2 c^2+2 a b c d-5 a^2 d^2}{2 a^2 c^3 (b c-a d) x}-\frac {d}{2 c (b c-a d) x^3 \left (c+d x^2\right )}+\frac {\int \frac {3 \left (2 b^3 c^3+2 a b^2 c^2 d+2 a^2 b c d^2-5 a^3 d^3\right )+3 b d \left (2 b^2 c^2+2 a b c d-5 a^2 d^2\right ) x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{6 a^2 c^3 (b c-a d)}\\ &=-\frac {2 b c-5 a d}{6 a c^2 (b c-a d) x^3}+\frac {2 b^2 c^2+2 a b c d-5 a^2 d^2}{2 a^2 c^3 (b c-a d) x}-\frac {d}{2 c (b c-a d) x^3 \left (c+d x^2\right )}+\frac {b^4 \int \frac {1}{a+b x^2} \, dx}{a^2 (b c-a d)^2}-\frac {\left (d^3 (7 b c-5 a d)\right ) \int \frac {1}{c+d x^2} \, dx}{2 c^3 (b c-a d)^2}\\ &=-\frac {2 b c-5 a d}{6 a c^2 (b c-a d) x^3}+\frac {2 b^2 c^2+2 a b c d-5 a^2 d^2}{2 a^2 c^3 (b c-a d) x}-\frac {d}{2 c (b c-a d) x^3 \left (c+d x^2\right )}+\frac {b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} (b c-a d)^2}-\frac {d^{5/2} (7 b c-5 a d) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 c^{7/2} (b c-a d)^2}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 142, normalized size = 0.75 \[ \frac {b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} (a d-b c)^2}+\frac {2 a d+b c}{a^2 c^3 x}-\frac {d^{5/2} (7 b c-5 a d) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 c^{7/2} (b c-a d)^2}-\frac {d^3 x}{2 c^3 \left (c+d x^2\right ) (b c-a d)}-\frac {1}{3 a c^2 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*x^2)*(c + d*x^2)^2),x]

[Out]

-1/3*1/(a*c^2*x^3) + (b*c + 2*a*d)/(a^2*c^3*x) - (d^3*x)/(2*c^3*(b*c - a*d)*(c + d*x^2)) + (b^(7/2)*ArcTan[(Sq
rt[b]*x)/Sqrt[a]])/(a^(5/2)*(-(b*c) + a*d)^2) - (d^(5/2)*(7*b*c - 5*a*d)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(2*c^(7/
2)*(b*c - a*d)^2)

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fricas [A]  time = 1.92, size = 1281, normalized size = 6.78 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^2,x, algorithm="fricas")

[Out]

[-1/12*(4*a*b^2*c^4 - 8*a^2*b*c^3*d + 4*a^3*c^2*d^2 - 6*(2*b^3*c^3*d - 7*a^2*b*c*d^3 + 5*a^3*d^4)*x^4 - 4*(3*b
^3*c^4 - a*b^2*c^3*d - 7*a^2*b*c^2*d^2 + 5*a^3*c*d^3)*x^2 - 6*(b^3*c^3*d*x^5 + b^3*c^4*x^3)*sqrt(-b/a)*log((b*
x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) + 3*((7*a^2*b*c*d^3 - 5*a^3*d^4)*x^5 + (7*a^2*b*c^2*d^2 - 5*a^3*c*d^3
)*x^3)*sqrt(-d/c)*log((d*x^2 + 2*c*x*sqrt(-d/c) - c)/(d*x^2 + c)))/((a^2*b^2*c^5*d - 2*a^3*b*c^4*d^2 + a^4*c^3
*d^3)*x^5 + (a^2*b^2*c^6 - 2*a^3*b*c^5*d + a^4*c^4*d^2)*x^3), -1/6*(2*a*b^2*c^4 - 4*a^2*b*c^3*d + 2*a^3*c^2*d^
2 - 3*(2*b^3*c^3*d - 7*a^2*b*c*d^3 + 5*a^3*d^4)*x^4 - 2*(3*b^3*c^4 - a*b^2*c^3*d - 7*a^2*b*c^2*d^2 + 5*a^3*c*d
^3)*x^2 + 3*((7*a^2*b*c*d^3 - 5*a^3*d^4)*x^5 + (7*a^2*b*c^2*d^2 - 5*a^3*c*d^3)*x^3)*sqrt(d/c)*arctan(x*sqrt(d/
c)) - 3*(b^3*c^3*d*x^5 + b^3*c^4*x^3)*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/((a^2*b^2*c^
5*d - 2*a^3*b*c^4*d^2 + a^4*c^3*d^3)*x^5 + (a^2*b^2*c^6 - 2*a^3*b*c^5*d + a^4*c^4*d^2)*x^3), -1/12*(4*a*b^2*c^
4 - 8*a^2*b*c^3*d + 4*a^3*c^2*d^2 - 6*(2*b^3*c^3*d - 7*a^2*b*c*d^3 + 5*a^3*d^4)*x^4 - 4*(3*b^3*c^4 - a*b^2*c^3
*d - 7*a^2*b*c^2*d^2 + 5*a^3*c*d^3)*x^2 - 12*(b^3*c^3*d*x^5 + b^3*c^4*x^3)*sqrt(b/a)*arctan(x*sqrt(b/a)) + 3*(
(7*a^2*b*c*d^3 - 5*a^3*d^4)*x^5 + (7*a^2*b*c^2*d^2 - 5*a^3*c*d^3)*x^3)*sqrt(-d/c)*log((d*x^2 + 2*c*x*sqrt(-d/c
) - c)/(d*x^2 + c)))/((a^2*b^2*c^5*d - 2*a^3*b*c^4*d^2 + a^4*c^3*d^3)*x^5 + (a^2*b^2*c^6 - 2*a^3*b*c^5*d + a^4
*c^4*d^2)*x^3), -1/6*(2*a*b^2*c^4 - 4*a^2*b*c^3*d + 2*a^3*c^2*d^2 - 3*(2*b^3*c^3*d - 7*a^2*b*c*d^3 + 5*a^3*d^4
)*x^4 - 2*(3*b^3*c^4 - a*b^2*c^3*d - 7*a^2*b*c^2*d^2 + 5*a^3*c*d^3)*x^2 - 6*(b^3*c^3*d*x^5 + b^3*c^4*x^3)*sqrt
(b/a)*arctan(x*sqrt(b/a)) + 3*((7*a^2*b*c*d^3 - 5*a^3*d^4)*x^5 + (7*a^2*b*c^2*d^2 - 5*a^3*c*d^3)*x^3)*sqrt(d/c
)*arctan(x*sqrt(d/c)))/((a^2*b^2*c^5*d - 2*a^3*b*c^4*d^2 + a^4*c^3*d^3)*x^5 + (a^2*b^2*c^6 - 2*a^3*b*c^5*d + a
^4*c^4*d^2)*x^3)]

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giac [A]  time = 0.34, size = 165, normalized size = 0.87 \[ \frac {b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} \sqrt {a b}} - \frac {d^{3} x}{2 \, {\left (b c^{4} - a c^{3} d\right )} {\left (d x^{2} + c\right )}} - \frac {{\left (7 \, b c d^{3} - 5 \, a d^{4}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, {\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2}\right )} \sqrt {c d}} + \frac {3 \, b c x^{2} + 6 \, a d x^{2} - a c}{3 \, a^{2} c^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^2,x, algorithm="giac")

[Out]

b^4*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*d^3*x/((b*c^4 - a*c^3*d)*(d*
x^2 + c)) - 1/2*(7*b*c*d^3 - 5*a*d^4)*arctan(d*x/sqrt(c*d))/((b^2*c^5 - 2*a*b*c^4*d + a^2*c^3*d^2)*sqrt(c*d))
+ 1/3*(3*b*c*x^2 + 6*a*d*x^2 - a*c)/(a^2*c^3*x^3)

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maple [A]  time = 0.02, size = 191, normalized size = 1.01 \[ \frac {a \,d^{4} x}{2 \left (a d -b c \right )^{2} \left (d \,x^{2}+c \right ) c^{3}}+\frac {5 a \,d^{4} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \left (a d -b c \right )^{2} \sqrt {c d}\, c^{3}}+\frac {b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\left (a d -b c \right )^{2} \sqrt {a b}\, a^{2}}-\frac {b \,d^{3} x}{2 \left (a d -b c \right )^{2} \left (d \,x^{2}+c \right ) c^{2}}-\frac {7 b \,d^{3} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \left (a d -b c \right )^{2} \sqrt {c d}\, c^{2}}+\frac {2 d}{a \,c^{3} x}+\frac {b}{a^{2} c^{2} x}-\frac {1}{3 a \,c^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(b*x^2+a)/(d*x^2+c)^2,x)

[Out]

1/a^2*b^4/(a*d-b*c)^2/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)+1/2*d^4/c^3/(a*d-b*c)^2*x/(d*x^2+c)*a-1/2*d^3/c^2/
(a*d-b*c)^2*x/(d*x^2+c)*b+5/2*d^4/c^3/(a*d-b*c)^2/(c*d)^(1/2)*arctan(1/(c*d)^(1/2)*d*x)*a-7/2*d^3/c^2/(a*d-b*c
)^2/(c*d)^(1/2)*arctan(1/(c*d)^(1/2)*d*x)*b-1/3/a/c^2/x^3+2/a/c^3/x*d+1/a^2/c^2/x*b

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maxima [A]  time = 2.37, size = 236, normalized size = 1.25 \[ \frac {b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} \sqrt {a b}} - \frac {{\left (7 \, b c d^{3} - 5 \, a d^{4}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, {\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2}\right )} \sqrt {c d}} - \frac {2 \, a b c^{3} - 2 \, a^{2} c^{2} d - 3 \, {\left (2 \, b^{2} c^{2} d + 2 \, a b c d^{2} - 5 \, a^{2} d^{3}\right )} x^{4} - 2 \, {\left (3 \, b^{2} c^{3} + 2 \, a b c^{2} d - 5 \, a^{2} c d^{2}\right )} x^{2}}{6 \, {\left ({\left (a^{2} b c^{4} d - a^{3} c^{3} d^{2}\right )} x^{5} + {\left (a^{2} b c^{5} - a^{3} c^{4} d\right )} x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^2,x, algorithm="maxima")

[Out]

b^4*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*(7*b*c*d^3 - 5*a*d^4)*arctan
(d*x/sqrt(c*d))/((b^2*c^5 - 2*a*b*c^4*d + a^2*c^3*d^2)*sqrt(c*d)) - 1/6*(2*a*b*c^3 - 2*a^2*c^2*d - 3*(2*b^2*c^
2*d + 2*a*b*c*d^2 - 5*a^2*d^3)*x^4 - 2*(3*b^2*c^3 + 2*a*b*c^2*d - 5*a^2*c*d^2)*x^2)/((a^2*b*c^4*d - a^3*c^3*d^
2)*x^5 + (a^2*b*c^5 - a^3*c^4*d)*x^3)

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mupad [B]  time = 0.94, size = 469, normalized size = 2.48 \[ -\frac {\frac {1}{3\,a\,c}-\frac {x^2\,\left (5\,a\,d+3\,b\,c\right )}{3\,a^2\,c^2}+\frac {x^4\,\left (-5\,a^2\,d^3+2\,a\,b\,c\,d^2+2\,b^2\,c^2\,d\right )}{2\,a^2\,c^3\,\left (a\,d-b\,c\right )}}{d\,x^5+c\,x^3}-\frac {\mathrm {atan}\left (\frac {b\,c^7\,x\,{\left (-a^5\,b^7\right )}^{3/2}\,4{}\mathrm {i}+a^{12}\,b\,d^7\,x\,\sqrt {-a^5\,b^7}\,25{}\mathrm {i}+a^{10}\,b^3\,c^2\,d^5\,x\,\sqrt {-a^5\,b^7}\,49{}\mathrm {i}-a^{11}\,b^2\,c\,d^6\,x\,\sqrt {-a^5\,b^7}\,70{}\mathrm {i}}{-25\,a^{15}\,b^4\,d^7+70\,a^{14}\,b^5\,c\,d^6-49\,a^{13}\,b^6\,c^2\,d^5+4\,a^8\,b^{11}\,c^7}\right )\,\sqrt {-a^5\,b^7}\,1{}\mathrm {i}}{a^7\,d^2-2\,a^6\,b\,c\,d+a^5\,b^2\,c^2}-\frac {\mathrm {atan}\left (\frac {a^7\,d^3\,x\,{\left (-c^7\,d^5\right )}^{3/2}\,25{}\mathrm {i}+b^7\,c^{14}\,d\,x\,\sqrt {-c^7\,d^5}\,4{}\mathrm {i}-a^6\,b\,c\,d^2\,x\,{\left (-c^7\,d^5\right )}^{3/2}\,70{}\mathrm {i}+a^5\,b^2\,c^2\,d\,x\,{\left (-c^7\,d^5\right )}^{3/2}\,49{}\mathrm {i}}{25\,a^7\,c^{11}\,d^{10}-70\,a^6\,b\,c^{12}\,d^9+49\,a^5\,b^2\,c^{13}\,d^8-4\,b^7\,c^{18}\,d^3}\right )\,\left (5\,a\,d-7\,b\,c\right )\,\sqrt {-c^7\,d^5}\,1{}\mathrm {i}}{2\,\left (a^2\,c^7\,d^2-2\,a\,b\,c^8\,d+b^2\,c^9\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b*x^2)*(c + d*x^2)^2),x)

[Out]

- (1/(3*a*c) - (x^2*(5*a*d + 3*b*c))/(3*a^2*c^2) + (x^4*(2*b^2*c^2*d - 5*a^2*d^3 + 2*a*b*c*d^2))/(2*a^2*c^3*(a
*d - b*c)))/(c*x^3 + d*x^5) - (atan((b*c^7*x*(-a^5*b^7)^(3/2)*4i + a^12*b*d^7*x*(-a^5*b^7)^(1/2)*25i + a^10*b^
3*c^2*d^5*x*(-a^5*b^7)^(1/2)*49i - a^11*b^2*c*d^6*x*(-a^5*b^7)^(1/2)*70i)/(4*a^8*b^11*c^7 - 25*a^15*b^4*d^7 +
70*a^14*b^5*c*d^6 - 49*a^13*b^6*c^2*d^5))*(-a^5*b^7)^(1/2)*1i)/(a^7*d^2 + a^5*b^2*c^2 - 2*a^6*b*c*d) - (atan((
a^7*d^3*x*(-c^7*d^5)^(3/2)*25i + b^7*c^14*d*x*(-c^7*d^5)^(1/2)*4i - a^6*b*c*d^2*x*(-c^7*d^5)^(3/2)*70i + a^5*b
^2*c^2*d*x*(-c^7*d^5)^(3/2)*49i)/(25*a^7*c^11*d^10 - 4*b^7*c^18*d^3 - 70*a^6*b*c^12*d^9 + 49*a^5*b^2*c^13*d^8)
)*(5*a*d - 7*b*c)*(-c^7*d^5)^(1/2)*1i)/(2*(b^2*c^9 + a^2*c^7*d^2 - 2*a*b*c^8*d))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(b*x**2+a)/(d*x**2+c)**2,x)

[Out]

Timed out

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