3.250 \(\int \frac {1}{x^3 (d+e x^2) (a+c x^4)^2} \, dx\)

Optimal. Leaf size=236 \[ -\frac {c^{3/2} d \left (2 a e^2+c d^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 a^{5/2} \left (a e^2+c d^2\right )^2}-\frac {c^{3/2} d \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 a^{5/2} \left (a e^2+c d^2\right )}+\frac {c e \left (2 a e^2+c d^2\right ) \log \left (a+c x^4\right )}{4 a^2 \left (a e^2+c d^2\right )^2}-\frac {c \left (a e+c d x^2\right )}{4 a^2 \left (a+c x^4\right ) \left (a e^2+c d^2\right )}-\frac {e \log (x)}{a^2 d^2}-\frac {1}{2 a^2 d x^2}+\frac {e^5 \log \left (d+e x^2\right )}{2 d^2 \left (a e^2+c d^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1252, 894, 639, 205, 635, 260} \[ -\frac {c^{3/2} d \left (2 a e^2+c d^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 a^{5/2} \left (a e^2+c d^2\right )^2}-\frac {c^{3/2} d \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 a^{5/2} \left (a e^2+c d^2\right )}-\frac {c \left (a e+c d x^2\right )}{4 a^2 \left (a+c x^4\right ) \left (a e^2+c d^2\right )}+\frac {c e \left (2 a e^2+c d^2\right ) \log \left (a+c x^4\right )}{4 a^2 \left (a e^2+c d^2\right )^2}-\frac {e \log (x)}{a^2 d^2}-\frac {1}{2 a^2 d x^2}+\frac {e^5 \log \left (d+e x^2\right )}{2 d^2 \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(d + e*x^2)*(a + c*x^4)^2),x]

[Out]

-1/(2*a^2*d*x^2) - (c*(a*e + c*d*x^2))/(4*a^2*(c*d^2 + a*e^2)*(a + c*x^4)) - (c^(3/2)*d*ArcTan[(Sqrt[c]*x^2)/S
qrt[a]])/(4*a^(5/2)*(c*d^2 + a*e^2)) - (c^(3/2)*d*(c*d^2 + 2*a*e^2)*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(2*a^(5/2)*
(c*d^2 + a*e^2)^2) - (e*Log[x])/(a^2*d^2) + (e^5*Log[d + e*x^2])/(2*d^2*(c*d^2 + a*e^2)^2) + (c*e*(c*d^2 + 2*a
*e^2)*Log[a + c*x^4])/(4*a^2*(c*d^2 + a*e^2)^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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1252

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.43, size = 248, normalized size = 1.05 \[ \frac {1}{4} \left (\frac {c^{3/2} d \left (5 a e^2+3 c d^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{a^{5/2} \left (a e^2+c d^2\right )^2}+\frac {c^{3/2} d \left (5 a e^2+3 c d^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{a^{5/2} \left (a e^2+c d^2\right )^2}-\frac {c \left (a e+c d x^2\right )}{a^2 \left (a+c x^4\right ) \left (a e^2+c d^2\right )}+\frac {c \left (2 a e^3+c d^2 e\right ) \log \left (a+c x^4\right )}{a^2 \left (a e^2+c d^2\right )^2}-\frac {4 e \log (x)}{a^2 d^2}-\frac {2}{a^2 d x^2}+\frac {2 e^5 \log \left (d+e x^2\right )}{\left (a d e^2+c d^3\right )^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(d + e*x^2)*(a + c*x^4)^2),x]

[Out]

(-2/(a^2*d*x^2) - (c*(a*e + c*d*x^2))/(a^2*(c*d^2 + a*e^2)*(a + c*x^4)) + (c^(3/2)*d*(3*c*d^2 + 5*a*e^2)*ArcTa
n[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(a^(5/2)*(c*d^2 + a*e^2)^2) + (c^(3/2)*d*(3*c*d^2 + 5*a*e^2)*ArcTan[1 + (S
qrt[2]*c^(1/4)*x)/a^(1/4)])/(a^(5/2)*(c*d^2 + a*e^2)^2) - (4*e*Log[x])/(a^2*d^2) + (2*e^5*Log[d + e*x^2])/(c*d
^3 + a*d*e^2)^2 + (c*(c*d^2*e + 2*a*e^3)*Log[a + c*x^4])/(a^2*(c*d^2 + a*e^2)^2))/4

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fricas [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^3/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.35, size = 344, normalized size = 1.46 \[ \frac {{\left (c^{2} d^{2} e + 2 \, a c e^{3}\right )} \log \left (c x^{4} + a\right )}{4 \, {\left (a^{2} c^{2} d^{4} + 2 \, a^{3} c d^{2} e^{2} + a^{4} e^{4}\right )}} + \frac {e^{6} \log \left ({\left | x^{2} e + d \right |}\right )}{2 \, {\left (c^{2} d^{6} e + 2 \, a c d^{4} e^{3} + a^{2} d^{2} e^{5}\right )}} - \frac {{\left (3 \, c^{3} d^{3} + 5 \, a c^{2} d e^{2}\right )} \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{4 \, {\left (a^{2} c^{2} d^{4} + 2 \, a^{3} c d^{2} e^{2} + a^{4} e^{4}\right )} \sqrt {a c}} - \frac {9 \, c^{3} d^{5} x^{4} + 15 \, a c^{2} d^{3} x^{4} e^{2} - 2 \, a^{2} c x^{6} e^{5} + 3 \, a c^{2} d^{4} x^{2} e + 6 \, a^{2} c d x^{4} e^{4} + 6 \, a c^{2} d^{5} + 3 \, a^{2} c d^{2} x^{2} e^{3} + 12 \, a^{2} c d^{3} e^{2} - 2 \, a^{3} x^{2} e^{5} + 6 \, a^{3} d e^{4}}{12 \, {\left (a^{2} c^{2} d^{6} + 2 \, a^{3} c d^{4} e^{2} + a^{4} d^{2} e^{4}\right )} {\left (c x^{6} + a x^{2}\right )}} - \frac {e \log \left (x^{2}\right )}{2 \, a^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="giac")

[Out]

1/4*(c^2*d^2*e + 2*a*c*e^3)*log(c*x^4 + a)/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4) + 1/2*e^6*log(abs(x^2*e +
d))/(c^2*d^6*e + 2*a*c*d^4*e^3 + a^2*d^2*e^5) - 1/4*(3*c^3*d^3 + 5*a*c^2*d*e^2)*arctan(c*x^2/sqrt(a*c))/((a^2
*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4)*sqrt(a*c)) - 1/12*(9*c^3*d^5*x^4 + 15*a*c^2*d^3*x^4*e^2 - 2*a^2*c*x^6*e^
5 + 3*a*c^2*d^4*x^2*e + 6*a^2*c*d*x^4*e^4 + 6*a*c^2*d^5 + 3*a^2*c*d^2*x^2*e^3 + 12*a^2*c*d^3*e^2 - 2*a^3*x^2*e
^5 + 6*a^3*d*e^4)/((a^2*c^2*d^6 + 2*a^3*c*d^4*e^2 + a^4*d^2*e^4)*(c*x^6 + a*x^2)) - 1/2*e*log(x^2)/(a^2*d^2)

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maple [A]  time = 0.02, size = 332, normalized size = 1.41 \[ -\frac {c^{2} d \,e^{2} x^{2}}{4 \left (a \,e^{2}+c \,d^{2}\right )^{2} \left (c \,x^{4}+a \right ) a}-\frac {c^{3} d^{3} x^{2}}{4 \left (a \,e^{2}+c \,d^{2}\right )^{2} \left (c \,x^{4}+a \right ) a^{2}}-\frac {5 c^{2} d \,e^{2} \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{4 \left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {a c}\, a}-\frac {3 c^{3} d^{3} \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{4 \left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {a c}\, a^{2}}-\frac {c^{2} d^{2} e}{4 \left (a \,e^{2}+c \,d^{2}\right )^{2} \left (c \,x^{4}+a \right ) a}+\frac {c \,e^{3} \ln \left (c \,x^{4}+a \right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2} a}+\frac {c^{2} d^{2} e \ln \left (c \,x^{4}+a \right )}{4 \left (a \,e^{2}+c \,d^{2}\right )^{2} a^{2}}-\frac {c \,e^{3}}{4 \left (a \,e^{2}+c \,d^{2}\right )^{2} \left (c \,x^{4}+a \right )}+\frac {e^{5} \ln \left (e \,x^{2}+d \right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2} d^{2}}-\frac {e \ln \relax (x )}{a^{2} d^{2}}-\frac {1}{2 a^{2} d \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(e*x^2+d)/(c*x^4+a)^2,x)

[Out]

-1/2/a^2/d/x^2-e*ln(x)/a^2/d^2-1/4*c^2/(a*e^2+c*d^2)^2/a/(c*x^4+a)*x^2*d*e^2-1/4*c^3/(a*e^2+c*d^2)^2/a^2/(c*x^
4+a)*x^2*d^3-1/4*c/(a*e^2+c*d^2)^2/(c*x^4+a)*e^3-1/4*c^2/(a*e^2+c*d^2)^2/a/(c*x^4+a)*e*d^2+1/2*c/(a*e^2+c*d^2)
^2/a*ln(c*x^4+a)*e^3+1/4*c^2/(a*e^2+c*d^2)^2/a^2*ln(c*x^4+a)*e*d^2-5/4*c^2/(a*e^2+c*d^2)^2/a/(a*c)^(1/2)*arcta
n(1/(a*c)^(1/2)*c*x^2)*d*e^2-3/4*c^3/(a*e^2+c*d^2)^2/a^2/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x^2)*d^3+1/2*e^5*l
n(e*x^2+d)/d^2/(a*e^2+c*d^2)^2

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maxima [A]  time = 2.01, size = 278, normalized size = 1.18 \[ \frac {e^{5} \log \left (e x^{2} + d\right )}{2 \, {\left (c^{2} d^{6} + 2 \, a c d^{4} e^{2} + a^{2} d^{2} e^{4}\right )}} + \frac {{\left (c^{2} d^{2} e + 2 \, a c e^{3}\right )} \log \left (c x^{4} + a\right )}{4 \, {\left (a^{2} c^{2} d^{4} + 2 \, a^{3} c d^{2} e^{2} + a^{4} e^{4}\right )}} - \frac {{\left (3 \, c^{3} d^{3} + 5 \, a c^{2} d e^{2}\right )} \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{4 \, {\left (a^{2} c^{2} d^{4} + 2 \, a^{3} c d^{2} e^{2} + a^{4} e^{4}\right )} \sqrt {a c}} - \frac {a c d e x^{2} + {\left (3 \, c^{2} d^{2} + 2 \, a c e^{2}\right )} x^{4} + 2 \, a c d^{2} + 2 \, a^{2} e^{2}}{4 \, {\left ({\left (a^{2} c^{2} d^{3} + a^{3} c d e^{2}\right )} x^{6} + {\left (a^{3} c d^{3} + a^{4} d e^{2}\right )} x^{2}\right )}} - \frac {e \log \left (x^{2}\right )}{2 \, a^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.94, size = 1337, normalized size = 5.67 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + c*x^4)^2*(d + e*x^2)),x)

[Out]

(log(81*a^10*c^16*d^24*x^2 + 1024*a^22*c^4*e^24*x^2 - 81*a^3*c^11*d^24*(-a^5*c^3)^(3/2) + 1024*a^20*c^2*e^24*(
-a^5*c^3)^(1/2) - 14496*a^6*d^8*e^16*(-a^5*c^3)^(5/2) - 5120*a^14*d^2*e^22*(-a^5*c^3)^(3/2) + 11647*c^6*d^20*e
^4*(-a^5*c^3)^(5/2) + 1638*a^11*c^15*d^22*e^2*x^2 + 11647*a^12*c^14*d^20*e^4*x^2 + 43524*a^13*c^13*d^18*e^6*x^
2 + 97311*a^14*c^12*d^16*e^8*x^2 + 133334*a^15*c^11*d^14*e^10*x^2 + 103633*a^16*c^10*d^12*e^12*x^2 + 29456*a^1
7*c^9*d^10*e^14*x^2 - 14496*a^18*c^8*d^8*e^16*x^2 - 7984*a^19*c^7*d^6*e^18*x^2 + 5888*a^20*c^6*d^4*e^20*x^2 +
5120*a^21*c^5*d^2*e^22*x^2 + 43524*a*c^5*d^18*e^6*(-a^5*c^3)^(5/2) + 29456*a^5*c*d^10*e^14*(-a^5*c^3)^(5/2) -
5888*a^13*c*d^4*e^20*(-a^5*c^3)^(3/2) + 97311*a^2*c^4*d^16*e^8*(-a^5*c^3)^(5/2) + 133334*a^3*c^3*d^14*e^10*(-a
^5*c^3)^(5/2) + 103633*a^4*c^2*d^12*e^12*(-a^5*c^3)^(5/2) - 1638*a^4*c^10*d^22*e^2*(-a^5*c^3)^(3/2) + 7984*a^1
2*c^2*d^6*e^18*(-a^5*c^3)^(3/2))*(4*a^4*c*e^3 - 3*c*d^3*(-a^5*c^3)^(1/2) + 2*a^3*c^2*d^2*e - 5*a*d*e^2*(-a^5*c
^3)^(1/2)))/(8*(a^7*e^4 + a^5*c^2*d^4 + 2*a^6*c*d^2*e^2)) - (1/(2*a*d) + (c*e*x^2)/(4*a*(a*e^2 + c*d^2)) + (c*
x^4*(2*a*e^2 + 3*c*d^2))/(4*a^2*d*(a*e^2 + c*d^2)))/(a*x^2 + c*x^6) + (log(81*a^10*c^16*d^24*x^2 + 1024*a^22*c
^4*e^24*x^2 + 81*a^3*c^11*d^24*(-a^5*c^3)^(3/2) - 1024*a^20*c^2*e^24*(-a^5*c^3)^(1/2) + 14496*a^6*d^8*e^16*(-a
^5*c^3)^(5/2) + 5120*a^14*d^2*e^22*(-a^5*c^3)^(3/2) - 11647*c^6*d^20*e^4*(-a^5*c^3)^(5/2) + 1638*a^11*c^15*d^2
2*e^2*x^2 + 11647*a^12*c^14*d^20*e^4*x^2 + 43524*a^13*c^13*d^18*e^6*x^2 + 97311*a^14*c^12*d^16*e^8*x^2 + 13333
4*a^15*c^11*d^14*e^10*x^2 + 103633*a^16*c^10*d^12*e^12*x^2 + 29456*a^17*c^9*d^10*e^14*x^2 - 14496*a^18*c^8*d^8
*e^16*x^2 - 7984*a^19*c^7*d^6*e^18*x^2 + 5888*a^20*c^6*d^4*e^20*x^2 + 5120*a^21*c^5*d^2*e^22*x^2 - 43524*a*c^5
*d^18*e^6*(-a^5*c^3)^(5/2) - 29456*a^5*c*d^10*e^14*(-a^5*c^3)^(5/2) + 5888*a^13*c*d^4*e^20*(-a^5*c^3)^(3/2) -
97311*a^2*c^4*d^16*e^8*(-a^5*c^3)^(5/2) - 133334*a^3*c^3*d^14*e^10*(-a^5*c^3)^(5/2) - 103633*a^4*c^2*d^12*e^12
*(-a^5*c^3)^(5/2) + 1638*a^4*c^10*d^22*e^2*(-a^5*c^3)^(3/2) - 7984*a^12*c^2*d^6*e^18*(-a^5*c^3)^(3/2))*(4*a^4*
c*e^3 + 3*c*d^3*(-a^5*c^3)^(1/2) + 2*a^3*c^2*d^2*e + 5*a*d*e^2*(-a^5*c^3)^(1/2)))/(8*(a^7*e^4 + a^5*c^2*d^4 +
2*a^6*c*d^2*e^2)) + (e^5*log(d + e*x^2))/(2*c^2*d^6 + 2*a^2*d^2*e^4 + 4*a*c*d^4*e^2) - (e*log(x))/(a^2*d^2)

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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**3/(e*x**2+d)/(c*x**4+a)**2,x)

[Out]

Timed out

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