3.246 \(\int \frac {x^5}{(d+e x^2) (a+c x^4)^2} \, dx\)

Optimal. Leaf size=155 \[ -\frac {d^2 e \log \left (a+c x^4\right )}{4 \left (a e^2+c d^2\right )^2}+\frac {d^2 e \log \left (d+e x^2\right )}{2 \left (a e^2+c d^2\right )^2}+\frac {d \left (c d^2-a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 \sqrt {a} \sqrt {c} \left (a e^2+c d^2\right )^2}+\frac {-a e-c d x^2}{4 c \left (a+c x^4\right ) \left (a e^2+c d^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 153, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1252, 1647, 801, 635, 205, 260} \[ -\frac {a e+c d x^2}{4 c \left (a+c x^4\right ) \left (a e^2+c d^2\right )}+\frac {d^2 e \log \left (d+e x^2\right )}{2 \left (a e^2+c d^2\right )^2}-\frac {d^2 e \log \left (a+c x^4\right )}{4 \left (a e^2+c d^2\right )^2}+\frac {d \left (c d^2-a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 \sqrt {a} \sqrt {c} \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^5/((d + e*x^2)*(a + c*x^4)^2),x]

[Out]

-(a*e + c*d*x^2)/(4*c*(c*d^2 + a*e^2)*(a + c*x^4)) + (d*(c*d^2 - a*e^2)*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(4*Sqrt
[a]*Sqrt[c]*(c*d^2 + a*e^2)^2) + (d^2*e*Log[d + e*x^2])/(2*(c*d^2 + a*e^2)^2) - (d^2*e*Log[a + c*x^4])/(4*(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 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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]

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 120, normalized size = 0.77 \[ \frac {\frac {d \left (c d^2-a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {c}}-\frac {\left (a e^2+c d^2\right ) \left (a e+c d x^2\right )}{c \left (a+c x^4\right )}-d^2 e \log \left (a+c x^4\right )+2 d^2 e \log \left (d+e x^2\right )}{4 \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/((d + e*x^2)*(a + c*x^4)^2),x]

[Out]

(-(((c*d^2 + a*e^2)*(a*e + c*d*x^2))/(c*(a + c*x^4))) + (d*(c*d^2 - a*e^2)*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(Sqr
t[a]*Sqrt[c]) + 2*d^2*e*Log[d + e*x^2] - d^2*e*Log[a + c*x^4])/(4*(c*d^2 + a*e^2)^2)

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fricas [A]  time = 6.53, size = 487, normalized size = 3.14 \[ \left [-\frac {2 \, a^{2} c d^{2} e + 2 \, a^{3} e^{3} + 2 \, {\left (a c^{2} d^{3} + a^{2} c d e^{2}\right )} x^{2} - {\left (a c d^{3} - a^{2} d e^{2} + {\left (c^{2} d^{3} - a c d e^{2}\right )} x^{4}\right )} \sqrt {-a c} \log \left (\frac {c x^{4} + 2 \, \sqrt {-a c} x^{2} - a}{c x^{4} + a}\right ) + 2 \, {\left (a c^{2} d^{2} e x^{4} + a^{2} c d^{2} e\right )} \log \left (c x^{4} + a\right ) - 4 \, {\left (a c^{2} d^{2} e x^{4} + a^{2} c d^{2} e\right )} \log \left (e x^{2} + d\right )}{8 \, {\left (a^{2} c^{3} d^{4} + 2 \, a^{3} c^{2} d^{2} e^{2} + a^{4} c e^{4} + {\left (a c^{4} d^{4} + 2 \, a^{2} c^{3} d^{2} e^{2} + a^{3} c^{2} e^{4}\right )} x^{4}\right )}}, -\frac {a^{2} c d^{2} e + a^{3} e^{3} + {\left (a c^{2} d^{3} + a^{2} c d e^{2}\right )} x^{2} + {\left (a c d^{3} - a^{2} d e^{2} + {\left (c^{2} d^{3} - a c d e^{2}\right )} x^{4}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c}}{c x^{2}}\right ) + {\left (a c^{2} d^{2} e x^{4} + a^{2} c d^{2} e\right )} \log \left (c x^{4} + a\right ) - 2 \, {\left (a c^{2} d^{2} e x^{4} + a^{2} c d^{2} e\right )} \log \left (e x^{2} + d\right )}{4 \, {\left (a^{2} c^{3} d^{4} + 2 \, a^{3} c^{2} d^{2} e^{2} + a^{4} c e^{4} + {\left (a c^{4} d^{4} + 2 \, a^{2} c^{3} d^{2} e^{2} + a^{3} c^{2} e^{4}\right )} x^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.33, size = 220, normalized size = 1.42 \[ -\frac {d^{2} e \log \left (c x^{4} + a\right )}{4 \, {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}} + \frac {d^{2} e^{2} \log \left ({\left | x^{2} e + d \right |}\right )}{2 \, {\left (c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )}} + \frac {{\left (c d^{3} - a d e^{2}\right )} \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{4 \, {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {a c}} + \frac {c^{2} d^{2} x^{4} e - c^{2} d^{3} x^{2} - a c d x^{2} e^{2} - a^{2} e^{3}}{4 \, {\left (c^{3} d^{4} + 2 \, a c^{2} d^{2} e^{2} + a^{2} c e^{4}\right )} {\left (c x^{4} + a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 252, normalized size = 1.63 \[ -\frac {a d \,e^{2} x^{2}}{4 \left (a \,e^{2}+c \,d^{2}\right )^{2} \left (c \,x^{4}+a \right )}-\frac {c \,d^{3} x^{2}}{4 \left (a \,e^{2}+c \,d^{2}\right )^{2} \left (c \,x^{4}+a \right )}-\frac {a d \,e^{2} \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{4 \left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {a c}}+\frac {c \,d^{3} \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{4 \left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {a c}}-\frac {a^{2} e^{3}}{4 \left (a \,e^{2}+c \,d^{2}\right )^{2} \left (c \,x^{4}+a \right ) c}-\frac {a \,d^{2} e}{4 \left (a \,e^{2}+c \,d^{2}\right )^{2} \left (c \,x^{4}+a \right )}-\frac {d^{2} e \ln \left (c \,x^{4}+a \right )}{4 \left (a \,e^{2}+c \,d^{2}\right )^{2}}+\frac {d^{2} e \ln \left (e \,x^{2}+d \right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(e*x^2+d)/(c*x^4+a)^2,x)

[Out]

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

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maxima [A]  time = 2.08, size = 192, normalized size = 1.24 \[ -\frac {d^{2} e \log \left (c x^{4} + a\right )}{4 \, {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}} + \frac {d^{2} e \log \left (e x^{2} + d\right )}{2 \, {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}} + \frac {{\left (c d^{3} - a d e^{2}\right )} \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{4 \, {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {a c}} - \frac {c d x^{2} + a e}{4 \, {\left (a c^{2} d^{2} + a^{2} c e^{2} + {\left (c^{3} d^{2} + a c^{2} e^{2}\right )} x^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.52, size = 528, normalized size = 3.41 \[ \frac {\ln \left (a^4\,e^8\,\sqrt {-a\,c}+c^4\,d^8\,\sqrt {-a\,c}+70\,d^4\,e^4\,{\left (-a\,c\right )}^{5/2}+c^5\,d^8\,x^2+a^4\,c\,e^8\,x^2-36\,a^2\,d^2\,e^6\,{\left (-a\,c\right )}^{3/2}-36\,c^2\,d^6\,e^2\,{\left (-a\,c\right )}^{3/2}+70\,a^2\,c^3\,d^4\,e^4\,x^2+36\,a^3\,c^2\,d^2\,e^6\,x^2+36\,a\,c^4\,d^6\,e^2\,x^2\right )\,\left (c\,\left (\frac {d^3\,\sqrt {-a\,c}}{8}-\frac {a\,d^2\,e}{4}\right )-\frac {a\,d\,e^2\,\sqrt {-a\,c}}{8}\right )}{a^3\,c\,e^4+2\,a^2\,c^2\,d^2\,e^2+a\,c^3\,d^4}-\frac {\frac {d\,x^2}{4\,\left (c\,d^2+a\,e^2\right )}+\frac {a\,e}{4\,c\,\left (c\,d^2+a\,e^2\right )}}{c\,x^4+a}-\frac {\ln \left (c^5\,d^8\,x^2-c^4\,d^8\,\sqrt {-a\,c}-70\,d^4\,e^4\,{\left (-a\,c\right )}^{5/2}-a^4\,e^8\,\sqrt {-a\,c}+a^4\,c\,e^8\,x^2+36\,a^2\,d^2\,e^6\,{\left (-a\,c\right )}^{3/2}+36\,c^2\,d^6\,e^2\,{\left (-a\,c\right )}^{3/2}+70\,a^2\,c^3\,d^4\,e^4\,x^2+36\,a^3\,c^2\,d^2\,e^6\,x^2+36\,a\,c^4\,d^6\,e^2\,x^2\right )\,\left (c\,\left (\frac {d^3\,\sqrt {-a\,c}}{8}+\frac {a\,d^2\,e}{4}\right )-\frac {a\,d\,e^2\,\sqrt {-a\,c}}{8}\right )}{a^3\,c\,e^4+2\,a^2\,c^2\,d^2\,e^2+a\,c^3\,d^4}+\frac {d^2\,e\,\ln \left (e\,x^2+d\right )}{2\,\left (a^2\,e^4+2\,a\,c\,d^2\,e^2+c^2\,d^4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/((a + c*x^4)^2*(d + e*x^2)),x)

[Out]

(log(a^4*e^8*(-a*c)^(1/2) + c^4*d^8*(-a*c)^(1/2) + 70*d^4*e^4*(-a*c)^(5/2) + c^5*d^8*x^2 + a^4*c*e^8*x^2 - 36*
a^2*d^2*e^6*(-a*c)^(3/2) - 36*c^2*d^6*e^2*(-a*c)^(3/2) + 70*a^2*c^3*d^4*e^4*x^2 + 36*a^3*c^2*d^2*e^6*x^2 + 36*
a*c^4*d^6*e^2*x^2)*(c*((d^3*(-a*c)^(1/2))/8 - (a*d^2*e)/4) - (a*d*e^2*(-a*c)^(1/2))/8))/(a*c^3*d^4 + a^3*c*e^4
+ 2*a^2*c^2*d^2*e^2) - ((d*x^2)/(4*(a*e^2 + c*d^2)) + (a*e)/(4*c*(a*e^2 + c*d^2)))/(a + c*x^4) - (log(c^5*d^8
*x^2 - c^4*d^8*(-a*c)^(1/2) - 70*d^4*e^4*(-a*c)^(5/2) - a^4*e^8*(-a*c)^(1/2) + a^4*c*e^8*x^2 + 36*a^2*d^2*e^6*
(-a*c)^(3/2) + 36*c^2*d^6*e^2*(-a*c)^(3/2) + 70*a^2*c^3*d^4*e^4*x^2 + 36*a^3*c^2*d^2*e^6*x^2 + 36*a*c^4*d^6*e^
2*x^2)*(c*((d^3*(-a*c)^(1/2))/8 + (a*d^2*e)/4) - (a*d*e^2*(-a*c)^(1/2))/8))/(a*c^3*d^4 + a^3*c*e^4 + 2*a^2*c^2
*d^2*e^2) + (d^2*e*log(d + e*x^2))/(2*(a^2*e^4 + c^2*d^4 + 2*a*c*d^2*e^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(x**5/(e*x**2+d)/(c*x**4+a)**2,x)

[Out]

Timed out

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