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

Optimal. Leaf size=158 \[ -\frac {\left (-a b e-2 a c d+b^2 d\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c \sqrt {b^2-4 a c} \left (a e^2-b d e+c d^2\right )}+\frac {d^2 \log \left (d+e x^2\right )}{2 e \left (a e^2-b d e+c d^2\right )}-\frac {(b d-a e) \log \left (a+b x^2+c x^4\right )}{4 c \left (a e^2-b d e+c d^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1251, 1628, 634, 618, 206, 628} \[ -\frac {\left (-a b e-2 a c d+b^2 d\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c \sqrt {b^2-4 a c} \left (a e^2-b d e+c d^2\right )}+\frac {d^2 \log \left (d+e x^2\right )}{2 e \left (a e^2-b d e+c d^2\right )}-\frac {(b d-a e) \log \left (a+b x^2+c x^4\right )}{4 c \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1251

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

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 139, normalized size = 0.88 \[ -\frac {\sqrt {4 a c-b^2} \left (e (b d-a e) \log \left (a+b x^2+c x^4\right )-2 c d^2 \log \left (d+e x^2\right )\right )+2 e \left (a b e+2 a c d+b^2 (-d)\right ) \tan ^{-1}\left (\frac {b+2 c x^2}{\sqrt {4 a c-b^2}}\right )}{4 c e \sqrt {4 a c-b^2} \left (e (a e-b d)+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(2*e*(-(b^2*d) + 2*a*c*d + a*b*e)*ArcTan[(b + 2*c*x^2)/Sqrt[-b^2 + 4*a*c]] + Sqrt[-b^2 + 4*a*c]*(-2*c*d^2
*Log[d + e*x^2] + e*(b*d - a*e)*Log[a + b*x^2 + c*x^4]))/(c*Sqrt[-b^2 + 4*a*c]*e*(c*d^2 + e*(-(b*d) + a*e)))

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fricas [A]  time = 144.36, size = 421, normalized size = 2.66 \[ \left [\frac {2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} \log \left (e x^{2} + d\right ) + {\left (a b e^{2} - {\left (b^{2} - 2 \, a c\right )} d e\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c + {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) - {\left ({\left (b^{3} - 4 \, a b c\right )} d e - {\left (a b^{2} - 4 \, a^{2} c\right )} e^{2}\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{2} e - {\left (b^{3} c - 4 \, a b c^{2}\right )} d e^{2} + {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} e^{3}\right )}}, \frac {2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} \log \left (e x^{2} + d\right ) + 2 \, {\left (a b e^{2} - {\left (b^{2} - 2 \, a c\right )} d e\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) - {\left ({\left (b^{3} - 4 \, a b c\right )} d e - {\left (a b^{2} - 4 \, a^{2} c\right )} e^{2}\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{2} e - {\left (b^{3} c - 4 \, a b c^{2}\right )} d e^{2} + {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} e^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.84, size = 157, normalized size = 0.99 \[ \frac {d^{2} \log \left ({\left | x^{2} e + d \right |}\right )}{2 \, {\left (c d^{2} e - b d e^{2} + a e^{3}\right )}} - \frac {{\left (b d - a e\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (c^{2} d^{2} - b c d e + a c e^{2}\right )}} + \frac {{\left (b^{2} d - 2 \, a c d - a b e\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, {\left (c^{2} d^{2} - b c d e + a c e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 289, normalized size = 1.83 \[ -\frac {a b e \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{2 \left (a \,e^{2}-d e b +c \,d^{2}\right ) \sqrt {4 a c -b^{2}}\, c}-\frac {a d \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-d e b +c \,d^{2}\right ) \sqrt {4 a c -b^{2}}}+\frac {b^{2} d \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{2 \left (a \,e^{2}-d e b +c \,d^{2}\right ) \sqrt {4 a c -b^{2}}\, c}+\frac {a e \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{4 \left (a \,e^{2}-d e b +c \,d^{2}\right ) c}-\frac {b d \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{4 \left (a \,e^{2}-d e b +c \,d^{2}\right ) c}+\frac {d^{2} \ln \left (e \,x^{2}+d \right )}{2 \left (a \,e^{2}-d e b +c \,d^{2}\right ) e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 11.05, size = 1853, normalized size = 11.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d^2*log(d + e*x^2))/(2*a*e^3 - 2*b*d*e^2 + 2*c*d^2*e) + (log(4*a^2*e^4*(b^2 - 4*a*c)^(5/2) + 8*c^2*d^4*(b^2 -
4*a*c)^(5/2) + 5*d^2*e^2*(b^2 - 4*a*c)^(7/2) + 3*d*e^3*x^2*(b^2 - 4*a*c)^(7/2) - 16*a^3*b^3*c*e^4 + 64*a^4*b*
c^2*e^4 + 640*a^3*c^4*d^3*e - 384*a^4*c^3*d*e^3 - 4*a^2*b^2*e^4*(b^2 - 4*a*c)^(3/2) - 8*b^2*c^2*d^4*(b^2 - 4*a
*c)^(3/2) - 6*b^2*d^2*e^2*(b^2 - 4*a*c)^(5/2) + b^4*d^2*e^2*(b^2 - 4*a*c)^(3/2) - 256*a^2*c^5*d^4*x^2 - 128*a^
4*c^3*e^4*x^2 - 16*b^4*c^3*d^4*x^2 + 80*a^2*b^3*c^2*d^2*e^2 + 96*a^3*b^2*c^2*e^4*x^2 + 640*a^3*c^4*d^2*e^2*x^2
+ 4*b^3*c*d^3*e*(b^2 - 4*a*c)^(3/2) + 4*a*b*e^4*x^2*(b^2 - 4*a*c)^(5/2) + 48*a*b^4*c^2*d^3*e - 16*a*b^5*c*d^2
*e^2 - 4*a*b^3*e^4*x^2*(b^2 - 4*a*c)^(3/2) - 16*b*c^3*d^4*x^2*(b^2 - 4*a*c)^(3/2) - 6*b^2*d*e^3*x^2*(b^2 - 4*a
*c)^(5/2) + 3*b^4*d*e^3*x^2*(b^2 - 4*a*c)^(3/2) + 20*c^2*d^3*e*x^2*(b^2 - 4*a*c)^(5/2) - 352*a^2*b^2*c^3*d^3*e
- 64*a^3*b*c^3*d^2*e^2 + 96*a^3*b^2*c^2*d*e^3 + 128*a*b^2*c^4*d^4*x^2 - 16*a^2*b^4*c*e^4*x^2 + 32*b^5*c^2*d^3
*e*x^2 - 16*b^6*c*d^2*e^2*x^2 - 4*b*c*d^3*e*(b^2 - 4*a*c)^(5/2) - 480*a^2*b^2*c^3*d^2*e^2*x^2 - 12*b*c*d^2*e^2
*x^2*(b^2 - 4*a*c)^(5/2) - 240*a*b^3*c^3*d^3*e*x^2 + 448*a^2*b*c^4*d^3*e*x^2 - 192*a^3*b*c^3*d*e^3*x^2 + 12*b^
2*c^2*d^3*e*x^2*(b^2 - 4*a*c)^(3/2) - 4*b^3*c*d^2*e^2*x^2*(b^2 - 4*a*c)^(3/2) + 144*a*b^4*c^2*d^2*e^2*x^2 + 48
*a^2*b^3*c^2*d*e^3*x^2)*((b^3*d)/4 + e*(a^2*c - (a*b^2)/4 + (a*b*(b^2 - 4*a*c)^(1/2))/4) - (b^2*d*(b^2 - 4*a*c
)^(1/2))/4 + (a*c*d*(b^2 - 4*a*c)^(1/2))/2 - a*b*c*d))/(4*a*c^3*d^2 + 4*a^2*c^2*e^2 - b^2*c^2*d^2 + b^3*c*d*e
- a*b^2*c*e^2 - 4*a*b*c^2*d*e) - (log(4*a^2*e^4*(b^2 - 4*a*c)^(5/2) + 8*c^2*d^4*(b^2 - 4*a*c)^(5/2) + 5*d^2*e^
2*(b^2 - 4*a*c)^(7/2) + 3*d*e^3*x^2*(b^2 - 4*a*c)^(7/2) + 16*a^3*b^3*c*e^4 - 64*a^4*b*c^2*e^4 - 640*a^3*c^4*d^
3*e + 384*a^4*c^3*d*e^3 - 4*a^2*b^2*e^4*(b^2 - 4*a*c)^(3/2) - 8*b^2*c^2*d^4*(b^2 - 4*a*c)^(3/2) - 6*b^2*d^2*e^
2*(b^2 - 4*a*c)^(5/2) + b^4*d^2*e^2*(b^2 - 4*a*c)^(3/2) + 256*a^2*c^5*d^4*x^2 + 128*a^4*c^3*e^4*x^2 + 16*b^4*c
^3*d^4*x^2 - 80*a^2*b^3*c^2*d^2*e^2 - 96*a^3*b^2*c^2*e^4*x^2 - 640*a^3*c^4*d^2*e^2*x^2 + 4*b^3*c*d^3*e*(b^2 -
4*a*c)^(3/2) + 4*a*b*e^4*x^2*(b^2 - 4*a*c)^(5/2) - 48*a*b^4*c^2*d^3*e + 16*a*b^5*c*d^2*e^2 - 4*a*b^3*e^4*x^2*(
b^2 - 4*a*c)^(3/2) - 16*b*c^3*d^4*x^2*(b^2 - 4*a*c)^(3/2) - 6*b^2*d*e^3*x^2*(b^2 - 4*a*c)^(5/2) + 3*b^4*d*e^3*
x^2*(b^2 - 4*a*c)^(3/2) + 20*c^2*d^3*e*x^2*(b^2 - 4*a*c)^(5/2) + 352*a^2*b^2*c^3*d^3*e + 64*a^3*b*c^3*d^2*e^2
- 96*a^3*b^2*c^2*d*e^3 - 128*a*b^2*c^4*d^4*x^2 + 16*a^2*b^4*c*e^4*x^2 - 32*b^5*c^2*d^3*e*x^2 + 16*b^6*c*d^2*e^
2*x^2 - 4*b*c*d^3*e*(b^2 - 4*a*c)^(5/2) + 480*a^2*b^2*c^3*d^2*e^2*x^2 - 12*b*c*d^2*e^2*x^2*(b^2 - 4*a*c)^(5/2)
+ 240*a*b^3*c^3*d^3*e*x^2 - 448*a^2*b*c^4*d^3*e*x^2 + 192*a^3*b*c^3*d*e^3*x^2 + 12*b^2*c^2*d^3*e*x^2*(b^2 - 4
*a*c)^(3/2) - 4*b^3*c*d^2*e^2*x^2*(b^2 - 4*a*c)^(3/2) - 144*a*b^4*c^2*d^2*e^2*x^2 - 48*a^2*b^3*c^2*d*e^3*x^2)*
(e*((a*b^2)/4 - a^2*c + (a*b*(b^2 - 4*a*c)^(1/2))/4) - (b^3*d)/4 - (b^2*d*(b^2 - 4*a*c)^(1/2))/4 + (a*c*d*(b^2
- 4*a*c)^(1/2))/2 + a*b*c*d))/(4*a*c^3*d^2 + 4*a^2*c^2*e^2 - b^2*c^2*d^2 + b^3*c*d*e - a*b^2*c*e^2 - 4*a*b*c^
2*d*e)

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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+b*x**2+a),x)

[Out]

Timed out

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