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

Optimal. Leaf size=105 \[ \frac{d^2 \log \left (d+e x^2\right )}{2 e \left (a e^2+c d^2\right )}+\frac{a e \log \left (a+c x^4\right )}{4 c \left (a e^2+c d^2\right )}-\frac{\sqrt{a} d \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{2 \sqrt{c} \left (a e^2+c d^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.136745, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {1252, 1629, 635, 205, 260} \[ \frac{d^2 \log \left (d+e x^2\right )}{2 e \left (a e^2+c d^2\right )}+\frac{a e \log \left (a+c x^4\right )}{4 c \left (a e^2+c d^2\right )}-\frac{\sqrt{a} d \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{2 \sqrt{c} \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^5/((d + e*x^2)*(a + c*x^4)),x]

[Out]

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

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 1629

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0347916, size = 77, normalized size = 0.73 \[ \frac{-2 \sqrt{a} \sqrt{c} d e \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )+a e^2 \log \left (a+c x^4\right )+2 c d^2 \log \left (d+e x^2\right )}{4 a c e^3+4 c^2 d^2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 92, normalized size = 0.9 \begin{align*}{\frac{ae\ln \left ( c{x}^{4}+a \right ) }{4\,c \left ( a{e}^{2}+c{d}^{2} \right ) }}-{\frac{ad}{2\,a{e}^{2}+2\,c{d}^{2}}\arctan \left ({c{x}^{2}{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{{d}^{2}\ln \left ( e{x}^{2}+d \right ) }{2\,e \left ( a{e}^{2}+c{d}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 5.37463, size = 365, normalized size = 3.48 \begin{align*} \left [\frac{c d e \sqrt{-\frac{a}{c}} \log \left (\frac{c x^{4} - 2 \, c x^{2} \sqrt{-\frac{a}{c}} - a}{c x^{4} + a}\right ) + a e^{2} \log \left (c x^{4} + a\right ) + 2 \, c d^{2} \log \left (e x^{2} + d\right )}{4 \,{\left (c^{2} d^{2} e + a c e^{3}\right )}}, -\frac{2 \, c d e \sqrt{\frac{a}{c}} \arctan \left (\frac{c x^{2} \sqrt{\frac{a}{c}}}{a}\right ) - a e^{2} \log \left (c x^{4} + a\right ) - 2 \, c d^{2} \log \left (e x^{2} + d\right )}{4 \,{\left (c^{2} d^{2} e + a c e^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.1149, size = 122, normalized size = 1.16 \begin{align*} \frac{a e \log \left (c x^{4} + a\right )}{4 \,{\left (c^{2} d^{2} + a c e^{2}\right )}} + \frac{d^{2} \log \left ({\left | x^{2} e + d \right |}\right )}{2 \,{\left (c d^{2} e + a e^{3}\right )}} - \frac{a d \arctan \left (\frac{c x^{2}}{\sqrt{a c}}\right )}{2 \,{\left (c d^{2} + a e^{2}\right )} \sqrt{a c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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