3.230 \(\int \frac{x^7}{(d+e x^2) (a+c x^4)} \, dx\)

Optimal. Leaf size=118 \[ -\frac{a^{3/2} e \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{2 c^{3/2} \left (a e^2+c d^2\right )}-\frac{d^3 \log \left (d+e x^2\right )}{2 e^2 \left (a e^2+c d^2\right )}-\frac{a d \log \left (a+c x^4\right )}{4 c \left (a e^2+c d^2\right )}+\frac{x^2}{2 c e} \]

[Out]

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

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Rubi [A]  time = 0.151951, antiderivative size = 118, 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} \[ -\frac{a^{3/2} e \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{2 c^{3/2} \left (a e^2+c d^2\right )}-\frac{d^3 \log \left (d+e x^2\right )}{2 e^2 \left (a e^2+c d^2\right )}-\frac{a d \log \left (a+c x^4\right )}{4 c \left (a e^2+c d^2\right )}+\frac{x^2}{2 c e} \]

Antiderivative was successfully verified.

[In]

Int[x^7/((d + e*x^2)*(a + c*x^4)),x]

[Out]

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

Mathematica [A]  time = 0.0954535, size = 99, normalized size = 0.84 \[ \frac{-\frac{2 a^{3/2} e^3 \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{c^{3/2}}+\frac{e \left (2 x^2 \left (a e^2+c d^2\right )-a d e \log \left (a+c x^4\right )\right )}{c}-2 d^3 \log \left (d+e x^2\right )}{4 e^2 \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^7/((d + e*x^2)*(a + c*x^4)),x]

[Out]

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

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Maple [A]  time = 0.007, size = 108, normalized size = 0.9 \begin{align*}{\frac{{x}^{2}}{2\,ce}}-{\frac{ad\ln \left ( c{x}^{4}+a \right ) }{4\, \left ( a{e}^{2}+c{d}^{2} \right ) c}}-{\frac{{a}^{2}e}{ \left ( 2\,a{e}^{2}+2\,c{d}^{2} \right ) c}\arctan \left ({c{x}^{2}{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}-{\frac{{d}^{3}\ln \left ( e{x}^{2}+d \right ) }{2\,{e}^{2} \left ( a{e}^{2}+c{d}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(e*x^2+d)/(c*x^4+a),x)

[Out]

1/2*x^2/c/e-1/4*a*d*ln(c*x^4+a)/c/(a*e^2+c*d^2)-1/2*a^2/(a*e^2+c*d^2)/c*e/(a*c)^(1/2)*arctan(c*x^2/(a*c)^(1/2)
)-1/2*d^3*ln(e*x^2+d)/e^2/(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^7/(e*x^2+d)/(c*x^4+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 9.13428, size = 446, normalized size = 3.78 \begin{align*} \left [\frac{a e^{3} \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 d e^{2} \log \left (c x^{4} + a\right ) - 2 \, c d^{3} \log \left (e x^{2} + d\right ) + 2 \,{\left (c d^{2} e + a e^{3}\right )} x^{2}}{4 \,{\left (c^{2} d^{2} e^{2} + a c e^{4}\right )}}, -\frac{2 \, a e^{3} \sqrt{\frac{a}{c}} \arctan \left (\frac{c x^{2} \sqrt{\frac{a}{c}}}{a}\right ) + a d e^{2} \log \left (c x^{4} + a\right ) + 2 \, c d^{3} \log \left (e x^{2} + d\right ) - 2 \,{\left (c d^{2} e + a e^{3}\right )} x^{2}}{4 \,{\left (c^{2} d^{2} e^{2} + a c e^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(e*x^2+d)/(c*x^4+a),x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.08914, size = 142, normalized size = 1.2 \begin{align*} -\frac{d^{3} \log \left ({\left | x^{2} e + d \right |}\right )}{2 \,{\left (c d^{2} e^{2} + a e^{4}\right )}} - \frac{a^{2} \arctan \left (\frac{c x^{2}}{\sqrt{a c}}\right ) e}{2 \,{\left (c^{2} d^{2} + a c e^{2}\right )} \sqrt{a c}} + \frac{x^{2} e^{\left (-1\right )}}{2 \, c} - \frac{a d \log \left (c x^{4} + a\right )}{4 \,{\left (c^{2} d^{2} + a c e^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(e*x^2+d)/(c*x^4+a),x, algorithm="giac")

[Out]

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