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

Optimal. Leaf size=96 \[ \frac{e \log \left (d+e x^2\right )}{2 \left (a e^2+c d^2\right )}-\frac{e \log \left (a+c x^4\right )}{4 \left (a e^2+c d^2\right )}+\frac{\sqrt{c} d \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{2 \sqrt{a} \left (a e^2+c d^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.0636733, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1248, 706, 31, 635, 205, 260} \[ \frac{e \log \left (d+e x^2\right )}{2 \left (a e^2+c d^2\right )}-\frac{e \log \left (a+c x^4\right )}{4 \left (a e^2+c d^2\right )}+\frac{\sqrt{c} d \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{2 \sqrt{a} \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x/((d + e*x^2)*(a + c*x^4)),x]

[Out]

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

Rule 1248

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

Rule 706

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

Rule 31

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

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

Mathematica [A]  time = 0.0367347, size = 67, normalized size = 0.7 \[ \frac{\frac{2 \sqrt{c} d \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{\sqrt{a}}-e \log \left (a+c x^4\right )+2 e \log \left (d+e x^2\right )}{4 a e^2+4 c d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 83, normalized size = 0.9 \begin{align*} -{\frac{e\ln \left ( c{x}^{4}+a \right ) }{4\,a{e}^{2}+4\,c{d}^{2}}}+{\frac{cd}{2\,a{e}^{2}+2\,c{d}^{2}}\arctan \left ({c{x}^{2}{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{e\ln \left ( e{x}^{2}+d \right ) }{2\,a{e}^{2}+2\,c{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.81944, size = 319, normalized size = 3.32 \begin{align*} \left [\frac{d \sqrt{-\frac{c}{a}} \log \left (\frac{c x^{4} + 2 \, a x^{2} \sqrt{-\frac{c}{a}} - a}{c x^{4} + a}\right ) - e \log \left (c x^{4} + a\right ) + 2 \, e \log \left (e x^{2} + d\right )}{4 \,{\left (c d^{2} + a e^{2}\right )}}, -\frac{2 \, d \sqrt{\frac{c}{a}} \arctan \left (\frac{a \sqrt{\frac{c}{a}}}{c x^{2}}\right ) + e \log \left (c x^{4} + a\right ) - 2 \, e \log \left (e x^{2} + d\right )}{4 \,{\left (c d^{2} + a e^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.09925, size = 115, normalized size = 1.2 \begin{align*} \frac{c d \arctan \left (\frac{c x^{2}}{\sqrt{a c}}\right )}{2 \,{\left (c d^{2} + a e^{2}\right )} \sqrt{a c}} - \frac{e \log \left (c x^{4} + a\right )}{4 \,{\left (c d^{2} + a e^{2}\right )}} + \frac{e^{2} \log \left ({\left | x^{2} e + d \right |}\right )}{2 \,{\left (c d^{2} e + a e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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