3.232 \(\int \frac{x^3}{(d+e x^2) (a+c x^4)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0944008, antiderivative size = 96, 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, 801, 635, 205, 260} \[ -\frac{d \log \left (d+e x^2\right )}{2 \left (a e^2+c d^2\right )}+\frac{d \log \left (a+c x^4\right )}{4 \left (a e^2+c d^2\right )}+\frac{\sqrt{a} e \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{2 \sqrt{c} \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^3/((d + e*x^2)*(a + c*x^4)),x]

[Out]

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

Mathematica [A]  time = 0.0392544, size = 66, normalized size = 0.69 \[ \frac{d \log \left (a+c x^4\right )+\frac{2 \sqrt{a} e \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{\sqrt{c}}-2 d \log \left (d+e x^2\right )}{4 a e^2+4 c d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/((d + e*x^2)*(a + c*x^4)),x]

[Out]

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

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Maple [A]  time = 0.007, size = 83, normalized size = 0.9 \begin{align*}{\frac{d\ln \left ( c{x}^{4}+a \right ) }{4\,a{e}^{2}+4\,c{d}^{2}}}+{\frac{ae}{2\,a{e}^{2}+2\,c{d}^{2}}\arctan \left ({c{x}^{2}{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}-{\frac{d\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^3/(e*x^2+d)/(c*x^4+a),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.3, size = 315, normalized size = 3.28 \begin{align*} \left [\frac{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 ) + d \log \left (c x^{4} + a\right ) - 2 \, d \log \left (e x^{2} + d\right )}{4 \,{\left (c d^{2} + a e^{2}\right )}}, \frac{2 \, e \sqrt{\frac{a}{c}} \arctan \left (\frac{c x^{2} \sqrt{\frac{a}{c}}}{a}\right ) + d \log \left (c x^{4} + a\right ) - 2 \, d \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^3/(e*x^2+d)/(c*x^4+a),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 144.776, size = 932, normalized size = 9.71 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(e*x**2+d)/(c*x**4+a),x)

[Out]

-d*log(x**2 + (16*a**2*c*d**3*e**4/(a*e**2 + c*d**2)**2 - 2*a**2*d*e**4/(a*e**2 + c*d**2) + 32*a*c**2*d**5*e**
2/(a*e**2 + c*d**2)**2 - 4*a*c*d**3*e**2/(a*e**2 + c*d**2) + 3*a*d*e**2 + 16*c**3*d**7/(a*e**2 + c*d**2)**2 -
2*c**2*d**5/(a*e**2 + c*d**2) - 5*c*d**3)/(a*e**3 + 9*c*d**2*e))/(2*(a*e**2 + c*d**2)) + (d/(4*(a*e**2 + c*d**
2)) - e*sqrt(-a*c)/(4*c*(a*e**2 + c*d**2)))*log(x**2 + (64*a**2*c*d*e**4*(d/(4*(a*e**2 + c*d**2)) - e*sqrt(-a*
c)/(4*c*(a*e**2 + c*d**2)))**2 + 4*a**2*e**4*(d/(4*(a*e**2 + c*d**2)) - e*sqrt(-a*c)/(4*c*(a*e**2 + c*d**2)))
+ 128*a*c**2*d**3*e**2*(d/(4*(a*e**2 + c*d**2)) - e*sqrt(-a*c)/(4*c*(a*e**2 + c*d**2)))**2 + 8*a*c*d**2*e**2*(
d/(4*(a*e**2 + c*d**2)) - e*sqrt(-a*c)/(4*c*(a*e**2 + c*d**2))) + 3*a*d*e**2 + 64*c**3*d**5*(d/(4*(a*e**2 + c*
d**2)) - e*sqrt(-a*c)/(4*c*(a*e**2 + c*d**2)))**2 + 4*c**2*d**4*(d/(4*(a*e**2 + c*d**2)) - e*sqrt(-a*c)/(4*c*(
a*e**2 + c*d**2))) - 5*c*d**3)/(a*e**3 + 9*c*d**2*e)) + (d/(4*(a*e**2 + c*d**2)) + e*sqrt(-a*c)/(4*c*(a*e**2 +
c*d**2)))*log(x**2 + (64*a**2*c*d*e**4*(d/(4*(a*e**2 + c*d**2)) + e*sqrt(-a*c)/(4*c*(a*e**2 + c*d**2)))**2 +
4*a**2*e**4*(d/(4*(a*e**2 + c*d**2)) + e*sqrt(-a*c)/(4*c*(a*e**2 + c*d**2))) + 128*a*c**2*d**3*e**2*(d/(4*(a*e
**2 + c*d**2)) + e*sqrt(-a*c)/(4*c*(a*e**2 + c*d**2)))**2 + 8*a*c*d**2*e**2*(d/(4*(a*e**2 + c*d**2)) + e*sqrt(
-a*c)/(4*c*(a*e**2 + c*d**2))) + 3*a*d*e**2 + 64*c**3*d**5*(d/(4*(a*e**2 + c*d**2)) + e*sqrt(-a*c)/(4*c*(a*e**
2 + c*d**2)))**2 + 4*c**2*d**4*(d/(4*(a*e**2 + c*d**2)) + e*sqrt(-a*c)/(4*c*(a*e**2 + c*d**2))) - 5*c*d**3)/(a
*e**3 + 9*c*d**2*e))

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Giac [A]  time = 1.11379, size = 116, normalized size = 1.21 \begin{align*} -\frac{d e \log \left ({\left | x^{2} e + d \right |}\right )}{2 \,{\left (c d^{2} e + a e^{3}\right )}} + \frac{a \arctan \left (\frac{c x^{2}}{\sqrt{a c}}\right ) e}{2 \,{\left (c d^{2} + a e^{2}\right )} \sqrt{a c}} + \frac{d \log \left (c x^{4} + a\right )}{4 \,{\left (c d^{2} + a e^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(e*x^2+d)/(c*x^4+a),x, algorithm="giac")

[Out]

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