3.236 \(\int \frac{1}{x^5 (d+e x^2) (a+c x^4)} \, dx\)

Optimal. Leaf size=156 \[ \frac{c^2 d \log \left (a+c x^4\right )}{4 a^2 \left (a e^2+c d^2\right )}+\frac{c^{3/2} e \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{2 a^{3/2} \left (a e^2+c d^2\right )}-\frac{\log (x) \left (c d^2-a e^2\right )}{a^2 d^3}-\frac{e^4 \log \left (d+e x^2\right )}{2 d^3 \left (a e^2+c d^2\right )}+\frac{e}{2 a d^2 x^2}-\frac{1}{4 a d x^4} \]

[Out]

-1/(4*a*d*x^4) + e/(2*a*d^2*x^2) + (c^(3/2)*e*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(2*a^(3/2)*(c*d^2 + a*e^2)) - ((c
*d^2 - a*e^2)*Log[x])/(a^2*d^3) - (e^4*Log[d + e*x^2])/(2*d^3*(c*d^2 + a*e^2)) + (c^2*d*Log[a + c*x^4])/(4*a^2
*(c*d^2 + a*e^2))

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Rubi [A]  time = 0.183466, antiderivative size = 156, 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, 894, 635, 205, 260} \[ \frac{c^2 d \log \left (a+c x^4\right )}{4 a^2 \left (a e^2+c d^2\right )}+\frac{c^{3/2} e \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{2 a^{3/2} \left (a e^2+c d^2\right )}-\frac{\log (x) \left (c d^2-a e^2\right )}{a^2 d^3}-\frac{e^4 \log \left (d+e x^2\right )}{2 d^3 \left (a e^2+c d^2\right )}+\frac{e}{2 a d^2 x^2}-\frac{1}{4 a d x^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(d + e*x^2)*(a + c*x^4)),x]

[Out]

-1/(4*a*d*x^4) + e/(2*a*d^2*x^2) + (c^(3/2)*e*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(2*a^(3/2)*(c*d^2 + a*e^2)) - ((c
*d^2 - a*e^2)*Log[x])/(a^2*d^3) - (e^4*Log[d + e*x^2])/(2*d^3*(c*d^2 + a*e^2)) + (c^2*d*Log[a + c*x^4])/(4*a^2
*(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 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

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{1}{x^5 \left (d+e x^2\right ) \left (a+c x^4\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{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}{a d x^3}-\frac{e}{a d^2 x^2}+\frac{-c d^2+a e^2}{a^2 d^3 x}-\frac{e^5}{d^3 \left (c d^2+a e^2\right ) (d+e x)}+\frac{c^2 (a e+c d x)}{a^2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac{1}{4 a d x^4}+\frac{e}{2 a d^2 x^2}-\frac{\left (c d^2-a e^2\right ) \log (x)}{a^2 d^3}-\frac{e^4 \log \left (d+e x^2\right )}{2 d^3 \left (c d^2+a e^2\right )}+\frac{c^2 \operatorname{Subst}\left (\int \frac{a e+c d x}{a+c x^2} \, dx,x,x^2\right )}{2 a^2 \left (c d^2+a e^2\right )}\\ &=-\frac{1}{4 a d x^4}+\frac{e}{2 a d^2 x^2}-\frac{\left (c d^2-a e^2\right ) \log (x)}{a^2 d^3}-\frac{e^4 \log \left (d+e x^2\right )}{2 d^3 \left (c d^2+a e^2\right )}+\frac{\left (c^3 d\right ) \operatorname{Subst}\left (\int \frac{x}{a+c x^2} \, dx,x,x^2\right )}{2 a^2 \left (c d^2+a e^2\right )}+\frac{\left (c^2 e\right ) \operatorname{Subst}\left (\int \frac{1}{a+c x^2} \, dx,x,x^2\right )}{2 a \left (c d^2+a e^2\right )}\\ &=-\frac{1}{4 a d x^4}+\frac{e}{2 a d^2 x^2}+\frac{c^{3/2} e \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{2 a^{3/2} \left (c d^2+a e^2\right )}-\frac{\left (c d^2-a e^2\right ) \log (x)}{a^2 d^3}-\frac{e^4 \log \left (d+e x^2\right )}{2 d^3 \left (c d^2+a e^2\right )}+\frac{c^2 d \log \left (a+c x^4\right )}{4 a^2 \left (c d^2+a e^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.0946543, size = 209, normalized size = 1.34 \[ -\frac{a^2 d^2 e^2-2 a^2 d e^3 x^2+2 a^2 e^4 x^4 \log \left (d+e x^2\right )-4 a^2 e^4 x^4 \log (x)+2 \sqrt{a} c^{3/2} d^3 e x^4 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+2 \sqrt{a} c^{3/2} d^3 e x^4 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )-c^2 d^4 x^4 \log \left (a+c x^4\right )-2 a c d^3 e x^2+a c d^4+4 c^2 d^4 x^4 \log (x)}{4 a^2 d^3 x^4 \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(d + e*x^2)*(a + c*x^4)),x]

[Out]

-(a*c*d^4 + a^2*d^2*e^2 - 2*a*c*d^3*e*x^2 - 2*a^2*d*e^3*x^2 + 2*Sqrt[a]*c^(3/2)*d^3*e*x^4*ArcTan[1 - (Sqrt[2]*
c^(1/4)*x)/a^(1/4)] + 2*Sqrt[a]*c^(3/2)*d^3*e*x^4*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)] + 4*c^2*d^4*x^4*Log[
x] - 4*a^2*e^4*x^4*Log[x] + 2*a^2*e^4*x^4*Log[d + e*x^2] - c^2*d^4*x^4*Log[a + c*x^4])/(4*a^2*d^3*(c*d^2 + a*e
^2)*x^4)

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Maple [A]  time = 0.013, size = 145, normalized size = 0.9 \begin{align*}{\frac{{c}^{2}d\ln \left ( c{x}^{4}+a \right ) }{4\, \left ( a{e}^{2}+c{d}^{2} \right ){a}^{2}}}+{\frac{{c}^{2}e}{ \left ( 2\,a{e}^{2}+2\,c{d}^{2} \right ) a}\arctan \left ({c{x}^{2}{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}-{\frac{1}{4\,ad{x}^{4}}}+{\frac{\ln \left ( x \right ){e}^{2}}{{d}^{3}a}}-{\frac{\ln \left ( x \right ) c}{d{a}^{2}}}+{\frac{e}{2\,{d}^{2}a{x}^{2}}}-{\frac{{e}^{4}\ln \left ( e{x}^{2}+d \right ) }{2\,{d}^{3} \left ( a{e}^{2}+c{d}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(e*x^2+d)/(c*x^4+a),x)

[Out]

1/4*c^2*d*ln(c*x^4+a)/a^2/(a*e^2+c*d^2)+1/2*c^2/(a*e^2+c*d^2)/a*e/(a*c)^(1/2)*arctan(c*x^2/(a*c)^(1/2))-1/4/a/
d/x^4+1/d^3/a*ln(x)*e^2-1/d/a^2*ln(x)*c+1/2*e/a/d^2/x^2-1/2*e^4*ln(e*x^2+d)/d^3/(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(1/x^5/(e*x^2+d)/(c*x^4+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [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(1/x^5/(e*x^2+d)/(c*x^4+a),x, algorithm="fricas")

[Out]

Timed out

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

[Out]

Timed out

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Giac [A]  time = 1.09408, size = 227, normalized size = 1.46 \begin{align*} \frac{c^{2} d \log \left (c x^{4} + a\right )}{4 \,{\left (a^{2} c d^{2} + a^{3} e^{2}\right )}} + \frac{c^{2} \arctan \left (\frac{c x^{2}}{\sqrt{a c}}\right ) e}{2 \,{\left (a c d^{2} + a^{2} e^{2}\right )} \sqrt{a c}} - \frac{e^{5} \log \left ({\left | x^{2} e + d \right |}\right )}{2 \,{\left (c d^{5} e + a d^{3} e^{3}\right )}} - \frac{{\left (c d^{2} - a e^{2}\right )} \log \left (x^{2}\right )}{2 \, a^{2} d^{3}} + \frac{3 \, c d^{2} x^{4} - 3 \, a x^{4} e^{2} + 2 \, a d x^{2} e - a d^{2}}{4 \, a^{2} d^{3} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(e*x^2+d)/(c*x^4+a),x, algorithm="giac")

[Out]

1/4*c^2*d*log(c*x^4 + a)/(a^2*c*d^2 + a^3*e^2) + 1/2*c^2*arctan(c*x^2/sqrt(a*c))*e/((a*c*d^2 + a^2*e^2)*sqrt(a
*c)) - 1/2*e^5*log(abs(x^2*e + d))/(c*d^5*e + a*d^3*e^3) - 1/2*(c*d^2 - a*e^2)*log(x^2)/(a^2*d^3) + 1/4*(3*c*d
^2*x^4 - 3*a*x^4*e^2 + 2*a*d*x^2*e - a*d^2)/(a^2*d^3*x^4)