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

Optimal. Leaf size=114 \[ -\frac{e^2 \log \left (d+e x^2\right )}{2 d \left (a e^2+c d^2\right )}-\frac{c d \log \left (a+c x^4\right )}{4 a \left (a e^2+c d^2\right )}-\frac{\sqrt{c} e \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{2 \sqrt{a} \left (a e^2+c d^2\right )}+\frac{\log (x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.124906, antiderivative size = 114, 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} \[ -\frac{e^2 \log \left (d+e x^2\right )}{2 d \left (a e^2+c d^2\right )}-\frac{c d \log \left (a+c x^4\right )}{4 a \left (a e^2+c d^2\right )}-\frac{\sqrt{c} e \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{2 \sqrt{a} \left (a e^2+c d^2\right )}+\frac{\log (x)}{a d} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(d + e*x^2)*(a + c*x^4)),x]

[Out]

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

Mathematica [A]  time = 0.0710177, size = 134, normalized size = 1.18 \[ \frac{-c d^2 \log \left (a+c x^4\right )+2 \sqrt{a} \sqrt{c} d e \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+2 \sqrt{a} \sqrt{c} d e \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )-2 a e^2 \log \left (d+e x^2\right )+4 a e^2 \log (x)+4 c d^2 \log (x)}{4 a^2 d e^2+4 a c d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 101, normalized size = 0.9 \begin{align*} -{\frac{cd\ln \left ( c{x}^{4}+a \right ) }{4\, \left ( a{e}^{2}+c{d}^{2} \right ) a}}-{\frac{ec}{2\,a{e}^{2}+2\,c{d}^{2}}\arctan \left ({c{x}^{2}{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{\ln \left ( x \right ) }{ad}}-{\frac{{e}^{2}\ln \left ( e{x}^{2}+d \right ) }{2\,d \left ( a{e}^{2}+c{d}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 36.2875, size = 439, normalized size = 3.85 \begin{align*} \left [\frac{a d e \sqrt{-\frac{c}{a}} \log \left (\frac{c x^{4} - 2 \, a x^{2} \sqrt{-\frac{c}{a}} - a}{c x^{4} + a}\right ) - c d^{2} \log \left (c x^{4} + a\right ) - 2 \, a e^{2} \log \left (e x^{2} + d\right ) + 4 \,{\left (c d^{2} + a e^{2}\right )} \log \left (x\right )}{4 \,{\left (a c d^{3} + a^{2} d e^{2}\right )}}, \frac{2 \, a d e \sqrt{\frac{c}{a}} \arctan \left (\frac{a \sqrt{\frac{c}{a}}}{c x^{2}}\right ) - c d^{2} \log \left (c x^{4} + a\right ) - 2 \, a e^{2} \log \left (e x^{2} + d\right ) + 4 \,{\left (c d^{2} + a e^{2}\right )} \log \left (x\right )}{4 \,{\left (a c d^{3} + a^{2} d e^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.0956, size = 138, normalized size = 1.21 \begin{align*} -\frac{c d \log \left (c x^{4} + a\right )}{4 \,{\left (a c d^{2} + a^{2} e^{2}\right )}} - \frac{c \arctan \left (\frac{c x^{2}}{\sqrt{a c}}\right ) e}{2 \,{\left (c d^{2} + a e^{2}\right )} \sqrt{a c}} - \frac{e^{3} \log \left ({\left | x^{2} e + d \right |}\right )}{2 \,{\left (c d^{3} e + a d e^{3}\right )}} + \frac{\log \left (x^{2}\right )}{2 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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