3.173 \(\int \frac {1}{x (a^4+x^4)^3} \, dx\)

Optimal. Leaf size=54 \[ \frac {\log (x)}{a^{12}}+\frac {1}{8 a^4 \left (a^4+x^4\right )^2}-\frac {\log \left (a^4+x^4\right )}{4 a^{12}}+\frac {1}{4 a^8 \left (a^4+x^4\right )} \]

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Rubi [A]  time = 0.03, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {266, 44} \[ \frac {1}{4 a^8 \left (a^4+x^4\right )}+\frac {1}{8 a^4 \left (a^4+x^4\right )^2}-\frac {\log \left (a^4+x^4\right )}{4 a^{12}}+\frac {\log (x)}{a^{12}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a^4 + x^4)^3),x]

[Out]

1/(8*a^4*(a^4 + x^4)^2) + 1/(4*a^8*(a^4 + x^4)) + Log[x]/a^12 - Log[a^4 + x^4]/(4*a^12)

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {1}{x \left (a^4+x^4\right )^3} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{x \left (a^4+x\right )^3} \, dx,x,x^4\right )\\ &=\frac {1}{4} \operatorname {Subst}\left (\int \left (\frac {1}{a^{12} x}-\frac {1}{a^4 \left (a^4+x\right )^3}-\frac {1}{a^8 \left (a^4+x\right )^2}-\frac {1}{a^{12} \left (a^4+x\right )}\right ) \, dx,x,x^4\right )\\ &=\frac {1}{8 a^4 \left (a^4+x^4\right )^2}+\frac {1}{4 a^8 \left (a^4+x^4\right )}+\frac {\log (x)}{a^{12}}-\frac {\log \left (a^4+x^4\right )}{4 a^{12}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 46, normalized size = 0.85 \[ \frac {-2 \log \left (a^4+x^4\right )+\frac {3 a^8+2 a^4 x^4}{\left (a^4+x^4\right )^2}+8 \log (x)}{8 a^{12}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a^4 + x^4)^3),x]

[Out]

((3*a^8 + 2*a^4*x^4)/(a^4 + x^4)^2 + 8*Log[x] - 2*Log[a^4 + x^4])/(8*a^12)

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IntegrateAlgebraic [A]  time = 0.04, size = 49, normalized size = 0.91 \[ \frac {\log (x)}{a^{12}}-\frac {\log \left (a^4+x^4\right )}{4 a^{12}}+\frac {3 a^4+2 x^4}{8 a^8 \left (a^4+x^4\right )^2} \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x*(a^4 + x^4)^3),x]

[Out]

(3*a^4 + 2*x^4)/(8*a^8*(a^4 + x^4)^2) + Log[x]/a^12 - Log[a^4 + x^4]/(4*a^12)

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fricas [A]  time = 1.30, size = 81, normalized size = 1.50 \[ \frac {3 \, a^{8} + 2 \, a^{4} x^{4} - 2 \, {\left (a^{8} + 2 \, a^{4} x^{4} + x^{8}\right )} \log \left (a^{4} + x^{4}\right ) + 8 \, {\left (a^{8} + 2 \, a^{4} x^{4} + x^{8}\right )} \log \relax (x)}{8 \, {\left (a^{20} + 2 \, a^{16} x^{4} + a^{12} x^{8}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a^4+x^4)^3,x, algorithm="fricas")

[Out]

1/8*(3*a^8 + 2*a^4*x^4 - 2*(a^8 + 2*a^4*x^4 + x^8)*log(a^4 + x^4) + 8*(a^8 + 2*a^4*x^4 + x^8)*log(x))/(a^20 +
2*a^16*x^4 + a^12*x^8)

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giac [A]  time = 0.94, size = 56, normalized size = 1.04 \[ -\frac {\log \left (a^{4} + x^{4}\right )}{4 \, a^{12}} + \frac {\log \left (x^{4}\right )}{4 \, a^{12}} + \frac {6 \, a^{8} + 8 \, a^{4} x^{4} + 3 \, x^{8}}{8 \, {\left (a^{4} + x^{4}\right )}^{2} a^{12}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a^4+x^4)^3,x, algorithm="giac")

[Out]

-1/4*log(a^4 + x^4)/a^12 + 1/4*log(x^4)/a^12 + 1/8*(6*a^8 + 8*a^4*x^4 + 3*x^8)/((a^4 + x^4)^2*a^12)

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maple [A]  time = 0.27, size = 45, normalized size = 0.83




method result size



norman \(\frac {\frac {3}{8 a^{4}}+\frac {x^{4}}{4 a^{8}}}{\left (a^{4}+x^{4}\right )^{2}}+\frac {\ln \relax (x )}{a^{12}}-\frac {\ln \left (a^{4}+x^{4}\right )}{4 a^{12}}\) \(45\)
risch \(\frac {\frac {3}{8 a^{4}}+\frac {x^{4}}{4 a^{8}}}{\left (a^{4}+x^{4}\right )^{2}}+\frac {\ln \relax (x )}{a^{12}}-\frac {\ln \left (a^{4}+x^{4}\right )}{4 a^{12}}\) \(45\)
default \(-\frac {-\frac {a^{4}}{2 \left (a^{4}+x^{4}\right )}+\frac {\ln \left (a^{4}+x^{4}\right )}{2}-\frac {a^{8}}{4 \left (a^{4}+x^{4}\right )^{2}}}{2 a^{12}}+\frac {\ln \relax (x )}{a^{12}}\) \(52\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(a^4+x^4)^3,x,method=_RETURNVERBOSE)

[Out]

(3/8/a^4+1/4/a^8*x^4)/(a^4+x^4)^2+ln(x)/a^12-1/4*ln(a^4+x^4)/a^12

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maxima [A]  time = 0.45, size = 57, normalized size = 1.06 \[ \frac {3 \, a^{4} + 2 \, x^{4}}{8 \, {\left (a^{16} + 2 \, a^{12} x^{4} + a^{8} x^{8}\right )}} - \frac {\log \left (a^{4} + x^{4}\right )}{4 \, a^{12}} + \frac {\log \left (x^{4}\right )}{4 \, a^{12}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a^4+x^4)^3,x, algorithm="maxima")

[Out]

1/8*(3*a^4 + 2*x^4)/(a^16 + 2*a^12*x^4 + a^8*x^8) - 1/4*log(a^4 + x^4)/a^12 + 1/4*log(x^4)/a^12

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mupad [B]  time = 0.11, size = 52, normalized size = 0.96 \[ \frac {\ln \relax (x)}{a^{12}}+\frac {\frac {3}{8\,a^4}+\frac {x^4}{4\,a^8}}{a^8+2\,a^4\,x^4+x^8}-\frac {\ln \left (a^4+x^4\right )}{4\,a^{12}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a^4 + x^4)^3),x)

[Out]

log(x)/a^12 + (3/(8*a^4) + x^4/(4*a^8))/(a^8 + x^8 + 2*a^4*x^4) - log(a^4 + x^4)/(4*a^12)

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sympy [A]  time = 0.46, size = 51, normalized size = 0.94 \[ \frac {3 a^{4} + 2 x^{4}}{8 a^{16} + 16 a^{12} x^{4} + 8 a^{8} x^{8}} + \frac {\log {\relax (x )}}{a^{12}} - \frac {\log {\left (a^{4} + x^{4} \right )}}{4 a^{12}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a**4+x**4)**3,x)

[Out]

(3*a**4 + 2*x**4)/(8*a**16 + 16*a**12*x**4 + 8*a**8*x**8) + log(x)/a**12 - log(a**4 + x**4)/(4*a**12)

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