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

Optimal. Leaf size=54 \[ -\frac {\log \left (a+c x^4\right )}{4 a^3}+\frac {\log (x)}{a^3}+\frac {1}{4 a^2 \left (a+c x^4\right )}+\frac {1}{8 a \left (a+c x^4\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.04, 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} \[ \frac {1}{4 a^2 \left (a+c x^4\right )}-\frac {\log \left (a+c x^4\right )}{4 a^3}+\frac {\log (x)}{a^3}+\frac {1}{8 a \left (a+c x^4\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + c*x^4)^3),x]

[Out]

1/(8*a*(a + c*x^4)^2) + 1/(4*a^2*(a + c*x^4)) + Log[x]/a^3 - Log[a + c*x^4]/(4*a^3)

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

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Mathematica [A]  time = 0.04, size = 43, normalized size = 0.80 \[ \frac {\frac {a \left (3 a+2 c x^4\right )}{\left (a+c x^4\right )^2}-2 \log \left (a+c x^4\right )+8 \log (x)}{8 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + c*x^4)^3),x]

[Out]

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

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fricas [A]  time = 0.51, size = 90, normalized size = 1.67 \[ \frac {2 \, a c x^{4} + 3 \, a^{2} - 2 \, {\left (c^{2} x^{8} + 2 \, a c x^{4} + a^{2}\right )} \log \left (c x^{4} + a\right ) + 8 \, {\left (c^{2} x^{8} + 2 \, a c x^{4} + a^{2}\right )} \log \relax (x)}{8 \, {\left (a^{3} c^{2} x^{8} + 2 \, a^{4} c x^{4} + a^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*a*c*x^4 + 3*a^2 - 2*(c^2*x^8 + 2*a*c*x^4 + a^2)*log(c*x^4 + a) + 8*(c^2*x^8 + 2*a*c*x^4 + a^2)*log(x))/
(a^3*c^2*x^8 + 2*a^4*c*x^4 + a^5)

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giac [A]  time = 0.16, size = 59, normalized size = 1.09 \[ \frac {\log \left (x^{4}\right )}{4 \, a^{3}} - \frac {\log \left ({\left | c x^{4} + a \right |}\right )}{4 \, a^{3}} + \frac {3 \, c^{2} x^{8} + 8 \, a c x^{4} + 6 \, a^{2}}{8 \, {\left (c x^{4} + a\right )}^{2} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 49, normalized size = 0.91 \[ \frac {1}{8 \left (c \,x^{4}+a \right )^{2} a}+\frac {1}{4 \left (c \,x^{4}+a \right ) a^{2}}+\frac {\ln \relax (x )}{a^{3}}-\frac {\ln \left (c \,x^{4}+a \right )}{4 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(c*x^4+a)^3,x)

[Out]

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

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maxima [A]  time = 1.36, size = 60, normalized size = 1.11 \[ \frac {2 \, c x^{4} + 3 \, a}{8 \, {\left (a^{2} c^{2} x^{8} + 2 \, a^{3} c x^{4} + a^{4}\right )}} - \frac {\log \left (c x^{4} + a\right )}{4 \, a^{3}} + \frac {\log \left (x^{4}\right )}{4 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.04, size = 56, normalized size = 1.04 \[ \frac {\ln \relax (x)}{a^3}+\frac {\frac {3}{8\,a}+\frac {c\,x^4}{4\,a^2}}{a^2+2\,a\,c\,x^4+c^2\,x^8}-\frac {\ln \left (c\,x^4+a\right )}{4\,a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + c*x^4)^3),x)

[Out]

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

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sympy [A]  time = 0.78, size = 56, normalized size = 1.04 \[ \frac {3 a + 2 c x^{4}}{8 a^{4} + 16 a^{3} c x^{4} + 8 a^{2} c^{2} x^{8}} + \frac {\log {\relax (x )}}{a^{3}} - \frac {\log {\left (\frac {a}{c} + x^{4} \right )}}{4 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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