3.332 \(\int \frac{1}{x (a+b x^3)^2} \, dx\)

Optimal. Leaf size=38 \[ -\frac{\log \left (a+b x^3\right )}{3 a^2}+\frac{\log (x)}{a^2}+\frac{1}{3 a \left (a+b x^3\right )} \]

[Out]

1/(3*a*(a + b*x^3)) + Log[x]/a^2 - Log[a + b*x^3]/(3*a^2)

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Rubi [A]  time = 0.0258655, antiderivative size = 38, normalized size of antiderivative = 1., 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} \[ -\frac{\log \left (a+b x^3\right )}{3 a^2}+\frac{\log (x)}{a^2}+\frac{1}{3 a \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^3)^2),x]

[Out]

1/(3*a*(a + b*x^3)) + Log[x]/a^2 - Log[a + b*x^3]/(3*a^2)

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]]

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])

Rubi steps

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

Mathematica [A]  time = 0.01231, size = 33, normalized size = 0.87 \[ \frac{\frac{a}{a+b x^3}-\log \left (a+b x^3\right )+3 \log (x)}{3 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^3)^2),x]

[Out]

(a/(a + b*x^3) + 3*Log[x] - Log[a + b*x^3])/(3*a^2)

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Maple [A]  time = 0.01, size = 35, normalized size = 0.9 \begin{align*}{\frac{1}{3\,a \left ( b{x}^{3}+a \right ) }}+{\frac{\ln \left ( x \right ) }{{a}^{2}}}-{\frac{\ln \left ( b{x}^{3}+a \right ) }{3\,{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(b*x^3+a)^2,x)

[Out]

1/3/a/(b*x^3+a)+ln(x)/a^2-1/3*ln(b*x^3+a)/a^2

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Maxima [A]  time = 0.944508, size = 50, normalized size = 1.32 \begin{align*} \frac{1}{3 \,{\left (a b x^{3} + a^{2}\right )}} - \frac{\log \left (b x^{3} + a\right )}{3 \, a^{2}} + \frac{\log \left (x^{3}\right )}{3 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

1/3/(a*b*x^3 + a^2) - 1/3*log(b*x^3 + a)/a^2 + 1/3*log(x^3)/a^2

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Fricas [A]  time = 1.7356, size = 108, normalized size = 2.84 \begin{align*} -\frac{{\left (b x^{3} + a\right )} \log \left (b x^{3} + a\right ) - 3 \,{\left (b x^{3} + a\right )} \log \left (x\right ) - a}{3 \,{\left (a^{2} b x^{3} + a^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

-1/3*((b*x^3 + a)*log(b*x^3 + a) - 3*(b*x^3 + a)*log(x) - a)/(a^2*b*x^3 + a^3)

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Sympy [A]  time = 0.671214, size = 34, normalized size = 0.89 \begin{align*} \frac{1}{3 a^{2} + 3 a b x^{3}} + \frac{\log{\left (x \right )}}{a^{2}} - \frac{\log{\left (\frac{a}{b} + x^{3} \right )}}{3 a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x**3+a)**2,x)

[Out]

1/(3*a**2 + 3*a*b*x**3) + log(x)/a**2 - log(a/b + x**3)/(3*a**2)

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Giac [A]  time = 1.12494, size = 61, normalized size = 1.61 \begin{align*} -\frac{\log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{2}} + \frac{\log \left ({\left | x \right |}\right )}{a^{2}} + \frac{b x^{3} + 2 \, a}{3 \,{\left (b x^{3} + a\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x^3+a)^2,x, algorithm="giac")

[Out]

-1/3*log(abs(b*x^3 + a))/a^2 + log(abs(x))/a^2 + 1/3*(b*x^3 + 2*a)/((b*x^3 + a)*a^2)