3.4.44 \(\int \frac {1}{x (a+b x^3)^3} \, dx\) [344]

Optimal. Leaf size=54 \[ \frac {1}{6 a \left (a+b x^3\right )^2}+\frac {1}{3 a^2 \left (a+b x^3\right )}+\frac {\log (x)}{a^3}-\frac {\log \left (a+b x^3\right )}{3 a^3} \]

[Out]

1/6/a/(b*x^3+a)^2+1/3/a^2/(b*x^3+a)+ln(x)/a^3-1/3*ln(b*x^3+a)/a^3

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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 = {272, 46} \begin {gather*} -\frac {\log \left (a+b x^3\right )}{3 a^3}+\frac {\log (x)}{a^3}+\frac {1}{3 a^2 \left (a+b x^3\right )}+\frac {1}{6 a \left (a+b x^3\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^3)^3),x]

[Out]

1/(6*a*(a + b*x^3)^2) + 1/(3*a^2*(a + b*x^3)) + Log[x]/a^3 - Log[a + b*x^3]/(3*a^3)

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 272

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+b x^3\right )^3} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {1}{x (a+b x)^3} \, dx,x,x^3\right )\\ &=\frac {1}{3} \text {Subst}\left (\int \left (\frac {1}{a^3 x}-\frac {b}{a (a+b x)^3}-\frac {b}{a^2 (a+b x)^2}-\frac {b}{a^3 (a+b x)}\right ) \, dx,x,x^3\right )\\ &=\frac {1}{6 a \left (a+b x^3\right )^2}+\frac {1}{3 a^2 \left (a+b x^3\right )}+\frac {\log (x)}{a^3}-\frac {\log \left (a+b x^3\right )}{3 a^3}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 43, normalized size = 0.80 \begin {gather*} \frac {\frac {a \left (3 a+2 b x^3\right )}{\left (a+b x^3\right )^2}+6 \log (x)-2 \log \left (a+b x^3\right )}{6 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^3)^3),x]

[Out]

((a*(3*a + 2*b*x^3))/(a + b*x^3)^2 + 6*Log[x] - 2*Log[a + b*x^3])/(6*a^3)

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Maple [A]
time = 0.14, size = 59, normalized size = 1.09

method	result	size
risch	\(\frac {\frac {b \,x^{3}}{3 a^{2}}+\frac {1}{2 a}}{\left (b \,x^{3}+a \right )^{2}}+\frac {\ln \left (x \right )}{a^{3}}-\frac {\ln \left (b \,x^{3}+a \right )}{3 a^{3}}\)	\(46\)
norman	\(\frac {-\frac {2 b \,x^{3}}{3 a^{2}}-\frac {b^{2} x^{6}}{2 a^{3}}}{\left (b \,x^{3}+a \right )^{2}}+\frac {\ln \left (x \right )}{a^{3}}-\frac {\ln \left (b \,x^{3}+a \right )}{3 a^{3}}\)	\(52\)
default	\(-\frac {b \left (-\frac {a}{b \left (b \,x^{3}+a \right )}-\frac {a^{2}}{2 b \left (b \,x^{3}+a \right )^{2}}+\frac {\ln \left (b \,x^{3}+a \right )}{b}\right )}{3 a^{3}}+\frac {\ln \left (x \right )}{a^{3}}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 60, normalized size = 1.11 \begin {gather*} \frac {2 \, b x^{3} + 3 \, a}{6 \, {\left (a^{2} b^{2} x^{6} + 2 \, a^{3} b x^{3} + a^{4}\right )}} - \frac {\log \left (b x^{3} + a\right )}{3 \, a^{3}} + \frac {\log \left (x^{3}\right )}{3 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 90, normalized size = 1.67 \begin {gather*} \frac {2 \, a b x^{3} + 3 \, a^{2} - 2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )} \log \left (b x^{3} + a\right ) + 6 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )} \log \left (x\right )}{6 \, {\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{3} + a^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a*b*x^3 + 3*a^2 - 2*(b^2*x^6 + 2*a*b*x^3 + a^2)*log(b*x^3 + a) + 6*(b^2*x^6 + 2*a*b*x^3 + a^2)*log(x))/
(a^3*b^2*x^6 + 2*a^4*b*x^3 + a^5)

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Sympy [A]
time = 0.23, size = 56, normalized size = 1.04 \begin {gather*} \frac {3 a + 2 b x^{3}}{6 a^{4} + 12 a^{3} b x^{3} + 6 a^{2} b^{2} x^{6}} + \frac {\log {\left (x \right )}}{a^{3}} - \frac {\log {\left (\frac {a}{b} + x^{3} \right )}}{3 a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*a + 2*b*x**3)/(6*a**4 + 12*a**3*b*x**3 + 6*a**2*b**2*x**6) + log(x)/a**3 - log(a/b + x**3)/(3*a**3)

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Giac [A]
time = 1.55, size = 57, normalized size = 1.06 \begin {gather*} -\frac {\log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{3}} + \frac {\log \left ({\left | x \right |}\right )}{a^{3}} + \frac {3 \, b^{2} x^{6} + 8 \, a b x^{3} + 6 \, a^{2}}{6 \, {\left (b x^{3} + a\right )}^{2} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.07, size = 56, normalized size = 1.04 \begin {gather*} \frac {\ln \left (x\right )}{a^3}+\frac {\frac {1}{2\,a}+\frac {b\,x^3}{3\,a^2}}{a^2+2\,a\,b\,x^3+b^2\,x^6}-\frac {\ln \left (b\,x^3+a\right )}{3\,a^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*x^3)^3),x)

[Out]

log(x)/a^3 + (1/(2*a) + (b*x^3)/(3*a^2))/(a^2 + b^2*x^6 + 2*a*b*x^3) - log(a + b*x^3)/(3*a^3)

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