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

Optimal. Leaf size=22 \[ \frac {\log (x)}{a}-\frac {\log \left (a+b x^3\right )}{3 a} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {266, 36, 29, 31} \[ \frac {\log (x)}{a}-\frac {\log \left (a+b x^3\right )}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^3)),x]

[Out]

Log[x]/a - Log[a + b*x^3]/(3*a)

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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+b x^3\right )} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x (a+b x)} \, dx,x,x^3\right )\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^3\right )}{3 a}-\frac {b \operatorname {Subst}\left (\int \frac {1}{a+b x} \, dx,x,x^3\right )}{3 a}\\ &=\frac {\log (x)}{a}-\frac {\log \left (a+b x^3\right )}{3 a}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.00 \[ \frac {\log (x)}{a}-\frac {\log \left (a+b x^3\right )}{3 a} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^3)),x]

[Out]

Log[x]/a - Log[a + b*x^3]/(3*a)

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fricas [A]  time = 0.61, size = 18, normalized size = 0.82 \[ -\frac {\log \left (b x^{3} + a\right ) - 3 \, \log \relax (x)}{3 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(log(b*x^3 + a) - 3*log(x))/a

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giac [A]  time = 0.15, size = 22, normalized size = 1.00 \[ -\frac {\log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a} + \frac {\log \left ({\left | x \right |}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 21, normalized size = 0.95 \[ \frac {\ln \relax (x )}{a}-\frac {\ln \left (b \,x^{3}+a \right )}{3 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.32, size = 23, normalized size = 1.05 \[ -\frac {\log \left (b x^{3} + a\right )}{3 \, a} + \frac {\log \left (x^{3}\right )}{3 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.00, size = 18, normalized size = 0.82 \[ -\frac {\ln \left (b\,x^3+a\right )-3\,\ln \relax (x)}{3\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*x^3)),x)

[Out]

-(log(a + b*x^3) - 3*log(x))/(3*a)

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sympy [A]  time = 0.29, size = 15, normalized size = 0.68 \[ \frac {\log {\relax (x )}}{a} - \frac {\log {\left (\frac {a}{b} + x^{3} \right )}}{3 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)/a - log(a/b + x**3)/(3*a)

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