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

Optimal. Leaf size=115 \[ -\frac {\log \left (\sqrt [3]{b} \sqrt [3]{a x^3+b}+\left (a x^3+b\right )^{2/3}+b^{2/3}\right )}{6 \sqrt [3]{b}}+\frac {\log \left (\sqrt [3]{a x^3+b}-\sqrt [3]{b}\right )}{3 \sqrt [3]{b}}+\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{a x^3+b}}{\sqrt {3} \sqrt [3]{b}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}} \]

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Rubi [A]  time = 0.06, antiderivative size = 83, normalized size of antiderivative = 0.72, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {266, 55, 617, 204, 31} \begin {gather*} \frac {\log \left (\sqrt [3]{b}-\sqrt [3]{a x^3+b}\right )}{2 \sqrt [3]{b}}+\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{a x^3+b}+\sqrt [3]{b}}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {\log (x)}{2 \sqrt [3]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(b + a*x^3)^(1/3)),x]

[Out]

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

Rule 31

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

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 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]]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 70, normalized size = 0.61 \begin {gather*} \frac {3 \log \left (\sqrt [3]{b}-\sqrt [3]{a x^3+b}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a x^3+b}}{\sqrt [3]{b}}+1}{\sqrt {3}}\right )-3 \log (x)}{6 \sqrt [3]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(b + a*x^3)^(1/3)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.10, size = 115, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b+a x^3}}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt {3} \sqrt [3]{b}}+\frac {\log \left (-\sqrt [3]{b}+\sqrt [3]{b+a x^3}\right )}{3 \sqrt [3]{b}}-\frac {\log \left (b^{2/3}+\sqrt [3]{b} \sqrt [3]{b+a x^3}+\left (b+a x^3\right )^{2/3}\right )}{6 \sqrt [3]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 236, normalized size = 2.05 \begin {gather*} \left [\frac {3 \, \sqrt {\frac {1}{3}} b \sqrt {-\frac {1}{b^{\frac {2}{3}}}} \log \left (\frac {2 \, a x^{3} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (a x^{3} + b\right )}^{\frac {2}{3}} b^{\frac {2}{3}} - {\left (a x^{3} + b\right )}^{\frac {1}{3}} b - b^{\frac {4}{3}}\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} - 3 \, {\left (a x^{3} + b\right )}^{\frac {1}{3}} b^{\frac {2}{3}} + 3 \, b}{x^{3}}\right ) - b^{\frac {2}{3}} \log \left ({\left (a x^{3} + b\right )}^{\frac {2}{3}} + {\left (a x^{3} + b\right )}^{\frac {1}{3}} b^{\frac {1}{3}} + b^{\frac {2}{3}}\right ) + 2 \, b^{\frac {2}{3}} \log \left ({\left (a x^{3} + b\right )}^{\frac {1}{3}} - b^{\frac {1}{3}}\right )}{6 \, b}, \frac {6 \, \sqrt {\frac {1}{3}} b^{\frac {2}{3}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, {\left (a x^{3} + b\right )}^{\frac {1}{3}} + b^{\frac {1}{3}}\right )}}{b^{\frac {1}{3}}}\right ) - b^{\frac {2}{3}} \log \left ({\left (a x^{3} + b\right )}^{\frac {2}{3}} + {\left (a x^{3} + b\right )}^{\frac {1}{3}} b^{\frac {1}{3}} + b^{\frac {2}{3}}\right ) + 2 \, b^{\frac {2}{3}} \log \left ({\left (a x^{3} + b\right )}^{\frac {1}{3}} - b^{\frac {1}{3}}\right )}{6 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*sqrt(1/3)*b*sqrt(-1/b^(2/3))*log((2*a*x^3 + 3*sqrt(1/3)*(2*(a*x^3 + b)^(2/3)*b^(2/3) - (a*x^3 + b)^(1/
3)*b - b^(4/3))*sqrt(-1/b^(2/3)) - 3*(a*x^3 + b)^(1/3)*b^(2/3) + 3*b)/x^3) - b^(2/3)*log((a*x^3 + b)^(2/3) + (
a*x^3 + b)^(1/3)*b^(1/3) + b^(2/3)) + 2*b^(2/3)*log((a*x^3 + b)^(1/3) - b^(1/3)))/b, 1/6*(6*sqrt(1/3)*b^(2/3)*
arctan(sqrt(1/3)*(2*(a*x^3 + b)^(1/3) + b^(1/3))/b^(1/3)) - b^(2/3)*log((a*x^3 + b)^(2/3) + (a*x^3 + b)^(1/3)*
b^(1/3) + b^(2/3)) + 2*b^(2/3)*log((a*x^3 + b)^(1/3) - b^(1/3)))/b]

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giac [A]  time = 0.69, size = 87, normalized size = 0.76 \begin {gather*} \frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (a x^{3} + b\right )}^{\frac {1}{3}} + b^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{3 \, b^{\frac {1}{3}}} - \frac {\log \left ({\left (a x^{3} + b\right )}^{\frac {2}{3}} + {\left (a x^{3} + b\right )}^{\frac {1}{3}} b^{\frac {1}{3}} + b^{\frac {2}{3}}\right )}{6 \, b^{\frac {1}{3}}} + \frac {\log \left ({\left | {\left (a x^{3} + b\right )}^{\frac {1}{3}} - b^{\frac {1}{3}} \right |}\right )}{3 \, b^{\frac {1}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1}{x \left (a \,x^{3}+b \right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 86, normalized size = 0.75 \begin {gather*} \frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (a x^{3} + b\right )}^{\frac {1}{3}} + b^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{3 \, b^{\frac {1}{3}}} - \frac {\log \left ({\left (a x^{3} + b\right )}^{\frac {2}{3}} + {\left (a x^{3} + b\right )}^{\frac {1}{3}} b^{\frac {1}{3}} + b^{\frac {2}{3}}\right )}{6 \, b^{\frac {1}{3}}} + \frac {\log \left ({\left (a x^{3} + b\right )}^{\frac {1}{3}} - b^{\frac {1}{3}}\right )}{3 \, b^{\frac {1}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.05, size = 100, normalized size = 0.87 \begin {gather*} \frac {\ln \left ({\left (a\,x^3+b\right )}^{1/3}-b^{1/3}\right )}{3\,b^{1/3}}+\frac {\ln \left ({\left (a\,x^3+b\right )}^{1/3}-\frac {b^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,b^{1/3}}-\frac {\ln \left ({\left (a\,x^3+b\right )}^{1/3}-\frac {b^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,b^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(b + a*x^3)^(1/3)),x)

[Out]

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

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sympy [C]  time = 0.89, size = 37, normalized size = 0.32 \begin {gather*} - \frac {\Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{3}}} \right )}}{3 \sqrt [3]{a} x \Gamma \left (\frac {4}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(1/3)*hyper((1/3, 1/3), (4/3,), b*exp_polar(I*pi)/(a*x**3))/(3*a**(1/3)*x*gamma(4/3))

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