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

Optimal. Leaf size=79 \[ \frac{3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{2 \sqrt [3]{a}}+\frac{\sqrt{3} \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt [3]{a}}-\frac{\log (x)}{2 \sqrt [3]{a}} \]

[Out]

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

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Rubi [A]  time = 0.0250843, antiderivative size = 79, normalized size of antiderivative = 1., 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 = {55, 617, 204, 31} \[ \frac{3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{2 \sqrt [3]{a}}+\frac{\sqrt{3} \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt [3]{a}}-\frac{\log (x)}{2 \sqrt [3]{a}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x)^(1/3)),x]

[Out]

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

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 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]

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 31

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

Rubi steps

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

Mathematica [A]  time = 0.0229908, size = 66, normalized size = 0.84 \[ \frac{3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}+1}{\sqrt{3}}\right )-\log (x)}{2 \sqrt [3]{a}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x)^(1/3)),x]

[Out]

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

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Maple [A]  time = 0.004, size = 75, normalized size = 1. \begin{align*}{\ln \left ( \sqrt [3]{bx+a}-\sqrt [3]{a} \right ){\frac{1}{\sqrt [3]{a}}}}-{\frac{1}{2}\ln \left ( \left ( bx+a \right ) ^{{\frac{2}{3}}}+\sqrt [3]{a}\sqrt [3]{bx+a}+{a}^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{a}}}}+{\sqrt{3}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\frac{\sqrt [3]{bx+a}}{\sqrt [3]{a}}}+1 \right ) } \right ){\frac{1}{\sqrt [3]{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.85358, size = 647, normalized size = 8.19 \begin{align*} \left [\frac{\sqrt{3} a \sqrt{-\frac{1}{a^{\frac{2}{3}}}} \log \left (\frac{2 \, b x + \sqrt{3}{\left (2 \,{\left (b x + a\right )}^{\frac{2}{3}} a^{\frac{2}{3}} -{\left (b x + a\right )}^{\frac{1}{3}} a - a^{\frac{4}{3}}\right )} \sqrt{-\frac{1}{a^{\frac{2}{3}}}} - 3 \,{\left (b x + a\right )}^{\frac{1}{3}} a^{\frac{2}{3}} + 3 \, a}{x}\right ) - a^{\frac{2}{3}} \log \left ({\left (b x + a\right )}^{\frac{2}{3}} +{\left (b x + a\right )}^{\frac{1}{3}} a^{\frac{1}{3}} + a^{\frac{2}{3}}\right ) + 2 \, a^{\frac{2}{3}} \log \left ({\left (b x + a\right )}^{\frac{1}{3}} - a^{\frac{1}{3}}\right )}{2 \, a}, \frac{2 \, \sqrt{3} a^{\frac{2}{3}} \arctan \left (\frac{\sqrt{3}{\left (2 \,{\left (b x + a\right )}^{\frac{1}{3}} + a^{\frac{1}{3}}\right )}}{3 \, a^{\frac{1}{3}}}\right ) - a^{\frac{2}{3}} \log \left ({\left (b x + a\right )}^{\frac{2}{3}} +{\left (b x + a\right )}^{\frac{1}{3}} a^{\frac{1}{3}} + a^{\frac{2}{3}}\right ) + 2 \, a^{\frac{2}{3}} \log \left ({\left (b x + a\right )}^{\frac{1}{3}} - a^{\frac{1}{3}}\right )}{2 \, a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 2.48862, size = 155, normalized size = 1.96 \begin{align*} \frac{2 \log{\left (1 - \frac{\sqrt [3]{b} \sqrt [3]{\frac{a}{b} + x}}{\sqrt [3]{a}} \right )} \Gamma \left (\frac{2}{3}\right )}{3 \sqrt [3]{a} \Gamma \left (\frac{5}{3}\right )} + \frac{2 e^{\frac{2 i \pi }{3}} \log{\left (1 - \frac{\sqrt [3]{b} \sqrt [3]{\frac{a}{b} + x} e^{\frac{2 i \pi }{3}}}{\sqrt [3]{a}} \right )} \Gamma \left (\frac{2}{3}\right )}{3 \sqrt [3]{a} \Gamma \left (\frac{5}{3}\right )} + \frac{2 e^{- \frac{2 i \pi }{3}} \log{\left (1 - \frac{\sqrt [3]{b} \sqrt [3]{\frac{a}{b} + x} e^{\frac{4 i \pi }{3}}}{\sqrt [3]{a}} \right )} \Gamma \left (\frac{2}{3}\right )}{3 \sqrt [3]{a} \Gamma \left (\frac{5}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(2/3)/(3*a**(1/3)*gamma(5/3)) + 2*exp(2*I*pi/3)*log(1 - b**
(1/3)*(a/b + x)**(1/3)*exp_polar(2*I*pi/3)/a**(1/3))*gamma(2/3)/(3*a**(1/3)*gamma(5/3)) + 2*exp(-2*I*pi/3)*log
(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*I*pi/3)/a**(1/3))*gamma(2/3)/(3*a**(1/3)*gamma(5/3))

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Giac [A]  time = 2.1971, size = 104, normalized size = 1.32 \begin{align*} \frac{\sqrt{3} \arctan \left (\frac{\sqrt{3}{\left (2 \,{\left (b x + a\right )}^{\frac{1}{3}} + a^{\frac{1}{3}}\right )}}{3 \, a^{\frac{1}{3}}}\right )}{a^{\frac{1}{3}}} - \frac{\log \left ({\left (b x + a\right )}^{\frac{2}{3}} +{\left (b x + a\right )}^{\frac{1}{3}} a^{\frac{1}{3}} + a^{\frac{2}{3}}\right )}{2 \, a^{\frac{1}{3}}} + \frac{\log \left ({\left |{\left (b x + a\right )}^{\frac{1}{3}} - a^{\frac{1}{3}} \right |}\right )}{a^{\frac{1}{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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