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

Optimal. Leaf size=74 \[ -\frac {3 \log \left (\sqrt [3]{b^3 x-a^3}+a\right )}{2 a}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {a-2 \sqrt [3]{b^3 x-a^3}}{\sqrt {3} a}\right )}{a}+\frac {\log (x)}{2 a} \]

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Rubi [A]  time = 0.03, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {56, 617, 204, 31} \begin {gather*} -\frac {3 \log \left (\sqrt [3]{b^3 x-a^3}+a\right )}{2 a}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {a-2 \sqrt [3]{b^3 x-a^3}}{\sqrt {3} a}\right )}{a}+\frac {\log (x)}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(-a^3 + b^3*x)^(1/3)),x]

[Out]

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

Rule 31

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

Rule 56

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, Simp
[Log[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] && Ne
gQ[(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 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]{-a^3+b^3 x}} \, dx &=\frac {\log (x)}{2 a}+\frac {3}{2} \operatorname {Subst}\left (\int \frac {1}{a^2-a x+x^2} \, dx,x,\sqrt [3]{-a^3+b^3 x}\right )-\frac {3 \operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,\sqrt [3]{-a^3+b^3 x}\right )}{2 a}\\ &=\frac {\log (x)}{2 a}-\frac {3 \log \left (a+\sqrt [3]{-a^3+b^3 x}\right )}{2 a}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{-a^3+b^3 x}}{a}\right )}{a}\\ &=-\frac {\sqrt {3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{-a^3+b^3 x}}{a}}{\sqrt {3}}\right )}{a}+\frac {\log (x)}{2 a}-\frac {3 \log \left (a+\sqrt [3]{-a^3+b^3 x}\right )}{2 a}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 41, normalized size = 0.55 \begin {gather*} \frac {3 \left (b^3 x-a^3\right )^{2/3} \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};1-\frac {b^3 x}{a^3}\right )}{2 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(-a^3 + b^3*x)^(2/3)*Hypergeometric2F1[2/3, 1, 5/3, 1 - (b^3*x)/a^3])/(2*a^3)

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IntegrateAlgebraic [A]  time = 0.05, size = 111, normalized size = 1.50 \begin {gather*} -\frac {\log \left (\sqrt [3]{b^3 x-a^3}+a\right )}{a}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b^3 x-a^3}}{\sqrt {3} a}\right )}{a}+\frac {\log \left (-a \sqrt [3]{b^3 x-a^3}+\left (b^3 x-a^3\right )^{2/3}+a^2\right )}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.83, size = 93, normalized size = 1.26 \begin {gather*} \frac {2 \, \sqrt {3} \arctan \left (-\frac {\sqrt {3} a - 2 \, \sqrt {3} {\left (b^{3} x - a^{3}\right )}^{\frac {1}{3}}}{3 \, a}\right ) + \log \left (a^{2} - {\left (b^{3} x - a^{3}\right )}^{\frac {1}{3}} a + {\left (b^{3} x - a^{3}\right )}^{\frac {2}{3}}\right ) - 2 \, \log \left (a + {\left (b^{3} x - a^{3}\right )}^{\frac {1}{3}}\right )}{2 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.07, size = 95, normalized size = 1.28 \begin {gather*} \frac {\sqrt {3} \arctan \left (-\frac {\sqrt {3} {\left (a - 2 \, {\left (b^{3} x - a^{3}\right )}^{\frac {1}{3}}\right )}}{3 \, a}\right )}{a} + \frac {\log \left (a^{2} - {\left (b^{3} x - a^{3}\right )}^{\frac {1}{3}} a + {\left (b^{3} x - a^{3}\right )}^{\frac {2}{3}}\right )}{2 \, a} - \frac {\log \left ({\left | a + {\left (b^{3} x - a^{3}\right )}^{\frac {1}{3}} \right |}\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 97, normalized size = 1.31 \begin {gather*} \frac {\sqrt {3}\, \arctan \left (\frac {\left (-a +2 \left (b^{3} x -a^{3}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a}\right )}{a}-\frac {\ln \left (a +\left (b^{3} x -a^{3}\right )^{\frac {1}{3}}\right )}{a}+\frac {\ln \left (a^{2}-\left (b^{3} x -a^{3}\right )^{\frac {1}{3}} a +\left (b^{3} x -a^{3}\right )^{\frac {2}{3}}\right )}{2 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.90, size = 94, normalized size = 1.27 \begin {gather*} \frac {\sqrt {3} \arctan \left (-\frac {\sqrt {3} {\left (a - 2 \, {\left (b^{3} x - a^{3}\right )}^{\frac {1}{3}}\right )}}{3 \, a}\right )}{a} + \frac {\log \left (a^{2} - {\left (b^{3} x - a^{3}\right )}^{\frac {1}{3}} a + {\left (b^{3} x - a^{3}\right )}^{\frac {2}{3}}\right )}{2 \, a} - \frac {\log \left (a + {\left (b^{3} x - a^{3}\right )}^{\frac {1}{3}}\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.11, size = 112, normalized size = 1.51 \begin {gather*} -\frac {\ln \left (9\,a+9\,{\left (b^3\,x-a^3\right )}^{1/3}\right )}{a}-\frac {\ln \left (\frac {9\,a\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}+9\,{\left (b^3\,x-a^3\right )}^{1/3}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,a}+\frac {\ln \left (\frac {9\,a\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}+9\,{\left (b^3\,x-a^3\right )}^{1/3}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(b^3*x - a^3)^(1/3)),x)

[Out]

(log((9*a*(3^(1/2)*1i + 1)^2)/4 + 9*(b^3*x - a^3)^(1/3))*(3^(1/2)*1i + 1))/(2*a) - (log((9*a*(3^(1/2)*1i - 1)^
2)/4 + 9*(b^3*x - a^3)^(1/3))*(3^(1/2)*1i - 1))/(2*a) - log(9*a + 9*(b^3*x - a^3)^(1/3))/a

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sympy [C]  time = 1.82, size = 134, normalized size = 1.81 \begin {gather*} - \frac {e^{- \frac {i \pi }{3}} \log {\left (- \frac {a e^{\frac {i \pi }{3}}}{b \sqrt [3]{- \frac {a^{3}}{b^{3}} + x}} + 1 \right )} \Gamma \left (- \frac {1}{3}\right )}{3 a \Gamma \left (\frac {2}{3}\right )} + \frac {\log {\left (- \frac {a e^{i \pi }}{b \sqrt [3]{- \frac {a^{3}}{b^{3}} + x}} + 1 \right )} \Gamma \left (- \frac {1}{3}\right )}{3 a \Gamma \left (\frac {2}{3}\right )} - \frac {e^{\frac {i \pi }{3}} \log {\left (- \frac {a e^{\frac {5 i \pi }{3}}}{b \sqrt [3]{- \frac {a^{3}}{b^{3}} + x}} + 1 \right )} \Gamma \left (- \frac {1}{3}\right )}{3 a \Gamma \left (\frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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