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

Optimal. Leaf size=73 \[ \frac {3 \log \left (a-\sqrt [3]{a^3-b^3 x}\right )}{2 a}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{a^3-b^3 x}+a}{\sqrt {3} a}\right )}{a}-\frac {\log (x)}{2 a} \]

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Rubi [A]  time = 0.03, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {55, 617, 204, 31} \begin {gather*} \frac {3 \log \left (a-\sqrt [3]{a^3-b^3 x}\right )}{2 a}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{a^3-b^3 x}+a}{\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 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 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 [A]  time = 0.02, size = 68, normalized size = 0.93 \begin {gather*} \frac {3 \log \left (a-\sqrt [3]{a^3-b^3 x}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{a^3-b^3 x}+a}{\sqrt {3} a}\right )-\log (x)}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.05, size = 106, normalized size = 1.45 \begin {gather*} \frac {\log \left (a-\sqrt [3]{a^3-b^3 x}\right )}{a}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{a^3-b^3 x}}{\sqrt {3} a}+\frac {1}{\sqrt {3}}\right )}{a}-\frac {\log \left (a \sqrt [3]{a^3-b^3 x}+\left (a^3-b^3 x\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 - Log[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 = 0.87, size = 92, 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 = 0.77, size = 91, normalized size = 1.25 \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 = 0.00, size = 91, normalized size = 1.25 \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/a*ln((-b^3*x+a^3)^(1/3)-a)-1/2/a*ln((-b^3*x+a^3)^(2/3)+(-b^3*x+a^3)^(1/3)*a+a^2)+arctan(1/3*(a+2*(-b^3*x+a^3
)^(1/3))/a*3^(1/2))*3^(1/2)/a

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maxima [A]  time = 2.99, size = 90, normalized size = 1.23 \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 = 0.13, size = 108, normalized size = 1.48 \begin {gather*} \frac {\ln \left (9\,{\left (a^3-b^3\,x\right )}^{1/3}-9\,a\right )}{a}+\frac {\ln \left (9\,{\left (a^3-b^3\,x\right )}^{1/3}-\frac {9\,a\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,a}-\frac {\ln \left (9\,{\left (a^3-b^3\,x\right )}^{1/3}-\frac {9\,a\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}\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*(a^3 - b^3*x)^(1/3)),x)

[Out]

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

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sympy [C]  time = 1.89, size = 136, normalized size = 1.86 \begin {gather*} - \frac {e^{- \frac {2 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 {e^{- \frac {i \pi }{3}} \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 {\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(-2*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)) + exp(-
I*pi/3)*log(-a*exp_polar(I*pi)/(b*(-a**3/b**3 + x)**(1/3)) + 1)*gamma(-1/3)/(3*a*gamma(2/3)) - log(-a*exp_pola
r(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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