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

Optimal. Leaf size=123 \[ \frac {\log \left (-\sqrt [3]{b} \sqrt [3]{a x^3-b}+\left (a x^3-b\right )^{2/3}+b^{2/3}\right )}{6 \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a x^3-b}+\sqrt [3]{b}\right )}{3 \sqrt [3]{b}}-\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{a x^3-b}}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt {3} \sqrt [3]{b}} \]

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Rubi [A]  time = 0.07, antiderivative size = 86, normalized size of antiderivative = 0.70, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {266, 56, 617, 204, 31} \begin {gather*} -\frac {\log \left (\sqrt [3]{a x^3-b}+\sqrt [3]{b}\right )}{2 \sqrt [3]{b}}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a x^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 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 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 [C]  time = 0.01, size = 39, normalized size = 0.32 \begin {gather*} \frac {\left (a x^3-b\right )^{2/3} \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};1-\frac {a x^3}{b}\right )}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(-b + a*x^3)^(1/3)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.12, size = 123, 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 = 0.62, size = 308, normalized size = 2.50 \begin {gather*} \left [\frac {3 \, \sqrt {\frac {1}{3}} b \sqrt {\frac {\left (-b\right )^{\frac {1}{3}}}{b}} \log \left (\frac {2 \, a x^{3} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (a x^{3} - b\right )}^{\frac {2}{3}} \left (-b\right )^{\frac {2}{3}} + {\left (a x^{3} - b\right )}^{\frac {1}{3}} b + \left (-b\right )^{\frac {1}{3}} b\right )} \sqrt {\frac {\left (-b\right )^{\frac {1}{3}}}{b}} - 3 \, {\left (a x^{3} - b\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {2}{3}} - 3 \, b}{x^{3}}\right ) + \left (-b\right )^{\frac {2}{3}} \log \left ({\left (a x^{3} - b\right )}^{\frac {2}{3}} + {\left (a x^{3} - b\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} + \left (-b\right )^{\frac {2}{3}}\right ) - 2 \, \left (-b\right )^{\frac {2}{3}} \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{3}} - \left (-b\right )^{\frac {1}{3}}\right )}{6 \, b}, \frac {6 \, \sqrt {\frac {1}{3}} b \sqrt {-\frac {\left (-b\right )^{\frac {1}{3}}}{b}} \arctan \left (\sqrt {\frac {1}{3}} {\left (2 \, {\left (a x^{3} - b\right )}^{\frac {1}{3}} + \left (-b\right )^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-b\right )^{\frac {1}{3}}}{b}}\right ) + \left (-b\right )^{\frac {2}{3}} \log \left ({\left (a x^{3} - b\right )}^{\frac {2}{3}} + {\left (a x^{3} - b\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} + \left (-b\right )^{\frac {2}{3}}\right ) - 2 \, \left (-b\right )^{\frac {2}{3}} \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{3}} - \left (-b\right )^{\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((-b)^(1/3)/b)*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)^(1/3)*b)*sqrt((-b)^(1/3)/b) - 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*sqrt(-(-b)^(1/3)/b)*arctan(sqrt(1/3)*(2*(a*x^3 - b)^(1/3) + (-b)^(1/3))*sqrt(-(-b)^(1/
3)/b)) + (-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 = 2.06, size = 120, normalized size = 0.98 \begin {gather*} -\frac {\sqrt {3} \left (-b\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (a x^{3} - b\right )}^{\frac {1}{3}} + \left (-b\right )^{\frac {1}{3}}\right )}}{3 \, \left (-b\right )^{\frac {1}{3}}}\right )}{3 \, b} + \frac {\left (-b\right )^{\frac {2}{3}} \log \left ({\left (a x^{3} - b\right )}^{\frac {2}{3}} + {\left (a x^{3} - b\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} + \left (-b\right )^{\frac {2}{3}}\right )}{6 \, b} - \frac {\left (-b\right )^{\frac {2}{3}} \log \left ({\left | {\left (a x^{3} - b\right )}^{\frac {1}{3}} - \left (-b\right )^{\frac {1}{3}} \right |}\right )}{3 \, b} \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)*(-b)^(2/3)*arctan(1/3*sqrt(3)*(2*(a*x^3 - b)^(1/3) + (-b)^(1/3))/(-b)^(1/3))/b + 1/6*(-b)^(2/3)*l
og((a*x^3 - b)^(2/3) + (a*x^3 - b)^(1/3)*(-b)^(1/3) + (-b)^(2/3))/b - 1/3*(-b)^(2/3)*log(abs((a*x^3 - b)^(1/3)
- (-b)^(1/3)))/b

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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 = 1.45, size = 95, normalized size = 0.77 \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 = 1.04, size = 118, normalized size = 0.96 \begin {gather*} \frac {\ln \left ({\left (a\,x^3-b\right )}^{1/3}-{\left (-b\right )}^{1/3}\right )}{3\,{\left (-b\right )}^{1/3}}+\frac {\ln \left ({\left (a\,x^3-b\right )}^{1/3}-\frac {{\left (-b\right )}^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,{\left (-b\right )}^{1/3}}-\frac {\ln \left ({\left (a\,x^3-b\right )}^{1/3}-\frac {{\left (-b\right )}^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,{\left (-b\right )}^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a*x^3 - b)^(1/3)),x)

[Out]

log((a*x^3 - b)^(1/3) - (-b)^(1/3))/(3*(-b)^(1/3)) + (log((a*x^3 - b)^(1/3) - ((-b)^(1/3)*(3^(1/2)*1i - 1)^2)/
4)*(3^(1/2)*1i - 1))/(6*(-b)^(1/3)) - (log((a*x^3 - b)^(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 = 39, 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^{2 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(2*I*pi)/(a*x**3))/(3*a**(1/3)*x*gamma(4/3))

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