3.8.6 \(\int \frac {1}{x \sqrt [3]{a+b x^2}} \, dx\) [706]

Optimal. Leaf size=86 \[ \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x^2}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{a}}-\frac {\log (x)}{2 \sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{4 \sqrt [3]{a}} \]

[Out]

-1/2*ln(x)/a^(1/3)+3/4*ln(a^(1/3)-(b*x^2+a)^(1/3))/a^(1/3)+1/2*arctan(1/3*(a^(1/3)+2*(b*x^2+a)^(1/3))/a^(1/3)*
3^(1/2))*3^(1/2)/a^(1/3)

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Rubi [A]
time = 0.04, antiderivative size = 86, normalized size of antiderivative = 1.00, 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 = {272, 57, 631, 210, 31} \begin {gather*} \frac {\sqrt {3} \text {ArcTan}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{4 \sqrt [3]{a}}-\frac {\log (x)}{2 \sqrt [3]{a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^2)^(1/3)),x]

[Out]

(Sqrt[3]*ArcTan[(a^(1/3) + 2*(a + b*x^2)^(1/3))/(Sqrt[3]*a^(1/3))])/(2*a^(1/3)) - Log[x]/(2*a^(1/3)) + (3*Log[
a^(1/3) - (a + b*x^2)^(1/3)])/(4*a^(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 57

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 210

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

Rule 272

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 631

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+b x^2}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {1}{x \sqrt [3]{a+b x}} \, dx,x,x^2\right )\\ &=-\frac {\log (x)}{2 \sqrt [3]{a}}+\frac {3}{4} \text {Subst}\left (\int \frac {1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x^2}\right )-\frac {3 \text {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^2}\right )}{4 \sqrt [3]{a}}\\ &=-\frac {\log (x)}{2 \sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{4 \sqrt [3]{a}}-\frac {3 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}\right )}{2 \sqrt [3]{a}}\\ &=\frac {\sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{2 \sqrt [3]{a}}-\frac {\log (x)}{2 \sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{4 \sqrt [3]{a}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 103, normalized size = 1.20 \begin {gather*} \frac {2 \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+2 \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^2}\right )-\log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}\right )}{4 \sqrt [3]{a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^2)^(1/3)),x]

[Out]

(2*Sqrt[3]*ArcTan[(1 + (2*(a + b*x^2)^(1/3))/a^(1/3))/Sqrt[3]] + 2*Log[-a^(1/3) + (a + b*x^2)^(1/3)] - Log[a^(
2/3) + a^(1/3)*(a + b*x^2)^(1/3) + (a + b*x^2)^(2/3)])/(4*a^(1/3))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{x \left (b \,x^{2}+a \right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(b*x^2+a)^(1/3),x)

[Out]

int(1/x/(b*x^2+a)^(1/3),x)

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Maxima [A]
time = 0.51, size = 86, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{2 \, a^{\frac {1}{3}}} - \frac {\log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{4 \, a^{\frac {1}{3}}} + \frac {\log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{2 \, a^{\frac {1}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x^2+a)^(1/3),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.73, size = 235, normalized size = 2.73 \begin {gather*} \left [\frac {\sqrt {3} a \sqrt {-\frac {1}{a^{\frac {2}{3}}}} \log \left (\frac {2 \, b x^{2} + \sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {2}{3}} a^{\frac {2}{3}} - {\left (b x^{2} + a\right )}^{\frac {1}{3}} a - a^{\frac {4}{3}}\right )} \sqrt {-\frac {1}{a^{\frac {2}{3}}}} - 3 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {2}{3}} + 3 \, a}{x^{2}}\right ) - a^{\frac {2}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + 2 \, a^{\frac {2}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{4 \, a}, \frac {2 \, \sqrt {3} a^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + 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^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + 2 \, a^{\frac {2}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{4 \, a}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x^2+a)^(1/3),x, algorithm="fricas")

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 0.46, size = 41, normalized size = 0.48 \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 {a e^{i \pi }}{b x^{2}}} \right )}}{2 \sqrt [3]{b} x^{\frac {2}{3}} \Gamma \left (\frac {4}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x**2+a)**(1/3),x)

[Out]

-gamma(1/3)*hyper((1/3, 1/3), (4/3,), a*exp_polar(I*pi)/(b*x**2))/(2*b**(1/3)*x**(2/3)*gamma(4/3))

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Giac [A]
time = 1.61, size = 87, normalized size = 1.01 \begin {gather*} \frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{2 \, a^{\frac {1}{3}}} - \frac {\log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{4 \, a^{\frac {1}{3}}} + \frac {\log \left ({\left | {\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{2 \, a^{\frac {1}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x^2+a)^(1/3),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 4.83, size = 106, normalized size = 1.23 \begin {gather*} \frac {\ln \left (\frac {9\,{\left (b\,x^2+a\right )}^{1/3}}{4}-\frac {9\,a^{1/3}}{4}\right )}{2\,a^{1/3}}+\frac {\ln \left (\frac {9\,{\left (b\,x^2+a\right )}^{1/3}}{4}-\frac {9\,a^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{4\,a^{1/3}}-\frac {\ln \left (\frac {9\,{\left (b\,x^2+a\right )}^{1/3}}{4}-\frac {9\,a^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{4\,a^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*x^2)^(1/3)),x)

[Out]

log((9*(a + b*x^2)^(1/3))/4 - (9*a^(1/3))/4)/(2*a^(1/3)) + (log((9*(a + b*x^2)^(1/3))/4 - (9*a^(1/3)*(3^(1/2)*
1i - 1)^2)/16)*(3^(1/2)*1i - 1))/(4*a^(1/3)) - (log((9*(a + b*x^2)^(1/3))/4 - (9*a^(1/3)*(3^(1/2)*1i + 1)^2)/1
6)*(3^(1/2)*1i + 1))/(4*a^(1/3))

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