3.718 \(\int \frac {1}{x (a+b x^2)^{2/3}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.05, 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 = {266, 57, 617, 204, 31} \[ \frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{4 a^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 a^{2/3}}-\frac {\log (x)}{2 a^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^2)^(2/3)),x]

[Out]

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

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_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), 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 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 \left (a+b x^2\right )^{2/3}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{2/3}} \, dx,x,x^2\right )\\ &=-\frac {\log (x)}{2 a^{2/3}}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^2}\right )}{4 a^{2/3}}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x^2}\right )}{4 \sqrt [3]{a}}\\ &=-\frac {\log (x)}{2 a^{2/3}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{4 a^{2/3}}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}\right )}{2 a^{2/3}}\\ &=-\frac {\sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{2 a^{2/3}}-\frac {\log (x)}{2 a^{2/3}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{4 a^{2/3}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 101, normalized size = 1.17 \[ -\frac {\log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}\right )-2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )}{4 a^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^2)^(2/3)),x]

[Out]

-1/4*(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)])/a^(2/3)

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fricas [B]  time = 0.60, size = 123, normalized size = 1.43 \[ -\frac {2 \, \sqrt {3} {\left (a^{2}\right )}^{\frac {1}{6}} a \arctan \left (\frac {\sqrt {3} {\left (a^{2}\right )}^{\frac {1}{6}} {\left ({\left (a^{2}\right )}^{\frac {1}{3}} a + 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right )}}{3 \, a^{2}}\right ) + {\left (a^{2}\right )}^{\frac {2}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} a + {\left (a^{2}\right )}^{\frac {1}{3}} a + {\left (b x^{2} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right ) - 2 \, {\left (a^{2}\right )}^{\frac {2}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} a - {\left (a^{2}\right )}^{\frac {2}{3}}\right )}{4 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*sqrt(3)*(a^2)^(1/6)*a*arctan(1/3*sqrt(3)*(a^2)^(1/6)*((a^2)^(1/3)*a + 2*(b*x^2 + a)^(1/3)*(a^2)^(2/3))
/a^2) + (a^2)^(2/3)*log((b*x^2 + a)^(2/3)*a + (a^2)^(1/3)*a + (b*x^2 + a)^(1/3)*(a^2)^(2/3)) - 2*(a^2)^(2/3)*l
og((b*x^2 + a)^(1/3)*a - (a^2)^(2/3)))/a^2

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giac [A]  time = 1.13, size = 87, normalized size = 1.01 \[ -\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{2 \, a^{\frac {2}{3}}} - \frac {\log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{4 \, a^{\frac {2}{3}}} + \frac {\log \left ({\left | {\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{2 \, a^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {2}{3}} x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.09, size = 86, normalized size = 1.00 \[ -\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{2 \, a^{\frac {2}{3}}} - \frac {\log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{4 \, a^{\frac {2}{3}}} + \frac {\log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{2 \, a^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x^2+a)^(2/3),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^(2/3) - 1/4*log((b*x^2 + a)^(2/3) +
(b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(2/3) + 1/2*log((b*x^2 + a)^(1/3) - a^(1/3))/a^(2/3)

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mupad [B]  time = 4.84, size = 102, normalized size = 1.19 \[ \frac {\ln \left (\frac {9\,{\left (b\,x^2+a\right )}^{1/3}}{2}-\frac {9\,a^{1/3}}{2}\right )}{2\,a^{2/3}}+\frac {\ln \left (\frac {9\,a^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{4}-\frac {9\,{\left (b\,x^2+a\right )}^{1/3}}{2}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{4\,a^{2/3}}-\frac {\ln \left (\frac {9\,a^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{4}+\frac {9\,{\left (b\,x^2+a\right )}^{1/3}}{2}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{4\,a^{2/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*x^2)^(2/3)),x)

[Out]

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

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sympy [C]  time = 1.06, size = 41, normalized size = 0.48 \[ - \frac {\Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 b^{\frac {2}{3}} x^{\frac {4}{3}} \Gamma \left (\frac {5}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(2/3)*hyper((2/3, 2/3), (5/3,), a*exp_polar(I*pi)/(b*x**2))/(2*b**(2/3)*x**(4/3)*gamma(5/3))

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