\(\int \frac {1}{x (a+b x^{3/2})^{2/3}} \, dx\) [2279]

Optimal result

Integrand size = 17, antiderivative size = 85 \[ \int \frac {1}{x \left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {2 \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x^{3/2}}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3}}-\frac {\log (x)}{2 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^{3/2}}\right )}{a^{2/3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {272, 59, 631, 210, 31} \[ \int \frac {1}{x \left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {2 \arctan \left (\frac {2 \sqrt [3]{a+b x^{3/2}}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3}}+\frac {\log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^{3/2}}\right )}{a^{2/3}}-\frac {\log (x)}{2 a^{2/3}} \]

[In]

Int[1/(x*(a + b*x^(3/2))^(2/3)),x]

[Out]

(-2*ArcTan[(a^(1/3) + 2*(a + b*x^(3/2))^(1/3))/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(2/3)) - Log[x]/(2*a^(2/3)) + Lo
g[a^(1/3) - (a + b*x^(3/2))^(1/3)]/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 59

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.28 \[ \int \frac {1}{x \left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x^{3/2}}}{\sqrt [3]{a}}}{\sqrt {3}}\right )-2 \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^{3/2}}\right )+\log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^{3/2}}+\left (a+b x^{3/2}\right )^{2/3}\right )}{3 a^{2/3}} \]

[In]

Integrate[1/(x*(a + b*x^(3/2))^(2/3)),x]

[Out]

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

Maple [A] (verified)

Time = 5.74 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {2 \ln \left (\left (a +b \,x^{\frac {3}{2}}\right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )}{3 a^{\frac {2}{3}}}-\frac {\ln \left (\left (a +b \,x^{\frac {3}{2}}\right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (a +b \,x^{\frac {3}{2}}\right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{3 a^{\frac {2}{3}}}-\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (a +b \,x^{\frac {3}{2}}\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{3 a^{\frac {2}{3}}}\)	\(85\)
default	\(\frac {2 \ln \left (\left (a +b \,x^{\frac {3}{2}}\right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )}{3 a^{\frac {2}{3}}}-\frac {\ln \left (\left (a +b \,x^{\frac {3}{2}}\right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (a +b \,x^{\frac {3}{2}}\right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{3 a^{\frac {2}{3}}}-\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (a +b \,x^{\frac {3}{2}}\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{3 a^{\frac {2}{3}}}\)	\(85\)

[In]

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

[Out]

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

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{x \left (a+b x^{3/2}\right )^{2/3}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.67 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.48 \[ \int \frac {1}{x \left (a+b x^{3/2}\right )^{2/3}} \, dx=- \frac {2 \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{\frac {3}{2}}}} \right )}}{3 b^{\frac {2}{3}} x \Gamma \left (\frac {5}{3}\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.01 \[ \int \frac {1}{x \left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{3 \, a^{\frac {2}{3}}} - \frac {\log \left ({\left (b x^{\frac {3}{2}} + a\right )}^{\frac {2}{3}} + {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{3 \, a^{\frac {2}{3}}} + \frac {2 \, \log \left ({\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{3 \, a^{\frac {2}{3}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 1.60 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x \left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{3 \, a^{\frac {2}{3}}} - \frac {\log \left ({\left (b x^{\frac {3}{2}} + a\right )}^{\frac {2}{3}} + {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{3 \, a^{\frac {2}{3}}} + \frac {2 \, \log \left ({\left | {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{3 \, a^{\frac {2}{3}}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.01 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.20 \[ \int \frac {1}{x \left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {2\,\ln \left (6\,{\left (a+b\,x^{3/2}\right )}^{1/3}-6\,a^{1/3}\right )}{3\,a^{2/3}}+\frac {\ln \left (3\,a^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )-6\,{\left (a+b\,x^{3/2}\right )}^{1/3}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{3\,a^{2/3}}-\frac {\ln \left (3\,a^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )+6\,{\left (a+b\,x^{3/2}\right )}^{1/3}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{3\,a^{2/3}} \]

[In]

int(1/(x*(a + b*x^(3/2))^(2/3)),x)

[Out]

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