\(\int (a+\frac {b}{x^{3/2}})^{2/3} \, dx\) [2393]

Optimal result

Integrand size = 13, antiderivative size = 95 \[ \int \left (a+\frac {b}{x^{3/2}}\right )^{2/3} \, dx=\left (a+\frac {b}{x^{3/2}}\right )^{2/3} x-\frac {2 b^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b}}{\sqrt [3]{a+\frac {b}{x^{3/2}}} \sqrt {x}}}{\sqrt {3}}\right )}{\sqrt {3}}+b^{2/3} \log \left (\sqrt [3]{a+\frac {b}{x^{3/2}}}-\frac {\sqrt [3]{b}}{\sqrt {x}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {249, 342, 283, 245} \[ \int \left (a+\frac {b}{x^{3/2}}\right )^{2/3} \, dx=-\frac {2 b^{2/3} \arctan \left (\frac {\frac {2 \sqrt [3]{b}}{\sqrt {x} \sqrt [3]{a+\frac {b}{x^{3/2}}}}+1}{\sqrt {3}}\right )}{\sqrt {3}}+b^{2/3} \log \left (\sqrt [3]{a+\frac {b}{x^{3/2}}}-\frac {\sqrt [3]{b}}{\sqrt {x}}\right )+x \left (a+\frac {b}{x^{3/2}}\right )^{2/3} \]

[In]

Int[(a + b/x^(3/2))^(2/3),x]

[Out]

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

Rule 245

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rule 249

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x^(k - 1)*(a + b*
x^(k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, p}, x] && FractionQ[n]

Rule 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(280\) vs. \(2(95)=190\).

Time = 7.35 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.95 \[ \int \left (a+\frac {b}{x^{3/2}}\right )^{2/3} \, dx=-\frac {\left (a+\frac {b}{x^{3/2}}\right )^{2/3} \left (\sqrt [3]{b}-\sqrt [3]{b+a x^{3/2}}\right ) \left (b^{2/3}+\sqrt [3]{b} \sqrt [3]{b+a x^{3/2}}+\left (b+a x^{3/2}\right )^{2/3}\right )^2 \left (3 \left (b+a x^{3/2}\right )^{2/3}+2 \sqrt {3} b^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b+a x^{3/2}}}{\sqrt [3]{b}}}{\sqrt {3}}\right )+2 b^{2/3} \log \left (-\sqrt [3]{b}+\sqrt [3]{b+a x^{3/2}}\right )-b^{2/3} \log \left (b^{2/3}+\sqrt [3]{b} \sqrt [3]{b+a x^{3/2}}+\left (b+a x^{3/2}\right )^{2/3}\right )\right )}{3 a \sqrt {x} \sqrt [3]{b+a x^{3/2}} \left (b+a x^{3/2}+b^{2/3} \sqrt [3]{b+a x^{3/2}}+\sqrt [3]{b} \left (b+a x^{3/2}\right )^{2/3}\right )} \]

[In]

Integrate[(a + b/x^(3/2))^(2/3),x]

[Out]

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

Maple [F]

\[\int \left (a +\frac {b}{x^{\frac {3}{2}}}\right )^{\frac {2}{3}}d x\]

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

integrate((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 = 1.74 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.48 \[ \int \left (a+\frac {b}{x^{3/2}}\right )^{2/3} \, dx=- \frac {2 a^{\frac {2}{3}} x \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {2}{3} \\ \frac {1}{3} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{\frac {3}{2}}}} \right )}}{3 \Gamma \left (\frac {1}{3}\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.15 \[ \int \left (a+\frac {b}{x^{3/2}}\right )^{2/3} \, dx=\frac {2}{3} \, \sqrt {3} b^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (a + \frac {b}{x^{\frac {3}{2}}}\right )}^{\frac {1}{3}} \sqrt {x} + b^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right ) + {\left (a + \frac {b}{x^{\frac {3}{2}}}\right )}^{\frac {2}{3}} x - \frac {1}{3} \, b^{\frac {2}{3}} \log \left ({\left (a + \frac {b}{x^{\frac {3}{2}}}\right )}^{\frac {2}{3}} x + {\left (a + \frac {b}{x^{\frac {3}{2}}}\right )}^{\frac {1}{3}} b^{\frac {1}{3}} \sqrt {x} + b^{\frac {2}{3}}\right ) + \frac {2}{3} \, b^{\frac {2}{3}} \log \left ({\left (a + \frac {b}{x^{\frac {3}{2}}}\right )}^{\frac {1}{3}} \sqrt {x} - b^{\frac {1}{3}}\right ) \]

[In]

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

[Out]

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

Giac [F]

\[ \int \left (a+\frac {b}{x^{3/2}}\right )^{2/3} \, dx=\int { {\left (a + \frac {b}{x^{\frac {3}{2}}}\right )}^{\frac {2}{3}} \,d x } \]

[In]

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

[Out]

integrate((a + b/x^(3/2))^(2/3), x)

Mupad [B] (verification not implemented)

Time = 6.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.39 \[ \int \left (a+\frac {b}{x^{3/2}}\right )^{2/3} \, dx=\frac {x\,{\left (a+\frac {b}{x^{3/2}}\right )}^{2/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {2}{3},-\frac {2}{3};\ \frac {1}{3};\ -\frac {b}{a\,x^{3/2}}\right )}{{\left (\frac {b}{a\,x^{3/2}}+1\right )}^{2/3}} \]

[In]

int((a + b/x^(3/2))^(2/3),x)

[Out]

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