Integrand size = 13, antiderivative size = 95 \[ \int \left (a+\frac {b}{x^{3/2}}\right )^{2/3} \, dx=\left (a+\frac {b}{x^{3/2}}\right )^{2/3} x-\frac {2 b^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b}}{\sqrt [3]{a+\frac {b}{x^{3/2}}} \sqrt {x}}}{\sqrt {3}}\right )}{\sqrt {3}}+b^{2/3} \log \left (\sqrt [3]{a+\frac {b}{x^{3/2}}}-\frac {\sqrt [3]{b}}{\sqrt {x}}\right ) \] Output:
(a+b/x^(3/2))^(2/3)*x-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)+b^(2/3)*ln((a+b/x^(3/2))^(1/3)-b^(1/3)/x^(1/ 2))
Leaf count is larger than twice the leaf count of optimal. \(280\) vs. \(2(95)=190\).
Time = 6.02 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.95 \[ \int \left (a+\frac {b}{x^{3/2}}\right )^{2/3} \, dx=-\frac {\left (a+\frac {b}{x^{3/2}}\right )^{2/3} \left (\sqrt [3]{b}-\sqrt [3]{b+a x^{3/2}}\right ) \left (b^{2/3}+\sqrt [3]{b} \sqrt [3]{b+a x^{3/2}}+\left (b+a x^{3/2}\right )^{2/3}\right )^2 \left (3 \left (b+a x^{3/2}\right )^{2/3}+2 \sqrt {3} b^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b+a x^{3/2}}}{\sqrt [3]{b}}}{\sqrt {3}}\right )+2 b^{2/3} \log \left (-\sqrt [3]{b}+\sqrt [3]{b+a x^{3/2}}\right )-b^{2/3} \log \left (b^{2/3}+\sqrt [3]{b} \sqrt [3]{b+a x^{3/2}}+\left (b+a x^{3/2}\right )^{2/3}\right )\right )}{3 a \sqrt {x} \sqrt [3]{b+a x^{3/2}} \left (b+a x^{3/2}+b^{2/3} \sqrt [3]{b+a x^{3/2}}+\sqrt [3]{b} \left (b+a x^{3/2}\right )^{2/3}\right )} \] Input:
Integrate[(a + b/x^(3/2))^(2/3),x]
Output:
-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)))
Time = 0.37 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {774, 858, 809, 769}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \left (a+\frac {b}{x^{3/2}}\right )^{2/3} \, dx\) |
\(\Big \downarrow \) 774 |
\(\displaystyle 2 \int \left (a+\frac {b}{x^{3/2}}\right )^{2/3} \sqrt {x}d\sqrt {x}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -2 \int \frac {\left (b x^{3/2}+a\right )^{2/3}}{x^{3/2}}d\frac {1}{\sqrt {x}}\) |
\(\Big \downarrow \) 809 |
\(\displaystyle -2 \left (b \int \frac {1}{\sqrt [3]{b x^{3/2}+a}}d\frac {1}{\sqrt {x}}-\frac {\left (a+b x^{3/2}\right )^{2/3}}{2 x}\right )\) |
\(\Big \downarrow \) 769 |
\(\displaystyle -2 \left (b \left (\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b}}{\sqrt {x} \sqrt [3]{a+b x^{3/2}}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a+b x^{3/2}}-\frac {\sqrt [3]{b}}{\sqrt {x}}\right )}{2 \sqrt [3]{b}}\right )-\frac {\left (a+b x^{3/2}\right )^{2/3}}{2 x}\right )\) |
Input:
Int[(a + b/x^(3/2))^(2/3),x]
Output:
-2*(-1/2*(a + b*x^(3/2))^(2/3)/x + b*(ArcTan[(1 + (2*b^(1/3))/(Sqrt[x]*(a + b*x^(3/2))^(1/3)))/Sqrt[3]]/(Sqrt[3]*b^(1/3)) - Log[-(b^(1/3)/Sqrt[x]) + (a + b*x^(3/2))^(1/3)]/(2*b^(1/3))))
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]]/(Sqrt[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]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Simp[k Subst[Int[x^(k - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; Fre eQ[{a, b, p}, x] && FractionQ[n]
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] - Simp[b*n*(p/(c^n*(m + 1))) I nt[(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] && IntB inomialQ[a, b, c, n, m, p, x]
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] && Int egerQ[m]
\[\int \left (a +\frac {b}{x^{\frac {3}{2}}}\right )^{\frac {2}{3}}d x\]
Input:
int((a+b/x^(3/2))^(2/3),x)
Output:
int((a+b/x^(3/2))^(2/3),x)
Timed out. \[ \int \left (a+\frac {b}{x^{3/2}}\right )^{2/3} \, dx=\text {Timed out} \] Input:
integrate((a+b/x^(3/2))^(2/3),x, algorithm="fricas")
Output:
Timed out
Result contains complex when optimal does not.
Time = 1.93 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.48 \[ \int \left (a+\frac {b}{x^{3/2}}\right )^{2/3} \, dx=- \frac {2 a^{\frac {2}{3}} x \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {2}{3} \\ \frac {1}{3} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{\frac {3}{2}}}} \right )}}{3 \Gamma \left (\frac {1}{3}\right )} \] Input:
integrate((a+b/x**(3/2))**(2/3),x)
Output:
-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))
Time = 0.11 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.15 \[ \int \left (a+\frac {b}{x^{3/2}}\right )^{2/3} \, dx=\frac {2}{3} \, \sqrt {3} b^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (a + \frac {b}{x^{\frac {3}{2}}}\right )}^{\frac {1}{3}} \sqrt {x} + b^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right ) + {\left (a + \frac {b}{x^{\frac {3}{2}}}\right )}^{\frac {2}{3}} x - \frac {1}{3} \, b^{\frac {2}{3}} \log \left ({\left (a + \frac {b}{x^{\frac {3}{2}}}\right )}^{\frac {2}{3}} x + {\left (a + \frac {b}{x^{\frac {3}{2}}}\right )}^{\frac {1}{3}} b^{\frac {1}{3}} \sqrt {x} + b^{\frac {2}{3}}\right ) + \frac {2}{3} \, b^{\frac {2}{3}} \log \left ({\left (a + \frac {b}{x^{\frac {3}{2}}}\right )}^{\frac {1}{3}} \sqrt {x} - b^{\frac {1}{3}}\right ) \] Input:
integrate((a+b/x^(3/2))^(2/3),x, algorithm="maxima")
Output:
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))
\[ \int \left (a+\frac {b}{x^{3/2}}\right )^{2/3} \, dx=\int { {\left (a + \frac {b}{x^{\frac {3}{2}}}\right )}^{\frac {2}{3}} \,d x } \] Input:
integrate((a+b/x^(3/2))^(2/3),x, algorithm="giac")
Output:
integrate((a + b/x^(3/2))^(2/3), x)
Time = 0.56 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.39 \[ \int \left (a+\frac {b}{x^{3/2}}\right )^{2/3} \, dx=\frac {x\,{\left (a+\frac {b}{x^{3/2}}\right )}^{2/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {2}{3},-\frac {2}{3};\ \frac {1}{3};\ -\frac {b}{a\,x^{3/2}}\right )}{{\left (\frac {b}{a\,x^{3/2}}+1\right )}^{2/3}} \] Input:
int((a + b/x^(3/2))^(2/3),x)
Output:
(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)
\[ \int \left (a+\frac {b}{x^{3/2}}\right )^{2/3} \, dx=\left (\sqrt {x}\, a x +b \right )^{\frac {2}{3}}-\left (\int \frac {\left (\sqrt {x}\, a x +b \right )^{\frac {2}{3}}}{a^{2} x^{4}-b^{2} x}d x \right ) b^{2}+\left (\int \frac {\sqrt {x}\, \left (\sqrt {x}\, a x +b \right )^{\frac {2}{3}}}{a^{2} x^{3}-b^{2}}d x \right ) a b \] Input:
int((a+b/x^(3/2))^(2/3),x)
Output:
(sqrt(x)*a*x + b)**(2/3) - int((sqrt(x)*a*x + b)**(2/3)/(a**2*x**4 - b**2* x),x)*b**2 + int((sqrt(x)*(sqrt(x)*a*x + b)**(2/3))/(a**2*x**3 - b**2),x)* a*b