Integrand size = 11, antiderivative size = 50 \[ \int \left (a+\frac {b}{x^{3/2}}\right )^p \, dx=\left (a+\frac {b}{x^{3/2}}\right )^p \left (1+\frac {b}{a x^{3/2}}\right )^{-p} x \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},-p,\frac {1}{3},-\frac {b}{a x^{3/2}}\right ) \] Output:
(a+b/x^(3/2))^p*x*hypergeom([-2/3, -p],[1/3],-b/a/x^(3/2))/((1+b/a/x^(3/2) )^p)
Time = 0.22 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int \left (a+\frac {b}{x^{3/2}}\right )^p \, dx=\left (a+\frac {b}{x^{3/2}}\right )^p \left (1+\frac {b}{a x^{3/2}}\right )^{-p} x \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},-p,\frac {1}{3},-\frac {b}{a x^{3/2}}\right ) \] Input:
Integrate[(a + b/x^(3/2))^p,x]
Output:
((a + b/x^(3/2))^p*x*Hypergeometric2F1[-2/3, -p, 1/3, -(b/(a*x^(3/2)))])/( 1 + b/(a*x^(3/2)))^p
Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {774, 858, 889, 888}
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.
\(\displaystyle \int \left (a+\frac {b}{x^{3/2}}\right )^p \, dx\) |
\(\Big \downarrow \) 774 |
\(\displaystyle 2 \int \left (a+\frac {b}{x^{3/2}}\right )^p \sqrt {x}d\sqrt {x}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -2 \int \frac {\left (b x^{3/2}+a\right )^p}{x^{3/2}}d\frac {1}{\sqrt {x}}\) |
\(\Big \downarrow \) 889 |
\(\displaystyle -2 \left (a+b x^{3/2}\right )^p \left (\frac {b x^{3/2}}{a}+1\right )^{-p} \int \frac {\left (\frac {b x^{3/2}}{a}+1\right )^p}{x^{3/2}}d\frac {1}{\sqrt {x}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {\left (a+b x^{3/2}\right )^p \left (\frac {b x^{3/2}}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},-p,\frac {1}{3},-\frac {b x^{3/2}}{a}\right )}{x}\) |
Input:
Int[(a + b/x^(3/2))^p,x]
Output:
((a + b*x^(3/2))^p*Hypergeometric2F1[-2/3, -p, 1/3, -((b*x^(3/2))/a)])/(x* (1 + (b*x^(3/2))/a)^p)
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[(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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^I ntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]) Int[(c*x) ^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0 ] && !(ILtQ[p, 0] || GtQ[a, 0])
\[\int \left (a +\frac {b}{x^{\frac {3}{2}}}\right )^{p}d x\]
Input:
int((a+b/x^(3/2))^p,x)
Output:
int((a+b/x^(3/2))^p,x)
\[ \int \left (a+\frac {b}{x^{3/2}}\right )^p \, dx=\int { {\left (a + \frac {b}{x^{\frac {3}{2}}}\right )}^{p} \,d x } \] Input:
integrate((a+b/x^(3/2))^p,x, algorithm="fricas")
Output:
integral(((a*x^2 + b*sqrt(x))/x^2)^p, x)
Timed out. \[ \int \left (a+\frac {b}{x^{3/2}}\right )^p \, dx=\text {Timed out} \] Input:
integrate((a+b/x**(3/2))**p,x)
Output:
Timed out
\[ \int \left (a+\frac {b}{x^{3/2}}\right )^p \, dx=\int { {\left (a + \frac {b}{x^{\frac {3}{2}}}\right )}^{p} \,d x } \] Input:
integrate((a+b/x^(3/2))^p,x, algorithm="maxima")
Output:
integrate((a + b/x^(3/2))^p, x)
\[ \int \left (a+\frac {b}{x^{3/2}}\right )^p \, dx=\int { {\left (a + \frac {b}{x^{\frac {3}{2}}}\right )}^{p} \,d x } \] Input:
integrate((a+b/x^(3/2))^p,x, algorithm="giac")
Output:
integrate((a + b/x^(3/2))^p, x)
Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.82 \[ \int \left (a+\frac {b}{x^{3/2}}\right )^p \, dx=\frac {x\,{\left (a+\frac {b}{x^{3/2}}\right )}^p\,{{}}_2{\mathrm {F}}_1\left (-\frac {2}{3},-p;\ \frac {1}{3};\ -\frac {b}{a\,x^{3/2}}\right )}{{\left (\frac {b}{a\,x^{3/2}}+1\right )}^p} \] Input:
int((a + b/x^(3/2))^p,x)
Output:
(x*(a + b/x^(3/2))^p*hypergeom([-2/3, -p], 1/3, -b/(a*x^(3/2))))/(b/(a*x^( 3/2)) + 1)^p
\[ \int \left (a+\frac {b}{x^{3/2}}\right )^p \, dx=\frac {2 \left (\sqrt {x}\, a x +b \right )^{p} x -3 x^{\frac {3 p}{2}} \left (\int \frac {\left (\sqrt {x}\, a x +b \right )^{p}}{x^{\frac {3 p}{2}} a^{2} x^{3}-x^{\frac {3 p}{2}} b^{2}}d x \right ) b^{2} p +3 x^{\frac {3 p}{2}} \left (\int \frac {\sqrt {x}\, \left (\sqrt {x}\, a x +b \right )^{p} x}{x^{\frac {3 p}{2}} a^{2} x^{3}-x^{\frac {3 p}{2}} b^{2}}d x \right ) a b p}{2 x^{\frac {3 p}{2}}} \] Input:
int((a+b/x^(3/2))^p,x)
Output:
(2*(sqrt(x)*a*x + b)**p*x - 3*x**((3*p)/2)*int((sqrt(x)*a*x + b)**p/(x**(( 3*p)/2)*a**2*x**3 - x**((3*p)/2)*b**2),x)*b**2*p + 3*x**((3*p)/2)*int((sqr t(x)*(sqrt(x)*a*x + b)**p*x)/(x**((3*p)/2)*a**2*x**3 - x**((3*p)/2)*b**2), x)*a*b*p)/(2*x**((3*p)/2))