\(\int \frac {1}{x^5 (a+b x^{3/2})^{2/3}} \, dx\) [2284]

Optimal result

Integrand size = 17, antiderivative size = 57 \[ \int \frac {1}{x^5 \left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {\left (1+\frac {b x^{3/2}}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {8}{3},\frac {2}{3},-\frac {5}{3},-\frac {b x^{3/2}}{a}\right )}{4 x^4 \left (a+b x^{3/2}\right )^{2/3}} \]

[Out]

-1/4*(1+b*x^(3/2)/a)^(2/3)*hypergeom([-8/3, 2/3],[-5/3],-b*x^(3/2)/a)/x^4/(a+b*x^(3/2))^(2/3)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {348, 372, 371} \[ \int \frac {1}{x^5 \left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {\left (\frac {b x^{3/2}}{a}+1\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {8}{3},\frac {2}{3},-\frac {5}{3},-\frac {b x^{3/2}}{a}\right )}{4 x^4 \left (a+b x^{3/2}\right )^{2/3}} \]

[In]

Int[1/(x^5*(a + b*x^(3/2))^(2/3)),x]

[Out]

-1/4*((1 + (b*x^(3/2))/a)^(2/3)*Hypergeometric2F1[-8/3, 2/3, -5/3, -((b*x^(3/2))/a)])/(x^4*(a + b*x^(3/2))^(2/
3))

Rule 348

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[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])

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {1}{x^9 \left (a+b x^3\right )^{2/3}} \, dx,x,\sqrt {x}\right ) \\ & = \frac {\left (2 \left (1+\frac {b x^{3/2}}{a}\right )^{2/3}\right ) \text {Subst}\left (\int \frac {1}{x^9 \left (1+\frac {b x^3}{a}\right )^{2/3}} \, dx,x,\sqrt {x}\right )}{\left (a+b x^{3/2}\right )^{2/3}} \\ & = -\frac {\left (1+\frac {b x^{3/2}}{a}\right )^{2/3} \, _2F_1\left (-\frac {8}{3},\frac {2}{3};-\frac {5}{3};-\frac {b x^{3/2}}{a}\right )}{4 x^4 \left (a+b x^{3/2}\right )^{2/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^5 \left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {\left (1+\frac {b x^{3/2}}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {8}{3},\frac {2}{3},-\frac {5}{3},-\frac {b x^{3/2}}{a}\right )}{4 x^4 \left (a+b x^{3/2}\right )^{2/3}} \]

[In]

Integrate[1/(x^5*(a + b*x^(3/2))^(2/3)),x]

[Out]

-1/4*((1 + (b*x^(3/2))/a)^(2/3)*Hypergeometric2F1[-8/3, 2/3, -5/3, -((b*x^(3/2))/a)])/(x^4*(a + b*x^(3/2))^(2/
3))

Maple [F]

\[\int \frac {1}{x^{5} \left (a +b \,x^{\frac {3}{2}}\right )^{\frac {2}{3}}}d x\]

[In]

int(1/x^5/(a+b*x^(3/2))^(2/3),x)

[Out]

int(1/x^5/(a+b*x^(3/2))^(2/3),x)

Fricas [F]

\[ \int \frac {1}{x^5 \left (a+b x^{3/2}\right )^{2/3}} \, dx=\int { \frac {1}{{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {2}{3}} x^{5}} \,d x } \]

[In]

integrate(1/x^5/(a+b*x^(3/2))^(2/3),x, algorithm="fricas")

[Out]

integral((b*x^(3/2) + a)^(1/3)*(b*x^(3/2) - a)/(b^2*x^8 - a^2*x^5), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.84 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^5 \left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {2 \Gamma \left (- \frac {8}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {8}{3}, \frac {2}{3} \\ - \frac {5}{3} \end {matrix}\middle | {\frac {b x^{\frac {3}{2}} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} x^{4} \Gamma \left (- \frac {5}{3}\right )} \]

[In]

integrate(1/x**5/(a+b*x**(3/2))**(2/3),x)

[Out]

2*gamma(-8/3)*hyper((-8/3, 2/3), (-5/3,), b*x**(3/2)*exp_polar(I*pi)/a)/(3*a**(2/3)*x**4*gamma(-5/3))

Maxima [F]

\[ \int \frac {1}{x^5 \left (a+b x^{3/2}\right )^{2/3}} \, dx=\int { \frac {1}{{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {2}{3}} x^{5}} \,d x } \]

[In]

integrate(1/x^5/(a+b*x^(3/2))^(2/3),x, algorithm="maxima")

[Out]

integrate(1/((b*x^(3/2) + a)^(2/3)*x^5), x)

Giac [F]

\[ \int \frac {1}{x^5 \left (a+b x^{3/2}\right )^{2/3}} \, dx=\int { \frac {1}{{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {2}{3}} x^{5}} \,d x } \]

[In]

integrate(1/x^5/(a+b*x^(3/2))^(2/3),x, algorithm="giac")

[Out]

integrate(1/((b*x^(3/2) + a)^(2/3)*x^5), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^5 \left (a+b x^{3/2}\right )^{2/3}} \, dx=\int \frac {1}{x^5\,{\left (a+b\,x^{3/2}\right )}^{2/3}} \,d x \]

[In]

int(1/(x^5*(a + b*x^(3/2))^(2/3)),x)

[Out]

int(1/(x^5*(a + b*x^(3/2))^(2/3)), x)