\(\int \frac {1}{(b \sin (e+f x))^{5/3}} \, dx\) [321]

Optimal result

Integrand size = 12, antiderivative size = 58 \[ \int \frac {1}{(b \sin (e+f x))^{5/3}} \, dx=-\frac {3 \cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\sin ^2(e+f x)\right )}{2 b f \sqrt {\cos ^2(e+f x)} (b \sin (e+f x))^{2/3}} \]

[Out]

-3/2*cos(f*x+e)*hypergeom([-1/3, 1/2],[2/3],sin(f*x+e)^2)/b/f/(b*sin(f*x+e))^(2/3)/(cos(f*x+e)^2)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2722} \[ \int \frac {1}{(b \sin (e+f x))^{5/3}} \, dx=-\frac {3 \cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\sin ^2(e+f x)\right )}{2 b f \sqrt {\cos ^2(e+f x)} (b \sin (e+f x))^{2/3}} \]

[In]

Int[(b*Sin[e + f*x])^(-5/3),x]

[Out]

(-3*Cos[e + f*x]*Hypergeometric2F1[-1/3, 1/2, 2/3, Sin[e + f*x]^2])/(2*b*f*Sqrt[Cos[e + f*x]^2]*(b*Sin[e + f*x
])^(2/3))

Rule 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps \begin{align*} \text {integral}& = -\frac {3 \cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\sin ^2(e+f x)\right )}{2 b f \sqrt {\cos ^2(e+f x)} (b \sin (e+f x))^{2/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(b \sin (e+f x))^{5/3}} \, dx=-\frac {3 \sqrt {\cos ^2(e+f x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\sin ^2(e+f x)\right ) \tan (e+f x)}{2 f (b \sin (e+f x))^{5/3}} \]

[In]

Integrate[(b*Sin[e + f*x])^(-5/3),x]

[Out]

(-3*Sqrt[Cos[e + f*x]^2]*Hypergeometric2F1[-1/3, 1/2, 2/3, Sin[e + f*x]^2]*Tan[e + f*x])/(2*f*(b*Sin[e + f*x])
^(5/3))

Maple [F]

\[\int \frac {1}{\left (b \sin \left (f x +e \right )\right )^{\frac {5}{3}}}d x\]

[In]

int(1/(b*sin(f*x+e))^(5/3),x)

[Out]

int(1/(b*sin(f*x+e))^(5/3),x)

Fricas [F]

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

[In]

integrate(1/(b*sin(f*x+e))^(5/3),x, algorithm="fricas")

[Out]

integral(-(b*sin(f*x + e))^(1/3)/(b^2*cos(f*x + e)^2 - b^2), x)

Sympy [F]

\[ \int \frac {1}{(b \sin (e+f x))^{5/3}} \, dx=\int \frac {1}{\left (b \sin {\left (e + f x \right )}\right )^{\frac {5}{3}}}\, dx \]

[In]

integrate(1/(b*sin(f*x+e))**(5/3),x)

[Out]

Integral((b*sin(e + f*x))**(-5/3), x)

Maxima [F]

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

[In]

integrate(1/(b*sin(f*x+e))^(5/3),x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e))^(-5/3), x)

Giac [F]

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

[In]

integrate(1/(b*sin(f*x+e))^(5/3),x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e))^(-5/3), x)

Mupad [F(-1)]

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

[In]

int(1/(b*sin(e + f*x))^(5/3),x)

[Out]

int(1/(b*sin(e + f*x))^(5/3), x)