\(\int \frac {\sec ^4(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx\) [323]

Optimal result

Integrand size = 21, antiderivative size = 58 \[ \int \frac {\sec ^4(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx=-\frac {3 \sqrt {\cos ^2(e+f x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {5}{2},\frac {2}{3},\sin ^2(e+f x)\right ) \sec (e+f x)}{2 b f (b \sin (e+f x))^{2/3}} \]

[Out]

-3/2*hypergeom([-1/3, 5/2],[2/3],sin(f*x+e)^2)*sec(f*x+e)*(cos(f*x+e)^2)^(1/2)/b/f/(b*sin(f*x+e))^(2/3)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2657} \[ \int \frac {\sec ^4(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx=-\frac {3 \sqrt {\cos ^2(e+f x)} \sec (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {5}{2},\frac {2}{3},\sin ^2(e+f x)\right )}{2 b f (b \sin (e+f x))^{2/3}} \]

[In]

Int[Sec[e + f*x]^4/(b*Sin[e + f*x])^(5/3),x]

[Out]

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

Rule 2657

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b^(2*IntPart[
(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^
2)^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {3 \sqrt {\cos ^2(e+f x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {5}{2},\frac {2}{3},\sin ^2(e+f x)\right ) \sec (e+f x)}{2 b f (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 \[ \int \frac {\sec ^4(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx=-\frac {3 \sqrt {\cos ^2(e+f x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {5}{2},\frac {2}{3},\sin ^2(e+f x)\right ) \tan (e+f x)}{2 f (b \sin (e+f x))^{5/3}} \]

[In]

Integrate[Sec[e + f*x]^4/(b*Sin[e + f*x])^(5/3),x]

[Out]

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

Maple [F]

\[\int \frac {\sec ^{4}\left (f x +e \right )}{\left (b \sin \left (f x +e \right )\right )^{\frac {5}{3}}}d x\]

[In]

int(sec(f*x+e)^4/(b*sin(f*x+e))^(5/3),x)

[Out]

int(sec(f*x+e)^4/(b*sin(f*x+e))^(5/3),x)

Fricas [F]

\[ \int \frac {\sec ^4(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx=\int { \frac {\sec \left (f x + e\right )^{4}}{\left (b \sin \left (f x + e\right )\right )^{\frac {5}{3}}} \,d x } \]

[In]

integrate(sec(f*x+e)^4/(b*sin(f*x+e))^(5/3),x, algorithm="fricas")

[Out]

integral(-(b*sin(f*x + e))^(1/3)*sec(f*x + e)^4/(b^2*cos(f*x + e)^2 - b^2), x)

Sympy [F]

\[ \int \frac {\sec ^4(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx=\int \frac {\sec ^{4}{\left (e + f x \right )}}{\left (b \sin {\left (e + f x \right )}\right )^{\frac {5}{3}}}\, dx \]

[In]

integrate(sec(f*x+e)**4/(b*sin(f*x+e))**(5/3),x)

[Out]

Integral(sec(e + f*x)**4/(b*sin(e + f*x))**(5/3), x)

Maxima [F(-1)]

Timed out. \[ \int \frac {\sec ^4(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx=\text {Timed out} \]

[In]

integrate(sec(f*x+e)^4/(b*sin(f*x+e))^(5/3),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {\sec ^4(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx=\int { \frac {\sec \left (f x + e\right )^{4}}{\left (b \sin \left (f x + e\right )\right )^{\frac {5}{3}}} \,d x } \]

[In]

integrate(sec(f*x+e)^4/(b*sin(f*x+e))^(5/3),x, algorithm="giac")

[Out]

integrate(sec(f*x + e)^4/(b*sin(f*x + e))^(5/3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^4(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx=\int \frac {1}{{\cos \left (e+f\,x\right )}^4\,{\left (b\,\sin \left (e+f\,x\right )\right )}^{5/3}} \,d x \]

[In]

int(1/(cos(e + f*x)^4*(b*sin(e + f*x))^(5/3)),x)

[Out]

int(1/(cos(e + f*x)^4*(b*sin(e + f*x))^(5/3)), x)