\(\int \frac {(a \sin (e+f x))^m}{\sqrt {b \tan (e+f x)}} \, dx\) [172]

Optimal result

Integrand size = 23, antiderivative size = 79 \[ \int \frac {(a \sin (e+f x))^m}{\sqrt {b \tan (e+f x)}} \, dx=\frac {2 \sqrt [4]{\cos ^2(e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{4} (1+2 m),\frac {1}{4} (5+2 m),\sin ^2(e+f x)\right ) (a \sin (e+f x))^m \sqrt {b \tan (e+f x)}}{b f (1+2 m)} \]

[Out]

2*(cos(f*x+e)^2)^(1/4)*hypergeom([1/4, 1/4+1/2*m],[5/4+1/2*m],sin(f*x+e)^2)*(a*sin(f*x+e))^m*(b*tan(f*x+e))^(1
/2)/b/f/(1+2*m)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2682, 2657} \[ \int \frac {(a \sin (e+f x))^m}{\sqrt {b \tan (e+f x)}} \, dx=\frac {2 \sqrt [4]{\cos ^2(e+f x)} \sqrt {b \tan (e+f x)} (a \sin (e+f x))^m \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{4} (2 m+1),\frac {1}{4} (2 m+5),\sin ^2(e+f x)\right )}{b f (2 m+1)} \]

[In]

Int[(a*Sin[e + f*x])^m/Sqrt[b*Tan[e + f*x]],x]

[Out]

(2*(Cos[e + f*x]^2)^(1/4)*Hypergeometric2F1[1/4, (1 + 2*m)/4, (5 + 2*m)/4, Sin[e + f*x]^2]*(a*Sin[e + f*x])^m*
Sqrt[b*Tan[e + f*x]])/(b*f*(1 + 2*m))

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]

Rule 2682

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[a*Cos[e + f*
x]^(n + 1)*((b*Tan[e + f*x])^(n + 1)/(b*(a*Sin[e + f*x])^(n + 1))), Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^
n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[n]

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

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.10 \[ \int \frac {(a \sin (e+f x))^m}{\sqrt {b \tan (e+f x)}} \, dx=\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {2+m}{2},\frac {1}{4} (1+2 m),\frac {1}{4} (5+2 m),-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{m/2} (a \sin (e+f x))^m \sqrt {b \tan (e+f x)}}{b f (1+2 m)} \]

[In]

Integrate[(a*Sin[e + f*x])^m/Sqrt[b*Tan[e + f*x]],x]

[Out]

(2*Hypergeometric2F1[(2 + m)/2, (1 + 2*m)/4, (5 + 2*m)/4, -Tan[e + f*x]^2]*(Sec[e + f*x]^2)^(m/2)*(a*Sin[e + f
*x])^m*Sqrt[b*Tan[e + f*x]])/(b*f*(1 + 2*m))

Maple [F]

\[\int \frac {\left (\sin \left (f x +e \right ) a \right )^{m}}{\sqrt {b \tan \left (f x +e \right )}}d x\]

[In]

int((sin(f*x+e)*a)^m/(b*tan(f*x+e))^(1/2),x)

[Out]

int((sin(f*x+e)*a)^m/(b*tan(f*x+e))^(1/2),x)

Fricas [F]

\[ \int \frac {(a \sin (e+f x))^m}{\sqrt {b \tan (e+f x)}} \, dx=\int { \frac {\left (a \sin \left (f x + e\right )\right )^{m}}{\sqrt {b \tan \left (f x + e\right )}} \,d x } \]

[In]

integrate((a*sin(f*x+e))^m/(b*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*tan(f*x + e))*(a*sin(f*x + e))^m/(b*tan(f*x + e)), x)

Sympy [F]

\[ \int \frac {(a \sin (e+f x))^m}{\sqrt {b \tan (e+f x)}} \, dx=\int \frac {\left (a \sin {\left (e + f x \right )}\right )^{m}}{\sqrt {b \tan {\left (e + f x \right )}}}\, dx \]

[In]

integrate((a*sin(f*x+e))**m/(b*tan(f*x+e))**(1/2),x)

[Out]

Integral((a*sin(e + f*x))**m/sqrt(b*tan(e + f*x)), x)

Maxima [F]

\[ \int \frac {(a \sin (e+f x))^m}{\sqrt {b \tan (e+f x)}} \, dx=\int { \frac {\left (a \sin \left (f x + e\right )\right )^{m}}{\sqrt {b \tan \left (f x + e\right )}} \,d x } \]

[In]

integrate((a*sin(f*x+e))^m/(b*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e))^m/sqrt(b*tan(f*x + e)), x)

Giac [F]

\[ \int \frac {(a \sin (e+f x))^m}{\sqrt {b \tan (e+f x)}} \, dx=\int { \frac {\left (a \sin \left (f x + e\right )\right )^{m}}{\sqrt {b \tan \left (f x + e\right )}} \,d x } \]

[In]

integrate((a*sin(f*x+e))^m/(b*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e))^m/sqrt(b*tan(f*x + e)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a \sin (e+f x))^m}{\sqrt {b \tan (e+f x)}} \, dx=\int \frac {{\left (a\,\sin \left (e+f\,x\right )\right )}^m}{\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \]

[In]

int((a*sin(e + f*x))^m/(b*tan(e + f*x))^(1/2),x)

[Out]

int((a*sin(e + f*x))^m/(b*tan(e + f*x))^(1/2), x)