\(\int \frac {(a \sin (e+f x))^m}{(b \tan (e+f x))^{3/2}} \, dx\) [173]

Optimal result

Integrand size = 23, antiderivative size = 79 \[ \int \frac {(a \sin (e+f x))^m}{(b \tan (e+f x))^{3/2}} \, dx=-\frac {2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{4} (-1+2 m),\frac {1}{4} (3+2 m),\sin ^2(e+f x)\right ) (a \sin (e+f x))^m}{b f (1-2 m) \sqrt [4]{\cos ^2(e+f x)} \sqrt {b \tan (e+f x)}} \] Output:

-2*hypergeom([-1/4, -1/4+1/2*m],[3/4+1/2*m],sin(f*x+e)^2)*(a*sin(f*x+e))^m 
/b/f/(1-2*m)/(cos(f*x+e)^2)^(1/4)/(b*tan(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 1.24 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.10 \[ \int \frac {(a \sin (e+f x))^m}{(b \tan (e+f x))^{3/2}} \, dx=\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {2+m}{2},\frac {1}{4} (-1+2 m),\frac {1}{4} (3+2 m),-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{m/2} (a \sin (e+f x))^m}{b f (-1+2 m) \sqrt {b \tan (e+f x)}} \] Input:

Integrate[(a*Sin[e + f*x])^m/(b*Tan[e + f*x])^(3/2),x]
 

Output:

(2*Hypergeometric2F1[(2 + m)/2, (-1 + 2*m)/4, (3 + 2*m)/4, -Tan[e + f*x]^2 
]*(Sec[e + f*x]^2)^(m/2)*(a*Sin[e + f*x])^m)/(b*f*(-1 + 2*m)*Sqrt[b*Tan[e 
+ f*x]])
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3082, 3042, 3057}

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.

\(\displaystyle \int \frac {(a \sin (e+f x))^m}{(b \tan (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x))^m}{(b \tan (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3082

\(\displaystyle \frac {a \sqrt {a \sin (e+f x)} \int \cos ^{\frac {3}{2}}(e+f x) (a \sin (e+f x))^{m-\frac {3}{2}}dx}{b \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \sqrt {a \sin (e+f x)} \int \cos (e+f x)^{3/2} (a \sin (e+f x))^{m-\frac {3}{2}}dx}{b \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}\)

\(\Big \downarrow \) 3057

\(\displaystyle -\frac {2 (a \sin (e+f x))^m \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{4} (2 m-1),\frac {1}{4} (2 m+3),\sin ^2(e+f x)\right )}{b f (1-2 m) \sqrt [4]{\cos ^2(e+f x)} \sqrt {b \tan (e+f x)}}\)

Input:

Int[(a*Sin[e + f*x])^m/(b*Tan[e + f*x])^(3/2),x]
 

Output:

(-2*Hypergeometric2F1[-1/4, (-1 + 2*m)/4, (3 + 2*m)/4, Sin[e + f*x]^2]*(a* 
Sin[e + f*x])^m)/(b*f*(1 - 2*m)*(Cos[e + f*x]^2)^(1/4)*Sqrt[b*Tan[e + f*x] 
])
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3057

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*Frac 
Part[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^2)^Fr 
acPart[(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 3082

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[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]
 

Maple [F]

\[\int \frac {\left (a \sin \left (f x +e \right )\right )^{m}}{\left (b \tan \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]

Input:

int((a*sin(f*x+e))^m/(b*tan(f*x+e))^(3/2),x)
 

Output:

int((a*sin(f*x+e))^m/(b*tan(f*x+e))^(3/2),x)
 

Fricas [F]

\[ \int \frac {(a \sin (e+f x))^m}{(b \tan (e+f x))^{3/2}} \, dx=\int { \frac {\left (a \sin \left (f x + e\right )\right )^{m}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((a*sin(f*x+e))^m/(b*tan(f*x+e))^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {(a \sin (e+f x))^m}{(b \tan (e+f x))^{3/2}} \, dx=\int \frac {\left (a \sin {\left (e + f x \right )}\right )^{m}}{\left (b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((a*sin(f*x+e))**m/(b*tan(f*x+e))**(3/2),x)
 

Output:

Integral((a*sin(e + f*x))**m/(b*tan(e + f*x))**(3/2), x)
 

Maxima [F]

\[ \int \frac {(a \sin (e+f x))^m}{(b \tan (e+f x))^{3/2}} \, dx=\int { \frac {\left (a \sin \left (f x + e\right )\right )^{m}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((a*sin(f*x+e))^m/(b*tan(f*x+e))^(3/2),x, algorithm="maxima")
 

Output:

integrate((a*sin(f*x + e))^m/(b*tan(f*x + e))^(3/2), x)
 

Giac [F]

\[ \int \frac {(a \sin (e+f x))^m}{(b \tan (e+f x))^{3/2}} \, dx=\int { \frac {\left (a \sin \left (f x + e\right )\right )^{m}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((a*sin(f*x+e))^m/(b*tan(f*x+e))^(3/2),x, algorithm="giac")
 

Output:

integrate((a*sin(f*x + e))^m/(b*tan(f*x + e))^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a \sin (e+f x))^m}{(b \tan (e+f x))^{3/2}} \, dx=\int \frac {{\left (a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:

int((a*sin(e + f*x))^m/(b*tan(e + f*x))^(3/2),x)
 

Output:

int((a*sin(e + f*x))^m/(b*tan(e + f*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {(a \sin (e+f x))^m}{(b \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {b}\, a^{m} \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \sin \left (f x +e \right )^{m}}{\tan \left (f x +e \right )^{2}}d x \right )}{b^{2}} \] Input:

int((a*sin(f*x+e))^m/(b*tan(f*x+e))^(3/2),x)
 

Output:

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