\(\int \frac {(b \csc (e+f x))^m}{(d \tan (e+f x))^{3/2}} \, dx\) [386]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.10 \[ \int \frac {(b \csc (e+f x))^m}{(d \tan (e+f x))^{3/2}} \, dx=-\frac {2 (b \csc (e+f x))^m \operatorname {Hypergeometric2F1}\left (\frac {1}{4} (-1-2 m),1-\frac {m}{2},\frac {1}{4} (3-2 m),-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{-m/2}}{d f (1+2 m) \sqrt {d \tan (e+f x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3098, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3098

\(\displaystyle \left (\frac {\sin (e+f x)}{b}\right )^m (b \csc (e+f x))^m \int \frac {\left (\frac {\sin (e+f x)}{b}\right )^{-m}}{(d \tan (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \left (\frac {\sin (e+f x)}{b}\right )^m (b \csc (e+f x))^m \int \frac {\left (\frac {\sin (e+f x)}{b}\right )^{-m}}{(d \tan (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3082

\(\displaystyle \frac {\left (\frac {\sin (e+f x)}{b}\right )^{m+\frac {1}{2}} (b \csc (e+f x))^m \int \cos ^{\frac {3}{2}}(e+f x) \left (\frac {\sin (e+f x)}{b}\right )^{-m-\frac {3}{2}}dx}{b d \sqrt {\cos (e+f x)} \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\left (\frac {\sin (e+f x)}{b}\right )^{m+\frac {1}{2}} (b \csc (e+f x))^m \int \cos (e+f x)^{3/2} \left (\frac {\sin (e+f x)}{b}\right )^{-m-\frac {3}{2}}dx}{b d \sqrt {\cos (e+f x)} \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 3057

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

Input:

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

Output:

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

rule 3098

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[(a*Csc[e + f*x])^FracPart[m]*(Sin[e + f*x]/a)^FracPar 
t[m]   Int[(b*Tan[e + f*x])^n/(Sin[e + f*x]/a)^m, x], x] /; FreeQ[{a, b, e, 
 f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

integrate((b*csc(f*x+e))**m/(d*tan(f*x+e))**(3/2),x)
 

Output:

Integral((b*csc(e + f*x))**m/(d*tan(e + f*x))**(3/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(d)*b**m*int((sqrt(tan(e + f*x))*csc(e + f*x)**m)/tan(e + f*x)**2,x)) 
/d**2