\(\int \frac {(a \cos (e+f x))^m}{(b \csc (e+f x))^{5/2}} \, dx\) [294]

Optimal result

Integrand size = 23, antiderivative size = 78 \[ \int \frac {(a \cos (e+f x))^m}{(b \csc (e+f x))^{5/2}} \, dx=-\frac {(a \cos (e+f x))^{1+m} \sqrt {b \csc (e+f x)} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(e+f x)\right ) \sqrt [4]{\sin ^2(e+f x)}}{a b^3 f (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 11.91 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.60 \[ \int \frac {(a \cos (e+f x))^m}{(b \csc (e+f x))^{5/2}} \, dx=\frac {2 (a \cos (e+f x))^m (1+2 \cos (2 (e+f x))) \left (-\cot ^2(e+f x)\right )^{\frac {1-m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4} (-5-2 m),\frac {1-m}{2},\frac {1}{4} (-1-2 m),\csc ^2(e+f x)\right ) \tan (e+f x)}{b^2 f (5+2 m) \sqrt {b \csc (e+f x)} \left (-4+3 \csc ^2(e+f x)\right )} \] Input:

Integrate[(a*Cos[e + f*x])^m/(b*Csc[e + f*x])^(5/2),x]
 

Output:

(2*(a*Cos[e + f*x])^m*(1 + 2*Cos[2*(e + f*x)])*(-Cot[e + f*x]^2)^((1 - m)/ 
2)*Hypergeometric2F1[(-5 - 2*m)/4, (1 - m)/2, (-1 - 2*m)/4, Csc[e + f*x]^2 
]*Tan[e + f*x])/(b^2*f*(5 + 2*m)*Sqrt[b*Csc[e + f*x]]*(-4 + 3*Csc[e + f*x] 
^2))
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 78, 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, 3066, 3042, 3056}

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.

\(\displaystyle \int \frac {(a \cos (e+f x))^m}{(b \csc (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (e+f x+\frac {\pi }{2}\right )\right )^m}{\left (-b \sec \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2}}dx\)

\(\Big \downarrow \) 3066

\(\displaystyle \frac {\int (a \cos (e+f x))^m (b \sin (e+f x))^{5/2}dx}{b^2 (b \sin (e+f x))^{3/2} (b \csc (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (a \cos (e+f x))^m (b \sin (e+f x))^{5/2}dx}{b^2 (b \sin (e+f x))^{3/2} (b \csc (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3056

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

Input:

Int[(a*Cos[e + f*x])^m/(b*Csc[e + f*x])^(5/2),x]
 

Output:

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

Defintions of rubi rules used

rule 3042

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

rule 3056

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

rule 3066

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[(1/b^2)*(b*Cos[e + f*x])^(n + 1)*(b*Sec[e + f*x])^(n 
+ 1)   Int[(a*Sin[e + f*x])^m/(b*Cos[e + f*x])^n, x], x] /; FreeQ[{a, b, e, 
 f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && LtQ[n, 1]
 

Maple [F]

\[\int \frac {\left (\cos \left (f x +e \right ) a \right )^{m}}{\left (b \csc \left (f x +e \right )\right )^{\frac {5}{2}}}d x\]

Input:

int((cos(f*x+e)*a)^m/(b*csc(f*x+e))^(5/2),x)
 

Output:

int((cos(f*x+e)*a)^m/(b*csc(f*x+e))^(5/2),x)
 

Fricas [F]

\[ \int \frac {(a \cos (e+f x))^m}{(b \csc (e+f x))^{5/2}} \, dx=\int { \frac {\left (a \cos \left (f x + e\right )\right )^{m}}{\left (b \csc \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((a*cos(f*x+e))^m/(b*csc(f*x+e))^(5/2),x, algorithm="fricas")
 

Output:

integral(sqrt(b*csc(f*x + e))*(a*cos(f*x + e))^m/(b^3*csc(f*x + e)^3), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a \cos (e+f x))^m}{(b \csc (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((a*cos(f*x+e))**m/(b*csc(f*x+e))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a \cos (e+f x))^m}{(b \csc (e+f x))^{5/2}} \, dx=\int { \frac {\left (a \cos \left (f x + e\right )\right )^{m}}{\left (b \csc \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((a*cos(f*x+e))^m/(b*csc(f*x+e))^(5/2),x, algorithm="maxima")
 

Output:

integrate((a*cos(f*x + e))^m/(b*csc(f*x + e))^(5/2), x)
 

Giac [F]

\[ \int \frac {(a \cos (e+f x))^m}{(b \csc (e+f x))^{5/2}} \, dx=\int { \frac {\left (a \cos \left (f x + e\right )\right )^{m}}{\left (b \csc \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((a*cos(f*x+e))^m/(b*csc(f*x+e))^(5/2),x, algorithm="giac")
 

Output:

integrate((a*cos(f*x + e))^m/(b*csc(f*x + e))^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a \cos (e+f x))^m}{(b \csc (e+f x))^{5/2}} \, dx=\int \frac {{\left (a\,\cos \left (e+f\,x\right )\right )}^m}{{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^{5/2}} \,d x \] Input:

int((a*cos(e + f*x))^m/(b/sin(e + f*x))^(5/2),x)
 

Output:

int((a*cos(e + f*x))^m/(b/sin(e + f*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {(a \cos (e+f x))^m}{(b \csc (e+f x))^{5/2}} \, dx=\frac {\sqrt {b}\, a^{m} \left (\int \frac {\sqrt {\csc \left (f x +e \right )}\, \cos \left (f x +e \right )^{m}}{\csc \left (f x +e \right )^{3}}d x \right )}{b^{3}} \] Input:

int((a*cos(f*x+e))^m/(b*csc(f*x+e))^(5/2),x)
 

Output:

(sqrt(b)*a**m*int((sqrt(csc(e + f*x))*cos(e + f*x)**m)/csc(e + f*x)**3,x)) 
/b**3