\(\int \frac {(a \cos (e+f x))^m}{\sqrt {b \csc (e+f x)}} \, dx\) [292]

Optimal result

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

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 12.15 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.88 \[ \int \frac {(a \cos (e+f x))^m}{\sqrt {b \csc (e+f x)}} \, dx=\frac {14 b \operatorname {AppellF1}\left (\frac {3}{4},-m,\frac {3}{2}+m,\frac {7}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (a \cos (e+f x))^m}{3 f (b \csc (e+f x))^{3/2} \left (7 \operatorname {AppellF1}\left (\frac {3}{4},-m,\frac {3}{2}+m,\frac {7}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 \left (2 m \operatorname {AppellF1}\left (\frac {7}{4},1-m,\frac {3}{2}+m,\frac {11}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+(3+2 m) \operatorname {AppellF1}\left (\frac {7}{4},-m,\frac {5}{2}+m,\frac {11}{4},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )} \] Input:

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

Output:

(14*b*AppellF1[3/4, -m, 3/2 + m, 7/4, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2 
]^2]*(a*Cos[e + f*x])^m)/(3*f*(b*Csc[e + f*x])^(3/2)*(7*AppellF1[3/4, -m, 
3/2 + m, 7/4, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*(2*m*AppellF1[7 
/4, 1 - m, 3/2 + m, 11/4, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (3 + 
2*m)*AppellF1[7/4, -m, 5/2 + m, 11/4, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2 
]^2])*Tan[(e + f*x)/2]^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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3066

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3056

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

Input:

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

Output:

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

integrate((a*cos(f*x+e))**m/(b*csc(f*x+e))**(1/2),x)
 

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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