\(\int \cos ^4(e+f x) (b \csc (e+f x))^n \, dx\) [295]

Optimal result

Integrand size = 19, antiderivative size = 72 \[ \int \cos ^4(e+f x) (b \csc (e+f x))^n \, dx=\frac {b \cos (e+f x) (b \csc (e+f x))^{-1+n} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(e+f x)\right )}{f (1-n) \sqrt {\cos ^2(e+f x)}} \] Output:

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(246\) vs. \(2(72)=144\).

Time = 0.86 (sec) , antiderivative size = 246, normalized size of antiderivative = 3.42 \[ \int \cos ^4(e+f x) (b \csc (e+f x))^n \, dx=-\frac {2 (b \csc (e+f x))^n \left (\operatorname {Hypergeometric2F1}\left (1-n,\frac {1}{2}-\frac {n}{2},\frac {3}{2}-\frac {n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-8 \left (\operatorname {Hypergeometric2F1}\left (2-n,\frac {1}{2}-\frac {n}{2},\frac {3}{2}-\frac {n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-3 \operatorname {Hypergeometric2F1}\left (3-n,\frac {1}{2}-\frac {n}{2},\frac {3}{2}-\frac {n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+4 \operatorname {Hypergeometric2F1}\left (4-n,\frac {1}{2}-\frac {n}{2},\frac {3}{2}-\frac {n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 \operatorname {Hypergeometric2F1}\left (5-n,\frac {1}{2}-\frac {n}{2},\frac {3}{2}-\frac {n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )\right ) \sec ^2\left (\frac {1}{2} (e+f x)\right )^{-n} \tan \left (\frac {1}{2} (e+f x)\right )}{f (-1+n)} \] Input:

Integrate[Cos[e + f*x]^4*(b*Csc[e + f*x])^n,x]
 

Output:

(-2*(b*Csc[e + f*x])^n*(Hypergeometric2F1[1 - n, 1/2 - n/2, 3/2 - n/2, -Ta 
n[(e + f*x)/2]^2] - 8*(Hypergeometric2F1[2 - n, 1/2 - n/2, 3/2 - n/2, -Tan 
[(e + f*x)/2]^2] - 3*Hypergeometric2F1[3 - n, 1/2 - n/2, 3/2 - n/2, -Tan[( 
e + f*x)/2]^2] + 4*Hypergeometric2F1[4 - n, 1/2 - n/2, 3/2 - n/2, -Tan[(e 
+ f*x)/2]^2] - 2*Hypergeometric2F1[5 - n, 1/2 - n/2, 3/2 - n/2, -Tan[(e + 
f*x)/2]^2]))*Tan[(e + f*x)/2])/(f*(-1 + n)*(Sec[(e + f*x)/2]^2)^n)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3111, 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.

\(\displaystyle \int \cos ^4(e+f x) (b \csc (e+f x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(b \csc (e+f x))^n}{\sec (e+f x)^4}dx\)

\(\Big \downarrow \) 3111

\(\displaystyle b^2 (b \sin (e+f x))^{n-1} (b \csc (e+f x))^{n-1} \int \cos ^4(e+f x) (b \sin (e+f x))^{-n}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle b^2 (b \sin (e+f x))^{n-1} (b \csc (e+f x))^{n-1} \int \cos (e+f x)^4 (b \sin (e+f x))^{-n}dx\)

\(\Big \downarrow \) 3057

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

Input:

Int[Cos[e + f*x]^4*(b*Csc[e + f*x])^n,x]
 

Output:

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

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 3111

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

Maple [F]

\[\int \cos \left (f x +e \right )^{4} \left (b \csc \left (f x +e \right )\right )^{n}d x\]

Input:

int(cos(f*x+e)^4*(b*csc(f*x+e))^n,x)
 

Output:

int(cos(f*x+e)^4*(b*csc(f*x+e))^n,x)
 

Fricas [F]

\[ \int \cos ^4(e+f x) (b \csc (e+f x))^n \, dx=\int { \left (b \csc \left (f x + e\right )\right )^{n} \cos \left (f x + e\right )^{4} \,d x } \] Input:

integrate(cos(f*x+e)^4*(b*csc(f*x+e))^n,x, algorithm="fricas")
 

Output:

integral((b*csc(f*x + e))^n*cos(f*x + e)^4, x)
 

Sympy [F]

\[ \int \cos ^4(e+f x) (b \csc (e+f x))^n \, dx=\int \left (b \csc {\left (e + f x \right )}\right )^{n} \cos ^{4}{\left (e + f x \right )}\, dx \] Input:

integrate(cos(f*x+e)**4*(b*csc(f*x+e))**n,x)
 

Output:

Integral((b*csc(e + f*x))**n*cos(e + f*x)**4, x)
 

Maxima [F]

\[ \int \cos ^4(e+f x) (b \csc (e+f x))^n \, dx=\int { \left (b \csc \left (f x + e\right )\right )^{n} \cos \left (f x + e\right )^{4} \,d x } \] Input:

integrate(cos(f*x+e)^4*(b*csc(f*x+e))^n,x, algorithm="maxima")
 

Output:

integrate((b*csc(f*x + e))^n*cos(f*x + e)^4, x)
 

Giac [F]

\[ \int \cos ^4(e+f x) (b \csc (e+f x))^n \, dx=\int { \left (b \csc \left (f x + e\right )\right )^{n} \cos \left (f x + e\right )^{4} \,d x } \] Input:

integrate(cos(f*x+e)^4*(b*csc(f*x+e))^n,x, algorithm="giac")
 

Output:

integrate((b*csc(f*x + e))^n*cos(f*x + e)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^4(e+f x) (b \csc (e+f x))^n \, dx=\int {\cos \left (e+f\,x\right )}^4\,{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^n \,d x \] Input:

int(cos(e + f*x)^4*(b/sin(e + f*x))^n,x)
 

Output:

int(cos(e + f*x)^4*(b/sin(e + f*x))^n, x)
 

Reduce [F]

\[ \int \cos ^4(e+f x) (b \csc (e+f x))^n \, dx=b^{n} \left (\int \csc \left (f x +e \right )^{n} \cos \left (f x +e \right )^{4}d x \right ) \] Input:

int(cos(f*x+e)^4*(b*csc(f*x+e))^n,x)
 

Output:

b**n*int(csc(e + f*x)**n*cos(e + f*x)**4,x)