Integrand size = 23, antiderivative size = 76 \[ \int (a \cos (e+f x))^m (b \csc (e+f x))^{5/2} \, dx=-\frac {b (a \cos (e+f x))^{1+m} (b \csc (e+f x))^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(e+f x)\right ) \sin ^2(e+f x)^{3/4}}{a f (1+m)} \] Output:
-b*(a*cos(f*x+e))^(1+m)*(b*csc(f*x+e))^(3/2)*hypergeom([7/4, 1/2+1/2*m],[3 /2+1/2*m],cos(f*x+e)^2)*(sin(f*x+e)^2)^(3/4)/a/f/(1+m)
Time = 3.76 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.24 \[ \int (a \cos (e+f x))^m (b \csc (e+f x))^{5/2} \, dx=\frac {2 a b (a \cos (e+f x))^{-1+m} \left (-\cot ^2(e+f x)\right )^{\frac {1-m}{2}} (b \csc (e+f x))^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4} (5-2 m),\frac {1-m}{2},\frac {1}{4} (9-2 m),\csc ^2(e+f x)\right )}{f (-5+2 m)} \] Input:
Integrate[(a*Cos[e + f*x])^m*(b*Csc[e + f*x])^(5/2),x]
Output:
(2*a*b*(a*Cos[e + f*x])^(-1 + m)*(-Cot[e + f*x]^2)^((1 - m)/2)*(b*Csc[e + f*x])^(3/2)*Hypergeometric2F1[(5 - 2*m)/4, (1 - m)/2, (9 - 2*m)/4, Csc[e + f*x]^2])/(f*(-5 + 2*m))
Time = 0.36 (sec) , antiderivative size = 76, 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, 3067, 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.
\(\displaystyle \int (b \csc (e+f x))^{5/2} (a \cos (e+f x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (-b \sec \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2} \left (a \sin \left (e+f x+\frac {\pi }{2}\right )\right )^mdx\) |
\(\Big \downarrow \) 3067 |
\(\displaystyle b^2 (b \sin (e+f x))^{3/2} (b \csc (e+f x))^{3/2} \int \frac {(a \cos (e+f x))^m}{(b \sin (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 (b \sin (e+f x))^{3/2} (b \csc (e+f x))^{3/2} \int \frac {(a \cos (e+f x))^m}{(b \sin (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 3056 |
\(\displaystyle -\frac {b \sin ^2(e+f x)^{3/4} (b \csc (e+f x))^{3/2} (a \cos (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(e+f x)\right )}{a f (m+1)}\) |
Input:
Int[(a*Cos[e + f*x])^m*(b*Csc[e + f*x])^(5/2),x]
Output:
-((b*(a*Cos[e + f*x])^(1 + m)*(b*Csc[e + f*x])^(3/2)*Hypergeometric2F1[7/4 , (1 + m)/2, (3 + m)/2, Cos[e + f*x]^2]*(Sin[e + f*x]^2)^(3/4))/(a*f*(1 + m)))
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]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[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]
\[\int \left (\cos \left (f x +e \right ) a \right )^{m} \left (b \csc \left (f x +e \right )\right )^{\frac {5}{2}}d x\]
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)
\[ \int (a \cos (e+f x))^m (b \csc (e+f x))^{5/2} \, dx=\int { \left (b \csc \left (f x + e\right )\right )^{\frac {5}{2}} \left (a \cos \left (f x + e\right )\right )^{m} \,d x } \] 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^2*csc(f*x + e)^2, x)
Timed out. \[ \int (a \cos (e+f x))^m (b \csc (e+f x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate((a*cos(f*x+e))**m*(b*csc(f*x+e))**(5/2),x)
Output:
Timed out
\[ \int (a \cos (e+f x))^m (b \csc (e+f x))^{5/2} \, dx=\int { \left (b \csc \left (f x + e\right )\right )^{\frac {5}{2}} \left (a \cos \left (f x + e\right )\right )^{m} \,d x } \] Input:
integrate((a*cos(f*x+e))^m*(b*csc(f*x+e))^(5/2),x, algorithm="maxima")
Output:
integrate((b*csc(f*x + e))^(5/2)*(a*cos(f*x + e))^m, x)
\[ \int (a \cos (e+f x))^m (b \csc (e+f x))^{5/2} \, dx=\int { \left (b \csc \left (f x + e\right )\right )^{\frac {5}{2}} \left (a \cos \left (f x + e\right )\right )^{m} \,d x } \] Input:
integrate((a*cos(f*x+e))^m*(b*csc(f*x+e))^(5/2),x, algorithm="giac")
Output:
integrate((b*csc(f*x + e))^(5/2)*(a*cos(f*x + e))^m, x)
Timed out. \[ \int (a \cos (e+f x))^m (b \csc (e+f x))^{5/2} \, dx=\int {\left (a\,\cos \left (e+f\,x\right )\right )}^m\,{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^{5/2} \,d x \] 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)
\[ \int (a \cos (e+f x))^m (b \csc (e+f x))^{5/2} \, dx=\sqrt {b}\, a^{m} \left (\int \sqrt {\csc \left (f x +e \right )}\, \cos \left (f x +e \right )^{m} \csc \left (f x +e \right )^{2}d x \right ) b^{2} \] 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)**2,x)*b** 2