Integrand size = 23, antiderivative size = 78 \[ \int (a \cos (e+f x))^m \sqrt {b \csc (e+f x)} \, dx=-\frac {(a \cos (e+f x))^{1+m} (b \csc (e+f x))^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(e+f x)\right ) \sin ^2(e+f x)^{3/4}}{a b f (1+m)} \] Output:
-(a*cos(f*x+e))^(1+m)*(b*csc(f*x+e))^(3/2)*hypergeom([3/4, 1/2+1/2*m],[3/2 +1/2*m],cos(f*x+e)^2)*(sin(f*x+e)^2)^(3/4)/a/b/f/(1+m)
Time = 2.09 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.23 \[ \int (a \cos (e+f x))^m \sqrt {b \csc (e+f x)} \, dx=\frac {2 (a \cos (e+f x))^m \left (-\cot ^2(e+f x)\right )^{\frac {1-m}{2}} \sqrt {b \csc (e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4} (1-2 m),\frac {1-m}{2},\frac {1}{4} (5-2 m),\csc ^2(e+f x)\right ) \tan (e+f x)}{f (-1+2 m)} \] Input:
Integrate[(a*Cos[e + f*x])^m*Sqrt[b*Csc[e + f*x]],x]
Output:
(2*(a*Cos[e + f*x])^m*(-Cot[e + f*x]^2)^((1 - m)/2)*Sqrt[b*Csc[e + f*x]]*H ypergeometric2F1[(1 - 2*m)/4, (1 - m)/2, (5 - 2*m)/4, Csc[e + f*x]^2]*Tan[ e + f*x])/(f*(-1 + 2*m))
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.
\(\displaystyle \int \sqrt {b \csc (e+f x)} (a \cos (e+f x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {-b \sec \left (e+f x+\frac {\pi }{2}\right )} \left (a \sin \left (e+f x+\frac {\pi }{2}\right )\right )^mdx\) |
\(\Big \downarrow \) 3066 |
\(\displaystyle \frac {(b \sin (e+f x))^{3/2} (b \csc (e+f x))^{3/2} \int \frac {(a \cos (e+f x))^m}{\sqrt {b \sin (e+f x)}}dx}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(b \sin (e+f x))^{3/2} (b \csc (e+f x))^{3/2} \int \frac {(a \cos (e+f x))^m}{\sqrt {b \sin (e+f x)}}dx}{b^2}\) |
\(\Big \downarrow \) 3056 |
\(\displaystyle -\frac {\sin ^2(e+f x)^{3/4} (b \csc (e+f x))^{3/2} (a \cos (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(e+f x)\right )}{a b f (m+1)}\) |
Input:
Int[(a*Cos[e + f*x])^m*Sqrt[b*Csc[e + f*x]],x]
Output:
-(((a*Cos[e + f*x])^(1 + m)*(b*Csc[e + f*x])^(3/2)*Hypergeometric2F1[3/4, (1 + m)/2, (3 + m)/2, Cos[e + f*x]^2]*(Sin[e + f*x]^2)^(3/4))/(a*b*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[(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]
\[\int \left (\cos \left (f x +e \right ) a \right )^{m} \sqrt {b \csc \left (f x +e \right )}d x\]
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)
\[ \int (a \cos (e+f x))^m \sqrt {b \csc (e+f x)} \, dx=\int { \sqrt {b \csc \left (f x + e\right )} \left (a \cos \left (f x + e\right )\right )^{m} \,d x } \] Input:
integrate((a*cos(f*x+e))^m*(b*csc(f*x+e))^(1/2),x, algorithm="fricas")
Output:
integral(sqrt(b*csc(f*x + e))*(a*cos(f*x + e))^m, x)
\[ \int (a \cos (e+f x))^m \sqrt {b \csc (e+f x)} \, dx=\int \left (a \cos {\left (e + f x \right )}\right )^{m} \sqrt {b \csc {\left (e + f x \right )}}\, dx \] 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)
\[ \int (a \cos (e+f x))^m \sqrt {b \csc (e+f x)} \, dx=\int { \sqrt {b \csc \left (f x + e\right )} \left (a \cos \left (f x + e\right )\right )^{m} \,d x } \] Input:
integrate((a*cos(f*x+e))^m*(b*csc(f*x+e))^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(b*csc(f*x + e))*(a*cos(f*x + e))^m, x)
\[ \int (a \cos (e+f x))^m \sqrt {b \csc (e+f x)} \, dx=\int { \sqrt {b \csc \left (f x + e\right )} \left (a \cos \left (f x + e\right )\right )^{m} \,d x } \] Input:
integrate((a*cos(f*x+e))^m*(b*csc(f*x+e))^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(b*csc(f*x + e))*(a*cos(f*x + e))^m, x)
Timed out. \[ \int (a \cos (e+f x))^m \sqrt {b \csc (e+f x)} \, dx=\int {\left (a\,\cos \left (e+f\,x\right )\right )}^m\,\sqrt {\frac {b}{\sin \left (e+f\,x\right )}} \,d x \] 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)
\[ \int (a \cos (e+f x))^m \sqrt {b \csc (e+f x)} \, dx=\sqrt {b}\, a^{m} \left (\int \sqrt {\csc \left (f x +e \right )}\, \cos \left (f x +e \right )^{m}d x \right ) \] 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,x)