\(\int \frac {(a \cos (e+f x))^m}{(b \csc (e+f x))^{3/2}} \, dx\) [293]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2666, 2656} \[ \int \frac {(a \cos (e+f x))^m}{(b \csc (e+f x))^{3/2}} \, dx=-\frac {(a \cos (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(e+f x)\right )}{a b f (m+1) \sqrt [4]{\sin ^2(e+f x)} \sqrt {b \csc (e+f x)}} \]

[In]

Int[(a*Cos[e + f*x])^m/(b*Csc[e + f*x])^(3/2),x]

[Out]

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

Rule 2656

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^(2*IntPar
t[(n - 1)/2] + 1))*(b*Sin[e + f*x])^(2*FracPart[(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, Cos[e + f*x]^2], x] /; FreeQ[{a
, b, e, f, m, n}, x] && SimplerQ[n, m]

Rule 2666

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(1/b^2)*(b*Co
s[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (a \cos (e+f x))^m (b \sin (e+f x))^{3/2} \, dx}{b^2 \sqrt {b \csc (e+f x)} \sqrt {b \sin (e+f x)}} \\ & = -\frac {(a \cos (e+f x))^{1+m} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(e+f x)\right )}{a b f (1+m) \sqrt {b \csc (e+f x)} \sqrt [4]{\sin ^2(e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 11.13 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.49 \[ \int \frac {(a \cos (e+f x))^m}{(b \csc (e+f x))^{3/2}} \, dx=\frac {2 a (a \cos (e+f x))^{-1+m} \cos (2 (e+f x)) \left (-\cot ^2(e+f x)\right )^{\frac {1-m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4} (-3-2 m),\frac {1-m}{2},\frac {1}{4} (1-2 m),\csc ^2(e+f x)\right )}{b f (3+2 m) \sqrt {b \csc (e+f x)} \left (-2+\csc ^2(e+f x)\right )} \]

[In]

Integrate[(a*Cos[e + f*x])^m/(b*Csc[e + f*x])^(3/2),x]

[Out]

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

Maple [F]

\[\int \frac {\left (\cos \left (f x +e \right ) a \right )^{m}}{\left (b \csc \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]

[In]

int((cos(f*x+e)*a)^m/(b*csc(f*x+e))^(3/2),x)

[Out]

int((cos(f*x+e)*a)^m/(b*csc(f*x+e))^(3/2),x)

Fricas [F]

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

[In]

integrate((a*cos(f*x+e))^m/(b*csc(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate((a*cos(f*x+e))**m/(b*csc(f*x+e))**(3/2),x)

[Out]

Integral((a*cos(e + f*x))**m/(b*csc(e + f*x))**(3/2), x)

Maxima [F]

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

[In]

integrate((a*cos(f*x+e))^m/(b*csc(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*cos(f*x + e))^m/(b*csc(f*x + e))^(3/2), x)

Giac [F]

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

[In]

integrate((a*cos(f*x+e))^m/(b*csc(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((a*cos(f*x + e))^m/(b*csc(f*x + e))^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a \cos (e+f x))^m}{(b \csc (e+f x))^{3/2}} \, dx=\int \frac {{\left (a\,\cos \left (e+f\,x\right )\right )}^m}{{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int((a*cos(e + f*x))^m/(b/sin(e + f*x))^(3/2),x)

[Out]

int((a*cos(e + f*x))^m/(b/sin(e + f*x))^(3/2), x)