3.294 \(\int \frac{(a \cos (e+f x))^m}{(b \csc (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=78 \[ -\frac{\sqrt [4]{\sin ^2(e+f x)} \sqrt{b \csc (e+f x)} (a \cos (e+f x))^{m+1} \, _2F_1\left (-\frac{3}{4},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(e+f x)\right )}{a b^3 f (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.111243, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2586, 2576} \[ -\frac{(a \cos (e+f x))^{m+1} \, _2F_1\left (-\frac{3}{4},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(e+f x)\right )}{a b f (m+1) \sin ^2(e+f x)^{3/4} (b \csc (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cos[e + f*x])^m/(b*Csc[e + f*x])^(5/2),x]

[Out]

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

Rule 2586

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(1*(b*Cos[e +
f*x])^(n + 1)*(b*Sec[e + f*x])^(n + 1))/b^2, 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]

Rule 2576

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)*Hypergeometric2F1[(1 + m)/
2, (1 - n)/2, (3 + m)/2, Cos[e + f*x]^2])/(a*f*(m + 1)*(Sin[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a,
b, e, f, m, n}, x] && SimplerQ[n, m]

Rubi steps

\begin{align*} \int \frac{(a \cos (e+f x))^m}{(b \csc (e+f x))^{5/2}} \, dx &=\frac{\int (a \cos (e+f x))^m (b \sin (e+f x))^{5/2} \, dx}{b^2 (b \csc (e+f x))^{3/2} (b \sin (e+f x))^{3/2}}\\ &=-\frac{(a \cos (e+f x))^{1+m} \, _2F_1\left (-\frac{3}{4},\frac{1+m}{2};\frac{3+m}{2};\cos ^2(e+f x)\right )}{a b f (1+m) (b \csc (e+f x))^{3/2} \sin ^2(e+f x)^{3/4}}\\ \end{align*}

Mathematica [A]  time = 1.11356, size = 125, normalized size = 1.6 \[ \frac{2 (2 \cos (2 (e+f x))+1) \tan (e+f x) \left (-\cot ^2(e+f x)\right )^{\frac{1-m}{2}} (a \cos (e+f x))^m \, _2F_1\left (\frac{1}{4} (-2 m-5),\frac{1-m}{2};\frac{1}{4} (-2 m-1);\csc ^2(e+f x)\right )}{b^2 f (2 m+5) \left (3 \csc ^2(e+f x)-4\right ) \sqrt{b \csc (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cos[e + f*x])^m/(b*Csc[e + f*x])^(5/2),x]

[Out]

(2*(a*Cos[e + f*x])^m*(1 + 2*Cos[2*(e + f*x)])*(-Cot[e + f*x]^2)^((1 - m)/2)*Hypergeometric2F1[(-5 - 2*m)/4, (
1 - m)/2, (-1 - 2*m)/4, Csc[e + f*x]^2]*Tan[e + f*x])/(b^2*f*(5 + 2*m)*Sqrt[b*Csc[e + f*x]]*(-4 + 3*Csc[e + f*
x]^2))

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Maple [F]  time = 0.318, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a\cos \left ( fx+e \right ) \right ) ^{m} \left ( b\csc \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(f*x+e))^m/(b*csc(f*x+e))^(5/2),x)

[Out]

int((a*cos(f*x+e))^m/(b*csc(f*x+e))^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a \cos \left (f x + e\right )\right )^{m}}{\left (b \csc \left (f x + e\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(f*x+e))^m/(b*csc(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*cos(f*x + e))^m/(b*csc(f*x + e))^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \csc \left (f x + e\right )} \left (a \cos \left (f x + e\right )\right )^{m}}{b^{3} \csc \left (f x + e\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(f*x+e))^m/(b*csc(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(f*x+e))**m/(b*csc(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a \cos \left (f x + e\right )\right )^{m}}{\left (b \csc \left (f x + e\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(f*x+e))^m/(b*csc(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((a*cos(f*x + e))^m/(b*csc(f*x + e))^(5/2), x)