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

Optimal. Leaf size=76 \[ -\frac{b \sin ^2(e+f x)^{3/4} (b \csc (e+f x))^{3/2} (a \cos (e+f x))^{m+1} \, _2F_1\left (\frac{7}{4},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(e+f x)\right )}{a f (m+1)} \]

[Out]

-((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)))

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Rubi [A]  time = 0.115817, antiderivative size = 76, 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 = {2587, 2576} \[ -\frac{b \sin ^2(e+f x)^{3/4} (b \csc (e+f x))^{3/2} (a \cos (e+f x))^{m+1} \, _2F_1\left (\frac{7}{4},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(e+f x)\right )}{a f (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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)))

Rule 2587

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[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]

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

Mathematica [A]  time = 2.45064, size = 94, normalized size = 1.24 \[ \frac{2 a b (b \csc (e+f x))^{3/2} \left (-\cot ^2(e+f x)\right )^{\frac{1-m}{2}} (a \cos (e+f x))^{m-1} \, _2F_1\left (\frac{1}{4} (5-2 m),\frac{1-m}{2};\frac{1}{4} (9-2 m);\csc ^2(e+f x)\right )}{f (2 m-5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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))

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Maple [F]  time = 0.333, 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 \left (b \csc \left (f x + e\right )\right )^{\frac{5}{2}} \left (a \cos \left (f x + e\right )\right )^{m}\,{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((b*csc(f*x + e))^(5/2)*(a*cos(f*x + e))^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{b \csc \left (f x + e\right )} \left (a \cos \left (f x + e\right )\right )^{m} b^{2} \csc \left (f x + e\right )^{2}, 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^2*csc(f*x + e)^2, 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 \left (b \csc \left (f x + e\right )\right )^{\frac{5}{2}} \left (a \cos \left (f x + e\right )\right )^{m}\,{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((b*csc(f*x + e))^(5/2)*(a*cos(f*x + e))^m, x)