3.294 \(\int \cos ^2(e+f x) (b \csc (e+f x))^n \, dx\)

Optimal. Leaf size=72 \[ \frac{b \cos (e+f x) (b \csc (e+f x))^{n-1} \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{1-n}{2},\frac{3-n}{2},\sin ^2(e+f x)\right )}{f (1-n) \sqrt{\cos ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.0796031, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2631, 2577} \[ \frac{b \cos (e+f x) (b \csc (e+f x))^{n-1} \, _2F_1\left (-\frac{1}{2},\frac{1-n}{2};\frac{3-n}{2};\sin ^2(e+f x)\right )}{f (1-n) \sqrt{\cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^2*(b*Csc[e + f*x])^n,x]

[Out]

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

Rule 2631

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^2*(a*Csc[e
+ f*x])^(m - 1)*(b*Sec[e + f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1)*(b*Cos[e + f*x])^(n + 1))/b^2, Int[1/((a*Si
n[e + f*x])^m*(b*Cos[e + f*x])^n), x], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !SimplerQ[-m, -n]

Rule 2577

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b^(2*IntPart
[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Sin[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rubi steps

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

Mathematica [B]  time = 0.456792, size = 165, normalized size = 2.29 \[ -\frac{2 \tan \left (\frac{1}{2} (e+f x)\right ) \sec ^2\left (\frac{1}{2} (e+f x)\right )^{-n} (b \csc (e+f x))^n \left (\text{Hypergeometric2F1}\left (1-n,\frac{1}{2}-\frac{n}{2},\frac{3}{2}-\frac{n}{2},-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-4 \text{Hypergeometric2F1}\left (2-n,\frac{1}{2}-\frac{n}{2},\frac{3}{2}-\frac{n}{2},-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )+4 \text{Hypergeometric2F1}\left (3-n,\frac{1}{2}-\frac{n}{2},\frac{3}{2}-\frac{n}{2},-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )}{f (n-1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^2*(b*Csc[e + f*x])^n,x]

[Out]

(-2*(b*Csc[e + f*x])^n*(Hypergeometric2F1[1 - n, 1/2 - n/2, 3/2 - n/2, -Tan[(e + f*x)/2]^2] - 4*Hypergeometric
2F1[2 - n, 1/2 - n/2, 3/2 - n/2, -Tan[(e + f*x)/2]^2] + 4*Hypergeometric2F1[3 - n, 1/2 - n/2, 3/2 - n/2, -Tan[
(e + f*x)/2]^2])*Tan[(e + f*x)/2])/(f*(-1 + n)*(Sec[(e + f*x)/2]^2)^n)

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Maple [F]  time = 0.946, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( b\csc \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^2*(b*csc(f*x+e))^n,x)

[Out]

int(cos(f*x+e)^2*(b*csc(f*x+e))^n,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 )^{n} \cos \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(b*csc(f*x+e))^n,x, algorithm="maxima")

[Out]

integrate((b*csc(f*x + e))^n*cos(f*x + e)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(b*csc(f*x+e))^n,x, algorithm="fricas")

[Out]

integral((b*csc(f*x + e))^n*cos(f*x + e)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \csc{\left (e + f x \right )}\right )^{n} \cos ^{2}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**2*(b*csc(f*x+e))**n,x)

[Out]

Integral((b*csc(e + f*x))**n*cos(e + f*x)**2, x)

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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 )^{n} \cos \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(b*csc(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((b*csc(f*x + e))^n*cos(f*x + e)^2, x)