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

Optimal. Leaf size=72 \[ \frac {b \cos (e+f x) (b \csc (e+f x))^{n-1} \, _2F_1\left (-\frac {1}{2},\frac {1-n}{2};\frac {3-n}{2};\sin ^2(e+f x)\right )}{f (1-n) \sqrt {\cos ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 72, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [B]  time = 0.48, size = 165, normalized size = 2.29 \[ -\frac {2 \tan \left (\frac {1}{2} (e+f x)\right ) \sec ^2\left (\frac {1}{2} (e+f x)\right )^{-n} (b \csc (e+f x))^n \left (\, _2F_1\left (1-n,\frac {1}{2}-\frac {n}{2};\frac {3}{2}-\frac {n}{2};-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-4 \, _2F_1\left (2-n,\frac {1}{2}-\frac {n}{2};\frac {3}{2}-\frac {n}{2};-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+4 \, _2F_1\left (3-n,\frac {1}{2}-\frac {n}{2};\frac {3}{2}-\frac {n}{2};-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )}{f (n-1)} \]

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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fricas [F]  time = 2.14, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (b \csc \left (f x + e\right )\right )^{n} \cos \left (f x + e\right )^{2}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \csc \left (f x + e\right )\right )^{n} \cos \left (f x + e\right )^{2}\,{d x} \]

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)

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maple [F]  time = 2.68, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{2}\left (f x +e \right )\right ) \left (b \csc \left (f x +e \right )\right )^{n}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \csc \left (f x + e\right )\right )^{n} \cos \left (f x + e\right )^{2}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (e+f\,x\right )}^2\,{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \csc {\left (e + f x \right )}\right )^{n} \cos ^{2}{\left (e + f x \right )}\, dx \]

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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