3.183 \(\int \csc ^3(e+f x) (b \tan (e+f x))^n \, dx\)

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

[Out]

-((Cos[e + f*x]*Hypergeometric2F1[(1 - n)/2, (4 - n)/2, (3 - n)/2, Cos[e + f*x]^2]*(b*Tan[e + f*x])^n)/(f*(1 -
n)*(Sin[e + f*x]^2)^(n/2)))

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Rubi [A]  time = 0.0818814, antiderivative size = 78, 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 = {2601, 2576} \[ -\frac{\cos (e+f x) \sin ^2(e+f x)^{-n/2} (b \tan (e+f x))^n \, _2F_1\left (\frac{1-n}{2},\frac{4-n}{2};\frac{3-n}{2};\cos ^2(e+f x)\right )}{f (1-n)} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^3*(b*Tan[e + f*x])^n,x]

[Out]

-((Cos[e + f*x]*Hypergeometric2F1[(1 - n)/2, (4 - n)/2, (3 - n)/2, Cos[e + f*x]^2]*(b*Tan[e + f*x])^n)/(f*(1 -
n)*(Sin[e + f*x]^2)^(n/2)))

Rule 2601

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(Cos[e + f*x
]^n*(b*Tan[e + f*x])^n)/(a*Sin[e + f*x])^n, Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

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

Mathematica [B]  time = 7.39898, size = 182, normalized size = 2.33 \[ \frac{\tan ^2\left (\frac{1}{2} (e+f x)\right ) (b \tan (e+f x))^n \left (\cos (e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right )\right )^n \left (n (n+2) \cot ^4\left (\frac{1}{2} (e+f x)\right ) \, _2F_1\left (\frac{n}{2}-1,n;\frac{n}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )+(n-2) \left (n \, _2F_1\left (\frac{n}{2}+1,n;\frac{n}{2}+2;\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )+2 (n+2) \cot ^2\left (\frac{1}{2} (e+f x)\right ) \, _2F_1\left (\frac{n}{2},n;\frac{n}{2}+1;\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )\right )}{4 f n \left (n^2-4\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[e + f*x]^3*(b*Tan[e + f*x])^n,x]

[Out]

((n*(2 + n)*Cot[(e + f*x)/2]^4*Hypergeometric2F1[-1 + n/2, n, n/2, Tan[(e + f*x)/2]^2] + (-2 + n)*(n*Hypergeom
etric2F1[1 + n/2, n, 2 + n/2, Tan[(e + f*x)/2]^2] + 2*(2 + n)*Cot[(e + f*x)/2]^2*Hypergeometric2F1[n/2, n, 1 +
n/2, Tan[(e + f*x)/2]^2]))*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^n*Tan[(e + f*x)/2]^2*(b*Tan[e + f*x])^n)/(4*f*n*
(-4 + n^2))

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Maple [F]  time = 0.278, size = 0, normalized size = 0. \begin{align*} \int \left ( \csc \left ( fx+e \right ) \right ) ^{3} \left ( b\tan \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^3*(b*tan(f*x+e))^n,x)

[Out]

int(csc(f*x+e)^3*(b*tan(f*x+e))^n,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \tan \left (f x + e\right )\right )^{n} \csc \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(b*tan(f*x+e))^n,x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e))^n*csc(f*x + e)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(b*tan(f*x+e))^n,x, algorithm="fricas")

[Out]

integral((b*tan(f*x + e))^n*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(csc(f*x+e)**3*(b*tan(f*x+e))**n,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \tan \left (f x + e\right )\right )^{n} \csc \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(b*tan(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e))^n*csc(f*x + e)^3, x)