3.284 \(\int (b \csc (e+f x))^n \sec ^5(e+f x) \, dx\)

Optimal. Leaf size=48 \[ \frac{(b \csc (e+f x))^{n+5} \text{Hypergeometric2F1}\left (3,\frac{n+5}{2},\frac{n+7}{2},\csc ^2(e+f x)\right )}{b^5 f (n+5)} \]

[Out]

((b*Csc[e + f*x])^(5 + n)*Hypergeometric2F1[3, (5 + n)/2, (7 + n)/2, Csc[e + f*x]^2])/(b^5*f*(5 + n))

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Rubi [A]  time = 0.0508996, antiderivative size = 48, 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 = {2621, 364} \[ \frac{(b \csc (e+f x))^{n+5} \, _2F_1\left (3,\frac{n+5}{2};\frac{n+7}{2};\csc ^2(e+f x)\right )}{b^5 f (n+5)} \]

Antiderivative was successfully verified.

[In]

Int[(b*Csc[e + f*x])^n*Sec[e + f*x]^5,x]

[Out]

((b*Csc[e + f*x])^(5 + n)*Hypergeometric2F1[3, (5 + n)/2, (7 + n)/2, Csc[e + f*x]^2])/(b^5*f*(5 + n))

Rule 2621

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int (b \csc (e+f x))^n \sec ^5(e+f x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^{4+n}}{\left (-1+\frac{x^2}{b^2}\right )^3} \, dx,x,b \csc (e+f x)\right )}{b^5 f}\\ &=\frac{(b \csc (e+f x))^{5+n} \, _2F_1\left (3,\frac{5+n}{2};\frac{7+n}{2};\csc ^2(e+f x)\right )}{b^5 f (5+n)}\\ \end{align*}

Mathematica [A]  time = 0.0512588, size = 51, normalized size = 1.06 \[ -\frac{b (b \csc (e+f x))^{n-1} \text{Hypergeometric2F1}\left (3,\frac{1-n}{2},\frac{3-n}{2},\sin ^2(e+f x)\right )}{f (n-1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Csc[e + f*x])^n*Sec[e + f*x]^5,x]

[Out]

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

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Maple [F]  time = 0.338, size = 0, normalized size = 0. \begin{align*} \int \left ( b\csc \left ( fx+e \right ) \right ) ^{n} \left ( \sec \left ( fx+e \right ) \right ) ^{5}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*csc(f*x+e))^n*sec(f*x+e)^5,x)

[Out]

int((b*csc(f*x+e))^n*sec(f*x+e)^5,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} \sec \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*csc(f*x+e))^n*sec(f*x+e)^5,x, algorithm="maxima")

[Out]

integrate((b*csc(f*x + e))^n*sec(f*x + e)^5, 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} \sec \left (f x + e\right )^{5}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*csc(f*x+e))^n*sec(f*x+e)^5,x, algorithm="fricas")

[Out]

integral((b*csc(f*x + e))^n*sec(f*x + e)^5, 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((b*csc(f*x+e))**n*sec(f*x+e)**5,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 )^{n} \sec \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*csc(f*x+e))^n*sec(f*x+e)^5,x, algorithm="giac")

[Out]

integrate((b*csc(f*x + e))^n*sec(f*x + e)^5, x)