3.299 \(\int \frac{(b \csc (e+f x))^n}{(c \sec (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=81 \[ \frac{b (b \csc (e+f x))^{n-1} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1-n}{2},\frac{3-n}{2},\sin ^2(e+f x)\right )}{c f (1-n) \sqrt [4]{\cos ^2(e+f x)} \sqrt{c \sec (e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.107255, antiderivative size = 81, 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 = {2631, 2577} \[ \frac{b (b \csc (e+f x))^{n-1} \, _2F_1\left (-\frac{1}{4},\frac{1-n}{2};\frac{3-n}{2};\sin ^2(e+f x)\right )}{c f (1-n) \sqrt [4]{\cos ^2(e+f x)} \sqrt{c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Csc[e + f*x])^n/(c*Sec[e + f*x])^(3/2),x]

[Out]

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

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

Mathematica [A]  time = 1.00577, size = 115, normalized size = 1.42 \[ -\frac{2 \cos (2 (e+f x)) \cot (e+f x) \sqrt{c \sec (e+f x)} \left (-\tan ^2(e+f x)\right )^{\frac{n+1}{2}} (b \csc (e+f x))^n \text{Hypergeometric2F1}\left (\frac{n+1}{2},\frac{1}{4} (2 n-3),\frac{1}{4} (2 n+1),\sec ^2(e+f x)\right )}{c^2 f (2 n-3) \left (\sec ^2(e+f x)-2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Csc[e + f*x])^n/(c*Sec[e + f*x])^(3/2),x]

[Out]

(-2*Cos[2*(e + f*x)]*Cot[e + f*x]*(b*Csc[e + f*x])^n*Hypergeometric2F1[(1 + n)/2, (-3 + 2*n)/4, (1 + 2*n)/4, S
ec[e + f*x]^2]*Sqrt[c*Sec[e + f*x]]*(-Tan[e + f*x]^2)^((1 + n)/2))/(c^2*f*(-3 + 2*n)*(-2 + Sec[e + f*x]^2))

________________________________________________________________________________________

Maple [F]  time = 0.136, size = 0, normalized size = 0. \begin{align*} \int{ \left ( b\csc \left ( fx+e \right ) \right ) ^{n} \left ( c\sec \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*csc(f*x+e))^n/(c*sec(f*x+e))^(3/2),x)

[Out]

int((b*csc(f*x+e))^n/(c*sec(f*x+e))^(3/2),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \csc \left (f x + e\right )\right )^{n}}{\left (c \sec \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*csc(f*x+e))^n/(c*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*csc(f*x + e))^n/(c*sec(f*x + e))^(3/2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*csc(f*x+e))^n/(c*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*sec(f*x + e))*(b*csc(f*x + e))^n/(c^2*sec(f*x + e)^2), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \csc{\left (e + f x \right )}\right )^{n}}{\left (c \sec{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*csc(f*x+e))**n/(c*sec(f*x+e))**(3/2),x)

[Out]

Integral((b*csc(e + f*x))**n/(c*sec(e + f*x))**(3/2), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \csc \left (f x + e\right )\right )^{n}}{\left (c \sec \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*csc(f*x+e))^n/(c*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((b*csc(f*x + e))^n/(c*sec(f*x + e))^(3/2), x)