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

Optimal. Leaf size=81 \[ \frac{b \cos ^2(e+f x)^{5/4} (c \sec (e+f x))^{5/2} (b \csc (e+f x))^{n-1} \text{Hypergeometric2F1}\left (\frac{5}{4},\frac{1-n}{2},\frac{3-n}{2},\sin ^2(e+f x)\right )}{c f (1-n)} \]

[Out]

(b*(Cos[e + f*x]^2)^(5/4)*(b*Csc[e + f*x])^(-1 + n)*Hypergeometric2F1[5/4, (1 - n)/2, (3 - n)/2, Sin[e + f*x]^
2]*(c*Sec[e + f*x])^(5/2))/(c*f*(1 - n))

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Rubi [A]  time = 0.114924, 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} \[ \frac{b \cos ^2(e+f x)^{5/4} (c \sec (e+f x))^{5/2} (b \csc (e+f x))^{n-1} \, _2F_1\left (\frac{5}{4},\frac{1-n}{2};\frac{3-n}{2};\sin ^2(e+f x)\right )}{c f (1-n)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(Cos[e + f*x]^2)^(5/4)*(b*Csc[e + f*x])^(-1 + n)*Hypergeometric2F1[5/4, (1 - n)/2, (3 - n)/2, Sin[e + f*x]^
2]*(c*Sec[e + f*x])^(5/2))/(c*f*(1 - n))

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

Mathematica [A]  time = 1.82734, size = 92, normalized size = 1.14 \[ \frac{2 \cot (e+f x) (c \sec (e+f x))^{3/2} \left (-\tan ^2(e+f x)\right )^{\frac{n+1}{2}} (b \csc (e+f x))^n \text{Hypergeometric2F1}\left (\frac{n+1}{2},\frac{1}{4} (2 n+3),\frac{1}{4} (2 n+7),\sec ^2(e+f x)\right )}{f (2 n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cot[e + f*x]*(b*Csc[e + f*x])^n*Hypergeometric2F1[(1 + n)/2, (3 + 2*n)/4, (7 + 2*n)/4, Sec[e + f*x]^2]*(c*S
ec[e + f*x])^(3/2)*(-Tan[e + f*x]^2)^((1 + n)/2))/(f*(3 + 2*n))

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Maple [F]  time = 0.159, 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)

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{c \sec \left (f x + e\right )} \left (b \csc \left (f x + e\right )\right )^{n} c \sec \left (f x + e\right ), 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*sec(f*x + e), 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*(c*sec(f*x+e))**(3/2),x)

[Out]

Timed out

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