3.298 \(\int \frac{(b \csc (e+f x))^n}{\sqrt{c \sec (e+f x)}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0958881, 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} \[ \frac{b \sqrt [4]{\cos ^2(e+f x)} \sqrt{c \sec (e+f x)} (b \csc (e+f x))^{n-1} \, _2F_1\left (\frac{1}{4},\frac{1-n}{2};\frac{3-n}{2};\sin ^2(e+f x)\right )}{c f (1-n)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(Cos[e + f*x]^2)^(1/4)*(b*Csc[e + f*x])^(-1 + n)*Hypergeometric2F1[1/4, (1 - n)/2, (3 - n)/2, Sin[e + f*x]^
2]*Sqrt[c*Sec[e + f*x]])/(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 \frac{(b \csc (e+f x))^n}{\sqrt{c \sec (e+f x)}} \, dx &=\frac{\left (b^2 \sqrt{c \cos (e+f x)} (b \csc (e+f x))^{-1+n} \sqrt{c \sec (e+f x)} (b \sin (e+f x))^{-1+n}\right ) \int \sqrt{c \cos (e+f x)} (b \sin (e+f x))^{-n} \, dx}{c^2}\\ &=\frac{b \sqrt [4]{\cos ^2(e+f x)} (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 ) \sqrt{c \sec (e+f x)}}{c f (1-n)}\\ \end{align*}

Mathematica [C]  time = 3.10075, size = 326, normalized size = 4.02 \[ -\frac{4 (n-3) \sin \left (\frac{1}{2} (e+f x)\right ) \cos ^3\left (\frac{1}{2} (e+f x)\right ) F_1\left (\frac{1}{2}-\frac{n}{2};-\frac{1}{2},\frac{3}{2}-n;\frac{3}{2}-\frac{n}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right ) (b \csc (e+f x))^n}{f (n-1) \sqrt{c \sec (e+f x)} \left (2 (3-2 n) \sin ^2\left (\frac{1}{2} (e+f x)\right ) F_1\left (\frac{3}{2}-\frac{n}{2};-\frac{1}{2},\frac{5}{2}-n;\frac{5}{2}-\frac{n}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )+2 (n-3) \cos ^2\left (\frac{1}{2} (e+f x)\right ) F_1\left (\frac{1}{2}-\frac{n}{2};-\frac{1}{2},\frac{3}{2}-n;\frac{3}{2}-\frac{n}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-(\cos (e+f x)-1) F_1\left (\frac{3}{2}-\frac{n}{2};\frac{1}{2},\frac{3}{2}-n;\frac{5}{2}-\frac{n}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-4*(-3 + n)*AppellF1[1/2 - n/2, -1/2, 3/2 - n, 3/2 - n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f
*x)/2]^3*(b*Csc[e + f*x])^n*Sin[(e + f*x)/2])/(f*(-1 + n)*Sqrt[c*Sec[e + f*x]]*(2*(-3 + n)*AppellF1[1/2 - n/2,
-1/2, 3/2 - n, 3/2 - n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^2 - AppellF1[3/2 - n/2, 1
/2, 3/2 - n, 5/2 - n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(-1 + Cos[e + f*x]) + 2*(3 - 2*n)*AppellF1[3/
2 - n/2, -1/2, 5/2 - n, 5/2 - n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sin[(e + f*x)/2]^2))

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Maple [F]  time = 0.144, size = 0, normalized size = 0. \begin{align*} \int{ \left ( b\csc \left ( fx+e \right ) \right ) ^{n}{\frac{1}{\sqrt{c\sec \left ( fx+e \right ) }}}}\, 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))^(1/2),x)

[Out]

int((b*csc(f*x+e))^n/(c*sec(f*x+e))^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \csc \left (f x + e\right )\right )^{n}}{\sqrt{c \sec \left (f x + e\right )}}\,{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))^(1/2),x, algorithm="maxima")

[Out]

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

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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 \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))^(1/2),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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \csc{\left (e + f x \right )}\right )^{n}}{\sqrt{c \sec{\left (e + f x \right )}}}\, 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))**(1/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \csc \left (f x + e\right )\right )^{n}}{\sqrt{c \sec \left (f x + e\right )}}\,{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))^(1/2),x, algorithm="giac")

[Out]

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