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

Optimal. Leaf size=81 \[ \frac {b \cos ^2(e+f x)^{3/4} (c \sec (e+f x))^{3/2} (b \csc (e+f x))^{n-1} \, _2F_1\left (\frac {3}{4},\frac {1-n}{2};\frac {3-n}{2};\sin ^2(e+f x)\right )}{c f (1-n)} \]

[Out]

b*(cos(f*x+e)^2)^(3/4)*(b*csc(f*x+e))^(-1+n)*hypergeom([3/4, 1/2-1/2*n],[3/2-1/2*n],sin(f*x+e)^2)*(c*sec(f*x+e
))^(3/2)/c/f/(1-n)

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Rubi [A]  time = 0.10, antiderivative size = 81, normalized size of antiderivative = 1.00, 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} \[ \frac {b \cos ^2(e+f x)^{3/4} (c \sec (e+f x))^{3/2} (b \csc (e+f x))^{n-1} \, _2F_1\left (\frac {3}{4},\frac {1-n}{2};\frac {3-n}{2};\sin ^2(e+f x)\right )}{c f (1-n)} \]

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

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]

Rubi steps

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

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Mathematica [A]  time = 1.72, size = 90, normalized size = 1.11 \[ \frac {2 \cot (e+f x) \sqrt {c \sec (e+f x)} \left (-\tan ^2(e+f x)\right )^{\frac {n+1}{2}} (b \csc (e+f x))^n \, _2F_1\left (\frac {n+1}{2},\frac {1}{4} (2 n+1);\frac {1}{4} (2 n+5);\sec ^2(e+f x)\right )}{2 f n+f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cot[e + f*x]*(b*Csc[e + f*x])^n*Hypergeometric2F1[(1 + n)/2, (1 + 2*n)/4, (5 + 2*n)/4, Sec[e + f*x]^2]*Sqrt
[c*Sec[e + f*x]]*(-Tan[e + f*x]^2)^((1 + n)/2))/(f + 2*f*n)

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fricas [F]  time = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {c \sec \left (f x + e\right )} \left (b \csc \left (f x + e\right )\right )^{n}, x\right ) \]

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, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c \sec \left (f x + e\right )} \left (b \csc \left (f x + e\right )\right )^{n}\,{d x} \]

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

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maple [F]  time = 0.92, size = 0, normalized size = 0.00 \[ \int \left (b \csc \left (f x +e \right )\right )^{n} \sqrt {c \sec \left (f x +e \right )}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c \sec \left (f x + e\right )} \left (b \csc \left (f x + e\right )\right )^{n}\,{d x} \]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {\frac {c}{\cos \left (e+f\,x\right )}}\,{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \csc {\left (e + f x \right )}\right )^{n} \sqrt {c \sec {\left (e + f x \right )}}\, dx \]

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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