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

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

[Out]

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

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Rubi [A]  time = 0.11, 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} \[ \frac {b (b \csc (e+f x))^{n-1} \, _2F_1\left (-\frac {1}{4},\frac {1-n}{2};\frac {3-n}{2};\sin ^2(e+f x)\right )}{c f (1-n) \sqrt [4]{\cos ^2(e+f x)} \sqrt {c \sec (e+f x)}} \]

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 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 \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*}

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Mathematica [A]  time = 1.02, size = 115, normalized size = 1.42 \[ -\frac {2 \cos (2 (e+f x)) \cot (e+f x) \sqrt {c \sec (e+f x)} \left (-\tan ^2(e+f x)\right )^{\frac {n+1}{2}} (b \csc (e+f x))^n \, _2F_1\left (\frac {n+1}{2},\frac {1}{4} (2 n-3);\frac {1}{4} (2 n+1);\sec ^2(e+f x)\right )}{c^2 f (2 n-3) \left (\sec ^2(e+f x)-2\right )} \]

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))

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fricas [F]  time = 1.20, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c \sec \left (f x + e\right )} \left (b \csc \left (f x + e\right )\right )^{n}}{c^{2} \sec \left (f x + e\right )^{2}}, x\right ) \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \csc \left (f x + e\right )\right )^{n}}{\left (c \sec \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

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)

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maple [F]  time = 0.73, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \csc \left (f x +e \right )\right )^{n}}{\left (c \sec \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \csc \left (f x + e\right )\right )^{n}}{\left (c \sec \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^n}{{\left (\frac {c}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \csc {\left (e + f x \right )}\right )^{n}}{\left (c \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

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)

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