3.3.92 \(\int (b \csc (e+f x))^n \sec ^2(e+f x) \, dx\) [292]

Optimal. Leaf size=72 \[ \frac {b \sqrt {\cos ^2(e+f x)} (b \csc (e+f x))^{-1+n} \, _2F_1\left (\frac {3}{2},\frac {1-n}{2};\frac {3-n}{2};\sin ^2(e+f x)\right ) \sec (e+f x)}{f (1-n)} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2711, 2657} \begin {gather*} \frac {b \sqrt {\cos ^2(e+f x)} \sec (e+f x) (b \csc (e+f x))^{n-1} \, _2F_1\left (\frac {3}{2},\frac {1-n}{2};\frac {3-n}{2};\sin ^2(e+f x)\right )}{f (1-n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2657

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)/(a*f*(m + 1)*(Cos[e + f*x]^
2)^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rule 2711

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^2/b^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), 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 \sec ^2(e+f x) \, dx &=\left (b^2 (b \csc (e+f x))^{-1+n} (b \sin (e+f x))^{-1+n}\right ) \int \sec ^2(e+f x) (b \sin (e+f x))^{-n} \, dx\\ &=\frac {b \sqrt {\cos ^2(e+f x)} (b \csc (e+f x))^{-1+n} \, _2F_1\left (\frac {3}{2},\frac {1-n}{2};\frac {3-n}{2};\sin ^2(e+f x)\right ) \sec (e+f x)}{f (1-n)}\\ \end {align*}

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Mathematica [A]
time = 0.49, size = 75, normalized size = 1.04 \begin {gather*} \frac {(b \csc (e+f x))^n \, _2F_1\left (\frac {1}{2}-\frac {n}{2},-\frac {n}{2};\frac {3}{2}-\frac {n}{2};-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{-n/2} \tan (e+f x)}{f (1-n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.20, size = 0, normalized size = 0.00 \[\int \left (b \csc \left (f x +e \right )\right )^{n} \left (\sec ^{2}\left (f x +e \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (b \csc {\left (e + f x \right )}\right )^{n} \sec ^{2}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^n}{{\cos \left (e+f\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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