3.42 \(\int (b \cot (e+f x))^n (a \sec (e+f x))^m \, dx\)

Optimal. Leaf size=90 \[ -\frac{\sin ^2(e+f x)^{\frac{n+1}{2}} (a \sec (e+f x))^m (b \cot (e+f x))^{n+1} \text{Hypergeometric2F1}\left (\frac{n+1}{2},\frac{1}{2} (-m+n+1),\frac{1}{2} (-m+n+3),\cos ^2(e+f x)\right )}{b f (-m+n+1)} \]

[Out]

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

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Rubi [A]  time = 0.153449, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2618, 2602, 2576} \[ -\frac{\sin ^2(e+f x)^{\frac{n+1}{2}} (a \sec (e+f x))^m (b \cot (e+f x))^{n+1} \, _2F_1\left (\frac{n+1}{2},\frac{1}{2} (-m+n+1);\frac{1}{2} (-m+n+3);\cos ^2(e+f x)\right )}{b f (-m+n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(b*Cot[e + f*x])^n*(a*Sec[e + f*x])^m,x]

[Out]

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

Rule 2618

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Csc[e + f*
x])^FracPart[m]*(Sin[e + f*x]/a)^FracPart[m], Int[(b*Tan[e + f*x])^n/(Sin[e + f*x]/a)^m, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]

Rule 2602

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Cos[e + f
*x]^(n + 1)*(b*Tan[e + f*x])^(n + 1))/(b*(a*Sin[e + f*x])^(n + 1)), Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^
n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[n]

Rule 2576

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^(2*IntPar
t[(n - 1)/2] + 1)*(b*Sin[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Cos[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/
2, (1 - n)/2, (3 + m)/2, Cos[e + f*x]^2])/(a*f*(m + 1)*(Sin[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a,
b, e, f, m, n}, x] && SimplerQ[n, m]

Rubi steps

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

Mathematica [A]  time = 0.44545, size = 83, normalized size = 0.92 \[ -\frac{b \sec ^2(e+f x)^{-m/2} (a \sec (e+f x))^m (b \cot (e+f x))^{n-1} \text{Hypergeometric2F1}\left (1-\frac{m}{2},\frac{1-n}{2},\frac{3-n}{2},-\tan ^2(e+f x)\right )}{f (n-1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Cot[e + f*x])^n*(a*Sec[e + f*x])^m,x]

[Out]

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

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Maple [F]  time = 1.068, size = 0, normalized size = 0. \begin{align*} \int \left ( b\cot \left ( fx+e \right ) \right ) ^{n} \left ( a\sec \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cot(f*x+e))^n*(a*sec(f*x+e))^m,x)

[Out]

int((b*cot(f*x+e))^n*(a*sec(f*x+e))^m,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \cot \left (f x + e\right )\right )^{n} \left (a \sec \left (f x + e\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cot(f*x+e))^n*(a*sec(f*x+e))^m,x, algorithm="maxima")

[Out]

integrate((b*cot(f*x + e))^n*(a*sec(f*x + e))^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (b \cot \left (f x + e\right )\right )^{n} \left (a \sec \left (f x + e\right )\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cot(f*x+e))^n*(a*sec(f*x+e))^m,x, algorithm="fricas")

[Out]

integral((b*cot(f*x + e))^n*(a*sec(f*x + e))^m, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \sec{\left (e + f x \right )}\right )^{m} \left (b \cot{\left (e + f x \right )}\right )^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cot(f*x+e))**n*(a*sec(f*x+e))**m,x)

[Out]

Integral((a*sec(e + f*x))**m*(b*cot(e + f*x))**n, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \cot \left (f x + e\right )\right )^{n} \left (a \sec \left (f x + e\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cot(f*x+e))^n*(a*sec(f*x+e))^m,x, algorithm="giac")

[Out]

integrate((b*cot(f*x + e))^n*(a*sec(f*x + e))^m, x)