\(\int (b \cot (e+f x))^n (a \sec (e+f x))^m \, dx\) [42]

Optimal result

Integrand size = 21, antiderivative size = 90 \[ \int (b \cot (e+f x))^n (a \sec (e+f x))^m \, dx=-\frac {(b \cot (e+f x))^{1+n} \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {1}{2} (1-m+n),\frac {1}{2} (3-m+n),\cos ^2(e+f x)\right ) (a \sec (e+f x))^m \sin ^2(e+f x)^{\frac {1+n}{2}}}{b f (1-m+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2698, 2682, 2656} \[ \int (b \cot (e+f x))^n (a \sec (e+f x))^m \, dx=-\frac {\sin ^2(e+f x)^{\frac {n+1}{2}} (a \sec (e+f x))^m (b \cot (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {n+1}{2},\frac {1}{2} (-m+n+1),\frac {1}{2} (-m+n+3),\cos ^2(e+f x)\right )}{b f (-m+n+1)} \]

[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 2656

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

Rule 2682

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 2698

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]

Rubi steps \begin{align*} \text {integral}& = \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} \operatorname {Hypergeometric2F1}\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] (verified)

Time = 0.81 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.92 \[ \int (b \cot (e+f x))^n (a \sec (e+f x))^m \, dx=-\frac {b (b \cot (e+f x))^{-1+n} \operatorname {Hypergeometric2F1}\left (1-\frac {m}{2},\frac {1-n}{2},\frac {3-n}{2},-\tan ^2(e+f x)\right ) (a \sec (e+f x))^m \sec ^2(e+f x)^{-m/2}}{f (-1+n)} \]

[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)))

Maple [F]

\[\int \left (b \cot \left (f x +e \right )\right )^{n} \left (a \sec \left (f x +e \right )\right )^{m}d x\]

[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)

Fricas [F]

\[ \int (b \cot (e+f x))^n (a \sec (e+f x))^m \, dx=\int { \left (b \cot \left (f x + e\right )\right )^{n} \left (a \sec \left (f x + e\right )\right )^{m} \,d x } \]

[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)

Sympy [F]

\[ \int (b \cot (e+f x))^n (a \sec (e+f x))^m \, dx=\int \left (a \sec {\left (e + f x \right )}\right )^{m} \left (b \cot {\left (e + f x \right )}\right )^{n}\, dx \]

[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)

Maxima [F]

\[ \int (b \cot (e+f x))^n (a \sec (e+f x))^m \, dx=\int { \left (b \cot \left (f x + e\right )\right )^{n} \left (a \sec \left (f x + e\right )\right )^{m} \,d x } \]

[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)

Giac [F]

\[ \int (b \cot (e+f x))^n (a \sec (e+f x))^m \, dx=\int { \left (b \cot \left (f x + e\right )\right )^{n} \left (a \sec \left (f x + e\right )\right )^{m} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int (b \cot (e+f x))^n (a \sec (e+f x))^m \, dx=\int {\left (b\,\mathrm {cot}\left (e+f\,x\right )\right )}^n\,{\left (\frac {a}{\cos \left (e+f\,x\right )}\right )}^m \,d x \]

[In]

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

[Out]

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