\(\int (b \sec (e+f x))^n (a \sin (e+f x))^m \, dx\) [491]

Optimal result

Integrand size = 21, antiderivative size = 92 \[ \int (b \sec (e+f x))^n (a \sin (e+f x))^m \, dx=-\frac {a b \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right ) (b \sec (e+f x))^{-1+n} (a \sin (e+f x))^{-1+m} \sin ^2(e+f x)^{\frac {1-m}{2}}}{f (1-n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2667, 2656} \[ \int (b \sec (e+f x))^n (a \sin (e+f x))^m \, dx=-\frac {a b \sin ^2(e+f x)^{\frac {1-m}{2}} (a \sin (e+f x))^{m-1} (b \sec (e+f x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right )}{f (1-n)} \]

[In]

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

[Out]

-((a*b*Hypergeometric2F1[(1 - m)/2, (1 - n)/2, (3 - n)/2, Cos[e + f*x]^2]*(b*Sec[e + f*x])^(-1 + n)*(a*Sin[e +
f*x])^(-1 + m)*(Sin[e + f*x]^2)^((1 - m)/2))/(f*(1 - 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 2667

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[b^2*(b*Cos[e
+ f*x])^(n - 1)*(b*Sec[e + f*x])^(n - 1), Int[(a*Sin[e + f*x])^m/(b*Cos[e + f*x])^n, x], x] /; FreeQ[{a, b, e,
f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \left (b^2 (b \cos (e+f x))^{-1+n} (b \sec (e+f x))^{-1+n}\right ) \int (b \cos (e+f x))^{-n} (a \sin (e+f x))^m \, dx \\ & = -\frac {a b \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right ) (b \sec (e+f x))^{-1+n} (a \sin (e+f x))^{-1+m} \sin ^2(e+f x)^{\frac {1-m}{2}}}{f (1-n)} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.43 (sec) , antiderivative size = 289, normalized size of antiderivative = 3.14 \[ \int (b \sec (e+f x))^n (a \sin (e+f x))^m \, dx=\frac {4 (3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},n,1+m-n,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^3\left (\frac {1}{2} (e+f x)\right ) (b \sec (e+f x))^n \sin \left (\frac {1}{2} (e+f x)\right ) (a \sin (e+f x))^m}{f (1+m) \left ((3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},n,1+m-n,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (1+\cos (e+f x))-4 \left ((1+m-n) \operatorname {AppellF1}\left (\frac {3+m}{2},n,2+m-n,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-n \operatorname {AppellF1}\left (\frac {3+m}{2},1+n,1+m-n,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

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

[Out]

(4*(3 + m)*AppellF1[(1 + m)/2, n, 1 + m - n, (3 + m)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)
/2]^3*(b*Sec[e + f*x])^n*Sin[(e + f*x)/2]*(a*Sin[e + f*x])^m)/(f*(1 + m)*((3 + m)*AppellF1[(1 + m)/2, n, 1 + m
- n, (3 + m)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(1 + Cos[e + f*x]) - 4*((1 + m - n)*AppellF1[(3 + m)
/2, n, 2 + m - n, (5 + m)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - n*AppellF1[(3 + m)/2, 1 + n, 1 + m - n
, (5 + m)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Sin[(e + f*x)/2]^2))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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