Integrand size = 19, antiderivative size = 89 \[ \int \sec ^n(e+f x) (a \sin (e+f x))^m \, dx=-\frac {a \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right ) \sec ^{-1+n}(e+f x) (a \sin (e+f x))^{-1+m} \sin ^2(e+f x)^{\frac {1-m}{2}}}{f (1-n)} \] Output:
-a*hypergeom([1/2-1/2*m, 1/2-1/2*n],[3/2-1/2*n],cos(f*x+e)^2)*sec(f*x+e)^( -1+n)*(a*sin(f*x+e))^(-1+m)*(sin(f*x+e)^2)^(1/2-1/2*m)/f/(1-n)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.56 (sec) , antiderivative size = 287, normalized size of antiderivative = 3.22 \[ \int \sec ^n(e+f x) (a \sin (e+f x))^m \, dx=\frac {4 (3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},n,1+m-n,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^3\left (\frac {1}{2} (e+f x)\right ) \sec ^n(e+f x) \sin \left (\frac {1}{2} (e+f x)\right ) (a \sin (e+f x))^m}{f (1+m) \left ((3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},n,1+m-n,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (1+\cos (e+f x))-4 \left ((1+m-n) \operatorname {AppellF1}\left (\frac {3+m}{2},n,2+m-n,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-n \operatorname {AppellF1}\left (\frac {3+m}{2},1+n,1+m-n,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )} \] Input:
Integrate[Sec[e + f*x]^n*(a*Sin[e + f*x])^m,x]
Output:
(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*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))
Time = 0.52 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3067, 3042, 3056}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sec ^n(e+f x) (a \sin (e+f x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (e+f x)^n (a \sin (e+f x))^mdx\) |
\(\Big \downarrow \) 3067 |
\(\displaystyle \cos ^n(e+f x) \sec ^n(e+f x) \int \cos ^{-n}(e+f x) (a \sin (e+f x))^mdx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos ^n(e+f x) \sec ^n(e+f x) \int \cos (e+f x)^{-n} (a \sin (e+f x))^mdx\) |
\(\Big \downarrow \) 3056 |
\(\displaystyle -\frac {a \sin ^2(e+f x)^{\frac {1-m}{2}} \sec ^{n-1}(e+f x) (a \sin (e+f x))^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right )}{f (1-n)}\) |
Input:
Int[Sec[e + f*x]^n*(a*Sin[e + f*x])^m,x]
Output:
-((a*Hypergeometric2F1[(1 - m)/2, (1 - n)/2, (3 - n)/2, Cos[e + f*x]^2]*Se c[e + f*x]^(-1 + n)*(a*Sin[e + f*x])^(-1 + m)*(Sin[e + f*x]^2)^((1 - m)/2) )/(f*(1 - n)))
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[(-b^(2*IntPart[(n - 1)/2] + 1))*(b*Sin[e + f*x])^(2*F racPart[(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, C os[e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x] && SimplerQ[n, m]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[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]
\[\int \sec \left (f x +e \right )^{n} \left (a \sin \left (f x +e \right )\right )^{m}d x\]
Input:
int(sec(f*x+e)^n*(a*sin(f*x+e))^m,x)
Output:
int(sec(f*x+e)^n*(a*sin(f*x+e))^m,x)
\[ \int \sec ^n(e+f x) (a \sin (e+f x))^m \, dx=\int { \left (a \sin \left (f x + e\right )\right )^{m} \sec \left (f x + e\right )^{n} \,d x } \] Input:
integrate(sec(f*x+e)^n*(a*sin(f*x+e))^m,x, algorithm="fricas")
Output:
integral((a*sin(f*x + e))^m*sec(f*x + e)^n, x)
\[ \int \sec ^n(e+f x) (a \sin (e+f x))^m \, dx=\int \left (a \sin {\left (e + f x \right )}\right )^{m} \sec ^{n}{\left (e + f x \right )}\, dx \] Input:
integrate(sec(f*x+e)**n*(a*sin(f*x+e))**m,x)
Output:
Integral((a*sin(e + f*x))**m*sec(e + f*x)**n, x)
\[ \int \sec ^n(e+f x) (a \sin (e+f x))^m \, dx=\int { \left (a \sin \left (f x + e\right )\right )^{m} \sec \left (f x + e\right )^{n} \,d x } \] Input:
integrate(sec(f*x+e)^n*(a*sin(f*x+e))^m,x, algorithm="maxima")
Output:
integrate((a*sin(f*x + e))^m*sec(f*x + e)^n, x)
\[ \int \sec ^n(e+f x) (a \sin (e+f x))^m \, dx=\int { \left (a \sin \left (f x + e\right )\right )^{m} \sec \left (f x + e\right )^{n} \,d x } \] Input:
integrate(sec(f*x+e)^n*(a*sin(f*x+e))^m,x, algorithm="giac")
Output:
integrate((a*sin(f*x + e))^m*sec(f*x + e)^n, x)
Timed out. \[ \int \sec ^n(e+f x) (a \sin (e+f x))^m \, dx=\int {\left (a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (\frac {1}{\cos \left (e+f\,x\right )}\right )}^n \,d x \] Input:
int((a*sin(e + f*x))^m*(1/cos(e + f*x))^n,x)
Output:
int((a*sin(e + f*x))^m*(1/cos(e + f*x))^n, x)
\[ \int \sec ^n(e+f x) (a \sin (e+f x))^m \, dx=a^{m} \left (\int \sin \left (f x +e \right )^{m} \sec \left (f x +e \right )^{n}d x \right ) \] Input:
int(sec(f*x+e)^n*(a*sin(f*x+e))^m,x)
Output:
a**m*int(sin(e + f*x)**m*sec(e + f*x)**n,x)