Integrand size = 17, antiderivative size = 76 \[ \int \sin (e+f x) (b \tan (e+f x))^n \, dx=\frac {\cos ^2(e+f x)^{\frac {1+n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(e+f x)\right ) \sin (e+f x) (b \tan (e+f x))^{1+n}}{b f (2+n)} \] Output:
(cos(f*x+e)^2)^(1/2+1/2*n)*hypergeom([1+1/2*n, 1/2+1/2*n],[2+1/2*n],sin(f* x+e)^2)*sin(f*x+e)*(b*tan(f*x+e))^(1+n)/b/f/(2+n)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 1.46 (sec) , antiderivative size = 252, normalized size of antiderivative = 3.32 \[ \int \sin (e+f x) (b \tan (e+f x))^n \, dx=\frac {8 (4+n) \operatorname {AppellF1}\left (1+\frac {n}{2},n,2,2+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^4\left (\frac {1}{2} (e+f x)\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right ) (b \tan (e+f x))^n}{f (2+n) \left (2 \left (2 \operatorname {AppellF1}\left (2+\frac {n}{2},n,3,3+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-n \operatorname {AppellF1}\left (2+\frac {n}{2},1+n,2,3+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) (-1+\cos (e+f x))+(4+n) \operatorname {AppellF1}\left (1+\frac {n}{2},n,2,2+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (1+\cos (e+f x))\right )} \] Input:
Integrate[Sin[e + f*x]*(b*Tan[e + f*x])^n,x]
Output:
(8*(4 + n)*AppellF1[1 + n/2, n, 2, 2 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^4*Sin[(e + f*x)/2]^2*(b*Tan[e + f*x])^n)/(f*(2 + n)*(2*(2*AppellF1[2 + n/2, n, 3, 3 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - n*AppellF1[2 + n/2, 1 + n, 2, 3 + n/2, Tan[(e + f*x)/2]^2, - Tan[(e + f*x)/2]^2])*(-1 + Cos[e + f*x]) + (4 + n)*AppellF1[1 + n/2, n, 2, 2 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(1 + Cos[e + f*x])))
Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3042, 3082, 3042, 3057}
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.
\(\displaystyle \int \sin (e+f x) (b \tan (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (e+f x) (b \tan (e+f x))^ndx\) |
\(\Big \downarrow \) 3082 |
\(\displaystyle \frac {\sin ^{-n-1}(e+f x) \cos ^{n+1}(e+f x) (b \tan (e+f x))^{n+1} \int \cos ^{-n}(e+f x) \sin ^{n+1}(e+f x)dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin ^{-n-1}(e+f x) \cos ^{n+1}(e+f x) (b \tan (e+f x))^{n+1} \int \cos (e+f x)^{-n} \sin (e+f x)^{n+1}dx}{b}\) |
\(\Big \downarrow \) 3057 |
\(\displaystyle \frac {\sin (e+f x) \cos ^2(e+f x)^{\frac {n+1}{2}} (b \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {n+1}{2},\frac {n+2}{2},\frac {n+4}{2},\sin ^2(e+f x)\right )}{b f (n+2)}\) |
Input:
Int[Sin[e + f*x]*(b*Tan[e + f*x])^n,x]
Output:
((Cos[e + f*x]^2)^((1 + n)/2)*Hypergeometric2F1[(1 + n)/2, (2 + n)/2, (4 + n)/2, Sin[e + f*x]^2]*Sin[e + f*x]*(b*Tan[e + f*x])^(1 + n))/(b*f*(2 + n) )
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*Frac Part[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^2)^Fr acPart[(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]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[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]
\[\int \sin \left (f x +e \right ) \left (b \tan \left (f x +e \right )\right )^{n}d x\]
Input:
int(sin(f*x+e)*(b*tan(f*x+e))^n,x)
Output:
int(sin(f*x+e)*(b*tan(f*x+e))^n,x)
\[ \int \sin (e+f x) (b \tan (e+f x))^n \, dx=\int { \left (b \tan \left (f x + e\right )\right )^{n} \sin \left (f x + e\right ) \,d x } \] Input:
integrate(sin(f*x+e)*(b*tan(f*x+e))^n,x, algorithm="fricas")
Output:
integral((b*tan(f*x + e))^n*sin(f*x + e), x)
\[ \int \sin (e+f x) (b \tan (e+f x))^n \, dx=\int \left (b \tan {\left (e + f x \right )}\right )^{n} \sin {\left (e + f x \right )}\, dx \] Input:
integrate(sin(f*x+e)*(b*tan(f*x+e))**n,x)
Output:
Integral((b*tan(e + f*x))**n*sin(e + f*x), x)
\[ \int \sin (e+f x) (b \tan (e+f x))^n \, dx=\int { \left (b \tan \left (f x + e\right )\right )^{n} \sin \left (f x + e\right ) \,d x } \] Input:
integrate(sin(f*x+e)*(b*tan(f*x+e))^n,x, algorithm="maxima")
Output:
integrate((b*tan(f*x + e))^n*sin(f*x + e), x)
\[ \int \sin (e+f x) (b \tan (e+f x))^n \, dx=\int { \left (b \tan \left (f x + e\right )\right )^{n} \sin \left (f x + e\right ) \,d x } \] Input:
integrate(sin(f*x+e)*(b*tan(f*x+e))^n,x, algorithm="giac")
Output:
integrate((b*tan(f*x + e))^n*sin(f*x + e), x)
Timed out. \[ \int \sin (e+f x) (b \tan (e+f x))^n \, dx=\int \sin \left (e+f\,x\right )\,{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^n \,d x \] Input:
int(sin(e + f*x)*(b*tan(e + f*x))^n,x)
Output:
int(sin(e + f*x)*(b*tan(e + f*x))^n, x)
\[ \int \sin (e+f x) (b \tan (e+f x))^n \, dx=b^{n} \left (\int \tan \left (f x +e \right )^{n} \sin \left (f x +e \right )d x \right ) \] Input:
int(sin(f*x+e)*(b*tan(f*x+e))^n,x)
Output:
b**n*int(tan(e + f*x)**n*sin(e + f*x),x)