\(\int \sin (e+f x) (b \tan (e+f x))^n \, dx\) [181]

Optimal result

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)
 

Mathematica [C] (warning: unable to verify)

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])))
 

Rubi [A] (verified)

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) 
)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3057

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]
 

rule 3082

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]
 

Maple [F]

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

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)
 

Fricas [F]

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

Sympy [F]

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

Maxima [F]

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

Giac [F]

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

Mupad [F(-1)]

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)
 

Reduce [F]

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