\(\int \sin (e+f x) (b (c \tan (e+f x))^n)^p \, dx\) [171]

Optimal result

Integrand size = 21, antiderivative size = 91 \[ \int \sin (e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\frac {\cos ^2(e+f x)^{\frac {1}{2} (1+n p)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1+n p),\frac {1}{2} (2+n p),\frac {1}{2} (4+n p),\sin ^2(e+f x)\right ) \sin (e+f x) \tan (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (2+n p)} \] Output:

(cos(f*x+e)^2)^(1/2*n*p+1/2)*hypergeom([1/2*n*p+1, 1/2*n*p+1/2],[1/2*n*p+2 
],sin(f*x+e)^2)*sin(f*x+e)*tan(f*x+e)*(b*(c*tan(f*x+e))^n)^p/f/(n*p+2)
 

Mathematica [C] (warning: unable to verify)

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

Time = 1.11 (sec) , antiderivative size = 284, normalized size of antiderivative = 3.12 \[ \int \sin (e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\frac {8 (4+n p) \operatorname {AppellF1}\left (1+\frac {n p}{2},n p,2,2+\frac {n p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^4\left (\frac {1}{2} (e+f x)\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (2+n p) \left (2 (4+n p) \operatorname {AppellF1}\left (1+\frac {n p}{2},n p,2,2+\frac {n p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right )+2 \left (2 \operatorname {AppellF1}\left (2+\frac {n p}{2},n p,3,3+\frac {n p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-n p \operatorname {AppellF1}\left (2+\frac {n p}{2},1+n p,2,3+\frac {n p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) (-1+\cos (e+f x))\right )} \] Input:

Integrate[Sin[e + f*x]*(b*(c*Tan[e + f*x])^n)^p,x]
 

Output:

(8*(4 + n*p)*AppellF1[1 + (n*p)/2, n*p, 2, 2 + (n*p)/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*(c*Tan[e + 
 f*x])^n)^p)/(f*(2 + n*p)*(2*(4 + n*p)*AppellF1[1 + (n*p)/2, n*p, 2, 2 + ( 
n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^2 + 2*(2 
*AppellF1[2 + (n*p)/2, n*p, 3, 3 + (n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + 
f*x)/2]^2] - n*p*AppellF1[2 + (n*p)/2, 1 + n*p, 2, 3 + (n*p)/2, Tan[(e + f 
*x)/2]^2, -Tan[(e + f*x)/2]^2])*(-1 + Cos[e + f*x])))
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4142, 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.

\(\displaystyle \int \sin (e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (e+f x) \left (b (c \tan (e+f x))^n\right )^pdx\)

\(\Big \downarrow \) 4142

\(\displaystyle (c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p \int \sin (e+f x) (c \tan (e+f x))^{n p}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle (c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p \int \sin (e+f x) (c \tan (e+f x))^{n p}dx\)

\(\Big \downarrow \) 3082

\(\displaystyle \sin ^{-n p}(e+f x) \cos ^{n p}(e+f x) \left (b (c \tan (e+f x))^n\right )^p \int \cos ^{-n p}(e+f x) \sin ^{n p+1}(e+f x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sin ^{-n p}(e+f x) \cos ^{n p}(e+f x) \left (b (c \tan (e+f x))^n\right )^p \int \cos (e+f x)^{-n p} \sin (e+f x)^{n p+1}dx\)

\(\Big \downarrow \) 3057

\(\displaystyle \frac {\sin (e+f x) \tan (e+f x) \cos ^2(e+f x)^{\frac {1}{2} (n p+1)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (n p+1),\frac {1}{2} (n p+2),\frac {1}{2} (n p+4),\sin ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+2)}\)

Input:

Int[Sin[e + f*x]*(b*(c*Tan[e + f*x])^n)^p,x]
 

Output:

((Cos[e + f*x]^2)^((1 + n*p)/2)*Hypergeometric2F1[(1 + n*p)/2, (2 + n*p)/2 
, (4 + n*p)/2, Sin[e + f*x]^2]*Sin[e + f*x]*Tan[e + f*x]*(b*(c*Tan[e + f*x 
])^n)^p)/(f*(2 + n*p))
 

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]
 

rule 4142

Int[(u_.)*((b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> S 
imp[b^IntPart[p]*((b*(c*Tan[e + f*x])^n)^FracPart[p]/(c*Tan[e + f*x])^(n*Fr 
acPart[p]))   Int[ActivateTrig[u]*(c*Tan[e + f*x])^(n*p), x], x] /; FreeQ[{ 
b, c, e, f, n, p}, x] &&  !IntegerQ[p] &&  !IntegerQ[n] && (EqQ[u, 1] || Ma 
tchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, 
 cos, tan, cot, sec, csc}, trig]])
 

Maple [F]

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

Input:

int(sin(f*x+e)*(b*(c*tan(f*x+e))^n)^p,x)
 

Output:

int(sin(f*x+e)*(b*(c*tan(f*x+e))^n)^p,x)
 

Fricas [F]

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

integrate(sin(f*x+e)*(b*(c*tan(f*x+e))^n)^p,x, algorithm="fricas")
 

Output:

integral(((c*tan(f*x + e))^n*b)^p*sin(f*x + e), x)
 

Sympy [F]

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

integrate(sin(f*x+e)*(b*(c*tan(f*x+e))**n)**p,x)
 

Output:

Integral((b*(c*tan(e + f*x))**n)**p*sin(e + f*x), x)
 

Maxima [F]

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

integrate(sin(f*x+e)*(b*(c*tan(f*x+e))^n)^p,x, algorithm="maxima")
 

Output:

integrate(((c*tan(f*x + e))^n*b)^p*sin(f*x + e), x)
 

Giac [F]

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

integrate(sin(f*x+e)*(b*(c*tan(f*x+e))^n)^p,x, algorithm="giac")
 

Output:

integrate(((c*tan(f*x + e))^n*b)^p*sin(f*x + e), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sin (e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int \sin \left (e+f\,x\right )\,{\left (b\,{\left (c\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\right )}^p \,d x \] Input:

int(sin(e + f*x)*(b*(c*tan(e + f*x))^n)^p,x)
 

Output:

int(sin(e + f*x)*(b*(c*tan(e + f*x))^n)^p, x)
 

Reduce [F]

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

int(sin(f*x+e)*(b*(c*tan(f*x+e))^n)^p,x)
 

Output:

c**(n*p)*b**p*int(tan(e + f*x)**(n*p)*sin(e + f*x),x)