\(\int (c (d \sin (e+f x))^p)^n (a+b \sin (e+f x)) \, dx\) [835]

Optimal result

Integrand size = 25, antiderivative size = 163 \[ \int \left (c (d \sin (e+f x))^p\right )^n (a+b \sin (e+f x)) \, dx=\frac {a \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1+n p),\frac {1}{2} (3+n p),\sin ^2(e+f x)\right ) \sin (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (1+n p) \sqrt {\cos ^2(e+f x)}}+\frac {b \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (2+n p),\frac {1}{2} (4+n p),\sin ^2(e+f x)\right ) \sin ^2(e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (2+n p) \sqrt {\cos ^2(e+f x)}} \] Output:

a*cos(f*x+e)*hypergeom([1/2, 1/2*n*p+1/2],[1/2*n*p+3/2],sin(f*x+e)^2)*sin( 
f*x+e)*(c*(d*sin(f*x+e))^p)^n/f/(n*p+1)/(cos(f*x+e)^2)^(1/2)+b*cos(f*x+e)* 
hypergeom([1/2, 1/2*n*p+1],[1/2*n*p+2],sin(f*x+e)^2)*sin(f*x+e)^2*(c*(d*si 
n(f*x+e))^p)^n/f/(n*p+2)/(cos(f*x+e)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.79 \[ \int \left (c (d \sin (e+f x))^p\right )^n (a+b \sin (e+f x)) \, dx=\frac {\sqrt {\cos ^2(e+f x)} \left (c (d \sin (e+f x))^p\right )^n \left (a (2+n p) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1+n p),\frac {1}{2} (3+n p),\sin ^2(e+f x)\right )+b (1+n p) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+\frac {n p}{2},2+\frac {n p}{2},\sin ^2(e+f x)\right ) \sin (e+f x)\right ) \tan (e+f x)}{f (1+n p) (2+n p)} \] Input:

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

Output:

(Sqrt[Cos[e + f*x]^2]*(c*(d*Sin[e + f*x])^p)^n*(a*(2 + n*p)*Hypergeometric 
2F1[1/2, (1 + n*p)/2, (3 + n*p)/2, Sin[e + f*x]^2] + b*(1 + n*p)*Hypergeom 
etric2F1[1/2, 1 + (n*p)/2, 2 + (n*p)/2, Sin[e + f*x]^2]*Sin[e + f*x])*Tan[ 
e + f*x])/(f*(1 + n*p)*(2 + n*p))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3305, 3042, 3227, 3042, 3122}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \sin (e+f x)) \left (c (d \sin (e+f x))^p\right )^ndx\)

\(\Big \downarrow \) 3305

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n \left (\frac {a \cos (e+f x) (d \sin (e+f x))^{n p+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (n p+1),\frac {1}{2} (n p+3),\sin ^2(e+f x)\right )}{d f (n p+1) \sqrt {\cos ^2(e+f x)}}+\frac {b \cos (e+f x) (d \sin (e+f x))^{n p+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (n p+2),\frac {1}{2} (n p+4),\sin ^2(e+f x)\right )}{d^2 f (n p+2) \sqrt {\cos ^2(e+f x)}}\right )\)

Input:

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

Output:

((c*(d*Sin[e + f*x])^p)^n*((a*Cos[e + f*x]*Hypergeometric2F1[1/2, (1 + n*p 
)/2, (3 + n*p)/2, Sin[e + f*x]^2]*(d*Sin[e + f*x])^(1 + n*p))/(d*f*(1 + n* 
p)*Sqrt[Cos[e + f*x]^2]) + (b*Cos[e + f*x]*Hypergeometric2F1[1/2, (2 + n*p 
)/2, (4 + n*p)/2, Sin[e + f*x]^2]*(d*Sin[e + f*x])^(2 + n*p))/(d^2*f*(2 + 
n*p)*Sqrt[Cos[e + f*x]^2])))/(d*Sin[e + f*x])^(n*p)
 

Defintions of rubi rules used

rule 3042

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

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3305

Int[((c_.)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(p_))^(n_)*((a_.) + (b_.)*sin[(e 
_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Simp[c^IntPart[n]*((c*(d*Sin[e + f*x 
])^p)^FracPart[n]/(d*Sin[e + f*x])^(p*FracPart[n]))   Int[(a + b*Sin[e + f* 
x])^m*(d*Sin[e + f*x])^(n*p), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, 
x] &&  !IntegerQ[n]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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