\(\int (c (d \sin (e+f x))^p)^n (a+a \sin (e+f x))^2 \, dx\) [828]

Optimal result

Integrand size = 27, antiderivative size = 222 \[ \int \left (c (d \sin (e+f x))^p\right )^n (a+a \sin (e+f x))^2 \, dx=-\frac {a^2 \cos (e+f x) \sin (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (2+n p)}+\frac {a^2 (3+2 n p) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1+n p),\frac {1}{2} (3+n p),\sin ^2(e+f x)\right ) \sin (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (1+n p) (2+n p) \sqrt {\cos ^2(e+f x)}}+\frac {2 a^2 \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (2+n p),\frac {1}{2} (4+n p),\sin ^2(e+f x)\right ) \sin ^2(e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (2+n p) \sqrt {\cos ^2(e+f x)}} \] Output:

-a^2*cos(f*x+e)*sin(f*x+e)*(c*(d*sin(f*x+e))^p)^n/f/(n*p+2)+a^2*(2*n*p+3)* 
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)/(n*p+2)/(cos(f*x+e)^2)^(1/2)+2*a^2*c 
os(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*sin(f*x+e))^p)^n/f/(n*p+2)/(cos(f*x+e)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.00 \[ \int \left (c (d \sin (e+f x))^p\right )^n (a+a \sin (e+f x))^2 \, dx=-\frac {a^2 \cos (e+f x) \sqrt {\cos ^2(e+f x)} \sin (e+f x) \left (c (d \sin (e+f x))^p\right )^n \left (\left (6+5 n p+n^2 p^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1+n p),\frac {1}{2} (3+n p),\sin ^2(e+f x)\right )+(1+n p) \sin (e+f x) \left (2 (3+n p) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+\frac {n p}{2},2+\frac {n p}{2},\sin ^2(e+f x)\right )+(2+n p) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (3+n p),\frac {1}{2} (5+n p),\sin ^2(e+f x)\right ) \sin (e+f x)\right )\right )}{f (1+n p) (2+n p) (3+n p) (-1+\sin (e+f x)) (1+\sin (e+f x))} \] Input:

Integrate[(c*(d*Sin[e + f*x])^p)^n*(a + a*Sin[e + f*x])^2,x]
 

Output:

-((a^2*Cos[e + f*x]*Sqrt[Cos[e + f*x]^2]*Sin[e + f*x]*(c*(d*Sin[e + f*x])^ 
p)^n*((6 + 5*n*p + n^2*p^2)*Hypergeometric2F1[1/2, (1 + n*p)/2, (3 + n*p)/ 
2, Sin[e + f*x]^2] + (1 + n*p)*Sin[e + f*x]*(2*(3 + n*p)*Hypergeometric2F1 
[1/2, 1 + (n*p)/2, 2 + (n*p)/2, Sin[e + f*x]^2] + (2 + n*p)*Hypergeometric 
2F1[1/2, (3 + n*p)/2, (5 + n*p)/2, Sin[e + f*x]^2]*Sin[e + f*x])))/(f*(1 + 
 n*p)*(2 + n*p)*(3 + n*p)*(-1 + Sin[e + f*x])*(1 + Sin[e + f*x])))
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3042, 3305, 3042, 3242, 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.

\(\displaystyle \int (a \sin (e+f x)+a)^2 \left (c (d \sin (e+f x))^p\right )^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x)+a)^2 \left (c (d \sin (e+f x))^p\right )^ndx\)

\(\Big \downarrow \) 3305

\(\displaystyle (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n \int (d \sin (e+f x))^{n p} (\sin (e+f x) a+a)^2dx\)

\(\Big \downarrow \) 3042

\(\displaystyle (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n \int (d \sin (e+f x))^{n p} (\sin (e+f x) a+a)^2dx\)

\(\Big \downarrow \) 3242

\(\displaystyle (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n \left (\frac {\int (d \sin (e+f x))^{n p} \left (d (2 n p+3) a^2+2 d (n p+2) \sin (e+f x) a^2\right )dx}{d (n p+2)}-\frac {a^2 \cos (e+f x) (d \sin (e+f x))^{n p+1}}{d f (n p+2)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n \left (\frac {\int (d \sin (e+f x))^{n p} \left (d (2 n p+3) a^2+2 d (n p+2) \sin (e+f x) a^2\right )dx}{d (n p+2)}-\frac {a^2 \cos (e+f x) (d \sin (e+f x))^{n p+1}}{d f (n p+2)}\right )\)

\(\Big \downarrow \) 3227

\(\displaystyle (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n \left (\frac {a^2 d (2 n p+3) \int (d \sin (e+f x))^{n p}dx+2 a^2 (n p+2) \int (d \sin (e+f x))^{n p+1}dx}{d (n p+2)}-\frac {a^2 \cos (e+f x) (d \sin (e+f x))^{n p+1}}{d f (n p+2)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n \left (\frac {a^2 d (2 n p+3) \int (d \sin (e+f x))^{n p}dx+2 a^2 (n p+2) \int (d \sin (e+f x))^{n p+1}dx}{d (n p+2)}-\frac {a^2 \cos (e+f x) (d \sin (e+f x))^{n p+1}}{d f (n p+2)}\right )\)

\(\Big \downarrow \) 3122

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

Input:

Int[(c*(d*Sin[e + f*x])^p)^n*(a + a*Sin[e + f*x])^2,x]
 

Output:

((c*(d*Sin[e + f*x])^p)^n*(-((a^2*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n*p)) 
/(d*f*(2 + n*p))) + ((a^2*(3 + 2*n*p)*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))/(f*( 
1 + n*p)*Sqrt[Cos[e + f*x]^2]) + (2*a^2*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 
*f*Sqrt[Cos[e + f*x]^2]))/(d*(2 + n*p))))/(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 3242

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m 
+ n))   Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c* 
(m - 2) + b^2*d*(n + 1) + a^2*d*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 
 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c 
 - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !LtQ[ 
n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[ 
c, 0]))
 

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]

\[\int \left (c \left (d \sin \left (f x +e \right )\right )^{p}\right )^{n} \left (a +a \sin \left (f x +e \right )\right )^{2}d x\]

Input:

int((c*(d*sin(f*x+e))^p)^n*(a+a*sin(f*x+e))^2,x)
 

Output:

int((c*(d*sin(f*x+e))^p)^n*(a+a*sin(f*x+e))^2,x)
 

Fricas [F]

\[ \int \left (c (d \sin (e+f x))^p\right )^n (a+a \sin (e+f x))^2 \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{2} \left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n} \,d x } \] Input:

integrate((c*(d*sin(f*x+e))^p)^n*(a+a*sin(f*x+e))^2,x, algorithm="fricas")
 

Output:

integral(-(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2)*((d*sin(f*x + 
e))^p*c)^n, x)
 

Sympy [F]

\[ \int \left (c (d \sin (e+f x))^p\right )^n (a+a \sin (e+f x))^2 \, dx=a^{2} \left (\int \left (c \left (d \sin {\left (e + f x \right )}\right )^{p}\right )^{n}\, dx + \int 2 \left (c \left (d \sin {\left (e + f x \right )}\right )^{p}\right )^{n} \sin {\left (e + f x \right )}\, dx + \int \left (c \left (d \sin {\left (e + f x \right )}\right )^{p}\right )^{n} \sin ^{2}{\left (e + f x \right )}\, dx\right ) \] Input:

integrate((c*(d*sin(f*x+e))**p)**n*(a+a*sin(f*x+e))**2,x)
 

Output:

a**2*(Integral((c*(d*sin(e + f*x))**p)**n, x) + Integral(2*(c*(d*sin(e + f 
*x))**p)**n*sin(e + f*x), x) + Integral((c*(d*sin(e + f*x))**p)**n*sin(e + 
 f*x)**2, x))
 

Maxima [F]

\[ \int \left (c (d \sin (e+f x))^p\right )^n (a+a \sin (e+f x))^2 \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{2} \left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n} \,d x } \] Input:

integrate((c*(d*sin(f*x+e))^p)^n*(a+a*sin(f*x+e))^2,x, algorithm="maxima")
 

Output:

integrate((a*sin(f*x + e) + a)^2*((d*sin(f*x + e))^p*c)^n, x)
 

Giac [F]

\[ \int \left (c (d \sin (e+f x))^p\right )^n (a+a \sin (e+f x))^2 \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{2} \left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n} \,d x } \] Input:

integrate((c*(d*sin(f*x+e))^p)^n*(a+a*sin(f*x+e))^2,x, algorithm="giac")
 

Output:

integrate((a*sin(f*x + e) + a)^2*((d*sin(f*x + e))^p*c)^n, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (c (d \sin (e+f x))^p\right )^n (a+a \sin (e+f x))^2 \, dx=\int {\left (c\,{\left (d\,\sin \left (e+f\,x\right )\right )}^p\right )}^n\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2 \,d x \] Input:

int((c*(d*sin(e + f*x))^p)^n*(a + a*sin(e + f*x))^2,x)
 

Output:

int((c*(d*sin(e + f*x))^p)^n*(a + a*sin(e + f*x))^2, x)
 

Reduce [F]

\[ \int \left (c (d \sin (e+f x))^p\right )^n (a+a \sin (e+f x))^2 \, dx=d^{n p} c^{n} a^{2} \left (\int \sin \left (f x +e \right )^{n p}d x +\int \sin \left (f x +e \right )^{n p} \sin \left (f x +e \right )^{2}d x +2 \left (\int \sin \left (f x +e \right )^{n p} \sin \left (f x +e \right )d x \right )\right ) \] Input:

int((c*(d*sin(f*x+e))^p)^n*(a+a*sin(f*x+e))^2,x)
 

Output:

d**(n*p)*c**n*a**2*(int(sin(e + f*x)**(n*p),x) + int(sin(e + f*x)**(n*p)*s 
in(e + f*x)**2,x) + 2*int(sin(e + f*x)**(n*p)*sin(e + f*x),x))