\(\int \cos ^2(e+f x) (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx\) [938]

Optimal result

Integrand size = 31, antiderivative size = 117 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=-\frac {4 \sqrt {2} a \operatorname {AppellF1}\left (\frac {3}{2},-\frac {3}{2},-n,\frac {5}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (1-\sin (e+f x)) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{3 f \sqrt {1+\sin (e+f x)}} \] Output:

-4/3*2^(1/2)*a*AppellF1(3/2,-n,-3/2,5/2,d*(1-sin(f*x+e))/(c+d),1/2-1/2*sin 
(f*x+e))*cos(f*x+e)*(1-sin(f*x+e))*(c+d*sin(f*x+e))^n/f/(1+sin(f*x+e))^(1/ 
2)/(((c+d*sin(f*x+e))/(c+d))^n)
 

Mathematica [F]

\[ \int \cos ^2(e+f x) (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int \cos ^2(e+f x) (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3042, 3347, 156, 155}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3347

\(\displaystyle \frac {a \cos (e+f x) \int \sqrt {1-\sin (e+f x)} (\sin (e+f x)+1)^{3/2} (c+d \sin (e+f x))^nd\sin (e+f x)}{f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}\)

\(\Big \downarrow \) 156

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

\(\Big \downarrow \) 155

\(\displaystyle -\frac {4 \sqrt {2} a (1-\sin (e+f x)) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \operatorname {AppellF1}\left (\frac {3}{2},-\frac {3}{2},-n,\frac {5}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{3 f \sqrt {\sin (e+f x)+1}}\)

Input:

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

Output:

(-4*Sqrt[2]*a*AppellF1[3/2, -3/2, -n, 5/2, (1 - Sin[e + f*x])/2, (d*(1 - S 
in[e + f*x]))/(c + d)]*Cos[e + f*x]*(1 - Sin[e + f*x])*(c + d*Sin[e + f*x] 
)^n)/(3*f*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n)
 

Defintions of rubi rules used

rule 155

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ 
(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, 
 m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 156

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p 
]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p])   Int[(a + b*x)^m*(c + d*x)^n*Si 
mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
& GtQ[Simplify[b/(b*c - a*d)], 0] &&  !GtQ[Simplify[b/(b*e - a*f)], 0]
 

rule 3042

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

rule 3347

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*g*((g 
*Cos[e + f*x])^(p - 1)/(f*(1 + Sin[e + f*x])^((p - 1)/2)*(1 - Sin[e + f*x]) 
^((p - 1)/2)))   Subst[Int[(1 + (d/c)*x)^((p + 1)/2)*(1 - (d/c)*x)^((p - 1) 
/2)*(a + b*x)^m, x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, p} 
, x] && NeQ[a^2 - b^2, 0] && EqQ[c^2 - d^2, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))*(c+d*sin(f*x+e))^n,x, algorithm="m 
axima")
 

Output:

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

Giac [F]

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

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))*(c+d*sin(f*x+e))^n,x, algorithm="g 
iac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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