\(\int \frac {\cos ^2(e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx\) [940]

Optimal result

Integrand size = 33, antiderivative size = 119 \[ \int \frac {\cos ^2(e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=-\frac {\operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{2},-n,\frac {5}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (1-\sin (e+f x)) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{3 \sqrt {2} a^2 f \sqrt {1+\sin (e+f x)}} \] Output:

-1/6*AppellF1(3/2,-n,3/2,5/2,d*(1-sin(f*x+e))/(c+d),1/2-1/2*sin(f*x+e))*co 
s(f*x+e)*(1-sin(f*x+e))*(c+d*sin(f*x+e))^n*2^(1/2)/a^2/f/(1+sin(f*x+e))^(1 
/2)/(((c+d*sin(f*x+e))/(c+d))^n)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3042, 3396, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3396

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

\(\Big \downarrow \) 156

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

\(\Big \downarrow \) 155

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

Input:

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

Output:

-1/3*(AppellF1[3/2, 3/2, -n, 5/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f* 
x]))/(c + d)]*Cos[e + f*x]*(1 - Sin[e + f*x])*(c + d*Sin[e + f*x])^n)/(Sqr 
t[2]*a^2*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 3396

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\cos ^2(e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=\frac {\int \frac {\left (\sin \left (f x +e \right ) d +c \right )^{n} \cos \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{2}+2 \sin \left (f x +e \right )+1}d x}{a^{2}} \] Input:

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

Output:

int(((sin(e + f*x)*d + c)**n*cos(e + f*x)**2)/(sin(e + f*x)**2 + 2*sin(e + 
 f*x) + 1),x)/a**2