\(\int \frac {\cos ^2(e+f x) (c+d \sin (e+f x))^n}{a+a \sin (e+f x)} \, dx\) [939]

Optimal result

Integrand size = 33, antiderivative size = 110 \[ \int \frac {\cos ^2(e+f x) (c+d \sin (e+f x))^n}{a+a \sin (e+f x)} \, dx=\frac {2 \sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n}}{a f \sqrt {1-\sin (e+f x)}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(229\) vs. \(2(110)=220\).

Time = 1.47 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.08 \[ \int \frac {\cos ^2(e+f x) (c+d \sin (e+f x))^n}{a+a \sin (e+f x)} \, dx=-\frac {\sec (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sqrt {-\frac {d (-1+\sin (e+f x))}{c+d}} (c+d \sin (e+f x))^{1+n} \left (-\left ((c+d) (2+n) \operatorname {AppellF1}\left (1+n,\frac {1}{2},\frac {1}{2},2+n,\frac {c+d \sin (e+f x)}{c-d},\frac {c+d \sin (e+f x)}{c+d}\right )\right )+(1+n) \operatorname {AppellF1}\left (2+n,\frac {1}{2},\frac {1}{2},3+n,\frac {c+d \sin (e+f x)}{c-d},\frac {c+d \sin (e+f x)}{c+d}\right ) (c+d \sin (e+f x))\right )}{a d (-c+d) f (1+n) (2+n) \sqrt {\frac {d (1+\sin (e+f x))}{-c+d}}} \] Input:

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

Output:

-((Sec[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*Sqrt[-((d*(-1 + Si 
n[e + f*x]))/(c + d))]*(c + d*Sin[e + f*x])^(1 + n)*(-((c + d)*(2 + n)*App 
ellF1[1 + n, 1/2, 1/2, 2 + n, (c + d*Sin[e + f*x])/(c - d), (c + d*Sin[e + 
 f*x])/(c + d)]) + (1 + n)*AppellF1[2 + n, 1/2, 1/2, 3 + n, (c + d*Sin[e + 
 f*x])/(c - d), (c + d*Sin[e + f*x])/(c + d)]*(c + d*Sin[e + f*x])))/(a*d* 
(-c + d)*f*(1 + n)*(2 + n)*Sqrt[(d*(1 + Sin[e + f*x]))/(-c + d)]))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 3393, 3042, 3234, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3393

\(\displaystyle \frac {\int (a-a \sin (e+f x)) (c+d \sin (e+f x))^ndx}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (a-a \sin (e+f x)) (c+d \sin (e+f x))^ndx}{a^2}\)

\(\Big \downarrow \) 3234

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

\(\Big \downarrow \) 156

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

\(\Big \downarrow \) 155

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

Input:

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

Output:

-1/3*(Sqrt[2]*AppellF1[3/2, 1/2, -n, 5/2, (1 - Sin[e + f*x])/2, (d*(1 - Si 
n[e + f*x]))/(c + d)]*Cos[e + f*x]*(1 - Sin[e + f*x])*(c + d*Sin[e + f*x]) 
^n)/(a*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 3234

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)]), x_Symbol] :> Simp[c*(Cos[e + f*x]/(f*Sqrt[1 + Sin[e + f*x]]*Sq 
rt[1 - Sin[e + f*x]]))   Subst[Int[(a + b*x)^m*(Sqrt[1 + (d/c)*x]/Sqrt[1 - 
(d/c)*x]), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && N 
eQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*m] && EqQ[c^2 - d^2, 
0]
 

rule 3393

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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