Integrand size = 33, antiderivative size = 121 \[ \int \frac {\cos ^4(e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=-\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {5}{2},\frac {1}{2},-n,\frac {7}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (1-\sin (e+f x))^2 (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{5 a^2 f \sqrt {1+\sin (e+f x)}} \]
-1/5*AppellF1(5/2,-n,1/2,7/2,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))^2*(c+d*sin(f*x+e))^n*2^(1/2)/a^2/f/(((c+d*sin(f*x+ e))/(c+d))^n)/(1+sin(f*x+e))^(1/2)
\[ \int \frac {\cos ^4(e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=\int \frac {\cos ^4(e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx \]
Time = 0.44 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, 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, 3263, 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.
| \(\displaystyle \int \frac {\cos ^4(e+f x) (c+d \sin (e+f x))^n}{(a \sin (e+f x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (e+f x)^4 (c+d \sin (e+f x))^n}{(a \sin (e+f x)+a)^2}dx\) |
| \(\Big \downarrow \) 3393 |
| \(\displaystyle \frac {\int (a-a \sin (e+f x))^2 (c+d \sin (e+f x))^ndx}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int (a-a \sin (e+f x))^2 (c+d \sin (e+f x))^ndx}{a^4}\) |
| \(\Big \downarrow \) 3263 |
| \(\displaystyle \frac {\cos (e+f x) \int \frac {(1-\sin (e+f x))^{3/2} (c+d \sin (e+f x))^n}{\sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{a^2 f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}\) |
| \(\Big \downarrow \) 156 |
| \(\displaystyle \frac {\cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \int \frac {(1-\sin (e+f x))^{3/2} \left (\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}\right )^n}{\sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{a^2 f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}\) |
| \(\Big \downarrow \) 155 |
| \(\displaystyle -\frac {\sqrt {2} (1-\sin (e+f x))^2 \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \operatorname {AppellF1}\left (\frac {5}{2},\frac {1}{2},-n,\frac {7}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{5 a^2 f \sqrt {\sin (e+f x)+1}}\) |
-1/5*(Sqrt[2]*AppellF1[5/2, 1/2, -n, 7/2, (1 - Sin[e + f*x])/2, (d*(1 - Si n[e + f*x]))/(c + d)]*Cos[e + f*x]*(1 - Sin[e + f*x])^2*(c + d*Sin[e + f*x ])^n)/(a^2*f*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n)
3.10.49.3.1 Defintions of rubi rules used
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])
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]
Int[((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 - 1/2)*((c + d *x)^n/Sqrt[1 - (b/a)*x]), x], x, Sin[e + f*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] & & IntegerQ[m]
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 ]
\[\int \frac {\left (\cos ^{4}\left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )^{n}}{\left (a +a \sin \left (f x +e \right )\right )^{2}}d x\]
\[ \int \frac {\cos ^4(e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=\int { \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{4}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
integral(-(d*sin(f*x + e) + c)^n*cos(f*x + e)^4/(a^2*cos(f*x + e)^2 - 2*a^ 2*sin(f*x + e) - 2*a^2), x)
Exception generated. \[ \int \frac {\cos ^4(e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
\[ \int \frac {\cos ^4(e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=\int { \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{4}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {\cos ^4(e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=\int { \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{4}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\cos ^4(e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^4\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]