Integrand size = 29, antiderivative size = 134 \[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{a d (1+n) \sqrt {\cos ^2(c+d x)}}-\frac {\cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{a d (2+n) \sqrt {\cos ^2(c+d x)}} \]
cos(d*x+c)*hypergeom([-1/2, 1/2+1/2*n],[3/2+1/2*n],sin(d*x+c)^2)*sin(d*x+c )^(1+n)/a/d/(1+n)/(cos(d*x+c)^2)^(1/2)-cos(d*x+c)*hypergeom([-1/2, 1+1/2*n ],[1/2*n+2],sin(d*x+c)^2)*sin(d*x+c)^(2+n)/a/d/(2+n)/(cos(d*x+c)^2)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(441\) vs. \(2(134)=268\).
Time = 11.84 (sec) , antiderivative size = 441, normalized size of antiderivative = 3.29 \[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {2^{1+n} \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2 \tan \left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^n \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^n \left (\frac {\operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},4+n,\frac {3+n}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{1+n}+\tan \left (\frac {1}{2} (c+d x)\right ) \left (-\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {2+n}{2},4+n,\frac {4+n}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{2+n}+\tan \left (\frac {1}{2} (c+d x)\right ) \left (-\frac {\operatorname {Hypergeometric2F1}\left (\frac {3+n}{2},4+n,\frac {5+n}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{3+n}+\frac {4 \operatorname {Hypergeometric2F1}\left (\frac {4+n}{2},4+n,\frac {6+n}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{4+n}-\frac {\operatorname {Hypergeometric2F1}\left (4+n,\frac {5+n}{2},\frac {7+n}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{5+n}-\frac {2 \operatorname {Hypergeometric2F1}\left (4+n,\frac {6+n}{2},\frac {8+n}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan ^3\left (\frac {1}{2} (c+d x)\right )}{6+n}+\frac {\operatorname {Hypergeometric2F1}\left (4+n,\frac {7+n}{2},\frac {9+n}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan ^4\left (\frac {1}{2} (c+d x)\right )}{7+n}\right )\right )\right )}{d (a+a \sin (c+d x))} \]
(2^(1 + n)*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*Tan[(c + d*x)/2]*(Tan[( c + d*x)/2]/(1 + Tan[(c + d*x)/2]^2))^n*(1 + Tan[(c + d*x)/2]^2)^n*(Hyperg eometric2F1[(1 + n)/2, 4 + n, (3 + n)/2, -Tan[(c + d*x)/2]^2]/(1 + n) + Ta n[(c + d*x)/2]*((-2*Hypergeometric2F1[(2 + n)/2, 4 + n, (4 + n)/2, -Tan[(c + d*x)/2]^2])/(2 + n) + Tan[(c + d*x)/2]*(-(Hypergeometric2F1[(3 + n)/2, 4 + n, (5 + n)/2, -Tan[(c + d*x)/2]^2]/(3 + n)) + (4*Hypergeometric2F1[(4 + n)/2, 4 + n, (6 + n)/2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2])/(4 + n) - (Hypergeometric2F1[4 + n, (5 + n)/2, (7 + n)/2, -Tan[(c + d*x)/2]^2]*Tan[ (c + d*x)/2]^2)/(5 + n) - (2*Hypergeometric2F1[4 + n, (6 + n)/2, (8 + n)/2 , -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^3)/(6 + n) + (Hypergeometric2F1[4 + n, (7 + n)/2, (9 + n)/2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^4)/(7 + n )))))/(d*(a + a*Sin[c + d*x]))
Time = 0.42 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3042, 3318, 3042, 3057}
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.
| \(\displaystyle \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^4 \sin (c+d x)^n}{a \sin (c+d x)+a}dx\) |
| \(\Big \downarrow \) 3318 |
| \(\displaystyle \frac {\int \cos ^2(c+d x) \sin ^n(c+d x)dx}{a}-\frac {\int \cos ^2(c+d x) \sin ^{n+1}(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \cos (c+d x)^2 \sin (c+d x)^ndx}{a}-\frac {\int \cos (c+d x)^2 \sin (c+d x)^{n+1}dx}{a}\) |
| \(\Big \downarrow \) 3057 |
| \(\displaystyle \frac {\cos (c+d x) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {n+1}{2},\frac {n+3}{2},\sin ^2(c+d x)\right )}{a d (n+1) \sqrt {\cos ^2(c+d x)}}-\frac {\cos (c+d x) \sin ^{n+2}(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {n+2}{2},\frac {n+4}{2},\sin ^2(c+d x)\right )}{a d (n+2) \sqrt {\cos ^2(c+d x)}}\) |
(Cos[c + d*x]*Hypergeometric2F1[-1/2, (1 + n)/2, (3 + n)/2, Sin[c + d*x]^2 ]*Sin[c + d*x]^(1 + n))/(a*d*(1 + n)*Sqrt[Cos[c + d*x]^2]) - (Cos[c + d*x] *Hypergeometric2F1[-1/2, (2 + n)/2, (4 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x ]^(2 + n))/(a*d*(2 + n)*Sqrt[Cos[c + d*x]^2])
3.5.92.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[b^(2*IntPart[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*Frac Part[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^2)^Fr acPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[ e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x]
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g^2/a Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[g^2/(b*d) Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0]
\[\int \frac {\left (\cos ^{4}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right )}{a +a \sin \left (d x +c \right )}d x\]
\[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{a+a \sin (c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4}}{a \sin \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{a+a \sin (c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4}}{a \sin \left (d x + c\right ) + a} \,d x } \]
\[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{a+a \sin (c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4}}{a \sin \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{a+a \sin (c+d x)} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^n}{a+a\,\sin \left (c+d\,x\right )} \,d x \]