Integrand size = 29, antiderivative size = 200 \[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{d (1+n) \sqrt {\cos ^2(c+d x)}}+\frac {2 a^2 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{d (2+n) \sqrt {\cos ^2(c+d x)}}+\frac {a^2 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3+n}{2},\frac {5+n}{2},\sin ^2(c+d x)\right ) \sin ^{3+n}(c+d x)}{d (3+n) \sqrt {\cos ^2(c+d x)}} \]
a^2*cos(d*x+c)*hypergeom([-3/2, 1/2+1/2*n],[3/2+1/2*n],sin(d*x+c)^2)*sin(d *x+c)^(1+n)/d/(1+n)/(cos(d*x+c)^2)^(1/2)+2*a^2*cos(d*x+c)*hypergeom([-3/2, 1+1/2*n],[1/2*n+2],sin(d*x+c)^2)*sin(d*x+c)^(2+n)/d/(2+n)/(cos(d*x+c)^2)^ (1/2)+a^2*cos(d*x+c)*hypergeom([-3/2, 3/2+1/2*n],[5/2+1/2*n],sin(d*x+c)^2) *sin(d*x+c)^(3+n)/d/(3+n)/(cos(d*x+c)^2)^(1/2)
Time = 0.23 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.82 \[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \sqrt {\cos ^2(c+d x)} \sec (c+d x) \sin ^{1+n}(c+d x) \left (\left (6+5 n+n^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right )+(1+n) \sin (c+d x) \left (2 (3+n) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right )+(2+n) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3+n}{2},\frac {5+n}{2},\sin ^2(c+d x)\right ) \sin (c+d x)\right )\right )}{d (1+n) (2+n) (3+n)} \]
(a^2*Sqrt[Cos[c + d*x]^2]*Sec[c + d*x]*Sin[c + d*x]^(1 + n)*((6 + 5*n + n^ 2)*Hypergeometric2F1[-3/2, (1 + n)/2, (3 + n)/2, Sin[c + d*x]^2] + (1 + n) *Sin[c + d*x]*(2*(3 + n)*Hypergeometric2F1[-3/2, (2 + n)/2, (4 + n)/2, Sin [c + d*x]^2] + (2 + n)*Hypergeometric2F1[-3/2, (3 + n)/2, (5 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x])))/(d*(1 + n)*(2 + n)*(3 + n))
Time = 0.46 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3042, 3352, 2009}
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.
\(\displaystyle \int \cos ^4(c+d x) (a \sin (c+d x)+a)^2 \sin ^n(c+d x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^4 (a \sin (c+d x)+a)^2 \sin (c+d x)^ndx\) |
\(\Big \downarrow \) 3352 |
\(\displaystyle \int \left (2 a^2 \cos ^4(c+d x) \sin ^{n+1}(c+d x)+a^2 \cos ^4(c+d x) \sin ^{n+2}(c+d x)+a^2 \cos ^4(c+d x) \sin ^n(c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \cos (c+d x) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {n+1}{2},\frac {n+3}{2},\sin ^2(c+d x)\right )}{d (n+1) \sqrt {\cos ^2(c+d x)}}+\frac {2 a^2 \cos (c+d x) \sin ^{n+2}(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {n+2}{2},\frac {n+4}{2},\sin ^2(c+d x)\right )}{d (n+2) \sqrt {\cos ^2(c+d x)}}+\frac {a^2 \cos (c+d x) \sin ^{n+3}(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {n+3}{2},\frac {n+5}{2},\sin ^2(c+d x)\right )}{d (n+3) \sqrt {\cos ^2(c+d x)}}\) |
(a^2*Cos[c + d*x]*Hypergeometric2F1[-3/2, (1 + n)/2, (3 + n)/2, Sin[c + d* x]^2]*Sin[c + d*x]^(1 + n))/(d*(1 + n)*Sqrt[Cos[c + d*x]^2]) + (2*a^2*Cos[ c + d*x]*Hypergeometric2F1[-3/2, (2 + n)/2, (4 + n)/2, Sin[c + d*x]^2]*Sin [c + d*x]^(2 + n))/(d*(2 + n)*Sqrt[Cos[c + d*x]^2]) + (a^2*Cos[c + d*x]*Hy pergeometric2F1[-3/2, (3 + n)/2, (5 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x]^( 3 + n))/(d*(3 + n)*Sqrt[Cos[c + d*x]^2])
3.5.90.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig [(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]
\[\int \left (\cos ^{4}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{2}d x\]
\[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^2 \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4} \,d x } \]
integral(-(a^2*cos(d*x + c)^6 - 2*a^2*cos(d*x + c)^4*sin(d*x + c) - 2*a^2* cos(d*x + c)^4)*sin(d*x + c)^n, x)
Timed out. \[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^2 \, dx=\text {Timed out} \]
\[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^2 \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4} \,d x } \]
\[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^2 \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4} \,d x } \]
Timed out. \[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^2 \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^n\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2 \,d x \]