Integrand size = 29, antiderivative size = 173 \[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\cos (c+d x) \sin ^{1+n}(c+d x)}{a^2 d (2+n)}+\frac {(3+2 n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{a^2 d (1+n) (2+n) \sqrt {\cos ^2(c+d x)}}-\frac {2 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{a^2 d (2+n) \sqrt {\cos ^2(c+d x)}} \]
-cos(d*x+c)*sin(d*x+c)^(1+n)/a^2/d/(2+n)+(3+2*n)*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^2/d/(1+n)/(2+n)/ (cos(d*x+c)^2)^(1/2)-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^2/d/(2+n)/(cos(d*x+c)^2)^(1/2)
Time = 4.01 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.80 \[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \sec ^2\left (\frac {1}{2} (c+d x)\right )^n \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4 \sin ^n(c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) \left (\frac {\operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},3+n,\frac {3+n}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{1+n}+\tan \left (\frac {1}{2} (c+d x)\right ) \left (-\frac {4 \operatorname {Hypergeometric2F1}\left (\frac {2+n}{2},3+n,\frac {4+n}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{2+n}+\tan \left (\frac {1}{2} (c+d x)\right ) \left (\frac {6 \operatorname {Hypergeometric2F1}\left (\frac {3+n}{2},3+n,\frac {5+n}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{3+n}-\frac {4 \operatorname {Hypergeometric2F1}\left (3+n,\frac {4+n}{2},\frac {6+n}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{4+n}+\frac {\operatorname {Hypergeometric2F1}\left (3+n,\frac {5+n}{2},\frac {7+n}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{5+n}\right )\right )\right )}{d (a+a \sin (c+d x))^2} \]
(2*(Sec[(c + d*x)/2]^2)^n*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*Sin[c + d*x]^n*Tan[(c + d*x)/2]*(Hypergeometric2F1[(1 + n)/2, 3 + n, (3 + n)/2, -T an[(c + d*x)/2]^2]/(1 + n) + Tan[(c + d*x)/2]*((-4*Hypergeometric2F1[(2 + n)/2, 3 + n, (4 + n)/2, -Tan[(c + d*x)/2]^2])/(2 + n) + Tan[(c + d*x)/2]*( (6*Hypergeometric2F1[(3 + n)/2, 3 + n, (5 + n)/2, -Tan[(c + d*x)/2]^2])/(3 + n) - (4*Hypergeometric2F1[3 + n, (4 + n)/2, (6 + n)/2, -Tan[(c + d*x)/2 ]^2]*Tan[(c + d*x)/2])/(4 + n) + (Hypergeometric2F1[3 + n, (5 + n)/2, (7 + n)/2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2)/(5 + n)))))/(d*(a + a*Sin[ c + d*x])^2)
Time = 0.60 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3042, 3348, 3042, 3242, 3042, 3227, 3042, 3122}
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.
| \(\displaystyle \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^4 \sin (c+d x)^n}{(a \sin (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3348 |
| \(\displaystyle \frac {\int \sin ^n(c+d x) (a-a \sin (c+d x))^2dx}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sin (c+d x)^n (a-a \sin (c+d x))^2dx}{a^4}\) |
| \(\Big \downarrow \) 3242 |
| \(\displaystyle \frac {\frac {\int \sin ^n(c+d x) \left (a^2 (2 n+3)-2 a^2 (n+2) \sin (c+d x)\right )dx}{n+2}-\frac {a^2 \cos (c+d x) \sin ^{n+1}(c+d x)}{d (n+2)}}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \sin (c+d x)^n \left (a^2 (2 n+3)-2 a^2 (n+2) \sin (c+d x)\right )dx}{n+2}-\frac {a^2 \cos (c+d x) \sin ^{n+1}(c+d x)}{d (n+2)}}{a^4}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {a^2 (2 n+3) \int \sin ^n(c+d x)dx-2 a^2 (n+2) \int \sin ^{n+1}(c+d x)dx}{n+2}-\frac {a^2 \cos (c+d x) \sin ^{n+1}(c+d x)}{d (n+2)}}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 (2 n+3) \int \sin (c+d x)^ndx-2 a^2 (n+2) \int \sin (c+d x)^{n+1}dx}{n+2}-\frac {a^2 \cos (c+d x) \sin ^{n+1}(c+d x)}{d (n+2)}}{a^4}\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \frac {\frac {\frac {a^2 (2 n+3) \cos (c+d x) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+1}{2},\frac {n+3}{2},\sin ^2(c+d x)\right )}{d (n+1) \sqrt {\cos ^2(c+d x)}}-\frac {2 a^2 \cos (c+d x) \sin ^{n+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+2}{2},\frac {n+4}{2},\sin ^2(c+d x)\right )}{d \sqrt {\cos ^2(c+d x)}}}{n+2}-\frac {a^2 \cos (c+d x) \sin ^{n+1}(c+d x)}{d (n+2)}}{a^4}\) |
(-((a^2*Cos[c + d*x]*Sin[c + d*x]^(1 + n))/(d*(2 + n))) + ((a^2*(3 + 2*n)* Cos[c + d*x]*Hypergeometric2F1[1/2, (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[1/2, (2 + n)/2, (4 + n)/2, Sin[c + d*x]^2]*Sin[c + d* x]^(2 + n))/(d*Sqrt[Cos[c + d*x]^2]))/(2 + n))/a^4
3.5.93.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[2*n]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c* (m - 2) + b^2*d*(n + 1) + a^2*d*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], 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] && GtQ[m, 1] && !LtQ[ n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[ c, 0]))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[a^(2*m) Int[(d* Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 0]
\[\int \frac {\left (\cos ^{4}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right )}{\left (a +a \sin \left (d x +c \right )\right )^{2}}d x\]
\[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
integral(-sin(d*x + c)^n*cos(d*x + c)^4/(a^2*cos(d*x + c)^2 - 2*a^2*sin(d* x + c) - 2*a^2), x)
Timed out. \[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^n}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2} \,d x \]