Integrand size = 29, antiderivative size = 88 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\sin ^{1+n}(c+d x)}{a^4 d (1+n)}-\frac {4 \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{a^4 d}+\frac {4 \sin ^{1+n}(c+d x)}{d \left (a^4+a^4 \sin (c+d x)\right )} \]
sin(d*x+c)^(1+n)/a^4/d/(1+n)-4*hypergeom([1, 1+n],[2+n],-sin(d*x+c))*sin(d *x+c)^(1+n)/a^4/d+4*sin(d*x+c)^(1+n)/d/(a^4+a^4*sin(d*x+c))
Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\sin ^{1+n}(c+d x) (5+4 n+\sin (c+d x)-4 (1+n) \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x)) (1+\sin (c+d x)))}{a^4 d (1+n) (1+\sin (c+d x))} \]
(Sin[c + d*x]^(1 + n)*(5 + 4*n + Sin[c + d*x] - 4*(1 + n)*Hypergeometric2F 1[1, 1 + n, 2 + n, -Sin[c + d*x]]*(1 + Sin[c + d*x])))/(a^4*d*(1 + n)*(1 + Sin[c + d*x]))
Time = 0.32 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3315, 100, 90, 74}
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.
\(\displaystyle \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a \sin (c+d x)+a)^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5 \sin (c+d x)^n}{(a \sin (c+d x)+a)^4}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \frac {\sin ^n(c+d x) (a-a \sin (c+d x))^2}{(\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {\frac {4 a^2 \sin ^{n+1}(c+d x)}{a \sin (c+d x)+a}-\int \frac {\sin ^n(c+d x) (a (4 n+3)-a \sin (c+d x))}{\sin (c+d x) a+a}d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {-4 a (n+1) \int \frac {\sin ^n(c+d x)}{\sin (c+d x) a+a}d(a \sin (c+d x))+\frac {4 a^2 \sin ^{n+1}(c+d x)}{a \sin (c+d x)+a}+\frac {a \sin ^{n+1}(c+d x)}{n+1}}{a^5 d}\) |
\(\Big \downarrow \) 74 |
\(\displaystyle \frac {\frac {4 a^2 \sin ^{n+1}(c+d x)}{a \sin (c+d x)+a}-4 a \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}(1,n+1,n+2,-\sin (c+d x))+\frac {a \sin ^{n+1}(c+d x)}{n+1}}{a^5 d}\) |
((a*Sin[c + d*x]^(1 + n))/(1 + n) - 4*a*Hypergeometric2F1[1, 1 + n, 2 + n, -Sin[c + d*x]]*Sin[c + d*x]^(1 + n) + (4*a^2*Sin[c + d*x]^(1 + n))/(a + a *Sin[c + d*x]))/(a^5*d)
3.6.71.3.1 Defintions of rubi rules used
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x )^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] && !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
\[\int \frac {\left (\cos ^{5}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right )}{\left (a +a \sin \left (d x +c \right )\right )^{4}}d x\]
\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]
integral(sin(d*x + c)^n*cos(d*x + c)^5/(a^4*cos(d*x + c)^4 - 8*a^4*cos(d*x + c)^2 + 8*a^4 - 4*(a^4*cos(d*x + c)^2 - 2*a^4)*sin(d*x + c)), x)
Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]
\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^5\,{\sin \left (c+d\,x\right )}^n}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4} \,d x \]