Integrand size = 29, antiderivative size = 92 \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^{1+n}(c+d x)}{a^2 d (1+n)}-\frac {2 \sin ^{2+n}(c+d x)}{a^2 d (2+n)}+\frac {2 \sin ^{4+n}(c+d x)}{a^2 d (4+n)}-\frac {\sin ^{5+n}(c+d x)}{a^2 d (5+n)} \] Output:
sin(d*x+c)^(1+n)/a^2/d/(1+n)-2*sin(d*x+c)^(2+n)/a^2/d/(2+n)+2*sin(d*x+c)^( 4+n)/a^2/d/(4+n)-sin(d*x+c)^(5+n)/a^2/d/(5+n)
Time = 0.38 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.27 \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^{1+n}(c+d x) \left (40+38 n+11 n^2+n^3-2 \left (20+29 n+10 n^2+n^3\right ) \sin (c+d x)+2 \left (10+17 n+8 n^2+n^3\right ) \sin ^3(c+d x)-\left (8+14 n+7 n^2+n^3\right ) \sin ^4(c+d x)\right )}{a^2 d (1+n) (2+n) (4+n) (5+n)} \] Input:
Integrate[(Cos[c + d*x]^7*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^2,x]
Output:
(Sin[c + d*x]^(1 + n)*(40 + 38*n + 11*n^2 + n^3 - 2*(20 + 29*n + 10*n^2 + n^3)*Sin[c + d*x] + 2*(10 + 17*n + 8*n^2 + n^3)*Sin[c + d*x]^3 - (8 + 14*n + 7*n^2 + n^3)*Sin[c + d*x]^4))/(a^2*d*(1 + n)*(2 + n)*(4 + n)*(5 + n))
Time = 0.37 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3042, 3315, 84, 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.
\(\displaystyle \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^7 \sin (c+d x)^n}{(a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \sin ^n(c+d x) (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 84 |
\(\displaystyle \frac {\int \left (a^4 \sin ^n(c+d x)-2 a^4 \sin ^{n+1}(c+d x)+2 a^4 \sin ^{n+3}(c+d x)-a^4 \sin ^{n+4}(c+d x)\right )d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {a^5 \sin ^{n+1}(c+d x)}{n+1}-\frac {2 a^5 \sin ^{n+2}(c+d x)}{n+2}+\frac {2 a^5 \sin ^{n+4}(c+d x)}{n+4}-\frac {a^5 \sin ^{n+5}(c+d x)}{n+5}}{a^7 d}\) |
Input:
Int[(Cos[c + d*x]^7*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^2,x]
Output:
((a^5*Sin[c + d*x]^(1 + n))/(1 + n) - (2*a^5*Sin[c + d*x]^(2 + n))/(2 + n) + (2*a^5*Sin[c + d*x]^(4 + n))/(4 + n) - (a^5*Sin[c + d*x]^(5 + n))/(5 + n))/(a^7*d)
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] && !(ILtQ[n + p + 2, 0 ] && GtQ[n + 2*p, 0])
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]
Time = 2.87 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.34
method | result | size |
derivativedivides | \(\frac {\sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (1+n \right )}-\frac {2 \sin \left (d x +c \right )^{2} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (2+n \right )}+\frac {2 \sin \left (d x +c \right )^{4} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (4+n \right )}-\frac {\sin \left (d x +c \right )^{5} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (5+n \right )}\) | \(123\) |
default | \(\frac {\sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (1+n \right )}-\frac {2 \sin \left (d x +c \right )^{2} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (2+n \right )}+\frac {2 \sin \left (d x +c \right )^{4} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (4+n \right )}-\frac {\sin \left (d x +c \right )^{5} {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{a^{2} d \left (5+n \right )}\) | \(123\) |
parallelrisch | \(-\frac {\sin \left (d x +c \right )^{n} \left (\left (-4 n^{3}-32 n^{2}-68 n -40\right ) \cos \left (4 d x +4 c \right )+\left (-5 n^{3}-35 n^{2}-70 n -40\right ) \sin \left (3 d x +3 c \right )+\left (n^{3}+7 n^{2}+14 n +8\right ) \sin \left (5 d x +5 c \right )+\left (-32 n^{2}-192 n -160\right ) \cos \left (2 d x +2 c \right )+\left (-6 n^{3}-106 n^{2}-468 n -560\right ) \sin \left (d x +c \right )+4 n^{3}+64 n^{2}+260 n +200\right )}{16 a^{2} \left (4+n \right ) \left (2+n \right ) \left (1+n \right ) d \left (5+n \right )}\) | \(167\) |
Input:
int(cos(d*x+c)^7*sin(d*x+c)^n/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/a^2/d/(1+n)*sin(d*x+c)*exp(n*ln(sin(d*x+c)))-2/a^2/d/(2+n)*sin(d*x+c)^2* exp(n*ln(sin(d*x+c)))+2/a^2/d/(4+n)*sin(d*x+c)^4*exp(n*ln(sin(d*x+c)))-1/a ^2/d/(5+n)*sin(d*x+c)^5*exp(n*ln(sin(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.84 \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\left (2 \, {\left (n^{3} + 8 \, n^{2} + 17 \, n + 10\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (n^{3} + 6 \, n^{2} + 5 \, n\right )} \cos \left (d x + c\right )^{2} - 4 \, n^{2} - {\left ({\left (n^{3} + 7 \, n^{2} + 14 \, n + 8\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (n^{3} + 7 \, n^{2} + 14 \, n + 8\right )} \cos \left (d x + c\right )^{2} - 4 \, n^{2} - 24 \, n - 32\right )} \sin \left (d x + c\right ) - 24 \, n - 20\right )} \sin \left (d x + c\right )^{n}}{a^{2} d n^{4} + 12 \, a^{2} d n^{3} + 49 \, a^{2} d n^{2} + 78 \, a^{2} d n + 40 \, a^{2} d} \] Input:
integrate(cos(d*x+c)^7*sin(d*x+c)^n/(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
(2*(n^3 + 8*n^2 + 17*n + 10)*cos(d*x + c)^4 - 2*(n^3 + 6*n^2 + 5*n)*cos(d* x + c)^2 - 4*n^2 - ((n^3 + 7*n^2 + 14*n + 8)*cos(d*x + c)^4 - 2*(n^3 + 7*n ^2 + 14*n + 8)*cos(d*x + c)^2 - 4*n^2 - 24*n - 32)*sin(d*x + c) - 24*n - 2 0)*sin(d*x + c)^n/(a^2*d*n^4 + 12*a^2*d*n^3 + 49*a^2*d*n^2 + 78*a^2*d*n + 40*a^2*d)
Timed out. \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**7*sin(d*x+c)**n/(a+a*sin(d*x+c))**2,x)
Output:
Timed out
Time = 0.06 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.37 \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {{\left ({\left (n^{3} + 7 \, n^{2} + 14 \, n + 8\right )} \sin \left (d x + c\right )^{5} - 2 \, {\left (n^{3} + 8 \, n^{2} + 17 \, n + 10\right )} \sin \left (d x + c\right )^{4} + 2 \, {\left (n^{3} + 10 \, n^{2} + 29 \, n + 20\right )} \sin \left (d x + c\right )^{2} - {\left (n^{3} + 11 \, n^{2} + 38 \, n + 40\right )} \sin \left (d x + c\right )\right )} \sin \left (d x + c\right )^{n}}{{\left (n^{4} + 12 \, n^{3} + 49 \, n^{2} + 78 \, n + 40\right )} a^{2} d} \] Input:
integrate(cos(d*x+c)^7*sin(d*x+c)^n/(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
-((n^3 + 7*n^2 + 14*n + 8)*sin(d*x + c)^5 - 2*(n^3 + 8*n^2 + 17*n + 10)*si n(d*x + c)^4 + 2*(n^3 + 10*n^2 + 29*n + 20)*sin(d*x + c)^2 - (n^3 + 11*n^2 + 38*n + 40)*sin(d*x + c))*sin(d*x + c)^n/((n^4 + 12*n^3 + 49*n^2 + 78*n + 40)*a^2*d)
Time = 0.14 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.02 \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {\sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{5}}{n + 5} - \frac {2 \, \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{4}}{n + 4} + \frac {2 \, \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{2}}{n + 2} - \frac {\sin \left (d x + c\right )^{n + 1}}{n + 1}}{a^{2} d} \] Input:
integrate(cos(d*x+c)^7*sin(d*x+c)^n/(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
-(sin(d*x + c)^n*sin(d*x + c)^5/(n + 5) - 2*sin(d*x + c)^n*sin(d*x + c)^4/ (n + 4) + 2*sin(d*x + c)^n*sin(d*x + c)^2/(n + 2) - sin(d*x + c)^(n + 1)/( n + 1))/(a^2*d)
Time = 34.32 (sec) , antiderivative size = 280, normalized size of antiderivative = 3.04 \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\sin \left (c+d\,x\right )}^n\,\left (560\,\sin \left (c+d\,x\right )-260\,n+160\,\cos \left (2\,c+2\,d\,x\right )+40\,\cos \left (4\,c+4\,d\,x\right )+40\,\sin \left (3\,c+3\,d\,x\right )-8\,\sin \left (5\,c+5\,d\,x\right )+468\,n\,\sin \left (c+d\,x\right )+192\,n\,\cos \left (2\,c+2\,d\,x\right )+68\,n\,\cos \left (4\,c+4\,d\,x\right )+70\,n\,\sin \left (3\,c+3\,d\,x\right )-14\,n\,\sin \left (5\,c+5\,d\,x\right )+106\,n^2\,\sin \left (c+d\,x\right )+6\,n^3\,\sin \left (c+d\,x\right )-64\,n^2-4\,n^3+32\,n^2\,\cos \left (2\,c+2\,d\,x\right )+32\,n^2\,\cos \left (4\,c+4\,d\,x\right )+4\,n^3\,\cos \left (4\,c+4\,d\,x\right )+35\,n^2\,\sin \left (3\,c+3\,d\,x\right )+5\,n^3\,\sin \left (3\,c+3\,d\,x\right )-7\,n^2\,\sin \left (5\,c+5\,d\,x\right )-n^3\,\sin \left (5\,c+5\,d\,x\right )-200\right )}{16\,a^2\,d\,\left (n^4+12\,n^3+49\,n^2+78\,n+40\right )} \] Input:
int((cos(c + d*x)^7*sin(c + d*x)^n)/(a + a*sin(c + d*x))^2,x)
Output:
(sin(c + d*x)^n*(560*sin(c + d*x) - 260*n + 160*cos(2*c + 2*d*x) + 40*cos( 4*c + 4*d*x) + 40*sin(3*c + 3*d*x) - 8*sin(5*c + 5*d*x) + 468*n*sin(c + d* x) + 192*n*cos(2*c + 2*d*x) + 68*n*cos(4*c + 4*d*x) + 70*n*sin(3*c + 3*d*x ) - 14*n*sin(5*c + 5*d*x) + 106*n^2*sin(c + d*x) + 6*n^3*sin(c + d*x) - 64 *n^2 - 4*n^3 + 32*n^2*cos(2*c + 2*d*x) + 32*n^2*cos(4*c + 4*d*x) + 4*n^3*c os(4*c + 4*d*x) + 35*n^2*sin(3*c + 3*d*x) + 5*n^3*sin(3*c + 3*d*x) - 7*n^2 *sin(5*c + 5*d*x) - n^3*sin(5*c + 5*d*x) - 200))/(16*a^2*d*(78*n + 49*n^2 + 12*n^3 + n^4 + 40))
\[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\int \frac {\cos \left (d x +c \right )^{7} \sin \left (d x +c \right )^{n}}{\left (\sin \left (d x +c \right ) a +a \right )^{2}}d x \] Input:
int(cos(d*x+c)^7*sin(d*x+c)^n/(a+a*sin(d*x+c))^2,x)
Output:
int(cos(d*x+c)^7*sin(d*x+c)^n/(a+a*sin(d*x+c))^2,x)