Integrand size = 29, antiderivative size = 109 \[ \int \frac {\cos ^7(c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^5(c+d x)}{5 a d}-\frac {\sin ^6(c+d x)}{6 a d}-\frac {2 \sin ^7(c+d x)}{7 a d}+\frac {\sin ^8(c+d x)}{4 a d}+\frac {\sin ^9(c+d x)}{9 a d}-\frac {\sin ^{10}(c+d x)}{10 a d} \]
1/5*sin(d*x+c)^5/a/d-1/6*sin(d*x+c)^6/a/d-2/7*sin(d*x+c)^7/a/d+1/4*sin(d*x +c)^8/a/d+1/9*sin(d*x+c)^9/a/d-1/10*sin(d*x+c)^10/a/d
Time = 0.12 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.62 \[ \int \frac {\cos ^7(c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^5(c+d x) \left (252-210 \sin (c+d x)-360 \sin ^2(c+d x)+315 \sin ^3(c+d x)+140 \sin ^4(c+d x)-126 \sin ^5(c+d x)\right )}{1260 a d} \]
(Sin[c + d*x]^5*(252 - 210*Sin[c + d*x] - 360*Sin[c + d*x]^2 + 315*Sin[c + d*x]^3 + 140*Sin[c + d*x]^4 - 126*Sin[c + d*x]^5))/(1260*a*d)
Time = 0.32 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3315, 27, 99, 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.
\(\displaystyle \int \frac {\sin ^4(c+d x) \cos ^7(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^4 \cos (c+d x)^7}{a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \sin ^4(c+d x) (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int a^4 \sin ^4(c+d x) (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2d(a \sin (c+d x))}{a^{11} d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (-\sin ^9(c+d x) a^9+\sin ^8(c+d x) a^9+2 \sin ^7(c+d x) a^9-2 \sin ^6(c+d x) a^9-\sin ^5(c+d x) a^9+\sin ^4(c+d x) a^9\right )d(a \sin (c+d x))}{a^{11} d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{10} a^{10} \sin ^{10}(c+d x)+\frac {1}{9} a^{10} \sin ^9(c+d x)+\frac {1}{4} a^{10} \sin ^8(c+d x)-\frac {2}{7} a^{10} \sin ^7(c+d x)-\frac {1}{6} a^{10} \sin ^6(c+d x)+\frac {1}{5} a^{10} \sin ^5(c+d x)}{a^{11} d}\) |
((a^10*Sin[c + d*x]^5)/5 - (a^10*Sin[c + d*x]^6)/6 - (2*a^10*Sin[c + d*x]^ 7)/7 + (a^10*Sin[c + d*x]^8)/4 + (a^10*Sin[c + d*x]^9)/9 - (a^10*Sin[c + d *x]^10)/10)/(a^11*d)
3.7.79.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -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]
Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.64
method | result | size |
derivativedivides | \(-\frac {\frac {\left (\sin ^{10}\left (d x +c \right )\right )}{10}-\frac {\left (\sin ^{9}\left (d x +c \right )\right )}{9}-\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{4}+\frac {2 \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}}{a d}\) | \(70\) |
default | \(-\frac {\frac {\left (\sin ^{10}\left (d x +c \right )\right )}{10}-\frac {\left (\sin ^{9}\left (d x +c \right )\right )}{9}-\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{4}+\frac {2 \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}}{a d}\) | \(70\) |
parallelrisch | \(\frac {\left (\sin \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )-5 \sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+10 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (880 \cos \left (2 d x +2 c \right )-63 \sin \left (5 d x +5 c \right )-420 \sin \left (d x +c \right )-315 \sin \left (3 d x +3 c \right )+140 \cos \left (4 d x +4 c \right )+996\right ) \left (\cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )+5 \cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+10 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{80640 a d}\) | \(127\) |
risch | \(\frac {3 \sin \left (d x +c \right )}{128 a d}+\frac {\cos \left (10 d x +10 c \right )}{5120 a d}+\frac {\sin \left (9 d x +9 c \right )}{2304 d a}+\frac {\sin \left (7 d x +7 c \right )}{1792 d a}-\frac {5 \cos \left (6 d x +6 c \right )}{3072 a d}-\frac {\sin \left (5 d x +5 c \right )}{320 d a}-\frac {\sin \left (3 d x +3 c \right )}{192 d a}+\frac {5 \cos \left (2 d x +2 c \right )}{512 a d}\) | \(135\) |
-1/a/d*(1/10*sin(d*x+c)^10-1/9*sin(d*x+c)^9-1/4*sin(d*x+c)^8+2/7*sin(d*x+c )^7+1/6*sin(d*x+c)^6-1/5*sin(d*x+c)^5)
Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^7(c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {126 \, \cos \left (d x + c\right )^{10} - 315 \, \cos \left (d x + c\right )^{8} + 210 \, \cos \left (d x + c\right )^{6} + 4 \, {\left (35 \, \cos \left (d x + c\right )^{8} - 50 \, \cos \left (d x + c\right )^{6} + 3 \, \cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right )}{1260 \, a d} \]
1/1260*(126*cos(d*x + c)^10 - 315*cos(d*x + c)^8 + 210*cos(d*x + c)^6 + 4* (35*cos(d*x + c)^8 - 50*cos(d*x + c)^6 + 3*cos(d*x + c)^4 + 4*cos(d*x + c) ^2 + 8)*sin(d*x + c))/(a*d)
Leaf count of result is larger than twice the leaf count of optimal. 2093 vs. \(2 (82) = 164\).
Time = 119.04 (sec) , antiderivative size = 2093, normalized size of antiderivative = 19.20 \[ \int \frac {\cos ^7(c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \]
Piecewise((2016*tan(c/2 + d*x/2)**15/(315*a*d*tan(c/2 + d*x/2)**20 + 3150* a*d*tan(c/2 + d*x/2)**18 + 14175*a*d*tan(c/2 + d*x/2)**16 + 37800*a*d*tan( c/2 + d*x/2)**14 + 66150*a*d*tan(c/2 + d*x/2)**12 + 79380*a*d*tan(c/2 + d* x/2)**10 + 66150*a*d*tan(c/2 + d*x/2)**8 + 37800*a*d*tan(c/2 + d*x/2)**6 + 14175*a*d*tan(c/2 + d*x/2)**4 + 3150*a*d*tan(c/2 + d*x/2)**2 + 315*a*d) - 3360*tan(c/2 + d*x/2)**14/(315*a*d*tan(c/2 + d*x/2)**20 + 3150*a*d*tan(c/ 2 + d*x/2)**18 + 14175*a*d*tan(c/2 + d*x/2)**16 + 37800*a*d*tan(c/2 + d*x/ 2)**14 + 66150*a*d*tan(c/2 + d*x/2)**12 + 79380*a*d*tan(c/2 + d*x/2)**10 + 66150*a*d*tan(c/2 + d*x/2)**8 + 37800*a*d*tan(c/2 + d*x/2)**6 + 14175*a*d *tan(c/2 + d*x/2)**4 + 3150*a*d*tan(c/2 + d*x/2)**2 + 315*a*d) - 1440*tan( c/2 + d*x/2)**13/(315*a*d*tan(c/2 + d*x/2)**20 + 3150*a*d*tan(c/2 + d*x/2) **18 + 14175*a*d*tan(c/2 + d*x/2)**16 + 37800*a*d*tan(c/2 + d*x/2)**14 + 6 6150*a*d*tan(c/2 + d*x/2)**12 + 79380*a*d*tan(c/2 + d*x/2)**10 + 66150*a*d *tan(c/2 + d*x/2)**8 + 37800*a*d*tan(c/2 + d*x/2)**6 + 14175*a*d*tan(c/2 + d*x/2)**4 + 3150*a*d*tan(c/2 + d*x/2)**2 + 315*a*d) + 6720*tan(c/2 + d*x/ 2)**12/(315*a*d*tan(c/2 + d*x/2)**20 + 3150*a*d*tan(c/2 + d*x/2)**18 + 141 75*a*d*tan(c/2 + d*x/2)**16 + 37800*a*d*tan(c/2 + d*x/2)**14 + 66150*a*d*t an(c/2 + d*x/2)**12 + 79380*a*d*tan(c/2 + d*x/2)**10 + 66150*a*d*tan(c/2 + d*x/2)**8 + 37800*a*d*tan(c/2 + d*x/2)**6 + 14175*a*d*tan(c/2 + d*x/2)**4 + 3150*a*d*tan(c/2 + d*x/2)**2 + 315*a*d) + 3520*tan(c/2 + d*x/2)**11/...
Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.63 \[ \int \frac {\cos ^7(c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {126 \, \sin \left (d x + c\right )^{10} - 140 \, \sin \left (d x + c\right )^{9} - 315 \, \sin \left (d x + c\right )^{8} + 360 \, \sin \left (d x + c\right )^{7} + 210 \, \sin \left (d x + c\right )^{6} - 252 \, \sin \left (d x + c\right )^{5}}{1260 \, a d} \]
-1/1260*(126*sin(d*x + c)^10 - 140*sin(d*x + c)^9 - 315*sin(d*x + c)^8 + 3 60*sin(d*x + c)^7 + 210*sin(d*x + c)^6 - 252*sin(d*x + c)^5)/(a*d)
Time = 0.32 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.63 \[ \int \frac {\cos ^7(c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {126 \, \sin \left (d x + c\right )^{10} - 140 \, \sin \left (d x + c\right )^{9} - 315 \, \sin \left (d x + c\right )^{8} + 360 \, \sin \left (d x + c\right )^{7} + 210 \, \sin \left (d x + c\right )^{6} - 252 \, \sin \left (d x + c\right )^{5}}{1260 \, a d} \]
-1/1260*(126*sin(d*x + c)^10 - 140*sin(d*x + c)^9 - 315*sin(d*x + c)^8 + 3 60*sin(d*x + c)^7 + 210*sin(d*x + c)^6 - 252*sin(d*x + c)^5)/(a*d)
Time = 10.16 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^7(c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^5}{5\,a}-\frac {{\sin \left (c+d\,x\right )}^6}{6\,a}-\frac {2\,{\sin \left (c+d\,x\right )}^7}{7\,a}+\frac {{\sin \left (c+d\,x\right )}^8}{4\,a}+\frac {{\sin \left (c+d\,x\right )}^9}{9\,a}-\frac {{\sin \left (c+d\,x\right )}^{10}}{10\,a}}{d} \]