Integrand size = 29, antiderivative size = 91 \[ \int \frac {\cos ^7(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\cos ^6(c+d x)}{6 a d}+\frac {\cos ^8(c+d x)}{8 a d}-\frac {\sin ^5(c+d x)}{5 a d}+\frac {2 \sin ^7(c+d x)}{7 a d}-\frac {\sin ^9(c+d x)}{9 a d} \] Output:
-1/6*cos(d*x+c)^6/a/d+1/8*cos(d*x+c)^8/a/d-1/5*sin(d*x+c)^5/a/d+2/7*sin(d* x+c)^7/a/d-1/9*sin(d*x+c)^9/a/d
Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.75 \[ \int \frac {\cos ^7(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^4(c+d x) \left (630-504 \sin (c+d x)-840 \sin ^2(c+d x)+720 \sin ^3(c+d x)+315 \sin ^4(c+d x)-280 \sin ^5(c+d x)\right )}{2520 a d} \] Input:
Integrate[(Cos[c + d*x]^7*Sin[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
Output:
(Sin[c + d*x]^4*(630 - 504*Sin[c + d*x] - 840*Sin[c + d*x]^2 + 720*Sin[c + d*x]^3 + 315*Sin[c + d*x]^4 - 280*Sin[c + d*x]^5))/(2520*a*d)
Time = 0.49 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.87, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 3314, 3042, 3044, 244, 2009, 3045, 244, 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.
\(\displaystyle \int \frac {\sin ^3(c+d x) \cos ^7(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^3 \cos (c+d x)^7}{a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3314 |
\(\displaystyle \frac {\int \cos ^5(c+d x) \sin ^3(c+d x)dx}{a}-\frac {\int \cos ^5(c+d x) \sin ^4(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \cos (c+d x)^5 \sin (c+d x)^3dx}{a}-\frac {\int \cos (c+d x)^5 \sin (c+d x)^4dx}{a}\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {\int \cos (c+d x)^5 \sin (c+d x)^3dx}{a}-\frac {\int \sin ^4(c+d x) \left (1-\sin ^2(c+d x)\right )^2d\sin (c+d x)}{a d}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \cos (c+d x)^5 \sin (c+d x)^3dx}{a}-\frac {\int \left (\sin ^8(c+d x)-2 \sin ^6(c+d x)+\sin ^4(c+d x)\right )d\sin (c+d x)}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\int \cos (c+d x)^5 \sin (c+d x)^3dx}{a}-\frac {\frac {1}{9} \sin ^9(c+d x)-\frac {2}{7} \sin ^7(c+d x)+\frac {1}{5} \sin ^5(c+d x)}{a d}\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {\int \cos ^5(c+d x) \left (1-\cos ^2(c+d x)\right )d\cos (c+d x)}{a d}-\frac {\frac {1}{9} \sin ^9(c+d x)-\frac {2}{7} \sin ^7(c+d x)+\frac {1}{5} \sin ^5(c+d x)}{a d}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {\int \left (\cos ^5(c+d x)-\cos ^7(c+d x)\right )d\cos (c+d x)}{a d}-\frac {\frac {1}{9} \sin ^9(c+d x)-\frac {2}{7} \sin ^7(c+d x)+\frac {1}{5} \sin ^5(c+d x)}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {1}{9} \sin ^9(c+d x)-\frac {2}{7} \sin ^7(c+d x)+\frac {1}{5} \sin ^5(c+d x)}{a d}-\frac {\frac {1}{6} \cos ^6(c+d x)-\frac {1}{8} \cos ^8(c+d x)}{a d}\) |
Input:
Int[(Cos[c + d*x]^7*Sin[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
Output:
-((Cos[c + d*x]^6/6 - Cos[c + d*x]^8/8)/(a*d)) - (Sin[c + d*x]^5/5 - (2*Si n[c + d*x]^7)/7 + Sin[c + d*x]^9/9)/(a*d)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ Symbol] :> Simp[1/(a*f) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a *Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(I ntegerQ[(m - 1)/2] && LtQ[0, m, n])
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> Simp[-(a*f)^(-1) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/(( a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/a Int[Cos[e + f *x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[1/(b*d) Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] & & IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n, -p]))
Time = 1.16 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.77
method | result | size |
derivativedivides | \(-\frac {\frac {\sin \left (d x +c \right )^{9}}{9}-\frac {\sin \left (d x +c \right )^{8}}{8}-\frac {2 \sin \left (d x +c \right )^{7}}{7}+\frac {\sin \left (d x +c \right )^{6}}{3}+\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{4}}{4}}{d a}\) | \(70\) |
default | \(-\frac {\frac {\sin \left (d x +c \right )^{9}}{9}-\frac {\sin \left (d x +c \right )^{8}}{8}-\frac {2 \sin \left (d x +c \right )^{7}}{7}+\frac {\sin \left (d x +c \right )^{6}}{3}+\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{4}}{4}}{d a}\) | \(70\) |
parallelrisch | \(\frac {7665+1008 \sin \left (5 d x +5 c \right )+315 \cos \left (8 d x +8 c \right )+840 \cos \left (6 d x +6 c \right )-180 \sin \left (7 d x +7 c \right )-7560 \cos \left (2 d x +2 c \right )-7560 \sin \left (d x +c \right )+1680 \sin \left (3 d x +3 c \right )-1260 \cos \left (4 d x +4 c \right )-140 \sin \left (9 d x +9 c \right )}{322560 d a}\) | \(107\) |
risch | \(-\frac {3 \sin \left (d x +c \right )}{128 a d}-\frac {\sin \left (9 d x +9 c \right )}{2304 d a}+\frac {\cos \left (8 d x +8 c \right )}{1024 a d}-\frac {\sin \left (7 d x +7 c \right )}{1792 d a}+\frac {\cos \left (6 d x +6 c \right )}{384 a d}+\frac {\sin \left (5 d x +5 c \right )}{320 d a}-\frac {\cos \left (4 d x +4 c \right )}{256 a d}+\frac {\sin \left (3 d x +3 c \right )}{192 d a}-\frac {3 \cos \left (2 d x +2 c \right )}{128 a d}\) | \(152\) |
Input:
int(cos(d*x+c)^7*sin(d*x+c)^3/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
-1/d/a*(1/9*sin(d*x+c)^9-1/8*sin(d*x+c)^8-2/7*sin(d*x+c)^7+1/3*sin(d*x+c)^ 6+1/5*sin(d*x+c)^5-1/4*sin(d*x+c)^4)
Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^7(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {315 \, \cos \left (d x + c\right )^{8} - 420 \, \cos \left (d x + c\right )^{6} - 8 \, {\left (35 \, \cos \left (d x + c\right )^{8} - 50 \, \cos \left (d x + c\right )^{6} + 3 \, \cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right )}{2520 \, a d} \] Input:
integrate(cos(d*x+c)^7*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
1/2520*(315*cos(d*x + c)^8 - 420*cos(d*x + c)^6 - 8*(35*cos(d*x + c)^8 - 5 0*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. 1906 vs. \(2 (68) = 136\).
Time = 81.22 (sec) , antiderivative size = 1906, normalized size of antiderivative = 20.95 \[ \int \frac {\cos ^7(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**7*sin(d*x+c)**3/(a+a*sin(d*x+c)),x)
Output:
Piecewise((1260*tan(c/2 + d*x/2)**14/(315*a*d*tan(c/2 + d*x/2)**18 + 2835* a*d*tan(c/2 + d*x/2)**16 + 11340*a*d*tan(c/2 + d*x/2)**14 + 26460*a*d*tan( c/2 + d*x/2)**12 + 39690*a*d*tan(c/2 + d*x/2)**10 + 39690*a*d*tan(c/2 + d* x/2)**8 + 26460*a*d*tan(c/2 + d*x/2)**6 + 11340*a*d*tan(c/2 + d*x/2)**4 + 2835*a*d*tan(c/2 + d*x/2)**2 + 315*a*d) - 2016*tan(c/2 + d*x/2)**13/(315*a *d*tan(c/2 + d*x/2)**18 + 2835*a*d*tan(c/2 + d*x/2)**16 + 11340*a*d*tan(c/ 2 + d*x/2)**14 + 26460*a*d*tan(c/2 + d*x/2)**12 + 39690*a*d*tan(c/2 + d*x/ 2)**10 + 39690*a*d*tan(c/2 + d*x/2)**8 + 26460*a*d*tan(c/2 + d*x/2)**6 + 1 1340*a*d*tan(c/2 + d*x/2)**4 + 2835*a*d*tan(c/2 + d*x/2)**2 + 315*a*d) - 4 20*tan(c/2 + d*x/2)**12/(315*a*d*tan(c/2 + d*x/2)**18 + 2835*a*d*tan(c/2 + d*x/2)**16 + 11340*a*d*tan(c/2 + d*x/2)**14 + 26460*a*d*tan(c/2 + d*x/2)* *12 + 39690*a*d*tan(c/2 + d*x/2)**10 + 39690*a*d*tan(c/2 + d*x/2)**8 + 264 60*a*d*tan(c/2 + d*x/2)**6 + 11340*a*d*tan(c/2 + d*x/2)**4 + 2835*a*d*tan( c/2 + d*x/2)**2 + 315*a*d) + 3456*tan(c/2 + d*x/2)**11/(315*a*d*tan(c/2 + d*x/2)**18 + 2835*a*d*tan(c/2 + d*x/2)**16 + 11340*a*d*tan(c/2 + d*x/2)**1 4 + 26460*a*d*tan(c/2 + d*x/2)**12 + 39690*a*d*tan(c/2 + d*x/2)**10 + 3969 0*a*d*tan(c/2 + d*x/2)**8 + 26460*a*d*tan(c/2 + d*x/2)**6 + 11340*a*d*tan( c/2 + d*x/2)**4 + 2835*a*d*tan(c/2 + d*x/2)**2 + 315*a*d) + 2520*tan(c/2 + d*x/2)**10/(315*a*d*tan(c/2 + d*x/2)**18 + 2835*a*d*tan(c/2 + d*x/2)**16 + 11340*a*d*tan(c/2 + d*x/2)**14 + 26460*a*d*tan(c/2 + d*x/2)**12 + 396...
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^7(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {280 \, \sin \left (d x + c\right )^{9} - 315 \, \sin \left (d x + c\right )^{8} - 720 \, \sin \left (d x + c\right )^{7} + 840 \, \sin \left (d x + c\right )^{6} + 504 \, \sin \left (d x + c\right )^{5} - 630 \, \sin \left (d x + c\right )^{4}}{2520 \, a d} \] Input:
integrate(cos(d*x+c)^7*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
-1/2520*(280*sin(d*x + c)^9 - 315*sin(d*x + c)^8 - 720*sin(d*x + c)^7 + 84 0*sin(d*x + c)^6 + 504*sin(d*x + c)^5 - 630*sin(d*x + c)^4)/(a*d)
Time = 0.14 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^7(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {280 \, \sin \left (d x + c\right )^{9} - 315 \, \sin \left (d x + c\right )^{8} - 720 \, \sin \left (d x + c\right )^{7} + 840 \, \sin \left (d x + c\right )^{6} + 504 \, \sin \left (d x + c\right )^{5} - 630 \, \sin \left (d x + c\right )^{4}}{2520 \, a d} \] Input:
integrate(cos(d*x+c)^7*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/2520*(280*sin(d*x + c)^9 - 315*sin(d*x + c)^8 - 720*sin(d*x + c)^7 + 84 0*sin(d*x + c)^6 + 504*sin(d*x + c)^5 - 630*sin(d*x + c)^4)/(a*d)
Time = 0.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^7(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^4}{4\,a}-\frac {{\sin \left (c+d\,x\right )}^5}{5\,a}-\frac {{\sin \left (c+d\,x\right )}^6}{3\,a}+\frac {2\,{\sin \left (c+d\,x\right )}^7}{7\,a}+\frac {{\sin \left (c+d\,x\right )}^8}{8\,a}-\frac {{\sin \left (c+d\,x\right )}^9}{9\,a}}{d} \] Input:
int((cos(c + d*x)^7*sin(c + d*x)^3)/(a + a*sin(c + d*x)),x)
Output:
(sin(c + d*x)^4/(4*a) - sin(c + d*x)^5/(5*a) - sin(c + d*x)^6/(3*a) + (2*s in(c + d*x)^7)/(7*a) + sin(c + d*x)^8/(8*a) - sin(c + d*x)^9/(9*a))/d
Time = 0.19 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^7(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin \left (d x +c \right )^{4} \left (-280 \sin \left (d x +c \right )^{5}+315 \sin \left (d x +c \right )^{4}+720 \sin \left (d x +c \right )^{3}-840 \sin \left (d x +c \right )^{2}-504 \sin \left (d x +c \right )+630\right )}{2520 a d} \] Input:
int(cos(d*x+c)^7*sin(d*x+c)^3/(a+a*sin(d*x+c)),x)
Output:
(sin(c + d*x)**4*( - 280*sin(c + d*x)**5 + 315*sin(c + d*x)**4 + 720*sin(c + d*x)**3 - 840*sin(c + d*x)**2 - 504*sin(c + d*x) + 630))/(2520*a*d)