Integrand size = 29, antiderivative size = 91 \[ \int \frac {\cos ^7(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\cos ^6(c+d x)}{6 a d}-\frac {\cos ^8(c+d x)}{8 a d}+\frac {\sin ^3(c+d x)}{3 a d}-\frac {2 \sin ^5(c+d x)}{5 a d}+\frac {\sin ^7(c+d x)}{7 a d} \] Output:
1/6*cos(d*x+c)^6/a/d-1/8*cos(d*x+c)^8/a/d+1/3*sin(d*x+c)^3/a/d-2/5*sin(d*x +c)^5/a/d+1/7*sin(d*x+c)^7/a/d
Time = 0.16 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.75 \[ \int \frac {\cos ^7(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^3(c+d x) \left (280-210 \sin (c+d x)-336 \sin ^2(c+d x)+280 \sin ^3(c+d x)+120 \sin ^4(c+d x)-105 \sin ^5(c+d x)\right )}{840 a d} \] Input:
Integrate[(Cos[c + d*x]^7*Sin[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
Output:
(Sin[c + d*x]^3*(280 - 210*Sin[c + d*x] - 336*Sin[c + d*x]^2 + 280*Sin[c + d*x]^3 + 120*Sin[c + d*x]^4 - 105*Sin[c + d*x]^5))/(840*a*d)
Time = 0.48 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.85, 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.
\(\displaystyle \int \frac {\sin ^2(c+d x) \cos ^7(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^2 \cos (c+d x)^7}{a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3314 |
\(\displaystyle \frac {\int \cos ^5(c+d x) \sin ^2(c+d x)dx}{a}-\frac {\int \cos ^5(c+d x) \sin ^3(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \cos (c+d x)^5 \sin (c+d x)^2dx}{a}-\frac {\int \cos (c+d x)^5 \sin (c+d x)^3dx}{a}\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {\int \sin ^2(c+d x) \left (1-\sin ^2(c+d x)\right )^2d\sin (c+d x)}{a d}-\frac {\int \cos (c+d x)^5 \sin (c+d x)^3dx}{a}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \left (\sin ^6(c+d x)-2 \sin ^4(c+d x)+\sin ^2(c+d x)\right )d\sin (c+d x)}{a d}-\frac {\int \cos (c+d x)^5 \sin (c+d x)^3dx}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{7} \sin ^7(c+d x)-\frac {2}{5} \sin ^5(c+d x)+\frac {1}{3} \sin ^3(c+d x)}{a d}-\frac {\int \cos (c+d x)^5 \sin (c+d x)^3dx}{a}\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle \frac {\int \cos ^5(c+d x) \left (1-\cos ^2(c+d x)\right )d\cos (c+d x)}{a d}+\frac {\frac {1}{7} \sin ^7(c+d x)-\frac {2}{5} \sin ^5(c+d x)+\frac {1}{3} \sin ^3(c+d x)}{a d}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \left (\cos ^5(c+d x)-\cos ^7(c+d x)\right )d\cos (c+d x)}{a d}+\frac {\frac {1}{7} \sin ^7(c+d x)-\frac {2}{5} \sin ^5(c+d x)+\frac {1}{3} \sin ^3(c+d x)}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{7} \sin ^7(c+d x)-\frac {2}{5} \sin ^5(c+d x)+\frac {1}{3} \sin ^3(c+d x)}{a d}+\frac {\frac {1}{6} \cos ^6(c+d x)-\frac {1}{8} \cos ^8(c+d x)}{a d}\) |
Input:
Int[(Cos[c + d*x]^7*Sin[c + d*x]^2)/(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]^3/3 - (2*Sin[c + d*x]^5)/5 + Sin[c + d*x]^7/7)/(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 = 1.09 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.77
method | result | size |
derivativedivides | \(-\frac {\frac {\sin \left (d x +c \right )^{8}}{8}-\frac {\sin \left (d x +c \right )^{7}}{7}-\frac {\sin \left (d x +c \right )^{6}}{3}+\frac {2 \sin \left (d x +c \right )^{5}}{5}+\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{3}}{3}}{d a}\) | \(70\) |
default | \(-\frac {\frac {\sin \left (d x +c \right )^{8}}{8}-\frac {\sin \left (d x +c \right )^{7}}{7}-\frac {\sin \left (d x +c \right )^{6}}{3}+\frac {2 \sin \left (d x +c \right )^{5}}{5}+\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{3}}{3}}{d a}\) | \(70\) |
parallelrisch | \(\frac {\left (-\sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (-105 \sin \left (5 d x +5 c \right )+1728 \cos \left (2 d x +2 c \right )-1050 \sin \left (d x +c \right )-595 \sin \left (3 d x +3 c \right )+240 \cos \left (4 d x +4 c \right )+2512\right ) \left (\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{26880 a d}\) | \(107\) |
risch | \(\frac {5 \sin \left (d x +c \right )}{64 a d}-\frac {\cos \left (8 d x +8 c \right )}{1024 a d}-\frac {\sin \left (7 d x +7 c \right )}{448 d a}-\frac {\cos \left (6 d x +6 c \right )}{384 a d}-\frac {3 \sin \left (5 d x +5 c \right )}{320 d a}+\frac {\cos \left (4 d x +4 c \right )}{256 a d}-\frac {\sin \left (3 d x +3 c \right )}{192 d a}+\frac {3 \cos \left (2 d x +2 c \right )}{128 a d}\) | \(135\) |
Input:
int(cos(d*x+c)^7*sin(d*x+c)^2/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
-1/d/a*(1/8*sin(d*x+c)^8-1/7*sin(d*x+c)^7-1/3*sin(d*x+c)^6+2/5*sin(d*x+c)^ 5+1/4*sin(d*x+c)^4-1/3*sin(d*x+c)^3)
Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^7(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {105 \, \cos \left (d x + c\right )^{8} - 140 \, \cos \left (d x + c\right )^{6} + 8 \, {\left (15 \, \cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right )}{840 \, a d} \] Input:
integrate(cos(d*x+c)^7*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
-1/840*(105*cos(d*x + c)^8 - 140*cos(d*x + c)^6 + 8*(15*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. 1719 vs. \(2 (68) = 136\).
Time = 53.49 (sec) , antiderivative size = 1719, normalized size of antiderivative = 18.89 \[ \int \frac {\cos ^7(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**7*sin(d*x+c)**2/(a+a*sin(d*x+c)),x)
Output:
Piecewise((280*tan(c/2 + d*x/2)**13/(105*a*d*tan(c/2 + d*x/2)**16 + 840*a* d*tan(c/2 + d*x/2)**14 + 2940*a*d*tan(c/2 + d*x/2)**12 + 5880*a*d*tan(c/2 + d*x/2)**10 + 7350*a*d*tan(c/2 + d*x/2)**8 + 5880*a*d*tan(c/2 + d*x/2)**6 + 2940*a*d*tan(c/2 + d*x/2)**4 + 840*a*d*tan(c/2 + d*x/2)**2 + 105*a*d) - 420*tan(c/2 + d*x/2)**12/(105*a*d*tan(c/2 + d*x/2)**16 + 840*a*d*tan(c/2 + d*x/2)**14 + 2940*a*d*tan(c/2 + d*x/2)**12 + 5880*a*d*tan(c/2 + d*x/2)** 10 + 7350*a*d*tan(c/2 + d*x/2)**8 + 5880*a*d*tan(c/2 + d*x/2)**6 + 2940*a* d*tan(c/2 + d*x/2)**4 + 840*a*d*tan(c/2 + d*x/2)**2 + 105*a*d) + 56*tan(c/ 2 + d*x/2)**11/(105*a*d*tan(c/2 + d*x/2)**16 + 840*a*d*tan(c/2 + d*x/2)**1 4 + 2940*a*d*tan(c/2 + d*x/2)**12 + 5880*a*d*tan(c/2 + d*x/2)**10 + 7350*a *d*tan(c/2 + d*x/2)**8 + 5880*a*d*tan(c/2 + d*x/2)**6 + 2940*a*d*tan(c/2 + d*x/2)**4 + 840*a*d*tan(c/2 + d*x/2)**2 + 105*a*d) + 560*tan(c/2 + d*x/2) **10/(105*a*d*tan(c/2 + d*x/2)**16 + 840*a*d*tan(c/2 + d*x/2)**14 + 2940*a *d*tan(c/2 + d*x/2)**12 + 5880*a*d*tan(c/2 + d*x/2)**10 + 7350*a*d*tan(c/2 + d*x/2)**8 + 5880*a*d*tan(c/2 + d*x/2)**6 + 2940*a*d*tan(c/2 + d*x/2)**4 + 840*a*d*tan(c/2 + d*x/2)**2 + 105*a*d) + 688*tan(c/2 + d*x/2)**9/(105*a *d*tan(c/2 + d*x/2)**16 + 840*a*d*tan(c/2 + d*x/2)**14 + 2940*a*d*tan(c/2 + d*x/2)**12 + 5880*a*d*tan(c/2 + d*x/2)**10 + 7350*a*d*tan(c/2 + d*x/2)** 8 + 5880*a*d*tan(c/2 + d*x/2)**6 + 2940*a*d*tan(c/2 + d*x/2)**4 + 840*a*d* tan(c/2 + d*x/2)**2 + 105*a*d) - 1400*tan(c/2 + d*x/2)**8/(105*a*d*tan(...
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^7(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {105 \, \sin \left (d x + c\right )^{8} - 120 \, \sin \left (d x + c\right )^{7} - 280 \, \sin \left (d x + c\right )^{6} + 336 \, \sin \left (d x + c\right )^{5} + 210 \, \sin \left (d x + c\right )^{4} - 280 \, \sin \left (d x + c\right )^{3}}{840 \, a d} \] Input:
integrate(cos(d*x+c)^7*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
-1/840*(105*sin(d*x + c)^8 - 120*sin(d*x + c)^7 - 280*sin(d*x + c)^6 + 336 *sin(d*x + c)^5 + 210*sin(d*x + c)^4 - 280*sin(d*x + c)^3)/(a*d)
Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^7(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {105 \, \sin \left (d x + c\right )^{8} - 120 \, \sin \left (d x + c\right )^{7} - 280 \, \sin \left (d x + c\right )^{6} + 336 \, \sin \left (d x + c\right )^{5} + 210 \, \sin \left (d x + c\right )^{4} - 280 \, \sin \left (d x + c\right )^{3}}{840 \, a d} \] Input:
integrate(cos(d*x+c)^7*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/840*(105*sin(d*x + c)^8 - 120*sin(d*x + c)^7 - 280*sin(d*x + c)^6 + 336 *sin(d*x + c)^5 + 210*sin(d*x + c)^4 - 280*sin(d*x + c)^3)/(a*d)
Time = 31.52 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^7(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^3}{3\,a}-\frac {{\sin \left (c+d\,x\right )}^4}{4\,a}-\frac {2\,{\sin \left (c+d\,x\right )}^5}{5\,a}+\frac {{\sin \left (c+d\,x\right )}^6}{3\,a}+\frac {{\sin \left (c+d\,x\right )}^7}{7\,a}-\frac {{\sin \left (c+d\,x\right )}^8}{8\,a}}{d} \] Input:
int((cos(c + d*x)^7*sin(c + d*x)^2)/(a + a*sin(c + d*x)),x)
Output:
(sin(c + d*x)^3/(3*a) - sin(c + d*x)^4/(4*a) - (2*sin(c + d*x)^5)/(5*a) + sin(c + d*x)^6/(3*a) + sin(c + d*x)^7/(7*a) - sin(c + d*x)^8/(8*a))/d
Time = 0.16 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^7(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin \left (d x +c \right )^{3} \left (-105 \sin \left (d x +c \right )^{5}+120 \sin \left (d x +c \right )^{4}+280 \sin \left (d x +c \right )^{3}-336 \sin \left (d x +c \right )^{2}-210 \sin \left (d x +c \right )+280\right )}{840 a d} \] Input:
int(cos(d*x+c)^7*sin(d*x+c)^2/(a+a*sin(d*x+c)),x)
Output:
(sin(c + d*x)**3*( - 105*sin(c + d*x)**5 + 120*sin(c + d*x)**4 + 280*sin(c + d*x)**3 - 336*sin(c + d*x)**2 - 210*sin(c + d*x) + 280))/(840*a*d)