Integrand size = 29, antiderivative size = 73 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^4(c+d x)}{4 a d}-\frac {\sin ^5(c+d x)}{5 a d}-\frac {\sin ^6(c+d x)}{6 a d}+\frac {\sin ^7(c+d x)}{7 a d} \] Output:
1/4*sin(d*x+c)^4/a/d-1/5*sin(d*x+c)^5/a/d-1/6*sin(d*x+c)^6/a/d+1/7*sin(d*x +c)^7/a/d
Time = 0.23 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.66 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^4(c+d x) \left (105-84 \sin (c+d x)-70 \sin ^2(c+d x)+60 \sin ^3(c+d x)\right )}{420 a d} \] Input:
Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
Output:
(Sin[c + d*x]^4*(105 - 84*Sin[c + d*x] - 70*Sin[c + d*x]^2 + 60*Sin[c + d* x]^3))/(420*a*d)
Time = 0.32 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.93, 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, 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.
\(\displaystyle \int \frac {\sin ^3(c+d x) \cos ^5(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^3 \cos (c+d x)^5}{a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \sin ^3(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int a^3 \sin ^3(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)d(a \sin (c+d x))}{a^8 d}\) |
\(\Big \downarrow \) 84 |
\(\displaystyle \frac {\int \left (\sin ^6(c+d x) a^6-\sin ^5(c+d x) a^6-\sin ^4(c+d x) a^6+\sin ^3(c+d x) a^6\right )d(a \sin (c+d x))}{a^8 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{7} a^7 \sin ^7(c+d x)-\frac {1}{6} a^7 \sin ^6(c+d x)-\frac {1}{5} a^7 \sin ^5(c+d x)+\frac {1}{4} a^7 \sin ^4(c+d x)}{a^8 d}\) |
Input:
Int[(Cos[c + d*x]^5*Sin[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
Output:
((a^7*Sin[c + d*x]^4)/4 - (a^7*Sin[c + d*x]^5)/5 - (a^7*Sin[c + d*x]^6)/6 + (a^7*Sin[c + d*x]^7)/7)/(a^8*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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 = 0.70 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67
method | result | size |
derivativedivides | \(\frac {\frac {\sin \left (d x +c \right )^{7}}{7}-\frac {\sin \left (d x +c \right )^{6}}{6}-\frac {\sin \left (d x +c \right )^{5}}{5}+\frac {\sin \left (d x +c \right )^{4}}{4}}{d a}\) | \(49\) |
default | \(\frac {\frac {\sin \left (d x +c \right )^{7}}{7}-\frac {\sin \left (d x +c \right )^{6}}{6}-\frac {\sin \left (d x +c \right )^{5}}{5}+\frac {\sin \left (d x +c \right )^{4}}{4}}{d a}\) | \(49\) |
parallelrisch | \(\frac {280+21 \sin \left (5 d x +5 c \right )+35 \cos \left (6 d x +6 c \right )-15 \sin \left (7 d x +7 c \right )-315 \cos \left (2 d x +2 c \right )-315 \sin \left (d x +c \right )+105 \sin \left (3 d x +3 c \right )}{6720 d a}\) | \(74\) |
risch | \(-\frac {3 \sin \left (d x +c \right )}{64 a d}-\frac {\sin \left (7 d x +7 c \right )}{448 d a}+\frac {\cos \left (6 d x +6 c \right )}{192 a d}+\frac {\sin \left (5 d x +5 c \right )}{320 d a}+\frac {\sin \left (3 d x +3 c \right )}{64 d a}-\frac {3 \cos \left (2 d x +2 c \right )}{64 a d}\) | \(101\) |
norman | \(\frac {\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d a}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d a}-\frac {12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5 d a}-\frac {12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{5 d a}-\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{15 d a}-\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{15 d a}+\frac {464 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{105 d a}+\frac {464 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{105 d a}+\frac {184 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{105 d a}+\frac {184 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{105 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{8} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\) | \(221\) |
Input:
int(cos(d*x+c)^5*sin(d*x+c)^3/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(1/7*sin(d*x+c)^7-1/6*sin(d*x+c)^6-1/5*sin(d*x+c)^5+1/4*sin(d*x+c)^4 )
Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {70 \, \cos \left (d x + c\right )^{6} - 105 \, \cos \left (d x + c\right )^{4} - 12 \, {\left (5 \, \cos \left (d x + c\right )^{6} - 8 \, \cos \left (d x + c\right )^{4} + \cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right )}{420 \, a d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
1/420*(70*cos(d*x + c)^6 - 105*cos(d*x + c)^4 - 12*(5*cos(d*x + c)^6 - 8*c os(d*x + c)^4 + cos(d*x + c)^2 + 2)*sin(d*x + c))/(a*d)
Leaf count of result is larger than twice the leaf count of optimal. 981 vs. \(2 (53) = 106\).
Time = 35.74 (sec) , antiderivative size = 981, normalized size of antiderivative = 13.44 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)**3/(a+a*sin(d*x+c)),x)
Output:
Piecewise((420*tan(c/2 + d*x/2)**10/(105*a*d*tan(c/2 + d*x/2)**14 + 735*a* d*tan(c/2 + d*x/2)**12 + 2205*a*d*tan(c/2 + d*x/2)**10 + 3675*a*d*tan(c/2 + d*x/2)**8 + 3675*a*d*tan(c/2 + d*x/2)**6 + 2205*a*d*tan(c/2 + d*x/2)**4 + 735*a*d*tan(c/2 + d*x/2)**2 + 105*a*d) - 672*tan(c/2 + d*x/2)**9/(105*a* d*tan(c/2 + d*x/2)**14 + 735*a*d*tan(c/2 + d*x/2)**12 + 2205*a*d*tan(c/2 + d*x/2)**10 + 3675*a*d*tan(c/2 + d*x/2)**8 + 3675*a*d*tan(c/2 + d*x/2)**6 + 2205*a*d*tan(c/2 + d*x/2)**4 + 735*a*d*tan(c/2 + d*x/2)**2 + 105*a*d) + 140*tan(c/2 + d*x/2)**8/(105*a*d*tan(c/2 + d*x/2)**14 + 735*a*d*tan(c/2 + d*x/2)**12 + 2205*a*d*tan(c/2 + d*x/2)**10 + 3675*a*d*tan(c/2 + d*x/2)**8 + 3675*a*d*tan(c/2 + d*x/2)**6 + 2205*a*d*tan(c/2 + d*x/2)**4 + 735*a*d*ta n(c/2 + d*x/2)**2 + 105*a*d) + 576*tan(c/2 + d*x/2)**7/(105*a*d*tan(c/2 + d*x/2)**14 + 735*a*d*tan(c/2 + d*x/2)**12 + 2205*a*d*tan(c/2 + d*x/2)**10 + 3675*a*d*tan(c/2 + d*x/2)**8 + 3675*a*d*tan(c/2 + d*x/2)**6 + 2205*a*d*t an(c/2 + d*x/2)**4 + 735*a*d*tan(c/2 + d*x/2)**2 + 105*a*d) + 140*tan(c/2 + d*x/2)**6/(105*a*d*tan(c/2 + d*x/2)**14 + 735*a*d*tan(c/2 + d*x/2)**12 + 2205*a*d*tan(c/2 + d*x/2)**10 + 3675*a*d*tan(c/2 + d*x/2)**8 + 3675*a*d*t an(c/2 + d*x/2)**6 + 2205*a*d*tan(c/2 + d*x/2)**4 + 735*a*d*tan(c/2 + d*x/ 2)**2 + 105*a*d) - 672*tan(c/2 + d*x/2)**5/(105*a*d*tan(c/2 + d*x/2)**14 + 735*a*d*tan(c/2 + d*x/2)**12 + 2205*a*d*tan(c/2 + d*x/2)**10 + 3675*a*d*t an(c/2 + d*x/2)**8 + 3675*a*d*tan(c/2 + d*x/2)**6 + 2205*a*d*tan(c/2 + ...
Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {60 \, \sin \left (d x + c\right )^{7} - 70 \, \sin \left (d x + c\right )^{6} - 84 \, \sin \left (d x + c\right )^{5} + 105 \, \sin \left (d x + c\right )^{4}}{420 \, a d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
1/420*(60*sin(d*x + c)^7 - 70*sin(d*x + c)^6 - 84*sin(d*x + c)^5 + 105*sin (d*x + c)^4)/(a*d)
Time = 0.16 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {60 \, \sin \left (d x + c\right )^{7} - 70 \, \sin \left (d x + c\right )^{6} - 84 \, \sin \left (d x + c\right )^{5} + 105 \, \sin \left (d x + c\right )^{4}}{420 \, a d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/420*(60*sin(d*x + c)^7 - 70*sin(d*x + c)^6 - 84*sin(d*x + c)^5 + 105*sin (d*x + c)^4)/(a*d)
Time = 0.07 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^4}{4\,a}-\frac {{\sin \left (c+d\,x\right )}^5}{5\,a}-\frac {{\sin \left (c+d\,x\right )}^6}{6\,a}+\frac {{\sin \left (c+d\,x\right )}^7}{7\,a}}{d} \] Input:
int((cos(c + d*x)^5*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/(6*a) + sin( c + d*x)^7/(7*a))/d
Time = 0.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.63 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin \left (d x +c \right )^{4} \left (60 \sin \left (d x +c \right )^{3}-70 \sin \left (d x +c \right )^{2}-84 \sin \left (d x +c \right )+105\right )}{420 a d} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)^3/(a+a*sin(d*x+c)),x)
Output:
(sin(c + d*x)**4*(60*sin(c + d*x)**3 - 70*sin(c + d*x)**2 - 84*sin(c + d*x ) + 105))/(420*a*d)