Integrand size = 29, antiderivative size = 102 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {4 \log (1+\sin (c+d x))}{a^3 d}+\frac {4 \sin (c+d x)}{a^3 d}-\frac {2 \sin ^2(c+d x)}{a^3 d}+\frac {4 \sin ^3(c+d x)}{3 a^3 d}-\frac {3 \sin ^4(c+d x)}{4 a^3 d}+\frac {\sin ^5(c+d x)}{5 a^3 d} \] Output:
-4*ln(1+sin(d*x+c))/a^3/d+4*sin(d*x+c)/a^3/d-2*sin(d*x+c)^2/a^3/d+4/3*sin( d*x+c)^3/a^3/d-3/4*sin(d*x+c)^4/a^3/d+1/5*sin(d*x+c)^5/a^3/d
Time = 1.01 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.70 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {45-3840 \log (1+\sin (c+d x))+3840 \sin (c+d x)-1920 \sin ^2(c+d x)+1280 \sin ^3(c+d x)-720 \sin ^4(c+d x)+192 \sin ^5(c+d x)}{960 a^3 d} \] Input:
Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^3,x]
Output:
(45 - 3840*Log[1 + Sin[c + d*x]] + 3840*Sin[c + d*x] - 1920*Sin[c + d*x]^2 + 1280*Sin[c + d*x]^3 - 720*Sin[c + d*x]^4 + 192*Sin[c + d*x]^5)/(960*a^3 *d)
Time = 0.34 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.91, 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.
\(\displaystyle \int \frac {\sin ^3(c+d x) \cos ^5(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^3 \cos (c+d x)^5}{(a \sin (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \frac {\sin ^3(c+d x) (a-a \sin (c+d x))^2}{\sin (c+d x) a+a}d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a^3 \sin ^3(c+d x) (a-a \sin (c+d x))^2}{\sin (c+d x) a+a}d(a \sin (c+d x))}{a^8 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (-\frac {4 a^5}{\sin (c+d x) a+a}+\sin ^4(c+d x) a^4-3 \sin ^3(c+d x) a^4+4 \sin ^2(c+d x) a^4-4 \sin (c+d x) a^4+4 a^4\right )d(a \sin (c+d x))}{a^8 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{5} a^5 \sin ^5(c+d x)-\frac {3}{4} a^5 \sin ^4(c+d x)+\frac {4}{3} a^5 \sin ^3(c+d x)-2 a^5 \sin ^2(c+d x)+4 a^5 \sin (c+d x)-4 a^5 \log (a \sin (c+d x)+a)}{a^8 d}\) |
Input:
Int[(Cos[c + d*x]^5*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^3,x]
Output:
(-4*a^5*Log[a + a*Sin[c + d*x]] + 4*a^5*Sin[c + d*x] - 2*a^5*Sin[c + d*x]^ 2 + (4*a^5*Sin[c + d*x]^3)/3 - (3*a^5*Sin[c + d*x]^4)/4 + (a^5*Sin[c + d*x ]^5)/5)/(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[((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 = 0.83 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.67
method | result | size |
derivativedivides | \(\frac {\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {3 \sin \left (d x +c \right )^{4}}{4}+\frac {4 \sin \left (d x +c \right )^{3}}{3}-2 \sin \left (d x +c \right )^{2}+4 \sin \left (d x +c \right )-4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\) | \(68\) |
default | \(\frac {\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {3 \sin \left (d x +c \right )^{4}}{4}+\frac {4 \sin \left (d x +c \right )^{3}}{3}-2 \sin \left (d x +c \right )^{2}+4 \sin \left (d x +c \right )-4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\) | \(68\) |
parallelrisch | \(\frac {1920 \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-3840 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-615-190 \sin \left (3 d x +3 c \right )+2460 \sin \left (d x +c \right )+6 \sin \left (5 d x +5 c \right )+660 \cos \left (2 d x +2 c \right )-45 \cos \left (4 d x +4 c \right )}{480 d \,a^{3}}\) | \(91\) |
risch | \(\frac {4 i x}{a^{3}}-\frac {41 i {\mathrm e}^{i \left (d x +c \right )}}{16 d \,a^{3}}+\frac {41 i {\mathrm e}^{-i \left (d x +c \right )}}{16 d \,a^{3}}+\frac {8 i c}{d \,a^{3}}-\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}+\frac {\sin \left (5 d x +5 c \right )}{80 d \,a^{3}}-\frac {3 \cos \left (4 d x +4 c \right )}{32 d \,a^{3}}-\frac {19 \sin \left (3 d x +3 c \right )}{48 d \,a^{3}}+\frac {11 \cos \left (2 d x +2 c \right )}{8 d \,a^{3}}\) | \(144\) |
Input:
int(cos(d*x+c)^5*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d/a^3*(1/5*sin(d*x+c)^5-3/4*sin(d*x+c)^4+4/3*sin(d*x+c)^3-2*sin(d*x+c)^2 +4*sin(d*x+c)-4*ln(1+sin(d*x+c)))
Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {45 \, \cos \left (d x + c\right )^{4} - 210 \, \cos \left (d x + c\right )^{2} - 4 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 26 \, \cos \left (d x + c\right )^{2} + 83\right )} \sin \left (d x + c\right ) + 240 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{60 \, a^{3} d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="frica s")
Output:
-1/60*(45*cos(d*x + c)^4 - 210*cos(d*x + c)^2 - 4*(3*cos(d*x + c)^4 - 26*c os(d*x + c)^2 + 83)*sin(d*x + c) + 240*log(sin(d*x + c) + 1))/(a^3*d)
Leaf count of result is larger than twice the leaf count of optimal. 2558 vs. \(2 (94) = 188\).
Time = 91.22 (sec) , antiderivative size = 2558, normalized size of antiderivative = 25.08 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)**3/(a+a*sin(d*x+c))**3,x)
Output:
Piecewise((-120*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**10/(15*a**3*d* tan(c/2 + d*x/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)**4 + 75*a**3*d*tan(c/2 + d*x/2)* *2 + 15*a**3*d) - 600*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**8/(15*a* *3*d*tan(c/2 + d*x/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan (c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)**4 + 75*a**3*d*tan(c/2 + d* x/2)**2 + 15*a**3*d) - 1200*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**6/ (15*a**3*d*tan(c/2 + d*x/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3 *d*tan(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)**4 + 75*a**3*d*tan(c/ 2 + d*x/2)**2 + 15*a**3*d) - 1200*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/ 2)**4/(15*a**3*d*tan(c/2 + d*x/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 15 0*a**3*d*tan(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)**4 + 75*a**3*d* tan(c/2 + d*x/2)**2 + 15*a**3*d) - 600*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**2/(15*a**3*d*tan(c/2 + d*x/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*tan(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)**4 + 75*a* *3*d*tan(c/2 + d*x/2)**2 + 15*a**3*d) - 120*log(tan(c/2 + d*x/2) + 1)/(15* a**3*d*tan(c/2 + d*x/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 150*a**3*d*t an(c/2 + d*x/2)**6 + 150*a**3*d*tan(c/2 + d*x/2)**4 + 75*a**3*d*tan(c/2 + d*x/2)**2 + 15*a**3*d) + 60*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)* *10/(15*a**3*d*tan(c/2 + d*x/2)**10 + 75*a**3*d*tan(c/2 + d*x/2)**8 + 1...
Time = 0.05 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.72 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {12 \, \sin \left (d x + c\right )^{5} - 45 \, \sin \left (d x + c\right )^{4} + 80 \, \sin \left (d x + c\right )^{3} - 120 \, \sin \left (d x + c\right )^{2} + 240 \, \sin \left (d x + c\right )}{a^{3}} - \frac {240 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}}}{60 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="maxim a")
Output:
1/60*((12*sin(d*x + c)^5 - 45*sin(d*x + c)^4 + 80*sin(d*x + c)^3 - 120*sin (d*x + c)^2 + 240*sin(d*x + c))/a^3 - 240*log(sin(d*x + c) + 1)/a^3)/d
Time = 0.14 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.04 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {4 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} d} + \frac {12 \, a^{12} d^{4} \sin \left (d x + c\right )^{5} - 45 \, a^{12} d^{4} \sin \left (d x + c\right )^{4} + 80 \, a^{12} d^{4} \sin \left (d x + c\right )^{3} - 120 \, a^{12} d^{4} \sin \left (d x + c\right )^{2} + 240 \, a^{12} d^{4} \sin \left (d x + c\right )}{60 \, a^{15} d^{5}} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="giac" )
Output:
-4*log(abs(sin(d*x + c) + 1))/(a^3*d) + 1/60*(12*a^12*d^4*sin(d*x + c)^5 - 45*a^12*d^4*sin(d*x + c)^4 + 80*a^12*d^4*sin(d*x + c)^3 - 120*a^12*d^4*si n(d*x + c)^2 + 240*a^12*d^4*sin(d*x + c))/(a^15*d^5)
Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {4\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^3}-\frac {4\,\sin \left (c+d\,x\right )}{a^3}+\frac {2\,{\sin \left (c+d\,x\right )}^2}{a^3}-\frac {4\,{\sin \left (c+d\,x\right )}^3}{3\,a^3}+\frac {3\,{\sin \left (c+d\,x\right )}^4}{4\,a^3}-\frac {{\sin \left (c+d\,x\right )}^5}{5\,a^3}}{d} \] Input:
int((cos(c + d*x)^5*sin(c + d*x)^3)/(a + a*sin(c + d*x))^3,x)
Output:
-((4*log(sin(c + d*x) + 1))/a^3 - (4*sin(c + d*x))/a^3 + (2*sin(c + d*x)^2 )/a^3 - (4*sin(c + d*x)^3)/(3*a^3) + (3*sin(c + d*x)^4)/(4*a^3) - sin(c + d*x)^5/(5*a^3))/d
Time = 0.15 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {240 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )-480 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+12 \sin \left (d x +c \right )^{5}-45 \sin \left (d x +c \right )^{4}+80 \sin \left (d x +c \right )^{3}-120 \sin \left (d x +c \right )^{2}+240 \sin \left (d x +c \right )+96}{60 a^{3} d} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x)
Output:
(240*log(tan((c + d*x)/2)**2 + 1) - 480*log(tan((c + d*x)/2) + 1) + 12*sin (c + d*x)**5 - 45*sin(c + d*x)**4 + 80*sin(c + d*x)**3 - 120*sin(c + d*x)* *2 + 240*sin(c + d*x) + 96)/(60*a**3*d)