Integrand size = 12, antiderivative size = 60 \[ \int \frac {\cos (x) \log (\sin (x))}{(1+\cos (x))^2} \, dx=-\frac {2 x}{3}-\frac {\sin (x)}{9 (1+\cos (x))^2}+\frac {8 \sin (x)}{9 (1+\cos (x))}-\frac {\log (\sin (x)) \sin (x)}{3 (1+\cos (x))^2}+\frac {2 \log (\sin (x)) \sin (x)}{3 (1+\cos (x))} \]
-2/3*x-1/9*sin(x)/(1+cos(x))^2+8/9*sin(x)/(1+cos(x))-1/3*ln(sin(x))*sin(x) /(1+cos(x))^2+2/3*ln(sin(x))*sin(x)/(1+cos(x))
Time = 0.11 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.93 \[ \int \frac {\cos (x) \log (\sin (x))}{(1+\cos (x))^2} \, dx=-\frac {1}{18} \sec ^3\left (\frac {x}{2}\right ) \left (9 x \cos \left (\frac {x}{2}\right )+3 x \cos \left (\frac {3 x}{2}\right )-(7+3 \log (\sin (x))+\cos (x) (8+6 \log (\sin (x)))) \sin \left (\frac {x}{2}\right )\right ) \]
-1/18*(Sec[x/2]^3*(9*x*Cos[x/2] + 3*x*Cos[(3*x)/2] - (7 + 3*Log[Sin[x]] + Cos[x]*(8 + 6*Log[Sin[x]]))*Sin[x/2]))
Time = 0.49 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {3034, 27, 3042, 3447, 3042, 3498, 27, 3042, 3214, 3042, 3127}
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.
| \(\displaystyle \int \frac {\cos (x) \log (\sin (x))}{(\cos (x)+1)^2} \, dx\) |
| \(\Big \downarrow \) 3034 |
| \(\displaystyle -\int \frac {\cos (x) (2 \cos (x)+1)}{3 (\cos (x)+1)^2}dx+\frac {2 \sin (x) \log (\sin (x))}{3 (\cos (x)+1)}-\frac {\sin (x) \log (\sin (x))}{3 (\cos (x)+1)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {1}{3} \int \frac {\cos (x) (2 \cos (x)+1)}{(\cos (x)+1)^2}dx+\frac {2 \sin (x) \log (\sin (x))}{3 (\cos (x)+1)}-\frac {\sin (x) \log (\sin (x))}{3 (\cos (x)+1)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {1}{3} \int \frac {\sin \left (x+\frac {\pi }{2}\right ) \left (2 \sin \left (x+\frac {\pi }{2}\right )+1\right )}{\left (\sin \left (x+\frac {\pi }{2}\right )+1\right )^2}dx+\frac {2 \sin (x) \log (\sin (x))}{3 (\cos (x)+1)}-\frac {\sin (x) \log (\sin (x))}{3 (\cos (x)+1)^2}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle -\frac {1}{3} \int \frac {2 \cos ^2(x)+\cos (x)}{(\cos (x)+1)^2}dx+\frac {2 \sin (x) \log (\sin (x))}{3 (\cos (x)+1)}-\frac {\sin (x) \log (\sin (x))}{3 (\cos (x)+1)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {1}{3} \int \frac {2 \sin \left (x+\frac {\pi }{2}\right )^2+\sin \left (x+\frac {\pi }{2}\right )}{\left (\sin \left (x+\frac {\pi }{2}\right )+1\right )^2}dx+\frac {2 \sin (x) \log (\sin (x))}{3 (\cos (x)+1)}-\frac {\sin (x) \log (\sin (x))}{3 (\cos (x)+1)^2}\) |
| \(\Big \downarrow \) 3498 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{3} \int \frac {2 (1-3 \cos (x))}{\cos (x)+1}dx-\frac {\sin (x)}{3 (\cos (x)+1)^2}\right )+\frac {2 \sin (x) \log (\sin (x))}{3 (\cos (x)+1)}-\frac {\sin (x) \log (\sin (x))}{3 (\cos (x)+1)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {2}{3} \int \frac {1-3 \cos (x)}{\cos (x)+1}dx-\frac {\sin (x)}{3 (\cos (x)+1)^2}\right )+\frac {2 \sin (x) \log (\sin (x))}{3 (\cos (x)+1)}-\frac {\sin (x) \log (\sin (x))}{3 (\cos (x)+1)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {2}{3} \int \frac {1-3 \sin \left (x+\frac {\pi }{2}\right )}{\sin \left (x+\frac {\pi }{2}\right )+1}dx-\frac {\sin (x)}{3 (\cos (x)+1)^2}\right )+\frac {2 \sin (x) \log (\sin (x))}{3 (\cos (x)+1)}-\frac {\sin (x) \log (\sin (x))}{3 (\cos (x)+1)^2}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{3} \left (\frac {2}{3} \left (4 \int \frac {1}{\cos (x)+1}dx-3 x\right )-\frac {\sin (x)}{3 (\cos (x)+1)^2}\right )+\frac {2 \sin (x) \log (\sin (x))}{3 (\cos (x)+1)}-\frac {\sin (x) \log (\sin (x))}{3 (\cos (x)+1)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {2}{3} \left (4 \int \frac {1}{\sin \left (x+\frac {\pi }{2}\right )+1}dx-3 x\right )-\frac {\sin (x)}{3 (\cos (x)+1)^2}\right )+\frac {2 \sin (x) \log (\sin (x))}{3 (\cos (x)+1)}-\frac {\sin (x) \log (\sin (x))}{3 (\cos (x)+1)^2}\) |
| \(\Big \downarrow \) 3127 |
| \(\displaystyle \frac {1}{3} \left (\frac {2}{3} \left (\frac {4 \sin (x)}{\cos (x)+1}-3 x\right )-\frac {\sin (x)}{3 (\cos (x)+1)^2}\right )+\frac {2 \sin (x) \log (\sin (x))}{3 (\cos (x)+1)}-\frac {\sin (x) \log (\sin (x))}{3 (\cos (x)+1)^2}\) |
-1/3*(Log[Sin[x]]*Sin[x])/(1 + Cos[x])^2 + (2*Log[Sin[x]]*Sin[x])/(3*(1 + Cos[x])) + (-1/3*Sin[x]/(1 + Cos[x])^2 + (2*(-3*x + (4*Sin[x])/(1 + Cos[x] )))/3)/3
3.7.44.3.1 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[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Simp[Log[u] w, x ] - Int[SimplifyIntegrand[w*(D[u, x]/u), x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b ^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b - a* B + b*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[1 /(a^2*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b *B - a*C) + b*C*(2*m + 1)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]
Time = 0.59 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.92
| method | result | size |
| parallelrisch | \(\frac {-6 x \cos \left (2 x \right )+\left (6 \ln \left (\sin \left (x \right )\right )+8\right ) \sin \left (2 x \right )-24 x \cos \left (x \right )+6 \ln \left (\sin \left (x \right )\right ) \sin \left (x \right )-18 x +14 \sin \left (x \right )}{9 \cos \left (2 x \right )+27+36 \cos \left (x \right )}\) | \(55\) |
| default | \(\frac {6 \ln \left (\frac {\sin \left (x \right )}{2}\right ) \cos \left (x \right ) \sin \left (x \right )+12 \arctan \left (-\csc \left (x \right )+\cot \left (x \right )\right ) \left (\cos ^{2}\left (x \right )\right )+6 \ln \left (2\right ) \cos \left (x \right ) \sin \left (x \right )+3 \ln \left (\frac {\sin \left (x \right )}{2}\right ) \sin \left (x \right )+24 \arctan \left (-\csc \left (x \right )+\cot \left (x \right )\right ) \cos \left (x \right )+3 \ln \left (2\right ) \sin \left (x \right )+8 \cos \left (x \right ) \sin \left (x \right )+12 \arctan \left (-\csc \left (x \right )+\cot \left (x \right )\right )+7 \sin \left (x \right )}{9 \left (\cos \left (x \right )+1\right )^{2}}\) | \(90\) |
| risch | \(\text {Expression too large to display}\) | \(598\) |
(-6*x*cos(2*x)+(6*ln(sin(x))+8)*sin(2*x)-24*x*cos(x)+6*ln(sin(x))*sin(x)-1 8*x+14*sin(x))/(9*cos(2*x)+27+36*cos(x))
Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.88 \[ \int \frac {\cos (x) \log (\sin (x))}{(1+\cos (x))^2} \, dx=-\frac {6 \, x \cos \left (x\right )^{2} - 3 \, {\left (2 \, \cos \left (x\right ) + 1\right )} \log \left (\sin \left (x\right )\right ) \sin \left (x\right ) + 12 \, x \cos \left (x\right ) - {\left (8 \, \cos \left (x\right ) + 7\right )} \sin \left (x\right ) + 6 \, x}{9 \, {\left (\cos \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 1\right )}} \]
-1/9*(6*x*cos(x)^2 - 3*(2*cos(x) + 1)*log(sin(x))*sin(x) + 12*x*cos(x) - ( 8*cos(x) + 7)*sin(x) + 6*x)/(cos(x)^2 + 2*cos(x) + 1)
Time = 2.71 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.47 \[ \int \frac {\cos (x) \log (\sin (x))}{(1+\cos (x))^2} \, dx=- \frac {2 x}{3} - \frac {\log {\left (\frac {\tan {\left (\frac {x}{2} \right )}}{\tan ^{2}{\left (\frac {x}{2} \right )} + 1} \right )} \tan ^{3}{\left (\frac {x}{2} \right )}}{6} + \frac {\log {\left (\frac {\tan {\left (\frac {x}{2} \right )}}{\tan ^{2}{\left (\frac {x}{2} \right )} + 1} \right )} \tan {\left (\frac {x}{2} \right )}}{2} - \frac {\log {\left (2 \right )} \tan ^{3}{\left (\frac {x}{2} \right )}}{6} - \frac {\tan ^{3}{\left (\frac {x}{2} \right )}}{18} + \frac {\log {\left (2 \right )} \tan {\left (\frac {x}{2} \right )}}{2} + \frac {5 \tan {\left (\frac {x}{2} \right )}}{6} \]
-2*x/3 - log(tan(x/2)/(tan(x/2)**2 + 1))*tan(x/2)**3/6 + log(tan(x/2)/(tan (x/2)**2 + 1))*tan(x/2)/2 - log(2)*tan(x/2)**3/6 - tan(x/2)**3/18 + log(2) *tan(x/2)/2 + 5*tan(x/2)/6
Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.43 \[ \int \frac {\cos (x) \log (\sin (x))}{(1+\cos (x))^2} \, dx=\frac {1}{6} \, {\left (\frac {3 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} - \frac {\sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}}\right )} \log \left (\frac {2 \, \sin \left (x\right )}{{\left (\frac {\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 1\right )} {\left (\cos \left (x\right ) + 1\right )}}\right ) + \frac {5 \, \sin \left (x\right )}{6 \, {\left (\cos \left (x\right ) + 1\right )}} - \frac {\sin \left (x\right )^{3}}{18 \, {\left (\cos \left (x\right ) + 1\right )}^{3}} - \frac {4}{3} \, \arctan \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1}\right ) \]
1/6*(3*sin(x)/(cos(x) + 1) - sin(x)^3/(cos(x) + 1)^3)*log(2*sin(x)/((sin(x )^2/(cos(x) + 1)^2 + 1)*(cos(x) + 1))) + 5/6*sin(x)/(cos(x) + 1) - 1/18*si n(x)^3/(cos(x) + 1)^3 - 4/3*arctan(sin(x)/(cos(x) + 1))
Time = 0.35 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.60 \[ \int \frac {\cos (x) \log (\sin (x))}{(1+\cos (x))^2} \, dx=-\frac {1}{18} \, \tan \left (\frac {1}{2} \, x\right )^{3} - \frac {1}{6} \, {\left (\tan \left (\frac {1}{2} \, x\right )^{3} - 3 \, \tan \left (\frac {1}{2} \, x\right )\right )} \log \left (\sin \left (x\right )\right ) - \frac {2}{3} \, x + \frac {5}{6} \, \tan \left (\frac {1}{2} \, x\right ) \]
Time = 0.94 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.73 \[ \int \frac {\cos (x) \log (\sin (x))}{(1+\cos (x))^2} \, dx=\frac {\frac {4\,\sin \left (2\,x\right )}{9}-\frac {\ln \left (-2\,{\sin \left (x\right )}^2+\sin \left (2\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{3}-\frac {14\,x}{3}+\frac {\ln \left (\sin \left (x\right )\right )\,7{}\mathrm {i}}{3}+\frac {7\,\sin \left (x\right )}{9}+\frac {\sin \left (2\,x\right )\,\ln \left (\sin \left (x\right )\right )}{3}-\frac {{\sin \left (x\right )}^2\,8{}\mathrm {i}}{9}+{\sin \left (\frac {x}{2}\right )}^2\,\left (\frac {16\,x}{3}+\frac {\ln \left (-2\,{\sin \left (x\right )}^2+\sin \left (2\,x\right )\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{3}-\frac {\ln \left (\sin \left (x\right )\right )\,8{}\mathrm {i}}{3}-\frac {32}{9}{}\mathrm {i}\right )+\frac {\ln \left (-2\,{\sin \left (x\right )}^2+\sin \left (2\,x\right )\,1{}\mathrm {i}\right )\,\left (2\,{\sin \left (x\right )}^2-1\right )\,1{}\mathrm {i}}{3}+\frac {\ln \left (\sin \left (x\right )\right )\,\sin \left (x\right )}{3}-\frac {\ln \left (\sin \left (x\right )\right )\,\left (2\,{\sin \left (x\right )}^2-1\right )\,1{}\mathrm {i}}{3}+\frac {2\,x\,\left (2\,{\sin \left (x\right )}^2-1\right )}{3}+\frac {32}{9}{}\mathrm {i}}{{\left (2\,{\sin \left (\frac {x}{2}\right )}^2-2\right )}^2} \]
((4*sin(2*x))/9 - (log(sin(2*x)*1i - 2*sin(x)^2)*7i)/3 - (14*x)/3 + (log(s in(x))*7i)/3 + (7*sin(x))/9 + (sin(2*x)*log(sin(x)))/3 - (sin(x)^2*8i)/9 + sin(x/2)^2*((16*x)/3 + (log(sin(2*x)*1i - 2*sin(x)^2)*8i)/3 - (log(sin(x) )*8i)/3 - 32i/9) + (log(sin(2*x)*1i - 2*sin(x)^2)*(2*sin(x)^2 - 1)*1i)/3 + (log(sin(x))*sin(x))/3 - (log(sin(x))*(2*sin(x)^2 - 1)*1i)/3 + (2*x*(2*si n(x)^2 - 1))/3 + 32i/9)/(2*sin(x/2)^2 - 2)^2
Time = 0.00 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.10 \[ \int \frac {\cos (x) \log (\sin (x))}{(1+\cos (x))^2} \, dx=-\frac {\mathrm {log}\left (\frac {2 \tan \left (\frac {x}{2}\right )}{\tan \left (\frac {x}{2}\right )^{2}+1}\right ) \tan \left (\frac {x}{2}\right )^{3}}{6}+\frac {\mathrm {log}\left (\frac {2 \tan \left (\frac {x}{2}\right )}{\tan \left (\frac {x}{2}\right )^{2}+1}\right ) \tan \left (\frac {x}{2}\right )}{2}-\frac {\tan \left (\frac {x}{2}\right )^{3}}{18}+\frac {5 \tan \left (\frac {x}{2}\right )}{6}-\frac {2 x}{3} \]