3.644 \(\int \frac{\cos (x) \log (\sin (x))}{(1+\cos (x))^2} \, dx\)

Optimal. Leaf size=60 \[ -\frac{2 x}{3}+\frac{8 \sin (x)}{9 (\cos (x)+1)}-\frac{\sin (x)}{9 (\cos (x)+1)^2}+\frac{2 \sin (x) \log (\sin (x))}{3 (\cos (x)+1)}-\frac{\sin (x) \log (\sin (x))}{3 (\cos (x)+1)^2} \]

[Out]

(-2*x)/3 - Sin[x]/(9*(1 + Cos[x])^2) + (8*Sin[x])/(9*(1 + Cos[x])) - (Log[Sin[x]]*Sin[x])/(3*(1 + Cos[x])^2) +
(2*Log[Sin[x]]*Sin[x])/(3*(1 + Cos[x]))

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Rubi [A]  time = 0.132164, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {2750, 2648, 2554, 12, 2968, 3019, 2735} \[ -\frac{2 x}{3}+\frac{8 \sin (x)}{9 (\cos (x)+1)}-\frac{\sin (x)}{9 (\cos (x)+1)^2}+\frac{2 \sin (x) \log (\sin (x))}{3 (\cos (x)+1)}-\frac{\sin (x) \log (\sin (x))}{3 (\cos (x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[x]*Log[Sin[x]])/(1 + Cos[x])^2,x]

[Out]

(-2*x)/3 - Sin[x]/(9*(1 + Cos[x])^2) + (8*Sin[x])/(9*(1 + Cos[x])) - (Log[Sin[x]]*Sin[x])/(3*(1 + Cos[x])^2) +
(2*Log[Sin[x]]*Sin[x])/(3*(1 + Cos[x]))

Rule 2750

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b
*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rule 2648

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]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2968

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]

Rule 3019

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] + D
ist[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]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(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]

Rubi steps

\begin{align*} \int \frac{\cos (x) \log (\sin (x))}{(1+\cos (x))^2} \, dx &=-\frac{\log (\sin (x)) \sin (x)}{3 (1+\cos (x))^2}+\frac{2 \log (\sin (x)) \sin (x)}{3 (1+\cos (x))}-\int \frac{\cos (x) (1+2 \cos (x))}{3 (1+\cos (x))^2} \, dx\\ &=-\frac{\log (\sin (x)) \sin (x)}{3 (1+\cos (x))^2}+\frac{2 \log (\sin (x)) \sin (x)}{3 (1+\cos (x))}-\frac{1}{3} \int \frac{\cos (x) (1+2 \cos (x))}{(1+\cos (x))^2} \, dx\\ &=-\frac{\log (\sin (x)) \sin (x)}{3 (1+\cos (x))^2}+\frac{2 \log (\sin (x)) \sin (x)}{3 (1+\cos (x))}-\frac{1}{3} \int \frac{\cos (x)+2 \cos ^2(x)}{(1+\cos (x))^2} \, dx\\ &=-\frac{\sin (x)}{9 (1+\cos (x))^2}-\frac{\log (\sin (x)) \sin (x)}{3 (1+\cos (x))^2}+\frac{2 \log (\sin (x)) \sin (x)}{3 (1+\cos (x))}+\frac{1}{9} \int \frac{2-6 \cos (x)}{1+\cos (x)} \, dx\\ &=-\frac{2 x}{3}-\frac{\sin (x)}{9 (1+\cos (x))^2}-\frac{\log (\sin (x)) \sin (x)}{3 (1+\cos (x))^2}+\frac{2 \log (\sin (x)) \sin (x)}{3 (1+\cos (x))}+\frac{8}{9} \int \frac{1}{1+\cos (x)} \, 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))}\\ \end{align*}

Mathematica [A]  time = 0.147594, size = 56, normalized size = 0.93 \[ -\frac{1}{18} \sec ^3\left (\frac{x}{2}\right ) \left (9 x \cos \left (\frac{x}{2}\right )+3 x \cos \left (\frac{3 x}{2}\right )-\sin \left (\frac{x}{2}\right ) (3 \log (\sin (x))+\cos (x) (6 \log (\sin (x))+8)+7)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[x]*Log[Sin[x]])/(1 + Cos[x])^2,x]

[Out]

-(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]))/18

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Maple [B]  time = 0.086, size = 106, normalized size = 1.8 \begin{align*} -{\frac{1}{9\, \left ( \sin \left ( x \right ) \right ) ^{3}} \left ( 12\,\sin \left ( x \right ) \left ( \cos \left ( x \right ) \right ) ^{2}\arctan \left ({\frac{\cos \left ( x \right ) -1}{\sin \left ( x \right ) }} \right ) -6\, \left ( \cos \left ( x \right ) \right ) ^{3}\ln \left ( 2 \right ) -6\, \left ( \cos \left ( x \right ) \right ) ^{3}\ln \left ( 1/2\,\sin \left ( x \right ) \right ) -8\, \left ( \cos \left ( x \right ) \right ) ^{3}+9\, \left ( \cos \left ( x \right ) \right ) ^{2}\ln \left ( 2 \right ) +9\, \left ( \cos \left ( x \right ) \right ) ^{2}\ln \left ( 1/2\,\sin \left ( x \right ) \right ) -12\,\arctan \left ({\frac{\cos \left ( x \right ) -1}{\sin \left ( x \right ) }} \right ) \sin \left ( x \right ) +9\, \left ( \cos \left ( x \right ) \right ) ^{2}+6\,\cos \left ( x \right ) -3\,\ln \left ( 2 \right ) -3\,\ln \left ( 1/2\,\sin \left ( x \right ) \right ) -7 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)*ln(sin(x))/(cos(x)+1)^2,x)

[Out]

-1/9*(12*sin(x)*cos(x)^2*arctan((cos(x)-1)/sin(x))-6*cos(x)^3*ln(2)-6*cos(x)^3*ln(1/2*sin(x))-8*cos(x)^3+9*cos
(x)^2*ln(2)+9*cos(x)^2*ln(1/2*sin(x))-12*arctan((cos(x)-1)/sin(x))*sin(x)+9*cos(x)^2+6*cos(x)-3*ln(2)-3*ln(1/2
*sin(x))-7)/sin(x)^3

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Maxima [A]  time = 1.43172, size = 116, normalized size = 1.93 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*log(sin(x))/(1+cos(x))^2,x, algorithm="maxima")

[Out]

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*sin(x)^3/(cos(x) + 1)^3 - 4/3*arctan(sin(x)/(cos(x) + 1))

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Fricas [A]  time = 2.49149, size = 174, normalized size = 2.9 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*log(sin(x))/(1+cos(x))^2,x, algorithm="fricas")

[Out]

-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)

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Sympy [A]  time = 11.4039, size = 107, normalized size = 1.78 \begin{align*} - \frac{2 x}{3} + \frac{\log{\left (\tan ^{2}{\left (\frac{x}{2} \right )} + 1 \right )} \tan ^{3}{\left (\frac{x}{2} \right )}}{6} - \frac{\log{\left (\tan ^{2}{\left (\frac{x}{2} \right )} + 1 \right )} \tan{\left (\frac{x}{2} \right )}}{2} - \frac{\log{\left (\tan{\left (\frac{x}{2} \right )} \right )} \tan ^{3}{\left (\frac{x}{2} \right )}}{6} + \frac{\log{\left (\tan{\left (\frac{x}{2} \right )} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*ln(sin(x))/(1+cos(x))**2,x)

[Out]

-2*x/3 + log(tan(x/2)**2 + 1)*tan(x/2)**3/6 - log(tan(x/2)**2 + 1)*tan(x/2)/2 - log(tan(x/2))*tan(x/2)**3/6 +
log(tan(x/2))*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

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Giac [A]  time = 1.21404, size = 49, normalized size = 0.82 \begin{align*} -\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*log(sin(x))/(1+cos(x))^2,x, algorithm="giac")

[Out]

-1/18*tan(1/2*x)^3 - 1/6*(tan(1/2*x)^3 - 3*tan(1/2*x))*log(sin(x)) - 2/3*x + 5/6*tan(1/2*x)