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

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

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Rubi [A]  time = 0.13, antiderivative size = 60, normalized size of antiderivative = 1.00, 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 12

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

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 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.15, 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]

-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]
))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos (x) \log (\sin (x))}{(1+\cos (x))^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [A]  time = 0.86, size = 53, normalized size = 0.88 \[ -\frac {6 \, x \cos \relax (x)^{2} - 3 \, {\left (2 \, \cos \relax (x) + 1\right )} \log \left (\sin \relax (x)\right ) \sin \relax (x) + 12 \, x \cos \relax (x) - {\left (8 \, \cos \relax (x) + 7\right )} \sin \relax (x) + 6 \, x}{9 \, {\left (\cos \relax (x)^{2} + 2 \, \cos \relax (x) + 1\right )}} \]

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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giac [A]  time = 1.09, size = 36, normalized size = 0.60 \[ -\frac {1}{18} \, \tan \left (\frac {1}{2} \, x\right )^{3} - \frac {1}{6} \, {\left (\tan \left (\frac {1}{2} \, x\right )^{3} - 3 \, \tan \left (\frac {1}{2} \, x\right )\right )} \log \left (\sin \relax (x)\right ) - \frac {2}{3} \, x + \frac {5}{6} \, \tan \left (\frac {1}{2} \, x\right ) \]

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)

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maple [B]  time = 0.44, size = 106, normalized size = 1.77

method	result	size
default	\(\frac {6 \left (\cos ^{3}\relax (x )\right ) \ln \relax (2)+6 \left (\cos ^{3}\relax (x )\right ) \ln \left (\frac {\sin \relax (x )}{2}\right )-12 \left (\cos ^{2}\relax (x )\right ) \sin \relax (x ) \arctan \left (\frac {-1+\cos \relax (x )}{\sin \relax (x )}\right )+8 \left (\cos ^{3}\relax (x )\right )-9 \left (\cos ^{2}\relax (x )\right ) \ln \relax (2)-9 \left (\cos ^{2}\relax (x )\right ) \ln \left (\frac {\sin \relax (x )}{2}\right )-9 \left (\cos ^{2}\relax (x )\right )+12 \arctan \left (\frac {-1+\cos \relax (x )}{\sin \relax (x )}\right ) \sin \relax (x )-6 \cos \relax (x )+3 \ln \relax (2)+3 \ln \left (\frac {\sin \relax (x )}{2}\right )+7}{9 \sin \relax (x )^{3}}\)	\(106\)
risch	\(-\frac {2 i \left (3 \,{\mathrm e}^{2 i x}+3 \,{\mathrm e}^{i x}+2\right ) \ln \left ({\mathrm e}^{i x}\right )}{3 \left ({\mathrm e}^{i x}+1\right )^{3}}+\frac {6 \pi -12 x +9 \pi \,{\mathrm e}^{2 i x}+16 i-9 \,{\mathrm e}^{i x} \pi \mathrm {csgn}\left (\sin \relax (x )\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-i x}\right )+9 \,{\mathrm e}^{i x} \pi \,\mathrm {csgn}\left (\sin \relax (x )\right ) \mathrm {csgn}\left (i \sin \relax (x )\right )^{2}-9 \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{2 i x}-1\right )\right ) \mathrm {csgn}\left (\sin \relax (x )\right )^{2} {\mathrm e}^{2 i x}-9 \pi \mathrm {csgn}\left (\sin \relax (x )\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-i x}\right ) {\mathrm e}^{2 i x}+9 \pi \,\mathrm {csgn}\left (\sin \relax (x )\right ) \mathrm {csgn}\left (i \sin \relax (x )\right )^{2} {\mathrm e}^{2 i x}-6 \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{2 i x}-1\right )\right ) \mathrm {csgn}\left (\sin \relax (x )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-i x}\right )-9 \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{2 i x}-1\right )\right ) \mathrm {csgn}\left (\sin \relax (x )\right )^{2} {\mathrm e}^{i x}-9 \pi \,\mathrm {csgn}\left (\sin \relax (x )\right ) \mathrm {csgn}\left (i \sin \relax (x )\right ) {\mathrm e}^{2 i x}-36 x \,{\mathrm e}^{i x}-6 \pi \,\mathrm {csgn}\left (\sin \relax (x )\right ) \mathrm {csgn}\left (i \sin \relax (x )\right )+6 \mathrm {csgn}\left (i \sin \relax (x )\right )^{2} \mathrm {csgn}\left (\sin \relax (x )\right ) \pi -6 \,\mathrm {csgn}\left (i {\mathrm e}^{-i x}\right ) \mathrm {csgn}\left (\sin \relax (x )\right )^{2} \pi -9 \,{\mathrm e}^{i x} \pi \mathrm {csgn}\left (\sin \relax (x )\right )^{3}+18 i {\mathrm e}^{2 i x}-9 \,{\mathrm e}^{i x} \pi \mathrm {csgn}\left (i \sin \relax (x )\right )^{2}+9 \,{\mathrm e}^{i x} \pi \mathrm {csgn}\left (i \sin \relax (x )\right )^{3}-18 i {\mathrm e}^{i x} \ln \relax (2)-36 x \,{\mathrm e}^{2 i x}-12 i \ln \relax (2)-12 x \,{\mathrm e}^{3 i x}+9 \,{\mathrm e}^{i x} \pi +30 i {\mathrm e}^{i x}+6 i \ln \left ({\mathrm e}^{2 i x}-1\right )-6 \mathrm {csgn}\left (i \sin \relax (x )\right )^{2} \pi +6 \mathrm {csgn}\left (i \sin \relax (x )\right )^{3} \pi -6 \mathrm {csgn}\left (\sin \relax (x )\right )^{3} \pi -9 \pi \mathrm {csgn}\left (\sin \relax (x )\right )^{3} {\mathrm e}^{2 i x}-6 i \ln \left ({\mathrm e}^{2 i x}-1\right ) {\mathrm e}^{3 i x}-9 \pi \mathrm {csgn}\left (i \sin \relax (x )\right )^{2} {\mathrm e}^{2 i x}-18 i \ln \relax (2) {\mathrm e}^{2 i x}+9 \pi \mathrm {csgn}\left (i \sin \relax (x )\right )^{3} {\mathrm e}^{2 i x}-6 \mathrm {csgn}\left (\sin \relax (x )\right )^{2} \mathrm {csgn}\left (i \left ({\mathrm e}^{2 i x}-1\right )\right ) \pi -9 \,{\mathrm e}^{i x} \pi \,\mathrm {csgn}\left (\sin \relax (x )\right ) \mathrm {csgn}\left (i \sin \relax (x )\right )-9 \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{2 i x}-1\right )\right ) \mathrm {csgn}\left (\sin \relax (x )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-i x}\right ) {\mathrm e}^{i x}-9 \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{2 i x}-1\right )\right ) \mathrm {csgn}\left (\sin \relax (x )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-i x}\right ) {\mathrm e}^{2 i x}}{9 \left ({\mathrm e}^{i x}+1\right )^{3}}\)	\(598\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.96, size = 86, normalized size = 1.43 \[ \frac {1}{6} \, {\left (\frac {3 \, \sin \relax (x)}{\cos \relax (x) + 1} - \frac {\sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}}\right )} \log \left (\frac {2 \, \sin \relax (x)}{{\left (\frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 1\right )} {\left (\cos \relax (x) + 1\right )}}\right ) + \frac {5 \, \sin \relax (x)}{6 \, {\left (\cos \relax (x) + 1\right )}} - \frac {\sin \relax (x)^{3}}{18 \, {\left (\cos \relax (x) + 1\right )}^{3}} - \frac {4}{3} \, \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right ) \]

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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mupad [B]  time = 0.79, size = 164, normalized size = 2.73 \[ \frac {\frac {4\,\sin \left (2\,x\right )}{9}-\frac {\ln \left (-2\,{\sin \relax (x)}^2+\sin \left (2\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{3}-\frac {14\,x}{3}+\frac {\ln \left (\sin \relax (x)\right )\,7{}\mathrm {i}}{3}+\frac {7\,\sin \relax (x)}{9}+\frac {\sin \left (2\,x\right )\,\ln \left (\sin \relax (x)\right )}{3}-\frac {{\sin \relax (x)}^2\,8{}\mathrm {i}}{9}+{\sin \left (\frac {x}{2}\right )}^2\,\left (\frac {16\,x}{3}+\frac {\ln \left (-2\,{\sin \relax (x)}^2+\sin \left (2\,x\right )\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{3}-\frac {\ln \left (\sin \relax (x)\right )\,8{}\mathrm {i}}{3}-\frac {32}{9}{}\mathrm {i}\right )+\frac {\ln \left (-2\,{\sin \relax (x)}^2+\sin \left (2\,x\right )\,1{}\mathrm {i}\right )\,\left (2\,{\sin \relax (x)}^2-1\right )\,1{}\mathrm {i}}{3}+\frac {\ln \left (\sin \relax (x)\right )\,\sin \relax (x)}{3}-\frac {\ln \left (\sin \relax (x)\right )\,\left (2\,{\sin \relax (x)}^2-1\right )\,1{}\mathrm {i}}{3}+\frac {2\,x\,\left (2\,{\sin \relax (x)}^2-1\right )}{3}+\frac {32}{9}{}\mathrm {i}}{{\left (2\,{\sin \left (\frac {x}{2}\right )}^2-2\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((4*sin(2*x))/9 - (log(sin(2*x)*1i - 2*sin(x)^2)*7i)/3 - (14*x)/3 + (log(sin(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*sin(x)^2 - 1))/3 + 32i/9)/(2*sin(x/2)^2 - 2)^2

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sympy [A]  time = 8.00, size = 107, normalized size = 1.78 \[ - \frac {2 x}{3} + \frac {\log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )} \tan ^{3}{\left (\frac {x}{2} \right )}}{6} - \frac {\log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )} \tan {\left (\frac {x}{2} \right )}}{2} - \frac {\log {\left (\tan {\left (\frac {x}{2} \right )} \right )} \tan ^{3}{\left (\frac {x}{2} \right )}}{6} + \frac {\log {\left (\tan {\left (\frac {x}{2} \right )} \right )} \tan {\left (\frac {x}{2} \right )}}{2} - \frac {\log {\relax (2 )} \tan ^{3}{\left (\frac {x}{2} \right )}}{6} - \frac {\tan ^{3}{\left (\frac {x}{2} \right )}}{18} + \frac {\log {\relax (2 )} \tan {\left (\frac {x}{2} \right )}}{2} + \frac {5 \tan {\left (\frac {x}{2} \right )}}{6} \]

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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