3.40 \(\int \frac {\log (\sin (x))}{1+\sin (x)} \, dx\)

Optimal. Leaf size=22 \[ -x-\tanh ^{-1}(\cos (x))-\frac {\cos (x) \log (\sin (x))}{\sin (x)+1} \]

[Out]

-x-arctanh(cos(x))-cos(x)*ln(sin(x))/(1+sin(x))

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Rubi [A]  time = 0.07, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2648, 2554, 2839, 3770, 8} \[ -x-\tanh ^{-1}(\cos (x))-\frac {\cos (x) \log (\sin (x))}{\sin (x)+1} \]

Antiderivative was successfully verified.

[In]

Int[Log[Sin[x]]/(1 + Sin[x]),x]

[Out]

-x - ArcTanh[Cos[x]] - (Cos[x]*Log[Sin[x]])/(1 + Sin[x])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\log (\sin (x))}{1+\sin (x)} \, dx &=-\frac {\cos (x) \log (\sin (x))}{1+\sin (x)}+\int \frac {\cos (x) \cot (x)}{1+\sin (x)} \, dx\\ &=-\frac {\cos (x) \log (\sin (x))}{1+\sin (x)}-\int 1 \, dx+\int \csc (x) \, dx\\ &=-x-\tanh ^{-1}(\cos (x))-\frac {\cos (x) \log (\sin (x))}{1+\sin (x)}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 39, normalized size = 1.77 \[ -x-2 \log \left (\cos \left (\frac {x}{2}\right )\right )+\frac {2 \sin \left (\frac {x}{2}\right ) \log (\sin (x))}{\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[Sin[x]]/(1 + Sin[x]),x]

[Out]

-x - 2*Log[Cos[x/2]] + (2*Log[Sin[x]]*Sin[x/2])/(Cos[x/2] + Sin[x/2])

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fricas [B]  time = 0.45, size = 93, normalized size = 4.23 \[ -\frac {4 \, {\left (\cos \relax (x) + \sin \relax (x) + 1\right )} \arctan \left (-\frac {\cos \relax (x) + \sin \relax (x) + 1}{\cos \relax (x) - \sin \relax (x) + 1}\right ) + 4 \, x \cos \relax (x) + {\left (\cos \relax (x) + \sin \relax (x) + 1\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) - {\left (\cos \relax (x) + \sin \relax (x) + 1\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + 2 \, {\left (\cos \relax (x) - \sin \relax (x) + 1\right )} \log \left (\sin \relax (x)\right ) + 4 \, x \sin \relax (x) + 4 \, x}{2 \, {\left (\cos \relax (x) + \sin \relax (x) + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(sin(x))/(1+sin(x)),x, algorithm="fricas")

[Out]

-1/2*(4*(cos(x) + sin(x) + 1)*arctan(-(cos(x) + sin(x) + 1)/(cos(x) - sin(x) + 1)) + 4*x*cos(x) + (cos(x) + si
n(x) + 1)*log(1/2*cos(x) + 1/2) - (cos(x) + sin(x) + 1)*log(-1/2*cos(x) + 1/2) + 2*(cos(x) - sin(x) + 1)*log(s
in(x)) + 4*x*sin(x) + 4*x)/(cos(x) + sin(x) + 1)

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giac [A]  time = 1.13, size = 36, normalized size = 1.64 \[ -x - \frac {2 \, \log \left (\sin \relax (x)\right )}{\tan \left (\frac {1}{2} \, x\right ) + 1} - 2 \, \log \left (\tan \left (\frac {1}{4} \, x\right )^{2} + 1\right ) + 2 \, \log \left ({\left | \tan \left (\frac {1}{4} \, x\right ) \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(sin(x))/(1+sin(x)),x, algorithm="giac")

[Out]

-x - 2*log(sin(x))/(tan(1/2*x) + 1) - 2*log(tan(1/4*x)^2 + 1) + 2*log(abs(tan(1/4*x)))

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maple [B]  time = 0.17, size = 54, normalized size = 2.45 \[ \ln \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )+\frac {-x \tan \left (\frac {x}{2}\right )+2 \ln \left (\frac {2 \tan \left (\frac {x}{2}\right )}{\tan ^{2}\left (\frac {x}{2}\right )+1}\right ) \tan \left (\frac {x}{2}\right )-x}{\tan \left (\frac {x}{2}\right )+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(sin(x))/(sin(x)+1),x)

[Out]

(-x-x*tan(1/2*x)+2*tan(1/2*x)*ln(2*tan(1/2*x)/(tan(1/2*x)^2+1)))/(tan(1/2*x)+1)+ln(tan(1/2*x)^2+1)

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maxima [B]  time = 0.97, size = 82, normalized size = 3.73 \[ -\frac {2 \, \log \left (\frac {2 \, \sin \relax (x)}{{\left (\frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 1\right )} {\left (\cos \relax (x) + 1\right )}}\right )}{\frac {\sin \relax (x)}{\cos \relax (x) + 1} + 1} - 2 \, \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right ) + 2 \, \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right ) - \log \left (\frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(sin(x))/(1+sin(x)),x, algorithm="maxima")

[Out]

-2*log(2*sin(x)/((sin(x)^2/(cos(x) + 1)^2 + 1)*(cos(x) + 1)))/(sin(x)/(cos(x) + 1) + 1) - 2*arctan(sin(x)/(cos
(x) + 1)) + 2*log(sin(x)/(cos(x) + 1)) - log(sin(x)^2/(cos(x) + 1)^2 + 1)

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mupad [B]  time = 0.38, size = 55, normalized size = 2.50 \[ -2\,x+\ln \left (2\,\sin \relax (x)-\cos \relax (x)\,2{}\mathrm {i}-2{}\mathrm {i}\right )\,\left (-1-\mathrm {i}\right )+\ln \left (2\,\sin \relax (x)-\cos \relax (x)\,2{}\mathrm {i}+2{}\mathrm {i}\right )\,\left (1-\mathrm {i}\right )-\frac {2\,\ln \left (\sin \relax (x)\right )}{\cos \relax (x)+\sin \relax (x)\,1{}\mathrm {i}+1{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(sin(x))/(sin(x) + 1),x)

[Out]

log(2*sin(x) - cos(x)*2i + 2i)*(1 - 1i) - log(2*sin(x) - cos(x)*2i - 2i)*(1 + 1i) - 2*x - (2*log(sin(x)))/(cos
(x) + sin(x)*1i + 1i)

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sympy [B]  time = 1.37, size = 97, normalized size = 4.41 \[ - \frac {x \tan {\left (\frac {x}{2} \right )}}{\tan {\left (\frac {x}{2} \right )} + 1} - \frac {x}{\tan {\left (\frac {x}{2} \right )} + 1} - \frac {\log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )} \tan {\left (\frac {x}{2} \right )}}{\tan {\left (\frac {x}{2} \right )} + 1} + \frac {\log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )}}{\tan {\left (\frac {x}{2} \right )} + 1} + \frac {2 \log {\left (\tan {\left (\frac {x}{2} \right )} \right )} \tan {\left (\frac {x}{2} \right )}}{\tan {\left (\frac {x}{2} \right )} + 1} + \frac {2 \log {\relax (2 )} \tan {\left (\frac {x}{2} \right )}}{\tan {\left (\frac {x}{2} \right )} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(sin(x))/(1+sin(x)),x)

[Out]

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

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