3.2.47 \(\int \frac {\sin (x)}{1+\cos (x)+\sin (x)} \, dx\) [147]

Optimal. Leaf size=30 \[ \frac {x}{2}-\frac {1}{2} \log (1+\cos (x)+\sin (x))-\frac {1}{2} \log \left (1+\tan \left (\frac {x}{2}\right )\right ) \]

[Out]

1/2*x-1/2*ln(1+cos(x)+sin(x))-1/2*ln(1+tan(1/2*x))

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Rubi [A]
time = 0.02, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3216, 3203, 31} \begin {gather*} \frac {x}{2}-\frac {1}{2} \log \left (\tan \left (\frac {x}{2}\right )+1\right )-\frac {1}{2} \log (\sin (x)+\cos (x)+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[x]/(1 + Cos[x] + Sin[x]),x]

[Out]

x/2 - Log[1 + Cos[x] + Sin[x]]/2 - Log[1 + Tan[x/2]]/2

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3203

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Module[{f = Free
Factors[Tan[(d + e*x)/2], x]}, Dist[2*(f/e), Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d +
e*x)/2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]

Rule 3216

Int[((A_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(
x_)]), x_Symbol] :> Simp[c*C*((d + e*x)/(e*(b^2 + c^2))), x] + (Dist[(A*(b^2 + c^2) - a*c*C)/(b^2 + c^2), Int[
1/(a + b*Cos[d + e*x] + c*Sin[d + e*x]), x], x] - Simp[b*C*(Log[a + b*Cos[d + e*x] + c*Sin[d + e*x]]/(e*(b^2 +
c^2))), x]) /; FreeQ[{a, b, c, d, e, A, C}, x] && NeQ[b^2 + c^2, 0] && NeQ[A*(b^2 + c^2) - a*c*C, 0]

Rubi steps

\begin {align*} \int \frac {\sin (x)}{1+\cos (x)+\sin (x)} \, dx &=\frac {x}{2}-\frac {1}{2} \log (1+\cos (x)+\sin (x))-\frac {1}{2} \int \frac {1}{1+\cos (x)+\sin (x)} \, dx\\ &=\frac {x}{2}-\frac {1}{2} \log (1+\cos (x)+\sin (x))-\text {Subst}\left (\int \frac {1}{2+2 x} \, dx,x,\tan \left (\frac {x}{2}\right )\right )\\ &=\frac {x}{2}-\frac {1}{2} \log (1+\cos (x)+\sin (x))-\frac {1}{2} \log \left (1+\tan \left (\frac {x}{2}\right )\right )\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 22, normalized size = 0.73 \begin {gather*} \frac {x}{2}-\log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]/(1 + Cos[x] + Sin[x]),x]

[Out]

x/2 - Log[Cos[x/2] + Sin[x/2]]

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Mathics [A]
time = 1.99, size = 24, normalized size = 0.80 \begin {gather*} \frac {x}{2}-\text {Log}\left [1+\text {Tan}\left [\frac {x}{2}\right ]\right ]+\frac {\text {Log}\left [\frac {2}{1+\text {Cos}\left [x\right ]}\right ]}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[Sin[x]/(1 + Sin[x] + Cos[x]),x]')

[Out]

x / 2 - Log[1 + Tan[x / 2]] + Log[2 / (1 + Cos[x])] / 2

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Maple [A]
time = 0.06, size = 27, normalized size = 0.90

method	result	size
risch	\(\frac {x}{2}+\frac {i x}{2}-\ln \left ({\mathrm e}^{i x}+i\right )\)	\(20\)
default	\(-\ln \left (1+\tan \left (\frac {x}{2}\right )\right )+\frac {\ln \left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )}{2}+\arctan \left (\tan \left (\frac {x}{2}\right )\right )\)	\(27\)
norman	\(\frac {\frac {x}{2}+\frac {x \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{2}}{1+\tan ^{2}\left (\frac {x}{2}\right )}-\ln \left (1+\tan \left (\frac {x}{2}\right )\right )+\frac {\ln \left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )}{2}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)/(1+cos(x)+sin(x)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.35, size = 41, normalized size = 1.37 \begin {gather*} \arctan \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1}\right ) - \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right ) + \frac {1}{2} \, \log \left (\frac {\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 11, normalized size = 0.37 \begin {gather*} \frac {1}{2} \, x - \frac {1}{2} \, \log \left (\sin \left (x\right ) + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x - 1/2*log(sin(x) + 1)

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Sympy [A]
time = 0.15, size = 22, normalized size = 0.73 \begin {gather*} \frac {x}{2} - \log {\left (\tan {\left (\frac {x}{2} \right )} + 1 \right )} + \frac {\log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.00, size = 35, normalized size = 1.17 \begin {gather*} 2 \left (-\frac {\ln \left |\tan \left (\frac {x}{2}\right )+1\right |}{2}+\frac {\ln \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )}{4}+\frac {x}{2\cdot 2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.31, size = 34, normalized size = 1.13 \begin {gather*} -\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )+\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-\mathrm {i}\right )\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )+\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)/(cos(x) + sin(x) + 1),x)

[Out]

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

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