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

Optimal. Leaf size=29 \[ -\log \left (\tan \left (\frac{x}{2}\right )+1\right )-\frac{\cos (x)-\sin (x)}{\sin (x)+\cos (x)+1} \]

[Out]

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

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Rubi [A]  time = 0.0201917, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {3129, 3124, 31} \[ -\log \left (\tan \left (\frac{x}{2}\right )+1\right )-\frac{\cos (x)-\sin (x)}{\sin (x)+\cos (x)+1} \]

Antiderivative was successfully verified.

[In]

Int[(1 + Cos[x] + Sin[x])^(-2),x]

[Out]

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

Rule 3129

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(n_), x_Symbol] :> Simp[((-(c*Cos[d
+ e*x]) + b*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1))/(e*(n + 1)*(a^2 - b^2 - c^2)), x] +
Dist[1/((n + 1)*(a^2 - b^2 - c^2)), Int[(a*(n + 1) - b*(n + 2)*Cos[d + e*x] - c*(n + 2)*Sin[d + e*x])*(a + b*C
os[d + e*x] + c*Sin[d + e*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && LtQ[n
, -1] && NeQ[n, -3/2]

Rule 3124

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 31

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

Rubi steps

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

Mathematica [A]  time = 0.0327142, size = 56, normalized size = 1.93 \[ \frac{1}{2} \tan \left (\frac{x}{2}\right )+\log \left (\cos \left (\frac{x}{2}\right )\right )+\frac{\sin \left (\frac{x}{2}\right )}{\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )}-\log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Cos[x] + Sin[x])^(-2),x]

[Out]

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

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Maple [A]  time = 0.039, size = 27, normalized size = 0.9 \begin{align*}{\frac{1}{2}\tan \left ({\frac{x}{2}} \right ) }- \left ( 1+\tan \left ({\frac{x}{2}} \right ) \right ) ^{-1}-\ln \left ( 1+\tan \left ({\frac{x}{2}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.30164, size = 54, normalized size = 1.86 \begin{align*} \frac{\sin \left (x\right )}{2 \,{\left (\cos \left (x\right ) + 1\right )}} - \frac{1}{\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1} - \log \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.29235, size = 182, normalized size = 6.28 \begin{align*} \frac{{\left (\cos \left (x\right ) + \sin \left (x\right ) + 1\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) -{\left (\cos \left (x\right ) + \sin \left (x\right ) + 1\right )} \log \left (\sin \left (x\right ) + 1\right ) - 2 \, \cos \left (x\right ) + 2 \, \sin \left (x\right )}{2 \,{\left (\cos \left (x\right ) + \sin \left (x\right ) + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.785549, size = 66, normalized size = 2.28 \begin{align*} - \frac{2 \log{\left (\tan{\left (\frac{x}{2} \right )} + 1 \right )} \tan{\left (\frac{x}{2} \right )}}{2 \tan{\left (\frac{x}{2} \right )} + 2} - \frac{2 \log{\left (\tan{\left (\frac{x}{2} \right )} + 1 \right )}}{2 \tan{\left (\frac{x}{2} \right )} + 2} + \frac{\tan ^{2}{\left (\frac{x}{2} \right )}}{2 \tan{\left (\frac{x}{2} \right )} + 2} - \frac{3}{2 \tan{\left (\frac{x}{2} \right )} + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12091, size = 41, normalized size = 1.41 \begin{align*} \frac{\tan \left (\frac{1}{2} \, x\right )}{\tan \left (\frac{1}{2} \, x\right ) + 1} - \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) + 1 \right |}\right ) + \frac{1}{2} \, \tan \left (\frac{1}{2} \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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