3.78 \(\int \frac{\csc (x)}{1+\tan (x)} \, dx\)

Optimal. Leaf size=26 \[ \frac{\tanh ^{-1}\left (\frac{\cos (x)-\sin (x)}{\sqrt{2}}\right )}{\sqrt{2}}-\tanh ^{-1}(\cos (x)) \]

[Out]

-ArcTanh[Cos[x]] + ArcTanh[(Cos[x] - Sin[x])/Sqrt[2]]/Sqrt[2]

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Rubi [A]  time = 0.0744782, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.556, Rules used = {3518, 3110, 3770, 3074, 206} \[ \frac{\tanh ^{-1}\left (\frac{\cos (x)-\sin (x)}{\sqrt{2}}\right )}{\sqrt{2}}-\tanh ^{-1}(\cos (x)) \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]/(1 + Tan[x]),x]

[Out]

-ArcTanh[Cos[x]] + ArcTanh[(Cos[x] - Sin[x])/Sqrt[2]]/Sqrt[2]

Rule 3518

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[(Sin[e + f*x]
^m*(a*Cos[e + f*x] + b*Sin[e + f*x])^n)/Cos[e + f*x]^n, x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rule 3110

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(cos[c + d*x]^m*sin[c + d*x]^n)/(a*cos[c + d*x] + b*sin[c + d
*x]), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IntegersQ[m, n]

Rule 3770

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

Rule 3074

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 0.0413579, size = 41, normalized size = 1.58 \[ \log \left (\sin \left (\frac{x}{2}\right )\right )-\log \left (\cos \left (\frac{x}{2}\right )\right )+(1+i) (-1)^{3/4} \tanh ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right )-1}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]/(1 + Tan[x]),x]

[Out]

(1 + I)*(-1)^(3/4)*ArcTanh[(-1 + Tan[x/2])/Sqrt[2]] - Log[Cos[x/2]] + Log[Sin[x/2]]

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Maple [A]  time = 0.025, size = 26, normalized size = 1. \begin{align*} -\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{4} \left ( 2\,\tan \left ( x/2 \right ) -2 \right ) } \right ) +\ln \left ( \tan \left ({\frac{x}{2}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)/(1+tan(x)),x)

[Out]

-2^(1/2)*arctanh(1/4*(2*tan(1/2*x)-2)*2^(1/2))+ln(tan(1/2*x))

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Maxima [B]  time = 1.85027, size = 68, normalized size = 2.62 \begin{align*} \frac{1}{2} \, \sqrt{2} \log \left (-\frac{\sqrt{2} - \frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1}{\sqrt{2} + \frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} - 1}\right ) + \log \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(csc(x)/(1+tan(x)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.35363, size = 203, normalized size = 7.81 \begin{align*} \frac{1}{4} \, \sqrt{2} \log \left (\frac{2 \,{\left (\sqrt{2} + \cos \left (x\right )\right )} \sin \left (x\right ) - 2 \, \sqrt{2} \cos \left (x\right ) - 3}{2 \, \cos \left (x\right ) \sin \left (x\right ) + 1}\right ) - \frac{1}{2} \, \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) + \frac{1}{2} \, \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(1+tan(x)),x, algorithm="fricas")

[Out]

1/4*sqrt(2)*log((2*(sqrt(2) + cos(x))*sin(x) - 2*sqrt(2)*cos(x) - 3)/(2*cos(x)*sin(x) + 1)) - 1/2*log(1/2*cos(
x) + 1/2) + 1/2*log(-1/2*cos(x) + 1/2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc{\left (x \right )}}{\tan{\left (x \right )} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(1+tan(x)),x)

[Out]

Integral(csc(x)/(tan(x) + 1), x)

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Giac [A]  time = 1.29647, size = 59, normalized size = 2.27 \begin{align*} \frac{1}{2} \, \sqrt{2} \log \left (\frac{{\left | -2 \, \sqrt{2} + 2 \, \tan \left (\frac{1}{2} \, x\right ) - 2 \right |}}{{\left | 2 \, \sqrt{2} + 2 \, \tan \left (\frac{1}{2} \, x\right ) - 2 \right |}}\right ) + \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(1+tan(x)),x, algorithm="giac")

[Out]

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