3.805 \(\int \csc (2 x) (\cos (x)+\sin (x)) \, dx\)

Optimal. Leaf size=15 \[ \frac {1}{2} \tanh ^{-1}(\sin (x))-\frac {1}{2} \tanh ^{-1}(\cos (x)) \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4401, 4287, 3770, 4288} \[ \frac {1}{2} \tanh ^{-1}(\sin (x))-\frac {1}{2} \tanh ^{-1}(\cos (x)) \]

Antiderivative was successfully verified.

[In]

Int[Csc[2*x]*(Cos[x] + Sin[x]),x]

[Out]

-ArcTanh[Cos[x]]/2 + ArcTanh[Sin[x]]/2

Rule 3770

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

Rule 4287

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_Symbol] :> Dist[2^p/e^p, Int[(e*Cos
[a + b*x])^(m + p)*Sin[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2]
&& IntegerQ[p]

Rule 4288

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

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rubi steps

\begin {align*} \int \csc (2 x) (\cos (x)+\sin (x)) \, dx &=\int (\cos (x) \csc (2 x)+\csc (2 x) \sin (x)) \, dx\\ &=\int \cos (x) \csc (2 x) \, dx+\int \csc (2 x) \sin (x) \, dx\\ &=\frac {1}{2} \int \csc (x) \, dx+\frac {1}{2} \int \sec (x) \, dx\\ &=-\frac {1}{2} \tanh ^{-1}(\cos (x))+\frac {1}{2} \tanh ^{-1}(\sin (x))\\ \end {align*}

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Mathematica [B]  time = 0.01, size = 61, normalized size = 4.07 \[ \frac {1}{2} \log \left (\sin \left (\frac {x}{2}\right )\right )-\frac {1}{2} \log \left (\cos \left (\frac {x}{2}\right )\right )-\frac {1}{2} \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+\frac {1}{2} \log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[2*x]*(Cos[x] + Sin[x]),x]

[Out]

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

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fricas [B]  time = 0.75, size = 35, normalized size = 2.33 \[ -\frac {1}{4} \, \log \left (-\frac {1}{2} \, {\left (\cos \relax (x) + 1\right )} \sin \relax (x) + \frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + \frac {1}{4} \, \log \left (-\frac {1}{2} \, {\left (\cos \relax (x) - 1\right )} \sin \relax (x) - \frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(2*x)*(cos(x)+sin(x)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.15, size = 29, normalized size = 1.93 \[ \frac {1}{2} \, \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right ) - \frac {1}{2} \, \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right ) + \frac {1}{2} \, \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(2*x)*(cos(x)+sin(x)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.26, size = 20, normalized size = 1.33 \[ \frac {\ln \left (\sec \relax (x )+\tan \relax (x )\right )}{2}+\frac {\ln \left (\csc \relax (x )-\cot \relax (x )\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(2*x)*(cos(x)+sin(x)),x)

[Out]

1/2*ln(sec(x)+tan(x))+1/2*ln(csc(x)-cot(x))

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maxima [B]  time = 0.42, size = 69, normalized size = 4.60 \[ -\frac {1}{4} \, \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} + 2 \, \cos \relax (x) + 1\right ) + \frac {1}{4} \, \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} - 2 \, \cos \relax (x) + 1\right ) + \frac {1}{4} \, \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} + 2 \, \sin \relax (x) + 1\right ) - \frac {1}{4} \, \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} - 2 \, \sin \relax (x) + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(2*x)*(cos(x)+sin(x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.11, size = 24, normalized size = 1.60 \[ \frac {\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+\mathrm {tan}\left (\frac {x}{2}\right )\right )}{2}-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(x) + sin(x))/sin(2*x),x)

[Out]

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

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sympy [B]  time = 1.86, size = 32, normalized size = 2.13 \[ - \frac {\log {\left (\sin {\relax (x )} - 1 \right )}}{4} + \frac {\log {\left (\sin {\relax (x )} + 1 \right )}}{4} + \frac {\log {\left (\cos {\relax (x )} - 1 \right )}}{4} - \frac {\log {\left (\cos {\relax (x )} + 1 \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(2*x)*(cos(x)+sin(x)),x)

[Out]

-log(sin(x) - 1)/4 + log(sin(x) + 1)/4 + log(cos(x) - 1)/4 - log(cos(x) + 1)/4

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