3.48 \(\int \frac{\cot ^{-1}(c x)}{1+x^2} \, dx\)

Optimal. Leaf size=183 \[ -\frac{1}{4} \text{PolyLog}\left (2,1+\frac{2 i (-c x+i)}{(1-c) (1-i x)}\right )+\frac{1}{4} \text{PolyLog}\left (2,1+\frac{2 i (c x+i)}{(c+1) (1-i x)}\right )+\frac{1}{2} i \tan ^{-1}(x) \log \left (1-\frac{i}{c x}\right )-\frac{1}{2} i \tan ^{-1}(x) \log \left (1+\frac{i}{c x}\right )-\frac{1}{2} i \tan ^{-1}(x) \log \left (-\frac{2 i (-c x+i)}{(1-c) (1-i x)}\right )+\frac{1}{2} i \tan ^{-1}(x) \log \left (-\frac{2 i (c x+i)}{(c+1) (1-i x)}\right ) \]

[Out]

(I/2)*ArcTan[x]*Log[1 - I/(c*x)] - (I/2)*ArcTan[x]*Log[1 + I/(c*x)] - (I/2)*ArcTan[x]*Log[((-2*I)*(I - c*x))/(
(1 - c)*(1 - I*x))] + (I/2)*ArcTan[x]*Log[((-2*I)*(I + c*x))/((1 + c)*(1 - I*x))] - PolyLog[2, 1 + ((2*I)*(I -
c*x))/((1 - c)*(1 - I*x))]/4 + PolyLog[2, 1 + ((2*I)*(I + c*x))/((1 + c)*(1 - I*x))]/4

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Rubi [A]  time = 0.464618, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 25, number of rules used = 13, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.083, Rules used = {4909, 203, 2470, 260, 6688, 12, 4876, 4848, 2391, 4856, 2402, 2315, 2447} \[ -\frac{1}{4} \text{PolyLog}\left (2,1+\frac{2 i (-c x+i)}{(1-c) (1-i x)}\right )+\frac{1}{4} \text{PolyLog}\left (2,1+\frac{2 i (c x+i)}{(c+1) (1-i x)}\right )+\frac{1}{2} i \tan ^{-1}(x) \log \left (1-\frac{i}{c x}\right )-\frac{1}{2} i \tan ^{-1}(x) \log \left (1+\frac{i}{c x}\right )-\frac{1}{2} i \tan ^{-1}(x) \log \left (-\frac{2 i (-c x+i)}{(1-c) (1-i x)}\right )+\frac{1}{2} i \tan ^{-1}(x) \log \left (-\frac{2 i (c x+i)}{(c+1) (1-i x)}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcCot[c*x]/(1 + x^2),x]

[Out]

(I/2)*ArcTan[x]*Log[1 - I/(c*x)] - (I/2)*ArcTan[x]*Log[1 + I/(c*x)] - (I/2)*ArcTan[x]*Log[((-2*I)*(I - c*x))/(
(1 - c)*(1 - I*x))] + (I/2)*ArcTan[x]*Log[((-2*I)*(I + c*x))/((1 + c)*(1 - I*x))] - PolyLog[2, 1 + ((2*I)*(I -
c*x))/((1 - c)*(1 - I*x))]/4 + PolyLog[2, 1 + ((2*I)*(I + c*x))/((1 + c)*(1 - I*x))]/4

Rule 4909

Int[ArcCot[(c_.)*(x_)]/((d_.) + (e_.)*(x_)^2), x_Symbol] :> Dist[I/2, Int[Log[1 - I/(c*x)]/(d + e*x^2), x], x]
- Dist[I/2, Int[Log[1 + I/(c*x)]/(d + e*x^2), x], x] /; FreeQ[{c, d, e}, x]

Rule 203

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

Rule 2470

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))/((f_) + (g_.)*(x_)^2), x_Symbol] :> With[{u = In
tHide[1/(f + g*x^2), x]}, Simp[u*(a + b*Log[c*(d + e*x^n)^p]), x] - Dist[b*e*n*p, Int[(u*x^(n - 1))/(d + e*x^n
), x], x]] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && IntegerQ[n]

Rule 260

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

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 4876

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[Ex
pandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p,
0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rule 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4856

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])*Log[2/(1 -
I*c*x)])/e, x] + (Dist[(b*c)/e, Int[Log[2/(1 - I*c*x)]/(1 + c^2*x^2), x], x] - Dist[(b*c)/e, Int[Log[(2*c*(d
+ e*x))/((c*d + I*e)*(1 - I*c*x))]/(1 + c^2*x^2), x], x] + Simp[((a + b*ArcTan[c*x])*Log[(2*c*(d + e*x))/((c*d
+ I*e)*(1 - I*c*x))])/e, x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 + e^2, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rubi steps

\begin{align*} \int \frac{\cot ^{-1}(c x)}{1+x^2} \, dx &=\frac{1}{2} i \int \frac{\log \left (1-\frac{i}{c x}\right )}{1+x^2} \, dx-\frac{1}{2} i \int \frac{\log \left (1+\frac{i}{c x}\right )}{1+x^2} \, dx\\ &=\frac{1}{2} i \tan ^{-1}(x) \log \left (1-\frac{i}{c x}\right )-\frac{1}{2} i \tan ^{-1}(x) \log \left (1+\frac{i}{c x}\right )+\frac{\int \frac{\tan ^{-1}(x)}{\left (1-\frac{i}{c x}\right ) x^2} \, dx}{2 c}+\frac{\int \frac{\tan ^{-1}(x)}{\left (1+\frac{i}{c x}\right ) x^2} \, dx}{2 c}\\ &=\frac{1}{2} i \tan ^{-1}(x) \log \left (1-\frac{i}{c x}\right )-\frac{1}{2} i \tan ^{-1}(x) \log \left (1+\frac{i}{c x}\right )+\frac{\int \frac{c \tan ^{-1}(x)}{x (-i+c x)} \, dx}{2 c}+\frac{\int \frac{c \tan ^{-1}(x)}{x (i+c x)} \, dx}{2 c}\\ &=\frac{1}{2} i \tan ^{-1}(x) \log \left (1-\frac{i}{c x}\right )-\frac{1}{2} i \tan ^{-1}(x) \log \left (1+\frac{i}{c x}\right )+\frac{1}{2} \int \frac{\tan ^{-1}(x)}{x (-i+c x)} \, dx+\frac{1}{2} \int \frac{\tan ^{-1}(x)}{x (i+c x)} \, dx\\ &=\frac{1}{2} i \tan ^{-1}(x) \log \left (1-\frac{i}{c x}\right )-\frac{1}{2} i \tan ^{-1}(x) \log \left (1+\frac{i}{c x}\right )+\frac{1}{2} \int \left (\frac{i \tan ^{-1}(x)}{x}-\frac{i c \tan ^{-1}(x)}{-i+c x}\right ) \, dx+\frac{1}{2} \int \left (-\frac{i \tan ^{-1}(x)}{x}+\frac{i c \tan ^{-1}(x)}{i+c x}\right ) \, dx\\ &=\frac{1}{2} i \tan ^{-1}(x) \log \left (1-\frac{i}{c x}\right )-\frac{1}{2} i \tan ^{-1}(x) \log \left (1+\frac{i}{c x}\right )-\frac{1}{2} (i c) \int \frac{\tan ^{-1}(x)}{-i+c x} \, dx+\frac{1}{2} (i c) \int \frac{\tan ^{-1}(x)}{i+c x} \, dx\\ &=\frac{1}{2} i \tan ^{-1}(x) \log \left (1-\frac{i}{c x}\right )-\frac{1}{2} i \tan ^{-1}(x) \log \left (1+\frac{i}{c x}\right )-\frac{1}{2} i \tan ^{-1}(x) \log \left (-\frac{2 i (i-c x)}{(1-c) (1-i x)}\right )+\frac{1}{2} i \tan ^{-1}(x) \log \left (-\frac{2 i (i+c x)}{(1+c) (1-i x)}\right )+\frac{1}{2} i \int \frac{\log \left (\frac{2 (-i+c x)}{(-i+i c) (1-i x)}\right )}{1+x^2} \, dx-\frac{1}{2} i \int \frac{\log \left (\frac{2 (i+c x)}{(i+i c) (1-i x)}\right )}{1+x^2} \, dx\\ &=\frac{1}{2} i \tan ^{-1}(x) \log \left (1-\frac{i}{c x}\right )-\frac{1}{2} i \tan ^{-1}(x) \log \left (1+\frac{i}{c x}\right )-\frac{1}{2} i \tan ^{-1}(x) \log \left (-\frac{2 i (i-c x)}{(1-c) (1-i x)}\right )+\frac{1}{2} i \tan ^{-1}(x) \log \left (-\frac{2 i (i+c x)}{(1+c) (1-i x)}\right )-\frac{1}{4} \text{Li}_2\left (1+\frac{2 i (i-c x)}{(1-c) (1-i x)}\right )+\frac{1}{4} \text{Li}_2\left (1+\frac{2 i (i+c x)}{(1+c) (1-i x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.0782909, size = 319, normalized size = 1.74 \[ -\frac{1}{4} \text{PolyLog}\left (2,\frac{i c (-x+i)}{1-c}\right )+\frac{1}{4} \text{PolyLog}\left (2,-\frac{i c (-x+i)}{c+1}\right )-\frac{1}{4} \text{PolyLog}\left (2,\frac{i c (x+i)}{1-c}\right )+\frac{1}{4} \text{PolyLog}\left (2,-\frac{i c (x+i)}{c+1}\right )-\frac{1}{4} \log (-x+i) \log \left (-\frac{i (-c x+i)}{1-c}\right )+\frac{1}{4} \log (x+i) \log \left (-\frac{i (-c x+i)}{c+1}\right )+\frac{1}{4} \log (-x+i) \log \left (-\frac{-c x+i}{c x}\right )-\frac{1}{4} \log (x+i) \log \left (-\frac{-c x+i}{c x}\right )-\frac{1}{4} \log (x+i) \log \left (-\frac{i (c x+i)}{1-c}\right )+\frac{1}{4} \log (-x+i) \log \left (-\frac{i (c x+i)}{c+1}\right )-\frac{1}{4} \log (-x+i) \log \left (\frac{c x+i}{c x}\right )+\frac{1}{4} \log (x+i) \log \left (\frac{c x+i}{c x}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCot[c*x]/(1 + x^2),x]

[Out]

-(Log[I - x]*Log[((-I)*(I - c*x))/(1 - c)])/4 + (Log[I + x]*Log[((-I)*(I - c*x))/(1 + c)])/4 + (Log[I - x]*Log
[-((I - c*x)/(c*x))])/4 - (Log[I + x]*Log[-((I - c*x)/(c*x))])/4 - (Log[I + x]*Log[((-I)*(I + c*x))/(1 - c)])/
4 + (Log[I - x]*Log[((-I)*(I + c*x))/(1 + c)])/4 - (Log[I - x]*Log[(I + c*x)/(c*x)])/4 + (Log[I + x]*Log[(I +
c*x)/(c*x)])/4 - PolyLog[2, (I*c*(I - x))/(1 - c)]/4 + PolyLog[2, ((-I)*c*(I - x))/(1 + c)]/4 - PolyLog[2, (I*
c*(I + x))/(1 - c)]/4 + PolyLog[2, ((-I)*c*(I + x))/(1 + c)]/4

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Maple [A]  time = 0.184, size = 304, normalized size = 1.7 \begin{align*} \arctan \left ( x \right ){\rm arccot} \left (cx\right )+\arctan \left ( cx \right ) \arctan \left ( x \right ) +{\frac{i}{2}}\arctan \left ( cx \right ) \ln \left ( 1-{\frac{ \left ( 1+c \right ) \left ( 1+icx \right ) ^{2}}{ \left ({c}^{2}{x}^{2}+1 \right ) \left ( 1-c \right ) }} \right ) +{\frac{ \left ( \arctan \left ( cx \right ) \right ) ^{2}}{2}}+{\frac{1}{4}{\it polylog} \left ( 2,{\frac{ \left ( 1+c \right ) \left ( 1+icx \right ) ^{2}}{ \left ({c}^{2}{x}^{2}+1 \right ) \left ( 1-c \right ) }} \right ) }-{\frac{{\frac{i}{2}}c\arctan \left ( cx \right ) }{1+c}\ln \left ( 1-{\frac{ \left ( -1+c \right ) \left ( 1+icx \right ) ^{2}}{ \left ({c}^{2}{x}^{2}+1 \right ) \left ( -c-1 \right ) }} \right ) }-{\frac{{\frac{i}{2}}\arctan \left ( cx \right ) }{1+c}\ln \left ( 1-{\frac{ \left ( -1+c \right ) \left ( 1+icx \right ) ^{2}}{ \left ({c}^{2}{x}^{2}+1 \right ) \left ( -c-1 \right ) }} \right ) }-{\frac{c \left ( \arctan \left ( cx \right ) \right ) ^{2}}{2\,c+2}}-{\frac{c}{4+4\,c}{\it polylog} \left ( 2,{\frac{ \left ( -1+c \right ) \left ( 1+icx \right ) ^{2}}{ \left ({c}^{2}{x}^{2}+1 \right ) \left ( -c-1 \right ) }} \right ) }-{\frac{ \left ( \arctan \left ( cx \right ) \right ) ^{2}}{2\,c+2}}-{\frac{1}{4+4\,c}{\it polylog} \left ( 2,{\frac{ \left ( -1+c \right ) \left ( 1+icx \right ) ^{2}}{ \left ({c}^{2}{x}^{2}+1 \right ) \left ( -c-1 \right ) }} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccot(c*x)/(x^2+1),x)

[Out]

arctan(x)*arccot(c*x)+arctan(c*x)*arctan(x)+1/2*I*arctan(c*x)*ln(1-(1+c)*(1+I*c*x)^2/(c^2*x^2+1)/(1-c))+1/2*ar
ctan(c*x)^2+1/4*polylog(2,(1+c)*(1+I*c*x)^2/(c^2*x^2+1)/(1-c))-1/2*I*c/(1+c)*arctan(c*x)*ln(1-(-1+c)*(1+I*c*x)
^2/(c^2*x^2+1)/(-c-1))-1/2*I/(1+c)*ln(1-(-1+c)*(1+I*c*x)^2/(c^2*x^2+1)/(-c-1))*arctan(c*x)-1/2*c/(1+c)*arctan(
c*x)^2-1/4*c/(1+c)*polylog(2,(-1+c)*(1+I*c*x)^2/(c^2*x^2+1)/(-c-1))-1/2/(1+c)*arctan(c*x)^2-1/4/(1+c)*polylog(
2,(-1+c)*(1+I*c*x)^2/(c^2*x^2+1)/(-c-1))

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Maxima [A]  time = 1.59556, size = 266, normalized size = 1.45 \begin{align*} -\frac{1}{8} \, c{\left (\frac{8 \, \arctan \left (c x\right ) \arctan \left (x\right )}{c} - \frac{4 \, \arctan \left (x\right ) \arctan \left (\frac{c x}{c + 1}, \frac{1}{c + 1}\right ) - 4 \, \arctan \left (x\right ) \arctan \left (\frac{c x}{c - 1}, -\frac{1}{c - 1}\right ) + \log \left (x^{2} + 1\right ) \log \left (\frac{c^{2} x^{2} + 1}{c^{2} + 2 \, c + 1}\right ) - \log \left (x^{2} + 1\right ) \log \left (\frac{c^{2} x^{2} + 1}{c^{2} - 2 \, c + 1}\right ) + 2 \,{\rm Li}_2\left (\frac{i \, c x + c}{c + 1}\right ) + 2 \,{\rm Li}_2\left (-\frac{i \, c x - c}{c + 1}\right ) - 2 \,{\rm Li}_2\left (\frac{i \, c x + c}{c - 1}\right ) - 2 \,{\rm Li}_2\left (-\frac{i \, c x - c}{c - 1}\right )}{c}\right )} + \operatorname{arccot}\left (c x\right ) \arctan \left (x\right ) + \arctan \left (c x\right ) \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(c*x)/(x^2+1),x, algorithm="maxima")

[Out]

-1/8*c*(8*arctan(c*x)*arctan(x)/c - (4*arctan(x)*arctan2(c*x/(c + 1), 1/(c + 1)) - 4*arctan(x)*arctan2(c*x/(c
- 1), -1/(c - 1)) + log(x^2 + 1)*log((c^2*x^2 + 1)/(c^2 + 2*c + 1)) - log(x^2 + 1)*log((c^2*x^2 + 1)/(c^2 - 2*
c + 1)) + 2*dilog((I*c*x + c)/(c + 1)) + 2*dilog(-(I*c*x - c)/(c + 1)) - 2*dilog((I*c*x + c)/(c - 1)) - 2*dilo
g(-(I*c*x - c)/(c - 1)))/c) + arccot(c*x)*arctan(x) + arctan(c*x)*arctan(x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{arccot}\left (c x\right )}{x^{2} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(c*x)/(x^2+1),x, algorithm="fricas")

[Out]

integral(arccot(c*x)/(x^2 + 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acot}{\left (c x \right )}}{x^{2} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acot(c*x)/(x**2+1),x)

[Out]

Integral(acot(c*x)/(x**2 + 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arccot}\left (c x\right )}{x^{2} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(c*x)/(x^2+1),x, algorithm="giac")

[Out]

integrate(arccot(c*x)/(x^2 + 1), x)