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

Optimal. Leaf size=212 \[ \frac{1}{4} \text{PolyLog}\left (2,1+\frac{2 i (-c x+i)}{(1-c) (1-i x)}\right )-\frac{1}{4} \text{PolyLog}\left (2,1+\frac{2 i (c x+i)}{(c+1) (1-i x)}\right )+\frac{1}{2} c \log \left (c^2 x^2+1\right )-c \log (x)-\frac{\cot ^{-1}(c x)}{x}-\frac{1}{2} i \tan ^{-1}(x) \log \left (1-\frac{i}{c x}\right )+\frac{1}{2} i \tan ^{-1}(x) \log \left (1+\frac{i}{c x}\right )+\frac{1}{2} i \tan ^{-1}(x) \log \left (-\frac{2 i (-c x+i)}{(1-c) (1-i x)}\right )-\frac{1}{2} i \tan ^{-1}(x) \log \left (-\frac{2 i (c x+i)}{(c+1) (1-i x)}\right ) \]

[Out]

-(ArcCot[c*x]/x) - (I/2)*ArcTan[x]*Log[1 - I/(c*x)] + (I/2)*ArcTan[x]*Log[1 + I/(c*x)] - c*Log[x] + (I/2)*ArcT
an[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))
] + (c*Log[1 + c^2*x^2])/2 + 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 = 0.503106, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 31, number of rules used = 19, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.267, Rules used = {4919, 4853, 266, 36, 29, 31, 4909, 203, 2470, 260, 6688, 12, 4876, 4848, 2391, 4856, 2402, 2315, 2447} \[ \frac{1}{4} \text{PolyLog}\left (2,1+\frac{2 i (-c x+i)}{(1-c) (1-i x)}\right )-\frac{1}{4} \text{PolyLog}\left (2,1+\frac{2 i (c x+i)}{(c+1) (1-i x)}\right )+\frac{1}{2} c \log \left (c^2 x^2+1\right )-c \log (x)-\frac{\cot ^{-1}(c x)}{x}-\frac{1}{2} i \tan ^{-1}(x) \log \left (1-\frac{i}{c x}\right )+\frac{1}{2} i \tan ^{-1}(x) \log \left (1+\frac{i}{c x}\right )+\frac{1}{2} i \tan ^{-1}(x) \log \left (-\frac{2 i (-c x+i)}{(1-c) (1-i x)}\right )-\frac{1}{2} i \tan ^{-1}(x) \log \left (-\frac{2 i (c x+i)}{(c+1) (1-i x)}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcCot[c*x]/(x^2*(1 + x^2)),x]

[Out]

-(ArcCot[c*x]/x) - (I/2)*ArcTan[x]*Log[1 - I/(c*x)] + (I/2)*ArcTan[x]*Log[1 + I/(c*x)] - c*Log[x] + (I/2)*ArcT
an[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))
] + (c*Log[1 + c^2*x^2])/2 + 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 4919

Int[(((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d,
Int[(f*x)^m*(a + b*ArcCot[c*x])^p, x], x] - Dist[e/(d*f^2), Int[((f*x)^(m + 2)*(a + b*ArcCot[c*x])^p)/(d + e*
x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

Rule 4853

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcCo
t[c*x])^p)/(d*(m + 1)), x] + Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCot[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 36

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

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)}{x^2 \left (1+x^2\right )} \, dx &=\int \frac{\cot ^{-1}(c x)}{x^2} \, dx-\int \frac{\cot ^{-1}(c x)}{1+x^2} \, dx\\ &=-\frac{\cot ^{-1}(c x)}{x}-\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-c \int \frac{1}{x \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{\cot ^{-1}(c x)}{x}-\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} c \operatorname{Subst}\left (\int \frac{1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{\cot ^{-1}(c x)}{x}-\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} c \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{2} c^3 \operatorname{Subst}\left (\int \frac{1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac{\cot ^{-1}(c x)}{x}-\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 )-c \log (x)+\frac{1}{2} c \log \left (1+c^2 x^2\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{\cot ^{-1}(c x)}{x}-\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 )-c \log (x)+\frac{1}{2} c \log \left (1+c^2 x^2\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{\cot ^{-1}(c x)}{x}-\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 )-c \log (x)+\frac{1}{2} c \log \left (1+c^2 x^2\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{\cot ^{-1}(c x)}{x}-\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 )-c \log (x)+\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} c \log \left (1+c^2 x^2\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{\cot ^{-1}(c x)}{x}-\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 )-c \log (x)+\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} c \log \left (1+c^2 x^2\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 = 0.0847863, size = 348, normalized size = 1.64 \[ \frac{1}{4} \text{PolyLog}\left (2,\frac{i c (-x+i)}{1-c}\right )-\frac{1}{4} \text{PolyLog}\left (2,-\frac{i c (-x+i)}{c+1}\right )+\frac{1}{4} \text{PolyLog}\left (2,\frac{i c (x+i)}{1-c}\right )-\frac{1}{4} \text{PolyLog}\left (2,-\frac{i c (x+i)}{c+1}\right )+\frac{1}{2} c \log \left (c^2 x^2+1\right )-c \log (x)+\frac{1}{4} \log (-x+i) \log \left (-\frac{i (-c x+i)}{1-c}\right )-\frac{1}{4} \log (x+i) \log \left (-\frac{i (-c x+i)}{c+1}\right )-\frac{1}{4} \log (-x+i) \log \left (-\frac{-c x+i}{c x}\right )+\frac{1}{4} \log (x+i) \log \left (-\frac{-c x+i}{c x}\right )+\frac{1}{4} \log (x+i) \log \left (-\frac{i (c x+i)}{1-c}\right )-\frac{1}{4} \log (-x+i) \log \left (-\frac{i (c x+i)}{c+1}\right )+\frac{1}{4} \log (-x+i) \log \left (\frac{c x+i}{c x}\right )-\frac{1}{4} \log (x+i) \log \left (\frac{c x+i}{c x}\right )-\frac{\cot ^{-1}(c x)}{x} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCot[c*x]/(x^2*(1 + x^2)),x]

[Out]

-(ArcCot[c*x]/x) - c*Log[x] + (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]*Lo
g[((-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 + (c*Log[1 + c^2*x^2])/2 + PolyLog[2, (I*c*(I - x))/(1 - c)]/4 - P
olyLog[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 = 0.207, size = 271, normalized size = 1.3 \begin{align*} -\arctan \left ( x \right ){\rm arccot} \left (cx\right )-{\frac{{\rm arccot} \left (cx\right )}{x}}+{\frac{c\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2}}-c\ln \left ( x \right ) +{\frac{{\frac{i}{2}}c\arctan \left ( x \right ) }{-1+c}\ln \left ( 1-{\frac{ \left ( 1+c \right ) \left ( 1+ix \right ) ^{2}}{ \left ({x}^{2}+1 \right ) \left ( -1+c \right ) }} \right ) }-{\frac{{\frac{i}{2}}\arctan \left ( x \right ) }{-1+c}\ln \left ( 1-{\frac{ \left ( 1+c \right ) \left ( 1+ix \right ) ^{2}}{ \left ({x}^{2}+1 \right ) \left ( -1+c \right ) }} \right ) }+{\frac{c \left ( \arctan \left ( x \right ) \right ) ^{2}}{-2+2\,c}}+{\frac{c}{-4+4\,c}{\it polylog} \left ( 2,{\frac{ \left ( 1+c \right ) \left ( 1+ix \right ) ^{2}}{ \left ({x}^{2}+1 \right ) \left ( -1+c \right ) }} \right ) }-{\frac{ \left ( \arctan \left ( x \right ) \right ) ^{2}}{-2+2\,c}}-{\frac{1}{-4+4\,c}{\it polylog} \left ( 2,{\frac{ \left ( 1+c \right ) \left ( 1+ix \right ) ^{2}}{ \left ({x}^{2}+1 \right ) \left ( -1+c \right ) }} \right ) }-{\frac{i}{2}}\arctan \left ( x \right ) \ln \left ( 1-{\frac{ \left ( -1+c \right ) \left ( 1+ix \right ) ^{2}}{ \left ({x}^{2}+1 \right ) \left ( 1+c \right ) }} \right ) -{\frac{ \left ( \arctan \left ( x \right ) \right ) ^{2}}{2}}-{\frac{1}{4}{\it polylog} \left ( 2,{\frac{ \left ( -1+c \right ) \left ( 1+ix \right ) ^{2}}{ \left ({x}^{2}+1 \right ) \left ( 1+c \right ) }} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.54924, size = 261, normalized size = 1.23 \begin{align*} -{\left (\frac{1}{x} + \arctan \left (x\right )\right )} \operatorname{arccot}\left (c x\right ) - \frac{1}{2} \, \arctan \left (x\right ) \arctan \left (\frac{c x}{c + 1}, \frac{1}{c + 1}\right ) + \frac{1}{2} \, \arctan \left (x\right ) \arctan \left (\frac{c x}{c - 1}, -\frac{1}{c - 1}\right ) + \frac{1}{2} \, c \log \left (c^{2} x^{2} + 1\right ) - c \log \left (x\right ) - \frac{1}{8} \, \log \left (x^{2} + 1\right ) \log \left (\frac{c^{2} x^{2} + 1}{c^{2} + 2 \, c + 1}\right ) + \frac{1}{8} \, \log \left (x^{2} + 1\right ) \log \left (\frac{c^{2} x^{2} + 1}{c^{2} - 2 \, c + 1}\right ) - \frac{1}{4} \,{\rm Li}_2\left (\frac{i \, c x + c}{c + 1}\right ) - \frac{1}{4} \,{\rm Li}_2\left (-\frac{i \, c x - c}{c + 1}\right ) + \frac{1}{4} \,{\rm Li}_2\left (\frac{i \, c x + c}{c - 1}\right ) + \frac{1}{4} \,{\rm Li}_2\left (-\frac{i \, c x - c}{c - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{4} + x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccot(c*x)/(x^4 + x^2), 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} \left (x^{2} + 1\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acot(c*x)/(x**2*(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 )}{{\left (x^{2} + 1\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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