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

Optimal. Leaf size=206 \[ \frac {\log \left (c^2 x^2+1\right )}{2 c}+\frac {1}{4} \text {Li}_2\left (\frac {2 i (i-c x)}{(1-c) (1-i x)}+1\right )-\frac {1}{4} \text {Li}_2\left (\frac {2 i (c x+i)}{(c+1) (1-i x)}+1\right )+x \cot ^{-1}(c x)-\frac {1}{2} i \tan ^{-1}(x) \log \left (1-\frac {i}{c x}\right )+\frac {1}{2} i \tan ^{-1}(x) \log \left (1+\frac {i}{c x}\right )+\frac {1}{2} i \tan ^{-1}(x) \log \left (-\frac {2 i (-c x+i)}{(1-c) (1-i x)}\right )-\frac {1}{2} i \tan ^{-1}(x) \log \left (-\frac {2 i (c x+i)}{(c+1) (1-i x)}\right ) \]

[Out]

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

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Rubi [A]  time = 0.58, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 28, number of rules used = 15, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {4917, 4847, 260, 4909, 203, 2470, 6688, 12, 4876, 4848, 2391, 4856, 2402, 2315, 2447} \[ \frac {1}{4} \text {PolyLog}\left (2,1+\frac {2 i (-c x+i)}{(1-c) (1-i x)}\right )-\frac {1}{4} \text {PolyLog}\left (2,1+\frac {2 i (c x+i)}{(c+1) (1-i x)}\right )+\frac {\log \left (c^2 x^2+1\right )}{2 c}+x \cot ^{-1}(c x)-\frac {1}{2} i \tan ^{-1}(x) \log \left (1-\frac {i}{c x}\right )+\frac {1}{2} i \tan ^{-1}(x) \log \left (1+\frac {i}{c x}\right )+\frac {1}{2} i \tan ^{-1}(x) \log \left (-\frac {2 i (-c x+i)}{(1-c) (1-i x)}\right )-\frac {1}{2} i \tan ^{-1}(x) \log \left (-\frac {2 i (c x+i)}{(c+1) (1-i x)}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(x^2*ArcCot[c*x])/(1 + x^2),x]

[Out]

x*ArcCot[c*x] - (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))] + Log[1 + c^
2*x^2]/(2*c) + 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 12

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

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 4847

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcCot[c*x])^p, x] + Dist[b*c*p, Int[
(x*(a + b*ArcCot[c*x])^(p - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

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 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 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 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 4917

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

Rule 6688

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

Rubi steps

\begin {align*} \int \frac {x^2 \cot ^{-1}(c x)}{1+x^2} \, dx &=\int \cot ^{-1}(c x) \, dx-\int \frac {\cot ^{-1}(c x)}{1+x^2} \, dx\\ &=x \cot ^{-1}(c 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 {x}{1+c^2 x^2} \, dx\\ &=x \cot ^{-1}(c 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 {\log \left (1+c^2 x^2\right )}{2 c}-\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}\\ &=x \cot ^{-1}(c 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 {\log \left (1+c^2 x^2\right )}{2 c}-\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}\\ &=x \cot ^{-1}(c 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 {\log \left (1+c^2 x^2\right )}{2 c}-\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\\ &=x \cot ^{-1}(c 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 {\log \left (1+c^2 x^2\right )}{2 c}-\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\\ &=x \cot ^{-1}(c 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 {\log \left (1+c^2 x^2\right )}{2 c}+\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\\ &=x \cot ^{-1}(c 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 (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 {\log \left (1+c^2 x^2\right )}{2 c}-\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\\ &=x \cot ^{-1}(c 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 (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 {\log \left (1+c^2 x^2\right )}{2 c}+\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*}

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Mathematica [B]  time = 1.77, size = 626, normalized size = 3.04 \[ \frac {\frac {1}{4} \sqrt {-c^2} \left (i \left (\text {Li}_2\left (\frac {\left (c^2+2 i \sqrt {-c^2}+1\right ) \left (c x+\sqrt {-c^2}\right )}{\left (c^2-1\right ) \left (\sqrt {-c^2}-c x\right )}\right )-\text {Li}_2\left (\frac {\left (c^2-2 i \sqrt {-c^2}+1\right ) \left (c x+\sqrt {-c^2}\right )}{\left (c^2-1\right ) \left (\sqrt {-c^2}-c x\right )}\right )\right )+2 \cos ^{-1}\left (\frac {c^2+1}{c^2-1}\right ) \tanh ^{-1}\left (\frac {\sqrt {-c^2}}{c x}\right )-4 \cot ^{-1}(c x) \tanh ^{-1}\left (\frac {c x}{\sqrt {-c^2}}\right )-\log \left (-\frac {2 \left (c^2+i \sqrt {-c^2}\right ) (c x-i)}{\left (c^2-1\right ) \left (\sqrt {-c^2}-c x\right )}\right ) \left (\cos ^{-1}\left (\frac {c^2+1}{c^2-1}\right )-2 i \tanh ^{-1}\left (\frac {\sqrt {-c^2}}{c x}\right )\right )-\log \left (\frac {2 i \left (\sqrt {-c^2}+i c^2\right ) (c x+i)}{\left (c^2-1\right ) \left (\sqrt {-c^2}-c x\right )}\right ) \left (\cos ^{-1}\left (\frac {c^2+1}{c^2-1}\right )+2 i \tanh ^{-1}\left (\frac {\sqrt {-c^2}}{c x}\right )\right )+\left (-2 i \tanh ^{-1}\left (\frac {\sqrt {-c^2}}{c x}\right )+2 i \tanh ^{-1}\left (\frac {c x}{\sqrt {-c^2}}\right )+\cos ^{-1}\left (\frac {c^2+1}{c^2-1}\right )\right ) \log \left (\frac {\sqrt {2} \sqrt {-c^2} e^{-i \cot ^{-1}(c x)}}{\sqrt {c^2-1} \sqrt {\left (c^2-1\right ) \cos \left (2 \cot ^{-1}(c x)\right )-c^2-1}}\right )+\left (2 i \tanh ^{-1}\left (\frac {\sqrt {-c^2}}{c x}\right )-2 i \tanh ^{-1}\left (\frac {c x}{\sqrt {-c^2}}\right )+\cos ^{-1}\left (\frac {c^2+1}{c^2-1}\right )\right ) \log \left (\frac {\sqrt {2} \sqrt {-c^2} e^{i \cot ^{-1}(c x)}}{\sqrt {c^2-1} \sqrt {\left (c^2-1\right ) \cos \left (2 \cot ^{-1}(c x)\right )-c^2-1}}\right )\right )-\log \left (\frac {1}{c x \sqrt {\frac {1}{c^2 x^2}+1}}\right )+c x \cot ^{-1}(c x)}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x*ArcCot[c*x] - Log[1/(c*Sqrt[1 + 1/(c^2*x^2)]*x)] + (Sqrt[-c^2]*(2*ArcCos[(1 + c^2)/(-1 + c^2)]*ArcTanh[Sq
rt[-c^2]/(c*x)] - 4*ArcCot[c*x]*ArcTanh[(c*x)/Sqrt[-c^2]] - (ArcCos[(1 + c^2)/(-1 + c^2)] - (2*I)*ArcTanh[Sqrt
[-c^2]/(c*x)])*Log[(-2*(c^2 + I*Sqrt[-c^2])*(-I + c*x))/((-1 + c^2)*(Sqrt[-c^2] - c*x))] - (ArcCos[(1 + c^2)/(
-1 + c^2)] + (2*I)*ArcTanh[Sqrt[-c^2]/(c*x)])*Log[((2*I)*(I*c^2 + Sqrt[-c^2])*(I + c*x))/((-1 + c^2)*(Sqrt[-c^
2] - c*x))] + (ArcCos[(1 + c^2)/(-1 + c^2)] - (2*I)*ArcTanh[Sqrt[-c^2]/(c*x)] + (2*I)*ArcTanh[(c*x)/Sqrt[-c^2]
])*Log[(Sqrt[2]*Sqrt[-c^2])/(Sqrt[-1 + c^2]*E^(I*ArcCot[c*x])*Sqrt[-1 - c^2 + (-1 + c^2)*Cos[2*ArcCot[c*x]]])]
+ (ArcCos[(1 + c^2)/(-1 + c^2)] + (2*I)*ArcTanh[Sqrt[-c^2]/(c*x)] - (2*I)*ArcTanh[(c*x)/Sqrt[-c^2]])*Log[(Sqr
t[2]*Sqrt[-c^2]*E^(I*ArcCot[c*x]))/(Sqrt[-1 + c^2]*Sqrt[-1 - c^2 + (-1 + c^2)*Cos[2*ArcCot[c*x]]])] + I*(-Poly
Log[2, ((1 + c^2 - (2*I)*Sqrt[-c^2])*(Sqrt[-c^2] + c*x))/((-1 + c^2)*(Sqrt[-c^2] - c*x))] + PolyLog[2, ((1 + c
^2 + (2*I)*Sqrt[-c^2])*(Sqrt[-c^2] + c*x))/((-1 + c^2)*(Sqrt[-c^2] - c*x))])))/4)/c

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2} \operatorname {arccot}\left (c x\right )}{x^{2} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \operatorname {arccot}\left (c x\right )}{x^{2} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.74, size = 265, normalized size = 1.29 \[ -\mathrm {arccot}\left (c x \right ) \arctan \relax (x )+x \,\mathrm {arccot}\left (c x \right )+\frac {\ln \left (c^{2} x^{2}+1\right )}{2 c}+\frac {i c \ln \left (1-\frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-1+c \right )}\right ) \arctan \relax (x )}{-2+2 c}-\frac {i \ln \left (1-\frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-1+c \right )}\right ) \arctan \relax (x )}{2 \left (-1+c \right )}+\frac {c \arctan \relax (x )^{2}}{-2+2 c}+\frac {c \polylog \left (2, \frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-1+c \right )}\right )}{-4+4 c}-\frac {\arctan \relax (x )^{2}}{2 \left (-1+c \right )}-\frac {\polylog \left (2, \frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (-1+c \right )}\right )}{4 \left (-1+c \right )}-\frac {i \arctan \relax (x ) \ln \left (1-\frac {\left (-1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (1+c \right )}\right )}{2}-\frac {\arctan \relax (x )^{2}}{2}-\frac {\polylog \left (2, \frac {\left (-1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (1+c \right )}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arccot(c*x)*arctan(x)+x*arccot(c*x)+1/2*ln(c^2*x^2+1)/c+1/2*I*c/(-1+c)*ln(1-(1+c)*(1+I*x)^2/(x^2+1)/(-1+c))*a
rctan(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)*pol
ylog(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 = 0.45, size = 189, normalized size = 0.92 \[ {\left (x - \arctan \relax (x)\right )} \operatorname {arccot}\left (c x\right ) - \frac {4 \, c \arctan \left (c x\right ) \arctan \relax (x) - 4 \, c \arctan \relax (x) \arctan \left (\frac {c x}{c - 1}, -\frac {1}{c - 1}\right ) + c \log \left (x^{2} + 1\right ) \log \left (\frac {c^{2} x^{2} + 1}{c^{2} + 2 \, c + 1}\right ) - c \log \left (x^{2} + 1\right ) \log \left (\frac {c^{2} x^{2} + 1}{c^{2} - 2 \, c + 1}\right ) + 2 \, c {\rm Li}_2\left (\frac {i \, c x + c}{c + 1}\right ) + 2 \, c {\rm Li}_2\left (-\frac {i \, c x - c}{c + 1}\right ) - 2 \, c {\rm Li}_2\left (\frac {i \, c x + c}{c - 1}\right ) - 2 \, c {\rm Li}_2\left (-\frac {i \, c x - c}{c - 1}\right ) - 4 \, \log \left (c^{2} x^{2} + 1\right )}{8 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2\,\mathrm {acot}\left (c\,x\right )}{x^2+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*acot(c*x))/(x^2 + 1),x)

[Out]

int((x^2*acot(c*x))/(x^2 + 1), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \operatorname {acot}{\left (c x \right )}}{x^{2} + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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