3.1.46 \(\int \frac {x^2 \cot ^{-1}(c x)}{1+x^2} \, dx\) [46]

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

[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*(c*x+I)/(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*(c*x+I)/(1+c)/(1-I*x))

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Rubi [A]
time = 0.44, 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 = {5037, 4931, 266, 5029, 209, 2520, 6820, 12, 4996, 4940, 2438, 4966, 2449, 2352, 2497} \begin {gather*} -\frac {1}{2} i \text {ArcTan}(x) \log \left (1-\frac {i}{c x}\right )+\frac {1}{2} i \text {ArcTan}(x) \log \left (1+\frac {i}{c x}\right )+\frac {1}{2} i \text {ArcTan}(x) \log \left (-\frac {2 i (-c x+i)}{(1-c) (1-i x)}\right )-\frac {1}{2} i \text {ArcTan}(x) \log \left (-\frac {2 i (c x+i)}{(c+1) (1-i x)}\right )+\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) \end {gather*}

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 209

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

Rule 266

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 2352

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

Rule 2438

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 2449

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 2497

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 2520

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 4931

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

Rule 4940

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 4966

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 4996

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 5029

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 5037

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 6820

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] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(626\) vs. \(2(206)=412\).
time = 1.12, size = 626, normalized size = 3.04 \begin {gather*} \frac {c x \cot ^{-1}(c x)-\log \left (\frac {1}{c \sqrt {1+\frac {1}{c^2 x^2}} x}\right )+\frac {1}{4} \sqrt {-c^2} \left (2 \text {ArcCos}\left (\frac {1+c^2}{-1+c^2}\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 )-\left (\text {ArcCos}\left (\frac {1+c^2}{-1+c^2}\right )-2 i \tanh ^{-1}\left (\frac {\sqrt {-c^2}}{c x}\right )\right ) \log \left (-\frac {2 \left (c^2+i \sqrt {-c^2}\right ) (-i+c x)}{\left (-1+c^2\right ) \left (\sqrt {-c^2}-c x\right )}\right )-\left (\text {ArcCos}\left (\frac {1+c^2}{-1+c^2}\right )+2 i \tanh ^{-1}\left (\frac {\sqrt {-c^2}}{c x}\right )\right ) \log \left (\frac {2 i \left (i c^2+\sqrt {-c^2}\right ) (i+c x)}{\left (-1+c^2\right ) \left (\sqrt {-c^2}-c x\right )}\right )+\left (\text {ArcCos}\left (\frac {1+c^2}{-1+c^2}\right )-2 i \tanh ^{-1}\left (\frac {\sqrt {-c^2}}{c x}\right )+2 i \tanh ^{-1}\left (\frac {c x}{\sqrt {-c^2}}\right )\right ) \log \left (\frac {\sqrt {2} \sqrt {-c^2} e^{-i \cot ^{-1}(c x)}}{\sqrt {-1+c^2} \sqrt {-1-c^2+\left (-1+c^2\right ) \cos \left (2 \cot ^{-1}(c x)\right )}}\right )+\left (\text {ArcCos}\left (\frac {1+c^2}{-1+c^2}\right )+2 i \tanh ^{-1}\left (\frac {\sqrt {-c^2}}{c x}\right )-2 i \tanh ^{-1}\left (\frac {c x}{\sqrt {-c^2}}\right )\right ) \log \left (\frac {\sqrt {2} \sqrt {-c^2} e^{i \cot ^{-1}(c x)}}{\sqrt {-1+c^2} \sqrt {-1-c^2+\left (-1+c^2\right ) \cos \left (2 \cot ^{-1}(c x)\right )}}\right )+i \left (-\text {PolyLog}\left (2,\frac {\left (1+c^2-2 i \sqrt {-c^2}\right ) \left (\sqrt {-c^2}+c x\right )}{\left (-1+c^2\right ) \left (\sqrt {-c^2}-c x\right )}\right )+\text {PolyLog}\left (2,\frac {\left (1+c^2+2 i \sqrt {-c^2}\right ) \left (\sqrt {-c^2}+c x\right )}{\left (-1+c^2\right ) \left (\sqrt {-c^2}-c x\right )}\right )\right )\right )}{c} \end {gather*}

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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Maple [A]
time = 0.23, size = 280, normalized size = 1.36

method	result	size
risch	\(\frac {\pi x}{2}+\frac {i \pi }{2 c}-\frac {i \ln \left (-i c x +1\right ) x}{2}+\frac {i \ln \left (i c x +1\right ) x}{2}-\frac {\pi \arctan \left (x \right )}{2}+\frac {\ln \left (c^{2} x^{2}+1\right )}{2 c}-\frac {1}{c}+\frac {\ln \left (\frac {-i c x -c}{-c -1}\right ) \ln \left (-i c x +1\right )}{4}+\frac {\dilog \left (\frac {-i c x -c}{-c -1}\right )}{4}-\frac {\ln \left (\frac {-i c x +c}{c -1}\right ) \ln \left (-i c x +1\right )}{4}-\frac {\dilog \left (\frac {-i c x +c}{c -1}\right )}{4}+\frac {\ln \left (\frac {i c x -c}{-c -1}\right ) \ln \left (i c x +1\right )}{4}+\frac {\dilog \left (\frac {i c x -c}{-c -1}\right )}{4}-\frac {\ln \left (\frac {i c x +c}{c -1}\right ) \ln \left (i c x +1\right )}{4}-\frac {\dilog \left (\frac {i c x +c}{c -1}\right )}{4}\)	\(238\)
derivativedivides	\(\frac {-\mathrm {arccot}\left (c x \right ) \arctan \left (x \right ) c^{3}+\mathrm {arccot}\left (c x \right ) c^{3} x +c^{3} \left (\frac {\ln \left (c^{2} x^{2}+1\right )}{2 c}-\frac {i \arctan \left (x \right ) \ln \left (1-\frac {\left (c -1\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (1+c \right )}\right )}{2}-\frac {\arctan \left (x \right )^{2}}{2}-\frac {\polylog \left (2, \frac {\left (c -1\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (1+c \right )}\right )}{4}+\frac {i c \ln \left (1-\frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (c -1\right )}\right ) \arctan \left (x \right )}{2 c -2}-\frac {i \ln \left (1-\frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (c -1\right )}\right ) \arctan \left (x \right )}{2 \left (c -1\right )}+\frac {c \arctan \left (x \right )^{2}}{2 c -2}+\frac {c \polylog \left (2, \frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (c -1\right )}\right )}{4 c -4}-\frac {\arctan \left (x \right )^{2}}{2 \left (c -1\right )}-\frac {\polylog \left (2, \frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (c -1\right )}\right )}{4 \left (c -1\right )}\right )}{c^{3}}\)	\(280\)
default	\(\frac {-\mathrm {arccot}\left (c x \right ) \arctan \left (x \right ) c^{3}+\mathrm {arccot}\left (c x \right ) c^{3} x +c^{3} \left (\frac {\ln \left (c^{2} x^{2}+1\right )}{2 c}-\frac {i \arctan \left (x \right ) \ln \left (1-\frac {\left (c -1\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (1+c \right )}\right )}{2}-\frac {\arctan \left (x \right )^{2}}{2}-\frac {\polylog \left (2, \frac {\left (c -1\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (1+c \right )}\right )}{4}+\frac {i c \ln \left (1-\frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (c -1\right )}\right ) \arctan \left (x \right )}{2 c -2}-\frac {i \ln \left (1-\frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (c -1\right )}\right ) \arctan \left (x \right )}{2 \left (c -1\right )}+\frac {c \arctan \left (x \right )^{2}}{2 c -2}+\frac {c \polylog \left (2, \frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (c -1\right )}\right )}{4 c -4}-\frac {\arctan \left (x \right )^{2}}{2 \left (c -1\right )}-\frac {\polylog \left (2, \frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (c -1\right )}\right )}{4 \left (c -1\right )}\right )}{c^{3}}\)	\(280\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^3*(-arccot(c*x)*arctan(x)*c^3+arccot(c*x)*c^3*x+c^3*(1/2*ln(c^2*x^2+1)/c-1/2*I*arctan(x)*ln(1-(c-1)*(1+I*x
)^2/(x^2+1)/(1+c))-1/2*arctan(x)^2-1/4*polylog(2,(c-1)*(1+I*x)^2/(x^2+1)/(1+c))+1/2*I*c/(c-1)*ln(1-(1+c)*(1+I*
x)^2/(x^2+1)/(c-1))*arctan(x)-1/2*I/(c-1)*ln(1-(1+c)*(1+I*x)^2/(x^2+1)/(c-1))*arctan(x)+1/2*c/(c-1)*arctan(x)^
2+1/4*c/(c-1)*polylog(2,(1+c)*(1+I*x)^2/(x^2+1)/(c-1))-1/2/(c-1)*arctan(x)^2-1/4/(c-1)*polylog(2,(1+c)*(1+I*x)
^2/(x^2+1)/(c-1))))

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Maxima [A]
time = 0.48, size = 189, normalized size = 0.92 \begin {gather*} {\left (x - \arctan \left (x\right )\right )} \operatorname {arccot}\left (c x\right ) - \frac {4 \, c \arctan \left (c x\right ) \arctan \left (x\right ) - 4 \, c \arctan \left (x\right ) \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} \end {gather*}

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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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