3.1.50 \(\int \frac {\cot ^{-1}(c x)}{x^2 (1+x^2)} \, dx\) [50]

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

[Out]

-arccot(c*x)/x-1/2*I*arctan(x)*ln(1-I/c/x)+1/2*I*arctan(x)*ln(1+I/c/x)-c*ln(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*c*ln(c^2*x^2+1)+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 = 0.37, antiderivative size = 212, normalized size of antiderivative = 1.00, 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 = {5039, 4947, 272, 36, 29, 31, 5029, 209, 2520, 266, 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 {1}{2} c \log \left (c^2 x^2+1\right )+\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 )-c \log (x)-\frac {\cot ^{-1}(c x)}{x} \end {gather*}

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 12

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

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 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 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 272

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 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 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 4947

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

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 5039

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 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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 \text {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 \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} c^3 \text {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*}

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Mathematica [A]
time = 0.07, size = 348, normalized size = 1.64 \begin {gather*} -\frac {\cot ^{-1}(c x)}{x}-c \log (x)+\frac {1}{4} \log (i-x) \log \left (-\frac {i (i-c x)}{1-c}\right )-\frac {1}{4} \log (i+x) \log \left (-\frac {i (i-c x)}{1+c}\right )-\frac {1}{4} \log (i-x) \log \left (-\frac {i-c x}{c x}\right )+\frac {1}{4} \log (i+x) \log \left (-\frac {i-c x}{c x}\right )+\frac {1}{4} \log (i+x) \log \left (-\frac {i (i+c x)}{1-c}\right )-\frac {1}{4} \log (i-x) \log \left (-\frac {i (i+c x)}{1+c}\right )+\frac {1}{4} \log (i-x) \log \left (\frac {i+c x}{c x}\right )-\frac {1}{4} \log (i+x) \log \left (\frac {i+c x}{c x}\right )+\frac {1}{2} c \log \left (1+c^2 x^2\right )+\frac {1}{4} \text {PolyLog}\left (2,\frac {i c (i-x)}{1-c}\right )-\frac {1}{4} \text {PolyLog}\left (2,-\frac {i c (i-x)}{1+c}\right )+\frac {1}{4} \text {PolyLog}\left (2,\frac {i c (i+x)}{1-c}\right )-\frac {1}{4} \text {PolyLog}\left (2,-\frac {i c (i+x)}{1+c}\right ) \end {gather*}

Antiderivative was successfully verified.

[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 = 0.12, size = 312, normalized size = 1.47

method	result	size
risch	\(-\frac {i \ln \left (i c x +1\right )}{2 x}-\frac {\pi \arctan \left (x \right )}{2}+\frac {i \ln \left (-i c x +1\right )}{2 x}-\frac {\pi }{2 x}+\frac {\ln \left (\frac {-i c x -c}{-c -1}\right ) \ln \left (-i c x +1\right )}{4}+\frac {\dilog \left (\frac {-i c x -c}{-c -1}\right )}{4}-\frac {\ln \left (\frac {-i c x +c}{c -1}\right ) \ln \left (-i c x +1\right )}{4}-\frac {\dilog \left (\frac {-i c x +c}{c -1}\right )}{4}-\frac {c \ln \left (-i c x \right )}{2}+\frac {c \ln \left (c^{2} x^{2}+1\right )}{2}+\frac {\ln \left (\frac {i c x -c}{-c -1}\right ) \ln \left (i c x +1\right )}{4}+\frac {\dilog \left (\frac {i c x -c}{-c -1}\right )}{4}-\frac {\ln \left (\frac {i c x +c}{c -1}\right ) \ln \left (i c x +1\right )}{4}-\frac {\dilog \left (\frac {i c x +c}{c -1}\right )}{4}-\frac {c \ln \left (i c x \right )}{2}\)	\(248\)
derivativedivides	\(c \left (-\frac {\mathrm {arccot}\left (c x \right )}{c x}-\frac {\mathrm {arccot}\left (c x \right ) \arctan \left (x \right )}{c}+c^{3} \left (\frac {\ln \left (c^{2} x^{2}+1\right )}{2 c^{3}}-\frac {\ln \left (x \right )}{c^{3}}-\frac {i \arctan \left (x \right ) \ln \left (1-\frac {\left (c -1\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (1+c \right )}\right )}{2 c^{4}}-\frac {\arctan \left (x \right )^{2}}{2 c^{4}}-\frac {\polylog \left (2, \frac {\left (c -1\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (1+c \right )}\right )}{4 c^{4}}+\frac {i \ln \left (1-\frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (c -1\right )}\right ) \arctan \left (x \right )}{2 c^{3} \left (c -1\right )}-\frac {i \ln \left (1-\frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (c -1\right )}\right ) \arctan \left (x \right )}{2 c^{4} \left (c -1\right )}+\frac {\arctan \left (x \right )^{2}}{2 c^{3} \left (c -1\right )}+\frac {\polylog \left (2, \frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (c -1\right )}\right )}{4 c^{3} \left (c -1\right )}-\frac {\arctan \left (x \right )^{2}}{2 c^{4} \left (c -1\right )}-\frac {\polylog \left (2, \frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (c -1\right )}\right )}{4 c^{4} \left (c -1\right )}\right )\right )\)	\(312\)
default	\(c \left (-\frac {\mathrm {arccot}\left (c x \right )}{c x}-\frac {\mathrm {arccot}\left (c x \right ) \arctan \left (x \right )}{c}+c^{3} \left (\frac {\ln \left (c^{2} x^{2}+1\right )}{2 c^{3}}-\frac {\ln \left (x \right )}{c^{3}}-\frac {i \arctan \left (x \right ) \ln \left (1-\frac {\left (c -1\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (1+c \right )}\right )}{2 c^{4}}-\frac {\arctan \left (x \right )^{2}}{2 c^{4}}-\frac {\polylog \left (2, \frac {\left (c -1\right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (1+c \right )}\right )}{4 c^{4}}+\frac {i \ln \left (1-\frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (c -1\right )}\right ) \arctan \left (x \right )}{2 c^{3} \left (c -1\right )}-\frac {i \ln \left (1-\frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (c -1\right )}\right ) \arctan \left (x \right )}{2 c^{4} \left (c -1\right )}+\frac {\arctan \left (x \right )^{2}}{2 c^{3} \left (c -1\right )}+\frac {\polylog \left (2, \frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (c -1\right )}\right )}{4 c^{3} \left (c -1\right )}-\frac {\arctan \left (x \right )^{2}}{2 c^{4} \left (c -1\right )}-\frac {\polylog \left (2, \frac {\left (1+c \right ) \left (i x +1\right )^{2}}{\left (x^{2}+1\right ) \left (c -1\right )}\right )}{4 c^{4} \left (c -1\right )}\right )\right )\)	\(312\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 183, normalized size = 0.86 \begin {gather*} -{\left (\frac {1}{x} + \arctan \left (x\right )\right )} \operatorname {arccot}\left (c x\right ) - \frac {1}{2} \, \arctan \left (c x\right ) \arctan \left (x\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 {gather*}

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

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 = 0.00, 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(arccot(c*x)/x^2/(x^2+1),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(c*x)/(x^2*(x^2 + 1)),x)

[Out]

int(acot(c*x)/(x^2*(x^2 + 1)), x)

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