3.1.48 \(\int \frac {\cot ^{-1}(c x)}{1+x^2} \, dx\) [48]

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

[Out]

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*a
rctan(x)*ln(-2*I*(c*x+I)/(1+c)/(1-I*x))-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 = 0.34, antiderivative size = 183, normalized size of antiderivative = 1.00, 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 = {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}{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 ) \end {gather*}

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 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 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 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)}{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*}

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Mathematica [A]
time = 0.06, size = 319, normalized size = 1.74 \begin {gather*} -\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}{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]/(1 + x^2),x]

[Out]

-1/4*(Log[I - x]*Log[((-I)*(I - c*x))/(1 - c)]) + (Log[I + x]*Log[((-I)*(I - c*x))/(1 + c)])/4 + (Log[I - x]*L
og[-((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 [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 334 vs. \(2 (153 ) = 306\).
time = 0.26, size = 335, normalized size = 1.83

method	result	size
risch	\(\frac {\pi \arctan \left (x \right )}{2}-\frac {\ln \left (\frac {-i c x -c}{-c -1}\right ) \ln \left (-i c x +1\right )}{4}-\frac {\dilog \left (\frac {-i c x -c}{-c -1}\right )}{4}+\frac {\ln \left (\frac {-i c x +c}{c -1}\right ) \ln \left (-i c x +1\right )}{4}+\frac {\dilog \left (\frac {-i c x +c}{c -1}\right )}{4}-\frac {\ln \left (\frac {i c x -c}{-c -1}\right ) \ln \left (i c x +1\right )}{4}-\frac {\dilog \left (\frac {i c x -c}{-c -1}\right )}{4}+\frac {\ln \left (\frac {i c x +c}{c -1}\right ) \ln \left (i c x +1\right )}{4}+\frac {\dilog \left (\frac {i c x +c}{c -1}\right )}{4}\)	\(183\)
derivativedivides	\(\frac {c \arctan \left (x \right ) \mathrm {arccot}\left (c x \right )+c^{2} \left (\frac {\arctan \left (c x \right ) \arctan \left (x \right )}{c}-\frac {-\frac {i c \arctan \left (c x \right ) \ln \left (1-\frac {\left (1+c \right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (1-c \right )}\right )}{2}-\frac {c \arctan \left (c x \right )^{2}}{2}-\frac {c \polylog \left (2, \frac {\left (1+c \right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (1-c \right )}\right )}{4}+\frac {i c^{2} \ln \left (1-\frac {\left (c -1\right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c -1\right )}\right ) \arctan \left (c x \right )}{2+2 c}+\frac {i c \arctan \left (c x \right ) \ln \left (1-\frac {\left (c -1\right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c -1\right )}\right )}{2+2 c}+\frac {c^{2} \arctan \left (c x \right )^{2}}{2+2 c}+\frac {c^{2} \polylog \left (2, \frac {\left (c -1\right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c -1\right )}\right )}{4+4 c}+\frac {c \arctan \left (c x \right )^{2}}{2+2 c}+\frac {c \polylog \left (2, \frac {\left (c -1\right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c -1\right )}\right )}{4+4 c}}{c^{2}}\right )}{c}\)	\(335\)
default	\(\frac {c \arctan \left (x \right ) \mathrm {arccot}\left (c x \right )+c^{2} \left (\frac {\arctan \left (c x \right ) \arctan \left (x \right )}{c}-\frac {-\frac {i c \arctan \left (c x \right ) \ln \left (1-\frac {\left (1+c \right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (1-c \right )}\right )}{2}-\frac {c \arctan \left (c x \right )^{2}}{2}-\frac {c \polylog \left (2, \frac {\left (1+c \right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (1-c \right )}\right )}{4}+\frac {i c^{2} \ln \left (1-\frac {\left (c -1\right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c -1\right )}\right ) \arctan \left (c x \right )}{2+2 c}+\frac {i c \arctan \left (c x \right ) \ln \left (1-\frac {\left (c -1\right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c -1\right )}\right )}{2+2 c}+\frac {c^{2} \arctan \left (c x \right )^{2}}{2+2 c}+\frac {c^{2} \polylog \left (2, \frac {\left (c -1\right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c -1\right )}\right )}{4+4 c}+\frac {c \arctan \left (c x \right )^{2}}{2+2 c}+\frac {c \polylog \left (2, \frac {\left (c -1\right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c -1\right )}\right )}{4+4 c}}{c^{2}}\right )}{c}\)	\(335\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 187, normalized size = 1.02 \begin {gather*} -\frac {1}{8} \, c {\left (\frac {8 \, \arctan \left (c x\right ) \arctan \left (x\right )}{c} - \frac {4 \, \arctan \left (c x\right ) \arctan \left (x\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 {gather*}

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(c*x)*arctan(x) - 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*dilog(-(I*c*x - c)/(c -
1)))/c) + arccot(c*x)*arctan(x) + arctan(c*x)*arctan(x)

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Fricas [F]
time = 0.00, 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+1),x, algorithm="fricas")

[Out]

integral(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 {\operatorname {acot}{\left (c x \right )}}{x^{2} + 1}\, dx \end {gather*}

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 = 0.00, 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+1),x, algorithm="giac")

[Out]

integrate(arccot(c*x)/(x^2 + 1), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\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(acot(c*x)/(x^2 + 1),x)

[Out]

int(acot(c*x)/(x^2 + 1), x)

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