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

Optimal. Leaf size=188 \[ -\frac {1}{2} i \text {Li}_2\left (1-\frac {2}{1-i c x}\right )+\frac {1}{4} i \text {Li}_2\left (1-\frac {2 i c (i-x)}{(1-c) (1-i c x)}\right )+\frac {1}{4} i \text {Li}_2\left (\frac {2 i c (x+i)}{(c+1) (1-i c x)}+1\right )+\log \left (\frac {2}{1-i c x}\right ) \left (-\cot ^{-1}(c x)\right )+\frac {1}{2} \log \left (\frac {2 i c (-x+i)}{(1-c) (1-i c x)}\right ) \cot ^{-1}(c x)+\frac {1}{2} \log \left (-\frac {2 i c (x+i)}{(c+1) (1-i c x)}\right ) \cot ^{-1}(c x) \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {4929, 4857, 2402, 2315, 2447} \[ -\frac {1}{2} i \text {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )+\frac {1}{4} i \text {PolyLog}\left (2,1-\frac {2 i c (-x+i)}{(1-c) (1-i c x)}\right )+\frac {1}{4} i \text {PolyLog}\left (2,1+\frac {2 i c (x+i)}{(c+1) (1-i c x)}\right )+\log \left (\frac {2}{1-i c x}\right ) \left (-\cot ^{-1}(c x)\right )+\frac {1}{2} \log \left (\frac {2 i c (-x+i)}{(1-c) (1-i c x)}\right ) \cot ^{-1}(c x)+\frac {1}{2} \log \left (-\frac {2 i c (x+i)}{(c+1) (1-i c x)}\right ) \cot ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[(x*ArcCot[c*x])/(1 + x^2),x]

[Out]

-(ArcCot[c*x]*Log[2/(1 - I*c*x)]) + (ArcCot[c*x]*Log[((2*I)*c*(I - x))/((1 - c)*(1 - I*c*x))])/2 + (ArcCot[c*x
]*Log[((-2*I)*c*(I + x))/((1 + c)*(1 - I*c*x))])/2 - (I/2)*PolyLog[2, 1 - 2/(1 - I*c*x)] + (I/4)*PolyLog[2, 1
- ((2*I)*c*(I - x))/((1 - c)*(1 - I*c*x))] + (I/4)*PolyLog[2, 1 + ((2*I)*c*(I + x))/((1 + c)*(1 - I*c*x))]

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 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 4857

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcCot[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*ArcCot[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 4929

Int[(((a_.) + ArcCot[(c_.)*(x_)]*(b_.))*(x_)^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[a
+ b*ArcCot[c*x], x^m/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IntegerQ[m] &&  !(EqQ[m, 1] && NeQ[a,
0])

Rubi steps

\begin {align*} \int \frac {x \cot ^{-1}(c x)}{1+x^2} \, dx &=\int \left (-\frac {\cot ^{-1}(c x)}{2 (i-x)}+\frac {\cot ^{-1}(c x)}{2 (i+x)}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {\cot ^{-1}(c x)}{i-x} \, dx\right )+\frac {1}{2} \int \frac {\cot ^{-1}(c x)}{i+x} \, dx\\ &=-\cot ^{-1}(c x) \log \left (\frac {2}{1-i c x}\right )+\frac {1}{2} \cot ^{-1}(c x) \log \left (\frac {2 i c (i-x)}{(1-c) (1-i c x)}\right )+\frac {1}{2} \cot ^{-1}(c x) \log \left (-\frac {2 i c (i+x)}{(1+c) (1-i c x)}\right )-2 \left (\frac {1}{2} c \int \frac {\log \left (\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx\right )+\frac {1}{2} c \int \frac {\log \left (\frac {2 c (i-x)}{(-i+i c) (1-i c x)}\right )}{1+c^2 x^2} \, dx+\frac {1}{2} c \int \frac {\log \left (\frac {2 c (i+x)}{(i+i c) (1-i c x)}\right )}{1+c^2 x^2} \, dx\\ &=-\cot ^{-1}(c x) \log \left (\frac {2}{1-i c x}\right )+\frac {1}{2} \cot ^{-1}(c x) \log \left (\frac {2 i c (i-x)}{(1-c) (1-i c x)}\right )+\frac {1}{2} \cot ^{-1}(c x) \log \left (-\frac {2 i c (i+x)}{(1+c) (1-i c x)}\right )+\frac {1}{4} i \text {Li}_2\left (1-\frac {2 i c (i-x)}{(1-c) (1-i c x)}\right )+\frac {1}{4} i \text {Li}_2\left (1+\frac {2 i c (i+x)}{(1+c) (1-i c x)}\right )-2 \left (\frac {1}{2} i \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-i c x}\right )\right )\\ &=-\cot ^{-1}(c x) \log \left (\frac {2}{1-i c x}\right )+\frac {1}{2} \cot ^{-1}(c x) \log \left (\frac {2 i c (i-x)}{(1-c) (1-i c x)}\right )+\frac {1}{2} \cot ^{-1}(c x) \log \left (-\frac {2 i c (i+x)}{(1+c) (1-i c x)}\right )-\frac {1}{2} i \text {Li}_2\left (1-\frac {2}{1-i c x}\right )+\frac {1}{4} i \text {Li}_2\left (1-\frac {2 i c (i-x)}{(1-c) (1-i c x)}\right )+\frac {1}{4} i \text {Li}_2\left (1+\frac {2 i c (i+x)}{(1+c) (1-i c x)}\right )\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 343, normalized size = 1.82 \[ -\frac {1}{4} i \text {Li}_2\left (\frac {i c (i-x)}{1-c}\right )+\frac {1}{4} i \text {Li}_2\left (-\frac {i c (i-x)}{c+1}\right )+\frac {1}{4} i \text {Li}_2\left (\frac {i c (x+i)}{1-c}\right )-\frac {1}{4} i \text {Li}_2\left (-\frac {i c (x+i)}{c+1}\right )-\frac {1}{4} i \log (-x+i) \log \left (-\frac {i (-c x+i)}{1-c}\right )-\frac {1}{4} i \log (x+i) \log \left (-\frac {i (-c x+i)}{c+1}\right )+\frac {1}{4} i \log (-x+i) \log \left (-\frac {-c x+i}{c x}\right )+\frac {1}{4} i \log (x+i) \log \left (-\frac {-c x+i}{c x}\right )+\frac {1}{4} i \log (x+i) \log \left (-\frac {i (c x+i)}{1-c}\right )+\frac {1}{4} i \log (-x+i) \log \left (-\frac {i (c x+i)}{c+1}\right )-\frac {1}{4} i \log (-x+i) \log \left (\frac {c x+i}{c x}\right )-\frac {1}{4} i \log (x+i) \log \left (\frac {c x+i}{c x}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-1/4*I)*Log[I - x]*Log[((-I)*(I - c*x))/(1 - c)] - (I/4)*Log[I + x]*Log[((-I)*(I - c*x))/(1 + c)] + (I/4)*Log
[I - x]*Log[-((I - c*x)/(c*x))] + (I/4)*Log[I + x]*Log[-((I - c*x)/(c*x))] + (I/4)*Log[I + x]*Log[((-I)*(I + c
*x))/(1 - c)] + (I/4)*Log[I - x]*Log[((-I)*(I + c*x))/(1 + c)] - (I/4)*Log[I - x]*Log[(I + c*x)/(c*x)] - (I/4)
*Log[I + x]*Log[(I + c*x)/(c*x)] - (I/4)*PolyLog[2, (I*c*(I - x))/(1 - c)] + (I/4)*PolyLog[2, ((-I)*c*(I - x))
/(1 + c)] + (I/4)*PolyLog[2, (I*c*(I + x))/(1 - c)] - (I/4)*PolyLog[2, ((-I)*c*(I + x))/(1 + c)]

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fricas [F]  time = 2.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x \operatorname {arccot}\left (c x\right )}{x^{2} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*arccot(c*x)/(x^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arccot(c*x)/(x^2 + 1), x)

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maple [A]  time = 0.32, size = 284, normalized size = 1.51 \[ \frac {\ln \left (c^{2} x^{2}+c^{2}\right ) \mathrm {arccot}\left (c x \right )}{2}+\frac {i \ln \left (c x -i\right ) \ln \left (\frac {i \left (c x -i\right )-c -1}{-c -1}\right )}{4}+\frac {i \ln \left (c x -i\right ) \ln \left (\frac {i \left (c x -i\right )+c -1}{-1+c}\right )}{4}-\frac {i \ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+c^{2}\right )}{4}+\frac {i \dilog \left (\frac {i \left (c x -i\right )-c -1}{-c -1}\right )}{4}+\frac {i \dilog \left (\frac {i \left (c x -i\right )+c -1}{-1+c}\right )}{4}-\frac {i \ln \left (c x +i\right ) \ln \left (\frac {-i \left (c x +i\right )-c -1}{-c -1}\right )}{4}-\frac {i \ln \left (c x +i\right ) \ln \left (\frac {-i \left (c x +i\right )+c -1}{-1+c}\right )}{4}+\frac {i \ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+c^{2}\right )}{4}-\frac {i \dilog \left (\frac {-i \left (c x +i\right )+c -1}{-1+c}\right )}{4}-\frac {i \dilog \left (\frac {-i \left (c x +i\right )-c -1}{-c -1}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*ln(c^2*x^2+c^2)*arccot(c*x)+1/4*I*ln(c*x-I)*ln((I*(c*x-I)-c-1)/(-c-1))+1/4*I*ln(c*x-I)*ln((I*(c*x-I)+c-1)/
(-1+c))-1/4*I*ln(c*x-I)*ln(c^2*x^2+c^2)+1/4*I*dilog((I*(c*x-I)-c-1)/(-c-1))+1/4*I*dilog((I*(c*x-I)+c-1)/(-1+c)
)-1/4*I*ln(I+c*x)*ln((-I*(I+c*x)-c-1)/(-c-1))-1/4*I*ln(I+c*x)*ln((-I*(I+c*x)+c-1)/(-1+c))+1/4*I*ln(I+c*x)*ln(c
^2*x^2+c^2)-1/4*I*dilog((-I*(I+c*x)+c-1)/(-1+c))-1/4*I*dilog((-I*(I+c*x)-c-1)/(-c-1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arccot(c*x)/(x^2 + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,\mathrm {acot}\left (c\,x\right )}{x^2+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*acot(c*x))/(x^2 + 1),x)

[Out]

int((x*acot(c*x))/(x^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*acot(c*x)/(x**2 + 1), x)

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