3.17 \(\int \cot ^{-1}(a x)^2 \, dx\)

Optimal. Leaf size=67 \[ \frac {i \text {Li}_2\left (1-\frac {2}{i a x+1}\right )}{a}+x \cot ^{-1}(a x)^2+\frac {i \cot ^{-1}(a x)^2}{a}-\frac {2 \log \left (\frac {2}{1+i a x}\right ) \cot ^{-1}(a x)}{a} \]

[Out]

I*arccot(a*x)^2/a+x*arccot(a*x)^2-2*arccot(a*x)*ln(2/(1+I*a*x))/a+I*polylog(2,1-2/(1+I*a*x))/a

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Rubi [A]  time = 0.07, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {4847, 4921, 4855, 2402, 2315} \[ \frac {i \text {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )}{a}+x \cot ^{-1}(a x)^2+\frac {i \cot ^{-1}(a x)^2}{a}-\frac {2 \log \left (\frac {2}{1+i a x}\right ) \cot ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[ArcCot[a*x]^2,x]

[Out]

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

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 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 4855

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

Rule 4921

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

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 56, normalized size = 0.84 \[ \frac {i \text {Li}_2\left (e^{2 i \cot ^{-1}(a x)}\right )+\cot ^{-1}(a x) \left ((a x+i) \cot ^{-1}(a x)-2 \log \left (1-e^{2 i \cot ^{-1}(a x)}\right )\right )}{a} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCot[a*x]^2,x]

[Out]

(ArcCot[a*x]*((I + a*x)*ArcCot[a*x] - 2*Log[1 - E^((2*I)*ArcCot[a*x])]) + I*PolyLog[2, E^((2*I)*ArcCot[a*x])])
/a

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\operatorname {arccot}\left (a x\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(a*x)^2,x, algorithm="fricas")

[Out]

integral(arccot(a*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(a*x)^2,x, algorithm="giac")

[Out]

integrate(arccot(a*x)^2, x)

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maple [B]  time = 0.42, size = 136, normalized size = 2.03 \[ x \mathrm {arccot}\left (a x \right )^{2}+\frac {i \mathrm {arccot}\left (a x \right )^{2}}{a}-\frac {2 \,\mathrm {arccot}\left (a x \right ) \ln \left (1+\frac {a x +i}{\sqrt {a^{2} x^{2}+1}}\right )}{a}-\frac {2 \,\mathrm {arccot}\left (a x \right ) \ln \left (1-\frac {a x +i}{\sqrt {a^{2} x^{2}+1}}\right )}{a}+\frac {2 i \polylog \left (2, -\frac {a x +i}{\sqrt {a^{2} x^{2}+1}}\right )}{a}+\frac {2 i \polylog \left (2, \frac {a x +i}{\sqrt {a^{2} x^{2}+1}}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccot(a*x)^2,x)

[Out]

x*arccot(a*x)^2+I*arccot(a*x)^2/a-2/a*arccot(a*x)*ln(1+(I+a*x)/(a^2*x^2+1)^(1/2))-2/a*arccot(a*x)*ln(1-(I+a*x)
/(a^2*x^2+1)^(1/2))+2*I/a*polylog(2,-(I+a*x)/(a^2*x^2+1)^(1/2))+2*I/a*polylog(2,(I+a*x)/(a^2*x^2+1)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, x \arctan \left (1, a x\right )^{2} + 12 \, a^{2} \int \frac {x^{2} \arctan \left (\frac {1}{a x}\right )^{2}}{16 \, {\left (a^{2} x^{2} + 1\right )}}\,{d x} + a^{2} \int \frac {x^{2} \log \left (a^{2} x^{2} + 1\right )^{2}}{16 \, {\left (a^{2} x^{2} + 1\right )}}\,{d x} + 4 \, a^{2} \int \frac {x^{2} \log \left (a^{2} x^{2} + 1\right )}{16 \, {\left (a^{2} x^{2} + 1\right )}}\,{d x} - \frac {1}{16} \, x \log \left (a^{2} x^{2} + 1\right )^{2} + \frac {\arctan \left (a x\right )^{3}}{4 \, a} + \frac {3 \, \arctan \left (a x\right )^{2} \arctan \left (\frac {1}{a x}\right )}{4 \, a} + \frac {3 \, \arctan \left (a x\right ) \arctan \left (\frac {1}{a x}\right )^{2}}{4 \, a} + 8 \, a \int \frac {x \arctan \left (\frac {1}{a x}\right )}{16 \, {\left (a^{2} x^{2} + 1\right )}}\,{d x} + \int \frac {\log \left (a^{2} x^{2} + 1\right )^{2}}{16 \, {\left (a^{2} x^{2} + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(a*x)^2,x, algorithm="maxima")

[Out]

1/4*x*arctan2(1, a*x)^2 + 12*a^2*integrate(1/16*x^2*arctan(1/(a*x))^2/(a^2*x^2 + 1), x) + a^2*integrate(1/16*x
^2*log(a^2*x^2 + 1)^2/(a^2*x^2 + 1), x) + 4*a^2*integrate(1/16*x^2*log(a^2*x^2 + 1)/(a^2*x^2 + 1), x) - 1/16*x
*log(a^2*x^2 + 1)^2 + 1/4*arctan(a*x)^3/a + 3/4*arctan(a*x)^2*arctan(1/(a*x))/a + 3/4*arctan(a*x)*arctan(1/(a*
x))^2/a + 8*a*integrate(1/16*x*arctan(1/(a*x))/(a^2*x^2 + 1), x) + integrate(1/16*log(a^2*x^2 + 1)^2/(a^2*x^2
+ 1), x)

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mupad [B]  time = 0.59, size = 55, normalized size = 0.82 \[ \frac {-2\,\ln \left (1-{\mathrm {e}}^{\mathrm {acot}\left (a\,x\right )\,2{}\mathrm {i}}\right )\,\mathrm {acot}\left (a\,x\right )+\mathrm {polylog}\left (2,{\mathrm {e}}^{\mathrm {acot}\left (a\,x\right )\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}+{\mathrm {acot}\left (a\,x\right )}^2\,1{}\mathrm {i}}{a}+x\,{\mathrm {acot}\left (a\,x\right )}^2 \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(a*x)^2,x)

[Out]

(polylog(2, exp(acot(a*x)*2i))*1i - 2*log(1 - exp(acot(a*x)*2i))*acot(a*x) + acot(a*x)^2*1i)/a + x*acot(a*x)^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acot(a*x)**2,x)

[Out]

Integral(acot(a*x)**2, x)

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