3.285 \(\int \frac{x \tan ^{-1}(a x)^2}{c+a^2 c x^2} \, dx\)

Optimal. Leaf size=102 \[ -\frac{\text{PolyLog}\left (3,1-\frac{2}{1+i a x}\right )}{2 a^2 c}-\frac{i \tan ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1+i a x}\right )}{a^2 c}-\frac{i \tan ^{-1}(a x)^3}{3 a^2 c}-\frac{\log \left (\frac{2}{1+i a x}\right ) \tan ^{-1}(a x)^2}{a^2 c} \]

[Out]

((-I/3)*ArcTan[a*x]^3)/(a^2*c) - (ArcTan[a*x]^2*Log[2/(1 + I*a*x)])/(a^2*c) - (I*ArcTan[a*x]*PolyLog[2, 1 - 2/
(1 + I*a*x)])/(a^2*c) - PolyLog[3, 1 - 2/(1 + I*a*x)]/(2*a^2*c)

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Rubi [A]  time = 0.146067, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4920, 4854, 4884, 4994, 6610} \[ -\frac{\text{PolyLog}\left (3,1-\frac{2}{1+i a x}\right )}{2 a^2 c}-\frac{i \tan ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1+i a x}\right )}{a^2 c}-\frac{i \tan ^{-1}(a x)^3}{3 a^2 c}-\frac{\log \left (\frac{2}{1+i a x}\right ) \tan ^{-1}(a x)^2}{a^2 c} \]

Antiderivative was successfully verified.

[In]

Int[(x*ArcTan[a*x]^2)/(c + a^2*c*x^2),x]

[Out]

((-I/3)*ArcTan[a*x]^3)/(a^2*c) - (ArcTan[a*x]^2*Log[2/(1 + I*a*x)])/(a^2*c) - (I*ArcTan[a*x]*PolyLog[2, 1 - 2/
(1 + I*a*x)])/(a^2*c) - PolyLog[3, 1 - 2/(1 + I*a*x)]/(2*a^2*c)

Rule 4920

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

Rule 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTan[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 4884

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

Rule 4994

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

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin{align*} \int \frac{x \tan ^{-1}(a x)^2}{c+a^2 c x^2} \, dx &=-\frac{i \tan ^{-1}(a x)^3}{3 a^2 c}-\frac{\int \frac{\tan ^{-1}(a x)^2}{i-a x} \, dx}{a c}\\ &=-\frac{i \tan ^{-1}(a x)^3}{3 a^2 c}-\frac{\tan ^{-1}(a x)^2 \log \left (\frac{2}{1+i a x}\right )}{a^2 c}+\frac{2 \int \frac{\tan ^{-1}(a x) \log \left (\frac{2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a c}\\ &=-\frac{i \tan ^{-1}(a x)^3}{3 a^2 c}-\frac{\tan ^{-1}(a x)^2 \log \left (\frac{2}{1+i a x}\right )}{a^2 c}-\frac{i \tan ^{-1}(a x) \text{Li}_2\left (1-\frac{2}{1+i a x}\right )}{a^2 c}+\frac{i \int \frac{\text{Li}_2\left (1-\frac{2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a c}\\ &=-\frac{i \tan ^{-1}(a x)^3}{3 a^2 c}-\frac{\tan ^{-1}(a x)^2 \log \left (\frac{2}{1+i a x}\right )}{a^2 c}-\frac{i \tan ^{-1}(a x) \text{Li}_2\left (1-\frac{2}{1+i a x}\right )}{a^2 c}-\frac{\text{Li}_3\left (1-\frac{2}{1+i a x}\right )}{2 a^2 c}\\ \end{align*}

Mathematica [A]  time = 0.0098974, size = 110, normalized size = 1.08 \[ -\frac{\text{PolyLog}\left (3,\frac{a x+i}{a x-i}\right )}{2 a^2 c}-\frac{i \tan ^{-1}(a x) \text{PolyLog}\left (2,\frac{a x+i}{a x-i}\right )}{a^2 c}-\frac{i \tan ^{-1}(a x)^3}{3 a^2 c}-\frac{\log \left (\frac{2 i}{-a x+i}\right ) \tan ^{-1}(a x)^2}{a^2 c} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*ArcTan[a*x]^2)/(c + a^2*c*x^2),x]

[Out]

((-I/3)*ArcTan[a*x]^3)/(a^2*c) - (ArcTan[a*x]^2*Log[(2*I)/(I - a*x)])/(a^2*c) - (I*ArcTan[a*x]*PolyLog[2, (I +
a*x)/(-I + a*x)])/(a^2*c) - PolyLog[3, (I + a*x)/(-I + a*x)]/(2*a^2*c)

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Maple [C]  time = 0.346, size = 897, normalized size = 8.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctan(a*x)^2/(a^2*c*x^2+c),x)

[Out]

1/2/a^2/c*arctan(a*x)^2*ln(a^2*x^2+1)-1/a^2/c*arctan(a*x)^2*ln((1+I*a*x)/(a^2*x^2+1)^(1/2))+1/3*I/a^2/c*arctan
(a*x)^3+1/2*I/a^2/c*arctan(a*x)^2*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1))*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)
^2+1/4*I/a^2/c*arctan(a*x)^2*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))^3+1/4*I/a^2/c*arctan(a*x)^2*Pi*csgn(I/((1+I*a*
x)^2/(a^2*x^2+1)+1)^2)*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x)^2/(a^2*x^2+1)
+1)^2)+1/4*I/a^2/c*arctan(a*x)^2*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^3-1/4*I/a^2/
c*arctan(a*x)^2*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1))^2*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)+1/4*I/a^2/c*arc
tan(a*x)^2*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))*csgn(I*(1+I*a*x)/(a^2*x^2+1)^(1/2))^2-1/4*I/a^2/c*arctan(a*x)^2*
Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^2-1/4*I/a^2/c
*arctan(a*x)^2*Pi*csgn(I/((1+I*a*x)^2/(a^2*x^2+1)+1)^2)*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x)^2/(a^2*x^2+1
)+1)^2)^2-1/2*I/a^2/c*arctan(a*x)^2*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))^2*csgn(I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-1
/a^2/c*arctan(a*x)^2*ln(2)-1/4*I/a^2/c*arctan(a*x)^2*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^3-1/2/a^2/c*poly
log(3,-(1+I*a*x)^2/(a^2*x^2+1))+I/a^2/c*arctan(a*x)*polylog(2,-(1+I*a*x)^2/(a^2*x^2+1))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \arctan \left (a x\right )^{2}}{a^{2} c x^{2} + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctan(a*x)^2/(a^2*c*x^2+c),x, algorithm="maxima")

[Out]

integrate(x*arctan(a*x)^2/(a^2*c*x^2 + c), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x \arctan \left (a x\right )^{2}}{a^{2} c x^{2} + c}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctan(a*x)^2/(a^2*c*x^2+c),x, algorithm="fricas")

[Out]

integral(x*arctan(a*x)^2/(a^2*c*x^2 + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atan(a*x)**2/(a**2*c*x**2+c),x)

[Out]

Integral(x*atan(a*x)**2/(a**2*x**2 + 1), x)/c

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \arctan \left (a x\right )^{2}}{a^{2} c x^{2} + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctan(a*x)^2/(a^2*c*x^2+c),x, algorithm="giac")

[Out]

integrate(x*arctan(a*x)^2/(a^2*c*x^2 + c), x)