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

Optimal. Leaf size=98 \[ \frac{i \text{PolyLog}\left (2,1-\frac{2}{1+i a x}\right )}{a^3 c}-\frac{\tan ^{-1}(a x)^3}{3 a^3 c}+\frac{x \tan ^{-1}(a x)^2}{a^2 c}+\frac{i \tan ^{-1}(a x)^2}{a^3 c}+\frac{2 \log \left (\frac{2}{1+i a x}\right ) \tan ^{-1}(a x)}{a^3 c} \]

[Out]

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

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Rubi [A]  time = 0.166925, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {4916, 4846, 4920, 4854, 2402, 2315, 4884} \[ \frac{i \text{PolyLog}\left (2,1-\frac{2}{1+i a x}\right )}{a^3 c}-\frac{\tan ^{-1}(a x)^3}{3 a^3 c}+\frac{x \tan ^{-1}(a x)^2}{a^2 c}+\frac{i \tan ^{-1}(a x)^2}{a^3 c}+\frac{2 \log \left (\frac{2}{1+i a x}\right ) \tan ^{-1}(a x)}{a^3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4916

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

Rule 4846

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

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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.172592, size = 69, normalized size = 0.7 \[ \frac{-i \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(a x)}\right )-\frac{1}{3} \tan ^{-1}(a x) \left (\tan ^{-1}(a x)^2+(-3 a x+3 i) \tan ^{-1}(a x)-6 \log \left (1+e^{2 i \tan ^{-1}(a x)}\right )\right )}{a^3 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(ArcTan[a*x]*((3*I - 3*a*x)*ArcTan[a*x] + ArcTan[a*x]^2 - 6*Log[1 + E^((2*I)*ArcTan[a*x])]))/3 - I*PolyLog[2
, -E^((2*I)*ArcTan[a*x])])/(a^3*c)

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Maple [B]  time = 0.093, size = 230, normalized size = 2.4 \begin{align*}{\frac{x \left ( \arctan \left ( ax \right ) \right ) ^{2}}{{a}^{2}c}}-{\frac{ \left ( \arctan \left ( ax \right ) \right ) ^{3}}{3\,{a}^{3}c}}-{\frac{\arctan \left ( ax \right ) \ln \left ({a}^{2}{x}^{2}+1 \right ) }{{a}^{3}c}}-{\frac{{\frac{i}{2}}\ln \left ( ax-i \right ) \ln \left ({a}^{2}{x}^{2}+1 \right ) }{{a}^{3}c}}+{\frac{{\frac{i}{4}} \left ( \ln \left ( ax-i \right ) \right ) ^{2}}{{a}^{3}c}}+{\frac{{\frac{i}{2}}\ln \left ( ax-i \right ) \ln \left ( -{\frac{i}{2}} \left ( ax+i \right ) \right ) }{{a}^{3}c}}+{\frac{{\frac{i}{2}}{\it dilog} \left ( -{\frac{i}{2}} \left ( ax+i \right ) \right ) }{{a}^{3}c}}+{\frac{{\frac{i}{2}}\ln \left ( ax+i \right ) \ln \left ({a}^{2}{x}^{2}+1 \right ) }{{a}^{3}c}}-{\frac{{\frac{i}{4}} \left ( \ln \left ( ax+i \right ) \right ) ^{2}}{{a}^{3}c}}-{\frac{{\frac{i}{2}}\ln \left ({\frac{i}{2}} \left ( ax-i \right ) \right ) \ln \left ( ax+i \right ) }{{a}^{3}c}}-{\frac{{\frac{i}{2}}{\it dilog} \left ({\frac{i}{2}} \left ( ax-i \right ) \right ) }{{a}^{3}c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arctan(a*x)^2/a^2/c-1/3*arctan(a*x)^3/a^3/c-1/a^3/c*arctan(a*x)*ln(a^2*x^2+1)-1/2*I/a^3/c*ln(a*x-I)*ln(a^2*x
^2+1)+1/4*I/a^3/c*ln(a*x-I)^2+1/2*I/a^3/c*ln(a*x-I)*ln(-1/2*I*(a*x+I))+1/2*I/a^3/c*dilog(-1/2*I*(a*x+I))+1/2*I
/a^3/c*ln(a*x+I)*ln(a^2*x^2+1)-1/4*I/a^3/c*ln(a*x+I)^2-1/2*I/a^3/c*ln(1/2*I*(a*x-I))*ln(a*x+I)-1/2*I/a^3/c*dil
og(1/2*I*(a*x-I))

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

integral(x^2*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^{2} \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**2*atan(a*x)**2/(a**2*c*x**2+c),x)

[Out]

Integral(x**2*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^{2} \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^2*arctan(a*x)^2/(a^2*c*x^2+c),x, algorithm="giac")

[Out]

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