3.290 \(\int \frac{\tan ^{-1}(a x)^2}{x^4 (c+a^2 c x^2)} \, dx\)

Optimal. Leaf size=166 \[ \frac{4 i a^3 \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )}{3 c}-\frac{a^2}{3 c x}+\frac{a^3 \tan ^{-1}(a x)^3}{3 c}+\frac{4 i a^3 \tan ^{-1}(a x)^2}{3 c}-\frac{a^3 \tan ^{-1}(a x)}{3 c}+\frac{a^2 \tan ^{-1}(a x)^2}{c x}-\frac{8 a^3 \log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)}{3 c}-\frac{a \tan ^{-1}(a x)}{3 c x^2}-\frac{\tan ^{-1}(a x)^2}{3 c x^3} \]

[Out]

-a^2/(3*c*x) - (a^3*ArcTan[a*x])/(3*c) - (a*ArcTan[a*x])/(3*c*x^2) + (((4*I)/3)*a^3*ArcTan[a*x]^2)/c - ArcTan[
a*x]^2/(3*c*x^3) + (a^2*ArcTan[a*x]^2)/(c*x) + (a^3*ArcTan[a*x]^3)/(3*c) - (8*a^3*ArcTan[a*x]*Log[2 - 2/(1 - I
*a*x)])/(3*c) + (((4*I)/3)*a^3*PolyLog[2, -1 + 2/(1 - I*a*x)])/c

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Rubi [A]  time = 0.435952, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {4918, 4852, 325, 203, 4924, 4868, 2447, 4884} \[ \frac{4 i a^3 \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )}{3 c}-\frac{a^2}{3 c x}+\frac{a^3 \tan ^{-1}(a x)^3}{3 c}+\frac{4 i a^3 \tan ^{-1}(a x)^2}{3 c}-\frac{a^3 \tan ^{-1}(a x)}{3 c}+\frac{a^2 \tan ^{-1}(a x)^2}{c x}-\frac{8 a^3 \log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)}{3 c}-\frac{a \tan ^{-1}(a x)}{3 c x^2}-\frac{\tan ^{-1}(a x)^2}{3 c x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a*x]^2/(x^4*(c + a^2*c*x^2)),x]

[Out]

-a^2/(3*c*x) - (a^3*ArcTan[a*x])/(3*c) - (a*ArcTan[a*x])/(3*c*x^2) + (((4*I)/3)*a^3*ArcTan[a*x]^2)/c - ArcTan[
a*x]^2/(3*c*x^3) + (a^2*ArcTan[a*x]^2)/(c*x) + (a^3*ArcTan[a*x]^3)/(3*c) - (8*a^3*ArcTan[a*x]*Log[2 - 2/(1 - I
*a*x)])/(3*c) + (((4*I)/3)*a^3*PolyLog[2, -1 + 2/(1 - I*a*x)])/c

Rule 4918

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

Rule 4852

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 4924

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

Rule 4868

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTan[c*x]
)^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[2 - 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 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 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{\tan ^{-1}(a x)^2}{x^4 \left (c+a^2 c x^2\right )} \, dx &=-\left (a^2 \int \frac{\tan ^{-1}(a x)^2}{x^2 \left (c+a^2 c x^2\right )} \, dx\right )+\frac{\int \frac{\tan ^{-1}(a x)^2}{x^4} \, dx}{c}\\ &=-\frac{\tan ^{-1}(a x)^2}{3 c x^3}+a^4 \int \frac{\tan ^{-1}(a x)^2}{c+a^2 c x^2} \, dx+\frac{(2 a) \int \frac{\tan ^{-1}(a x)}{x^3 \left (1+a^2 x^2\right )} \, dx}{3 c}-\frac{a^2 \int \frac{\tan ^{-1}(a x)^2}{x^2} \, dx}{c}\\ &=-\frac{\tan ^{-1}(a x)^2}{3 c x^3}+\frac{a^2 \tan ^{-1}(a x)^2}{c x}+\frac{a^3 \tan ^{-1}(a x)^3}{3 c}+\frac{(2 a) \int \frac{\tan ^{-1}(a x)}{x^3} \, dx}{3 c}-\frac{\left (2 a^3\right ) \int \frac{\tan ^{-1}(a x)}{x \left (1+a^2 x^2\right )} \, dx}{3 c}-\frac{\left (2 a^3\right ) \int \frac{\tan ^{-1}(a x)}{x \left (1+a^2 x^2\right )} \, dx}{c}\\ &=-\frac{a \tan ^{-1}(a x)}{3 c x^2}+\frac{4 i a^3 \tan ^{-1}(a x)^2}{3 c}-\frac{\tan ^{-1}(a x)^2}{3 c x^3}+\frac{a^2 \tan ^{-1}(a x)^2}{c x}+\frac{a^3 \tan ^{-1}(a x)^3}{3 c}+\frac{a^2 \int \frac{1}{x^2 \left (1+a^2 x^2\right )} \, dx}{3 c}-\frac{\left (2 i a^3\right ) \int \frac{\tan ^{-1}(a x)}{x (i+a x)} \, dx}{3 c}-\frac{\left (2 i a^3\right ) \int \frac{\tan ^{-1}(a x)}{x (i+a x)} \, dx}{c}\\ &=-\frac{a^2}{3 c x}-\frac{a \tan ^{-1}(a x)}{3 c x^2}+\frac{4 i a^3 \tan ^{-1}(a x)^2}{3 c}-\frac{\tan ^{-1}(a x)^2}{3 c x^3}+\frac{a^2 \tan ^{-1}(a x)^2}{c x}+\frac{a^3 \tan ^{-1}(a x)^3}{3 c}-\frac{8 a^3 \tan ^{-1}(a x) \log \left (2-\frac{2}{1-i a x}\right )}{3 c}-\frac{a^4 \int \frac{1}{1+a^2 x^2} \, dx}{3 c}+\frac{\left (2 a^4\right ) \int \frac{\log \left (2-\frac{2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{3 c}+\frac{\left (2 a^4\right ) \int \frac{\log \left (2-\frac{2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c}\\ &=-\frac{a^2}{3 c x}-\frac{a^3 \tan ^{-1}(a x)}{3 c}-\frac{a \tan ^{-1}(a x)}{3 c x^2}+\frac{4 i a^3 \tan ^{-1}(a x)^2}{3 c}-\frac{\tan ^{-1}(a x)^2}{3 c x^3}+\frac{a^2 \tan ^{-1}(a x)^2}{c x}+\frac{a^3 \tan ^{-1}(a x)^3}{3 c}-\frac{8 a^3 \tan ^{-1}(a x) \log \left (2-\frac{2}{1-i a x}\right )}{3 c}+\frac{4 i a^3 \text{Li}_2\left (-1+\frac{2}{1-i a x}\right )}{3 c}\\ \end{align*}

Mathematica [A]  time = 0.345505, size = 120, normalized size = 0.72 \[ \frac{a^3 \left (4 i \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(a x)}\right )-\frac{\frac{\left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2}{a^2 x^2}-4 \tan ^{-1}(a x)^2+1}{a x}+\tan ^{-1}(a x) \left (-\frac{a^2 x^2+1}{a^2 x^2}+\tan ^{-1}(a x) \left (\tan ^{-1}(a x)+4 i\right )-8 \log \left (1-e^{2 i \tan ^{-1}(a x)}\right )\right )\right )}{3 c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTan[a*x]^2/(x^4*(c + a^2*c*x^2)),x]

[Out]

(a^3*(-((1 - 4*ArcTan[a*x]^2 + ((1 + a^2*x^2)*ArcTan[a*x]^2)/(a^2*x^2))/(a*x)) + ArcTan[a*x]*(-((1 + a^2*x^2)/
(a^2*x^2)) + ArcTan[a*x]*(4*I + ArcTan[a*x]) - 8*Log[1 - E^((2*I)*ArcTan[a*x])]) + (4*I)*PolyLog[2, E^((2*I)*A
rcTan[a*x])]))/(3*c)

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Maple [B]  time = 0.108, size = 374, normalized size = 2.3 \begin{align*}{\frac{{a}^{3} \left ( \arctan \left ( ax \right ) \right ) ^{3}}{3\,c}}-{\frac{ \left ( \arctan \left ( ax \right ) \right ) ^{2}}{3\,c{x}^{3}}}+{\frac{{a}^{2} \left ( \arctan \left ( ax \right ) \right ) ^{2}}{cx}}+{\frac{4\,{a}^{3}\arctan \left ( ax \right ) \ln \left ({a}^{2}{x}^{2}+1 \right ) }{3\,c}}-{\frac{a\arctan \left ( ax \right ) }{3\,c{x}^{2}}}-{\frac{8\,{a}^{3}\arctan \left ( ax \right ) \ln \left ( ax \right ) }{3\,c}}-{\frac{{a}^{3}\arctan \left ( ax \right ) }{3\,c}}-{\frac{{a}^{2}}{3\,cx}}+{\frac{{\frac{4\,i}{3}}{a}^{3}{\it dilog} \left ( 1-iax \right ) }{c}}-{\frac{{\frac{2\,i}{3}}{a}^{3}\ln \left ( ax+i \right ) \ln \left ({a}^{2}{x}^{2}+1 \right ) }{c}}+{\frac{{\frac{2\,i}{3}}{a}^{3}\ln \left ({\frac{i}{2}} \left ( ax-i \right ) \right ) \ln \left ( ax+i \right ) }{c}}-{\frac{{\frac{i}{3}}{a}^{3} \left ( \ln \left ( ax-i \right ) \right ) ^{2}}{c}}-{\frac{{\frac{4\,i}{3}}{a}^{3}{\it dilog} \left ( 1+iax \right ) }{c}}-{\frac{{\frac{4\,i}{3}}{a}^{3}\ln \left ( ax \right ) \ln \left ( 1+iax \right ) }{c}}+{\frac{{\frac{2\,i}{3}}{a}^{3}\ln \left ( ax-i \right ) \ln \left ({a}^{2}{x}^{2}+1 \right ) }{c}}-{\frac{{\frac{2\,i}{3}}{a}^{3}\ln \left ( ax-i \right ) \ln \left ( -{\frac{i}{2}} \left ( ax+i \right ) \right ) }{c}}+{\frac{{\frac{4\,i}{3}}{a}^{3}\ln \left ( ax \right ) \ln \left ( 1-iax \right ) }{c}}-{\frac{{\frac{2\,i}{3}}{a}^{3}{\it dilog} \left ( -{\frac{i}{2}} \left ( ax+i \right ) \right ) }{c}}+{\frac{{\frac{2\,i}{3}}{a}^{3}{\it dilog} \left ({\frac{i}{2}} \left ( ax-i \right ) \right ) }{c}}+{\frac{{\frac{i}{3}}{a}^{3} \left ( \ln \left ( ax+i \right ) \right ) ^{2}}{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(a*x)^2/x^4/(a^2*c*x^2+c),x)

[Out]

1/3*a^3*arctan(a*x)^3/c-1/3*arctan(a*x)^2/c/x^3+a^2*arctan(a*x)^2/c/x+4/3*a^3/c*arctan(a*x)*ln(a^2*x^2+1)-1/3*
a*arctan(a*x)/c/x^2-8/3*a^3/c*arctan(a*x)*ln(a*x)-1/3*a^3*arctan(a*x)/c-1/3*a^2/c/x+4/3*I*a^3/c*dilog(1-I*a*x)
-2/3*I*a^3/c*ln(a*x+I)*ln(a^2*x^2+1)+2/3*I*a^3/c*ln(1/2*I*(a*x-I))*ln(a*x+I)-1/3*I*a^3/c*ln(a*x-I)^2-4/3*I*a^3
/c*dilog(1+I*a*x)-4/3*I*a^3/c*ln(a*x)*ln(1+I*a*x)+2/3*I*a^3/c*ln(a*x-I)*ln(a^2*x^2+1)-2/3*I*a^3/c*ln(a*x-I)*ln
(-1/2*I*(a*x+I))+4/3*I*a^3/c*ln(a*x)*ln(1-I*a*x)-2/3*I*a^3/c*dilog(-1/2*I*(a*x+I))+2/3*I*a^3/c*dilog(1/2*I*(a*
x-I))+1/3*I*a^3/c*ln(a*x+I)^2

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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(arctan(a*x)^2/x^4/(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{\arctan \left (a x\right )^{2}}{a^{2} c x^{6} + c x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^2/x^4/(a^2*c*x^2+c),x, algorithm="fricas")

[Out]

integral(arctan(a*x)^2/(a^2*c*x^6 + c*x^4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\operatorname{atan}^{2}{\left (a x \right )}}{a^{2} x^{6} + x^{4}}\, dx}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(a*x)**2/x**4/(a**2*c*x**2+c),x)

[Out]

Integral(atan(a*x)**2/(a**2*x**6 + x**4), x)/c

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^2/x^4/(a^2*c*x^2+c),x, algorithm="giac")

[Out]

integrate(arctan(a*x)^2/((a^2*c*x^2 + c)*x^4), x)