3.190 \(\int \frac{\tan ^{-1}(a x)}{x^3 (c+a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=156 \[ \frac{i a^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )}{c^2}+\frac{a^3 x}{4 c^2 \left (a^2 x^2+1\right )}-\frac{a^2 \tan ^{-1}(a x)}{2 c^2 \left (a^2 x^2+1\right )}+\frac{i a^2 \tan ^{-1}(a x)^2}{c^2}-\frac{a^2 \tan ^{-1}(a x)}{4 c^2}-\frac{2 a^2 \log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)}{c^2}-\frac{\tan ^{-1}(a x)}{2 c^2 x^2}-\frac{a}{2 c^2 x} \]

[Out]

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

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Rubi [A]  time = 0.405977, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 11, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.55, Rules used = {4966, 4918, 4852, 325, 203, 4924, 4868, 2447, 4930, 199, 205} \[ \frac{i a^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )}{c^2}+\frac{a^3 x}{4 c^2 \left (a^2 x^2+1\right )}-\frac{a^2 \tan ^{-1}(a x)}{2 c^2 \left (a^2 x^2+1\right )}+\frac{i a^2 \tan ^{-1}(a x)^2}{c^2}-\frac{a^2 \tan ^{-1}(a x)}{4 c^2}-\frac{2 a^2 \log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)}{c^2}-\frac{\tan ^{-1}(a x)}{2 c^2 x^2}-\frac{a}{2 c^2 x} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a*x]/(x^3*(c + a^2*c*x^2)^2),x]

[Out]

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

Rule 4966

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

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 4930

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\tan ^{-1}(a x)}{x^3 \left (c+a^2 c x^2\right )^2} \, dx &=-\left (a^2 \int \frac{\tan ^{-1}(a x)}{x \left (c+a^2 c x^2\right )^2} \, dx\right )+\frac{\int \frac{\tan ^{-1}(a x)}{x^3 \left (c+a^2 c x^2\right )} \, dx}{c}\\ &=a^4 \int \frac{x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx+\frac{\int \frac{\tan ^{-1}(a x)}{x^3} \, dx}{c^2}-2 \frac{a^2 \int \frac{\tan ^{-1}(a x)}{x \left (c+a^2 c x^2\right )} \, dx}{c}\\ &=-\frac{\tan ^{-1}(a x)}{2 c^2 x^2}-\frac{a^2 \tan ^{-1}(a x)}{2 c^2 \left (1+a^2 x^2\right )}+\frac{1}{2} a^3 \int \frac{1}{\left (c+a^2 c x^2\right )^2} \, dx+\frac{a \int \frac{1}{x^2 \left (1+a^2 x^2\right )} \, dx}{2 c^2}-2 \left (-\frac{i a^2 \tan ^{-1}(a x)^2}{2 c^2}+\frac{\left (i a^2\right ) \int \frac{\tan ^{-1}(a x)}{x (i+a x)} \, dx}{c^2}\right )\\ &=-\frac{a}{2 c^2 x}+\frac{a^3 x}{4 c^2 \left (1+a^2 x^2\right )}-\frac{\tan ^{-1}(a x)}{2 c^2 x^2}-\frac{a^2 \tan ^{-1}(a x)}{2 c^2 \left (1+a^2 x^2\right )}-\frac{a^3 \int \frac{1}{1+a^2 x^2} \, dx}{2 c^2}-2 \left (-\frac{i a^2 \tan ^{-1}(a x)^2}{2 c^2}+\frac{a^2 \tan ^{-1}(a x) \log \left (2-\frac{2}{1-i a x}\right )}{c^2}-\frac{a^3 \int \frac{\log \left (2-\frac{2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c^2}\right )+\frac{a^3 \int \frac{1}{c+a^2 c x^2} \, dx}{4 c}\\ &=-\frac{a}{2 c^2 x}+\frac{a^3 x}{4 c^2 \left (1+a^2 x^2\right )}-\frac{a^2 \tan ^{-1}(a x)}{4 c^2}-\frac{\tan ^{-1}(a x)}{2 c^2 x^2}-\frac{a^2 \tan ^{-1}(a x)}{2 c^2 \left (1+a^2 x^2\right )}-2 \left (-\frac{i a^2 \tan ^{-1}(a x)^2}{2 c^2}+\frac{a^2 \tan ^{-1}(a x) \log \left (2-\frac{2}{1-i a x}\right )}{c^2}-\frac{i a^2 \text{Li}_2\left (-1+\frac{2}{1-i a x}\right )}{2 c^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.391501, size = 93, normalized size = 0.6 \[ \frac{a^2 \left (8 i \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(a x)}\right )+\tan ^{-1}(a x) \left (-\frac{4}{a^2 x^2}-16 \log \left (1-e^{2 i \tan ^{-1}(a x)}\right )-2 \cos \left (2 \tan ^{-1}(a x)\right )-4\right )-\frac{4}{a x}+8 i \tan ^{-1}(a x)^2+\sin \left (2 \tan ^{-1}(a x)\right )\right )}{8 c^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTan[a*x]/(x^3*(c + a^2*c*x^2)^2),x]

[Out]

(a^2*(-4/(a*x) + (8*I)*ArcTan[a*x]^2 + ArcTan[a*x]*(-4 - 4/(a^2*x^2) - 2*Cos[2*ArcTan[a*x]] - 16*Log[1 - E^((2
*I)*ArcTan[a*x])]) + (8*I)*PolyLog[2, E^((2*I)*ArcTan[a*x])] + Sin[2*ArcTan[a*x]]))/(8*c^2)

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Maple [B]  time = 0.108, size = 369, normalized size = 2.4 \begin{align*}{\frac{{a}^{2}\arctan \left ( ax \right ) \ln \left ({a}^{2}{x}^{2}+1 \right ) }{{c}^{2}}}-{\frac{{a}^{2}\arctan \left ( ax \right ) }{2\,{c}^{2} \left ({a}^{2}{x}^{2}+1 \right ) }}-{\frac{\arctan \left ( ax \right ) }{2\,{c}^{2}{x}^{2}}}-2\,{\frac{{a}^{2}\arctan \left ( ax \right ) \ln \left ( ax \right ) }{{c}^{2}}}-{\frac{{\frac{i}{2}}{a}^{2}{\it dilog} \left ( -{\frac{i}{2}} \left ( ax+i \right ) \right ) }{{c}^{2}}}+{\frac{i{a}^{2}{\it dilog} \left ( 1-iax \right ) }{{c}^{2}}}-{\frac{{\frac{i}{2}}{a}^{2}\ln \left ( ax+i \right ) \ln \left ({a}^{2}{x}^{2}+1 \right ) }{{c}^{2}}}+{\frac{i{a}^{2}\ln \left ( ax \right ) \ln \left ( 1-iax \right ) }{{c}^{2}}}-{\frac{{\frac{i}{2}}{a}^{2}\ln \left ( ax-i \right ) \ln \left ( -{\frac{i}{2}} \left ( ax+i \right ) \right ) }{{c}^{2}}}+{\frac{{\frac{i}{2}}{a}^{2}{\it dilog} \left ({\frac{i}{2}} \left ( ax-i \right ) \right ) }{{c}^{2}}}+{\frac{{\frac{i}{2}}{a}^{2}\ln \left ( ax+i \right ) \ln \left ({\frac{i}{2}} \left ( ax-i \right ) \right ) }{{c}^{2}}}+{\frac{{\frac{i}{2}}{a}^{2}\ln \left ( ax-i \right ) \ln \left ({a}^{2}{x}^{2}+1 \right ) }{{c}^{2}}}+{\frac{{\frac{i}{4}}{a}^{2} \left ( \ln \left ( ax+i \right ) \right ) ^{2}}{{c}^{2}}}-{\frac{i{a}^{2}\ln \left ( ax \right ) \ln \left ( 1+iax \right ) }{{c}^{2}}}-{\frac{{\frac{i}{4}}{a}^{2} \left ( \ln \left ( ax-i \right ) \right ) ^{2}}{{c}^{2}}}-{\frac{i{a}^{2}{\it dilog} \left ( 1+iax \right ) }{{c}^{2}}}+{\frac{{a}^{3}x}{4\,{c}^{2} \left ({a}^{2}{x}^{2}+1 \right ) }}-{\frac{{a}^{2}\arctan \left ( ax \right ) }{4\,{c}^{2}}}-{\frac{a}{2\,{c}^{2}x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(a*x)/x^3/(a^2*c*x^2+c)^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )}^{2} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)/x^3/(a^2*c*x^2+c)^2,x, algorithm="maxima")

[Out]

integrate(arctan(a*x)/((a^2*c*x^2 + c)^2*x^3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\arctan \left (a x\right )}{a^{4} c^{2} x^{7} + 2 \, a^{2} c^{2} x^{5} + c^{2} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)/x^3/(a^2*c*x^2+c)^2,x, algorithm="fricas")

[Out]

integral(arctan(a*x)/(a^4*c^2*x^7 + 2*a^2*c^2*x^5 + c^2*x^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\operatorname{atan}{\left (a x \right )}}{a^{4} x^{7} + 2 a^{2} x^{5} + x^{3}}\, dx}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(a*x)/x**3/(a**2*c*x**2+c)**2,x)

[Out]

Integral(atan(a*x)/(a**4*x**7 + 2*a**2*x**5 + x**3), x)/c**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )}^{2} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)/x^3/(a^2*c*x^2+c)^2,x, algorithm="giac")

[Out]

integrate(arctan(a*x)/((a^2*c*x^2 + c)^2*x^3), x)