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

Optimal. Leaf size=113 \[ \frac{i a^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )}{2 c}+\frac{i a^2 \tan ^{-1}(a x)^2}{2 c}-\frac{a^2 \tan ^{-1}(a x)}{2 c}-\frac{a^2 \log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)}{c}-\frac{\tan ^{-1}(a x)}{2 c x^2}-\frac{a}{2 c x} \]

[Out]

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

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Rubi [A]  time = 0.162361, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {4918, 4852, 325, 203, 4924, 4868, 2447} \[ \frac{i a^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )}{2 c}+\frac{i a^2 \tan ^{-1}(a x)^2}{2 c}-\frac{a^2 \tan ^{-1}(a x)}{2 c}-\frac{a^2 \log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)}{c}-\frac{\tan ^{-1}(a x)}{2 c x^2}-\frac{a}{2 c x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [C]  time = 0.0606948, size = 142, normalized size = 1.26 \[ -\frac{a \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-a^2 x^2\right )}{2 c x}-\frac{a^2 \left (\frac{1}{2} i \text{PolyLog}(2,-i a x)-\frac{1}{2} i \text{PolyLog}(2,i a x)+\frac{1}{2} \left (i \text{PolyLog}\left (2,-\frac{a x+i}{-a x+i}\right )+2 \log \left (\frac{2 i}{-a x+i}\right ) \tan ^{-1}(a x)\right )+\frac{1}{2} i \tan ^{-1}(a x)^2\right )}{c}-\frac{\tan ^{-1}(a x)}{2 c x^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-ArcTan[a*x]/(2*c*x^2) - (a*Hypergeometric2F1[-1/2, 1, 1/2, -(a^2*x^2)])/(2*c*x) - (a^2*((I/2)*ArcTan[a*x]^2 +
(I/2)*PolyLog[2, (-I)*a*x] - (I/2)*PolyLog[2, I*a*x] + (2*ArcTan[a*x]*Log[(2*I)/(I - a*x)] + I*PolyLog[2, -((
I + a*x)/(I - a*x))])/2))/c

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Maple [B]  time = 0.095, size = 327, normalized size = 2.9 \begin{align*}{\frac{{a}^{2}\arctan \left ( ax \right ) \ln \left ({a}^{2}{x}^{2}+1 \right ) }{2\,c}}-{\frac{\arctan \left ( ax \right ) }{2\,c{x}^{2}}}-{\frac{{a}^{2}\arctan \left ( ax \right ) \ln \left ( ax \right ) }{c}}-{\frac{{\frac{i}{4}}{a}^{2}\ln \left ({a}^{2}{x}^{2}+1 \right ) \ln \left ( ax+i \right ) }{c}}-{\frac{{\frac{i}{4}}{a}^{2}\ln \left ( ax-i \right ) \ln \left ( -{\frac{i}{2}} \left ( ax+i \right ) \right ) }{c}}+{\frac{{\frac{i}{2}}{a}^{2}{\it dilog} \left ( 1-iax \right ) }{c}}-{\frac{{\frac{i}{2}}{a}^{2}\ln \left ( ax \right ) \ln \left ( 1+iax \right ) }{c}}+{\frac{{\frac{i}{4}}{a}^{2}\ln \left ({a}^{2}{x}^{2}+1 \right ) \ln \left ( ax-i \right ) }{c}}-{\frac{{\frac{i}{8}}{a}^{2} \left ( \ln \left ( ax-i \right ) \right ) ^{2}}{c}}-{\frac{{\frac{i}{4}}{a}^{2}{\it dilog} \left ( -{\frac{i}{2}} \left ( ax+i \right ) \right ) }{c}}+{\frac{{\frac{i}{4}}{a}^{2}\ln \left ( ax+i \right ) \ln \left ({\frac{i}{2}} \left ( ax-i \right ) \right ) }{c}}-{\frac{{a}^{2}\arctan \left ( ax \right ) }{2\,c}}-{\frac{a}{2\,cx}}+{\frac{{\frac{i}{2}}{a}^{2}\ln \left ( ax \right ) \ln \left ( 1-iax \right ) }{c}}-{\frac{{\frac{i}{2}}{a}^{2}{\it dilog} \left ( 1+iax \right ) }{c}}+{\frac{{\frac{i}{4}}{a}^{2}{\it dilog} \left ({\frac{i}{2}} \left ( ax-i \right ) \right ) }{c}}+{\frac{{\frac{i}{8}}{a}^{2} \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)/x^3/(a^2*c*x^2+c),x)

[Out]

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

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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 )} 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),x, algorithm="maxima")

[Out]

integrate(arctan(a*x)/((a^2*c*x^2 + c)*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^{2} c x^{5} + c 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),x, algorithm="fricas")

[Out]

integral(arctan(a*x)/(a^2*c*x^5 + c*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^{2} x^{5} + x^{3}}\, dx}{c} \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),x)

[Out]

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

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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 )} 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),x, algorithm="giac")

[Out]

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