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

Optimal. Leaf size=159 \[ -\frac{i \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )}{2 c^3}-\frac{11 a x}{32 c^3 \left (a^2 x^2+1\right )}-\frac{a x}{16 c^3 \left (a^2 x^2+1\right )^2}+\frac{\tan ^{-1}(a x)}{2 c^3 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac{i \tan ^{-1}(a x)^2}{2 c^3}-\frac{11 \tan ^{-1}(a x)}{32 c^3}+\frac{\log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)}{c^3} \]

[Out]

-(a*x)/(16*c^3*(1 + a^2*x^2)^2) - (11*a*x)/(32*c^3*(1 + a^2*x^2)) - (11*ArcTan[a*x])/(32*c^3) + ArcTan[a*x]/(4
*c^3*(1 + a^2*x^2)^2) + ArcTan[a*x]/(2*c^3*(1 + a^2*x^2)) - ((I/2)*ArcTan[a*x]^2)/c^3 + (ArcTan[a*x]*Log[2 - 2
/(1 - I*a*x)])/c^3 - ((I/2)*PolyLog[2, -1 + 2/(1 - I*a*x)])/c^3

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Rubi [A]  time = 0.283952, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {4966, 4924, 4868, 2447, 4930, 199, 205} \[ -\frac{i \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )}{2 c^3}-\frac{11 a x}{32 c^3 \left (a^2 x^2+1\right )}-\frac{a x}{16 c^3 \left (a^2 x^2+1\right )^2}+\frac{\tan ^{-1}(a x)}{2 c^3 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac{i \tan ^{-1}(a x)^2}{2 c^3}-\frac{11 \tan ^{-1}(a x)}{32 c^3}+\frac{\log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)}{c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*x)/(16*c^3*(1 + a^2*x^2)^2) - (11*a*x)/(32*c^3*(1 + a^2*x^2)) - (11*ArcTan[a*x])/(32*c^3) + ArcTan[a*x]/(4
*c^3*(1 + a^2*x^2)^2) + ArcTan[a*x]/(2*c^3*(1 + a^2*x^2)) - ((I/2)*ArcTan[a*x]^2)/c^3 + (ArcTan[a*x]*Log[2 - 2
/(1 - I*a*x)])/c^3 - ((I/2)*PolyLog[2, -1 + 2/(1 - I*a*x)])/c^3

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 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 \left (c+a^2 c x^2\right )^3} \, dx &=-\left (a^2 \int \frac{x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^3} \, dx\right )+\frac{\int \frac{\tan ^{-1}(a x)}{x \left (c+a^2 c x^2\right )^2} \, dx}{c}\\ &=\frac{\tan ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac{1}{4} a \int \frac{1}{\left (c+a^2 c x^2\right )^3} \, dx+\frac{\int \frac{\tan ^{-1}(a x)}{x \left (c+a^2 c x^2\right )} \, dx}{c^2}-\frac{a^2 \int \frac{x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{c}\\ &=-\frac{a x}{16 c^3 \left (1+a^2 x^2\right )^2}+\frac{\tan ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{\tan ^{-1}(a x)}{2 c^3 \left (1+a^2 x^2\right )}-\frac{i \tan ^{-1}(a x)^2}{2 c^3}+\frac{i \int \frac{\tan ^{-1}(a x)}{x (i+a x)} \, dx}{c^3}-\frac{(3 a) \int \frac{1}{\left (c+a^2 c x^2\right )^2} \, dx}{16 c}-\frac{a \int \frac{1}{\left (c+a^2 c x^2\right )^2} \, dx}{2 c}\\ &=-\frac{a x}{16 c^3 \left (1+a^2 x^2\right )^2}-\frac{11 a x}{32 c^3 \left (1+a^2 x^2\right )}+\frac{\tan ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{\tan ^{-1}(a x)}{2 c^3 \left (1+a^2 x^2\right )}-\frac{i \tan ^{-1}(a x)^2}{2 c^3}+\frac{\tan ^{-1}(a x) \log \left (2-\frac{2}{1-i a x}\right )}{c^3}-\frac{a \int \frac{\log \left (2-\frac{2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c^3}-\frac{(3 a) \int \frac{1}{c+a^2 c x^2} \, dx}{32 c^2}-\frac{a \int \frac{1}{c+a^2 c x^2} \, dx}{4 c^2}\\ &=-\frac{a x}{16 c^3 \left (1+a^2 x^2\right )^2}-\frac{11 a x}{32 c^3 \left (1+a^2 x^2\right )}-\frac{11 \tan ^{-1}(a x)}{32 c^3}+\frac{\tan ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{\tan ^{-1}(a x)}{2 c^3 \left (1+a^2 x^2\right )}-\frac{i \tan ^{-1}(a x)^2}{2 c^3}+\frac{\tan ^{-1}(a x) \log \left (2-\frac{2}{1-i a x}\right )}{c^3}-\frac{i \text{Li}_2\left (-1+\frac{2}{1-i a x}\right )}{2 c^3}\\ \end{align*}

Mathematica [A]  time = 0.221693, size = 90, normalized size = 0.57 \[ -\frac{64 i \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(a x)}\right )+64 i \tan ^{-1}(a x)^2+24 \sin \left (2 \tan ^{-1}(a x)\right )+\sin \left (4 \tan ^{-1}(a x)\right )-4 \tan ^{-1}(a x) \left (32 \log \left (1-e^{2 i \tan ^{-1}(a x)}\right )+12 \cos \left (2 \tan ^{-1}(a x)\right )+\cos \left (4 \tan ^{-1}(a x)\right )\right )}{128 c^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((64*I)*ArcTan[a*x]^2 - 4*ArcTan[a*x]*(12*Cos[2*ArcTan[a*x]] + Cos[4*ArcTan[a*x]] + 32*Log[1 - E^((2*I)*ArcTa
n[a*x])]) + (64*I)*PolyLog[2, E^((2*I)*ArcTan[a*x])] + 24*Sin[2*ArcTan[a*x]] + Sin[4*ArcTan[a*x]])/(128*c^3)

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Maple [B]  time = 0.102, size = 340, normalized size = 2.1 \begin{align*}{\frac{\arctan \left ( ax \right ) }{4\,{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{\arctan \left ( ax \right ) \ln \left ({a}^{2}{x}^{2}+1 \right ) }{2\,{c}^{3}}}+{\frac{\arctan \left ( ax \right ) }{2\,{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) }}+{\frac{\arctan \left ( ax \right ) \ln \left ( ax \right ) }{{c}^{3}}}-{\frac{11\,{a}^{3}{x}^{3}}{32\,{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{13\,ax}{32\,{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{11\,\arctan \left ( ax \right ) }{32\,{c}^{3}}}+{\frac{{\frac{i}{8}} \left ( \ln \left ( ax-i \right ) \right ) ^{2}}{{c}^{3}}}-{\frac{{\frac{i}{4}}\ln \left ( ax-i \right ) \ln \left ({a}^{2}{x}^{2}+1 \right ) }{{c}^{3}}}-{\frac{{\frac{i}{4}}\ln \left ( ax+i \right ) \ln \left ({\frac{i}{2}} \left ( ax-i \right ) \right ) }{{c}^{3}}}+{\frac{{\frac{i}{4}}\ln \left ( ax+i \right ) \ln \left ({a}^{2}{x}^{2}+1 \right ) }{{c}^{3}}}-{\frac{{\frac{i}{8}} \left ( \ln \left ( ax+i \right ) \right ) ^{2}}{{c}^{3}}}+{\frac{{\frac{i}{4}}\ln \left ( ax-i \right ) \ln \left ( -{\frac{i}{2}} \left ( ax+i \right ) \right ) }{{c}^{3}}}+{\frac{{\frac{i}{2}}{\it dilog} \left ( 1+iax \right ) }{{c}^{3}}}-{\frac{{\frac{i}{2}}\ln \left ( ax \right ) \ln \left ( 1-iax \right ) }{{c}^{3}}}-{\frac{{\frac{i}{2}}{\it dilog} \left ( 1-iax \right ) }{{c}^{3}}}+{\frac{{\frac{i}{2}}\ln \left ( ax \right ) \ln \left ( 1+iax \right ) }{{c}^{3}}}+{\frac{{\frac{i}{4}}{\it dilog} \left ( -{\frac{i}{2}} \left ( ax+i \right ) \right ) }{{c}^{3}}}-{\frac{{\frac{i}{4}}{\it dilog} \left ({\frac{i}{2}} \left ( ax-i \right ) \right ) }{{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*arctan(a*x)/c^3/(a^2*x^2+1)^2-1/2/c^3*arctan(a*x)*ln(a^2*x^2+1)+1/2*arctan(a*x)/c^3/(a^2*x^2+1)+1/c^3*arct
an(a*x)*ln(a*x)-11/32/c^3/(a^2*x^2+1)^2*x^3*a^3-13/32*a*x/c^3/(a^2*x^2+1)^2-11/32*arctan(a*x)/c^3+1/8*I/c^3*ln
(a*x-I)^2-1/4*I/c^3*ln(a*x-I)*ln(a^2*x^2+1)-1/4*I/c^3*ln(a*x+I)*ln(1/2*I*(a*x-I))+1/4*I/c^3*ln(a*x+I)*ln(a^2*x
^2+1)-1/8*I/c^3*ln(a*x+I)^2+1/4*I/c^3*ln(a*x-I)*ln(-1/2*I*(a*x+I))+1/2*I/c^3*dilog(1+I*a*x)-1/2*I/c^3*ln(a*x)*
ln(1-I*a*x)-1/2*I/c^3*dilog(1-I*a*x)+1/2*I/c^3*ln(a*x)*ln(1+I*a*x)+1/4*I/c^3*dilog(-1/2*I*(a*x+I))-1/4*I/c^3*d
ilog(1/2*I*(a*x-I))

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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 )}^{3} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arctan(a*x)/((a^2*c*x^2 + c)^3*x), 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^{6} c^{3} x^{7} + 3 \, a^{4} c^{3} x^{5} + 3 \, a^{2} c^{3} x^{3} + c^{3} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctan(a*x)/(a^6*c^3*x^7 + 3*a^4*c^3*x^5 + 3*a^2*c^3*x^3 + c^3*x), x)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RecursionError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RecursionError

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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 )}^{3} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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