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

Optimal. Leaf size=135 \[ -\frac{2}{3} i a^3 c \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )-\frac{a^2 c}{3 x}-\frac{2}{3} i a^3 c \tan ^{-1}(a x)^2-\frac{1}{3} a^3 c \tan ^{-1}(a x)-\frac{a^2 c \tan ^{-1}(a x)^2}{x}+\frac{4}{3} a^3 c \log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)-\frac{a c \tan ^{-1}(a x)}{3 x^2}-\frac{c \tan ^{-1}(a x)^2}{3 x^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.312138, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4950, 4852, 4918, 325, 203, 4924, 4868, 2447} \[ -\frac{2}{3} i a^3 c \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )-\frac{a^2 c}{3 x}-\frac{2}{3} i a^3 c \tan ^{-1}(a x)^2-\frac{1}{3} a^3 c \tan ^{-1}(a x)-\frac{a^2 c \tan ^{-1}(a x)^2}{x}+\frac{4}{3} a^3 c \log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)-\frac{a c \tan ^{-1}(a x)}{3 x^2}-\frac{c \tan ^{-1}(a x)^2}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4950

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

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

Mathematica [A]  time = 0.590859, size = 103, normalized size = 0.76 \[ \frac{c \left (-2 i a^3 x^3 \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(a x)}\right )-a^2 x^2+a x \tan ^{-1}(a x) \left (-a^2 x^2+4 a^2 x^2 \log \left (1-e^{2 i \tan ^{-1}(a x)}\right )-1\right )+(1-2 i a x) (a x-i)^2 \tan ^{-1}(a x)^2\right )}{3 x^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.096, size = 323, normalized size = 2.4 \begin{align*} -{\frac{{a}^{2}c \left ( \arctan \left ( ax \right ) \right ) ^{2}}{x}}-{\frac{c \left ( \arctan \left ( ax \right ) \right ) ^{2}}{3\,{x}^{3}}}-{\frac{2\,{a}^{3}c\arctan \left ( ax \right ) \ln \left ({a}^{2}{x}^{2}+1 \right ) }{3}}-{\frac{ac\arctan \left ( ax \right ) }{3\,{x}^{2}}}+{\frac{4\,{a}^{3}c\arctan \left ( ax \right ) \ln \left ( ax \right ) }{3}}-{\frac{{a}^{3}c\arctan \left ( ax \right ) }{3}}-{\frac{{a}^{2}c}{3\,x}}-{\frac{i}{3}}{a}^{3}c\ln \left ( ax+i \right ) \ln \left ({\frac{i}{2}} \left ( ax-i \right ) \right ) +{\frac{i}{3}}{a}^{3}c{\it dilog} \left ( -{\frac{i}{2}} \left ( ax+i \right ) \right ) -{\frac{i}{3}}{a}^{3}c\ln \left ({a}^{2}{x}^{2}+1 \right ) \ln \left ( ax-i \right ) -{\frac{i}{3}}{a}^{3}c{\it dilog} \left ({\frac{i}{2}} \left ( ax-i \right ) \right ) -{\frac{2\,i}{3}}{a}^{3}c{\it dilog} \left ( 1-iax \right ) +{\frac{2\,i}{3}}{a}^{3}c{\it dilog} \left ( 1+iax \right ) +{\frac{2\,i}{3}}{a}^{3}c\ln \left ( ax \right ) \ln \left ( 1+iax \right ) -{\frac{2\,i}{3}}{a}^{3}c\ln \left ( ax \right ) \ln \left ( 1-iax \right ) +{\frac{i}{3}}{a}^{3}c\ln \left ( ax-i \right ) \ln \left ( -{\frac{i}{2}} \left ( ax+i \right ) \right ) +{\frac{i}{6}}{a}^{3}c \left ( \ln \left ( ax-i \right ) \right ) ^{2}+{\frac{i}{3}}{a}^{3}c\ln \left ({a}^{2}{x}^{2}+1 \right ) \ln \left ( ax+i \right ) -{\frac{i}{6}}{a}^{3}c \left ( \ln \left ( ax+i \right ) \right ) ^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2*c*arctan(a*x)^2/x-1/3*c*arctan(a*x)^2/x^3-2/3*a^3*c*arctan(a*x)*ln(a^2*x^2+1)-1/3*a*c*arctan(a*x)/x^2+4/3
*a^3*c*arctan(a*x)*ln(a*x)-1/3*a^3*c*arctan(a*x)-1/3*a^2*c/x-1/3*I*a^3*c*ln(a*x+I)*ln(1/2*I*(a*x-I))+1/3*I*a^3
*c*dilog(-1/2*I*(a*x+I))-1/3*I*a^3*c*ln(a^2*x^2+1)*ln(a*x-I)-1/3*I*a^3*c*dilog(1/2*I*(a*x-I))-2/3*I*a^3*c*dilo
g(1-I*a*x)+2/3*I*a^3*c*dilog(1+I*a*x)+2/3*I*a^3*c*ln(a*x)*ln(1+I*a*x)-2/3*I*a^3*c*ln(a*x)*ln(1-I*a*x)+1/3*I*a^
3*c*ln(a*x-I)*ln(-1/2*I*(a*x+I))+1/6*I*a^3*c*ln(a*x-I)^2+1/3*I*a^3*c*ln(a^2*x^2+1)*ln(a*x+I)-1/6*I*a^3*c*ln(a*
x+I)^2

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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