\(\int \frac {\arctan (x) \log (1+x^2)}{x^4} \, dx\) [1283]

Optimal result

Integrand size = 12, antiderivative size = 81 \[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^4} \, dx=-\frac {2 \arctan (x)}{3 x}-\frac {\arctan (x)^2}{3}+\log (x)-\frac {1}{2} \log \left (1+x^2\right )-\frac {\log \left (1+x^2\right )}{6 x^2}-\frac {\arctan (x) \log \left (1+x^2\right )}{3 x^3}+\frac {1}{12} \log ^2\left (1+x^2\right )+\frac {\operatorname {PolyLog}\left (2,-x^2\right )}{6} \]

[Out]

-2/3*arctan(x)/x-1/3*arctan(x)^2+ln(x)-1/2*ln(x^2+1)-1/6*ln(x^2+1)/x^2-1/3*arctan(x)*ln(x^2+1)/x^3+1/12*ln(x^2
+1)^2+1/6*polylog(2,-x^2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.250, Rules used = {4946, 272, 46, 5137, 2525, 2457, 2442, 36, 29, 31, 2438, 2437, 2338, 5038, 5004} \[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^4} \, dx=-\frac {\arctan (x) \log \left (x^2+1\right )}{3 x^3}-\frac {1}{3} \arctan (x)^2-\frac {2 \arctan (x)}{3 x}+\frac {\operatorname {PolyLog}\left (2,-x^2\right )}{6}+\frac {1}{12} \log ^2\left (x^2+1\right )-\frac {\log \left (x^2+1\right )}{6 x^2}-\frac {1}{2} \log \left (x^2+1\right )+\log (x) \]

[In]

Int[(ArcTan[x]*Log[1 + x^2])/x^4,x]

[Out]

(-2*ArcTan[x])/(3*x) - ArcTan[x]^2/3 + Log[x] - Log[1 + x^2]/2 - Log[1 + x^2]/(6*x^2) - (ArcTan[x]*Log[1 + x^2
])/(3*x^3) + Log[1 + x^2]^2/12 + PolyLog[2, -x^2]/6

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2437

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[(f*(x/d))^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2442

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[(f + g*
x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/(g*(q + 1))), x] - Dist[b*e*(n/(g*(q + 1))), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 2457

Int[(Log[(c_.)*((d_) + (e_.)*(x_))]*(x_)^(m_.))/((f_) + (g_.)*(x_)), x_Symbol] :> Int[ExpandIntegrand[Log[c*(d
+ e*x)], x^m/(f + g*x), x], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[e*f - d*g, 0] && EqQ[c*d, 1] && IntegerQ[m
]

Rule 2525

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rule 4946

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

Rule 5004

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]

Rule 5038

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 5137

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.))*(x_)^(m_.), x_Symbol] :> Simp
[x^(m + 1)*(d + e*Log[f + g*x^2])*((a + b*ArcTan[c*x])/(m + 1)), x] + (-Dist[b*(c/(m + 1)), Int[x^(m + 1)*((d
+ e*Log[f + g*x^2])/(1 + c^2*x^2)), x], x] - Dist[2*e*(g/(m + 1)), Int[x^(m + 2)*((a + b*ArcTan[c*x])/(f + g*x
^2)), x], x]) /; FreeQ[{a, b, c, d, e, f, g}, x] && ILtQ[m/2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\arctan (x) \log \left (1+x^2\right )}{3 x^3}+\frac {1}{3} \int \frac {\log \left (1+x^2\right )}{x^3 \left (1+x^2\right )} \, dx+\frac {2}{3} \int \frac {\arctan (x)}{x^2 \left (1+x^2\right )} \, dx \\ & = -\frac {\arctan (x) \log \left (1+x^2\right )}{3 x^3}+\frac {1}{6} \text {Subst}\left (\int \frac {\log (1+x)}{x^2 (1+x)} \, dx,x,x^2\right )+\frac {2}{3} \int \frac {\arctan (x)}{x^2} \, dx-\frac {2}{3} \int \frac {\arctan (x)}{1+x^2} \, dx \\ & = -\frac {2 \arctan (x)}{3 x}-\frac {\arctan (x)^2}{3}-\frac {\arctan (x) \log \left (1+x^2\right )}{3 x^3}+\frac {1}{6} \text {Subst}\left (\int \left (\frac {\log (1+x)}{x^2}-\frac {\log (1+x)}{x}+\frac {\log (1+x)}{1+x}\right ) \, dx,x,x^2\right )+\frac {2}{3} \int \frac {1}{x \left (1+x^2\right )} \, dx \\ & = -\frac {2 \arctan (x)}{3 x}-\frac {\arctan (x)^2}{3}-\frac {\arctan (x) \log \left (1+x^2\right )}{3 x^3}+\frac {1}{6} \text {Subst}\left (\int \frac {\log (1+x)}{x^2} \, dx,x,x^2\right )-\frac {1}{6} \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,x^2\right )+\frac {1}{6} \text {Subst}\left (\int \frac {\log (1+x)}{1+x} \, dx,x,x^2\right )+\frac {1}{3} \text {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,x^2\right ) \\ & = -\frac {2 \arctan (x)}{3 x}-\frac {\arctan (x)^2}{3}-\frac {\log \left (1+x^2\right )}{6 x^2}-\frac {\arctan (x) \log \left (1+x^2\right )}{3 x^3}+\frac {\operatorname {PolyLog}\left (2,-x^2\right )}{6}+\frac {1}{6} \text {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,x^2\right )+\frac {1}{6} \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1+x^2\right )+\frac {1}{3} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{3} \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^2\right ) \\ & = -\frac {2 \arctan (x)}{3 x}-\frac {\arctan (x)^2}{3}+\frac {2 \log (x)}{3}-\frac {1}{3} \log \left (1+x^2\right )-\frac {\log \left (1+x^2\right )}{6 x^2}-\frac {\arctan (x) \log \left (1+x^2\right )}{3 x^3}+\frac {1}{12} \log ^2\left (1+x^2\right )+\frac {\operatorname {PolyLog}\left (2,-x^2\right )}{6}+\frac {1}{6} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{6} \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^2\right ) \\ & = -\frac {2 \arctan (x)}{3 x}-\frac {\arctan (x)^2}{3}+\log (x)-\frac {1}{2} \log \left (1+x^2\right )-\frac {\log \left (1+x^2\right )}{6 x^2}-\frac {\arctan (x) \log \left (1+x^2\right )}{3 x^3}+\frac {1}{12} \log ^2\left (1+x^2\right )+\frac {\operatorname {PolyLog}\left (2,-x^2\right )}{6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00 \[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^4} \, dx=-\frac {2 \arctan (x)}{3 x}-\frac {\arctan (x)^2}{3}+\log (x)-\frac {1}{2} \log \left (1+x^2\right )-\frac {\log \left (1+x^2\right )}{6 x^2}-\frac {\arctan (x) \log \left (1+x^2\right )}{3 x^3}+\frac {1}{12} \log ^2\left (1+x^2\right )+\frac {\operatorname {PolyLog}\left (2,-x^2\right )}{6} \]

[In]

Integrate[(ArcTan[x]*Log[1 + x^2])/x^4,x]

[Out]

(-2*ArcTan[x])/(3*x) - ArcTan[x]^2/3 + Log[x] - Log[1 + x^2]/2 - Log[1 + x^2]/(6*x^2) - (ArcTan[x]*Log[1 + x^2
])/(3*x^3) + Log[1 + x^2]^2/12 + PolyLog[2, -x^2]/6

Maple [F]

\[\int \frac {\arctan \left (x \right ) \ln \left (x^{2}+1\right )}{x^{4}}d x\]

[In]

int(arctan(x)*ln(x^2+1)/x^4,x)

[Out]

int(arctan(x)*ln(x^2+1)/x^4,x)

Fricas [F]

\[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^4} \, dx=\int { \frac {\arctan \left (x\right ) \log \left (x^{2} + 1\right )}{x^{4}} \,d x } \]

[In]

integrate(arctan(x)*log(x^2+1)/x^4,x, algorithm="fricas")

[Out]

integral(arctan(x)*log(x^2 + 1)/x^4, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 12.99 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.20 \[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^4} \, dx=\frac {2 \log {\left (x \right )}}{3} + \frac {\log {\left (2 x^{2} \right )}}{6} + \frac {\log {\left (x^{2} + 1 \right )}^{2}}{12} - \frac {\log {\left (x^{2} + 1 \right )}}{3} - \frac {\log {\left (2 x^{2} + 2 \right )}}{6} - \frac {\operatorname {atan}^{2}{\left (x \right )}}{3} + \frac {\operatorname {Li}_{2}\left (x^{2} e^{i \pi }\right )}{6} - \frac {2 \operatorname {atan}{\left (x \right )}}{3 x} - \frac {\log {\left (x^{2} + 1 \right )}}{6 x^{2}} - \frac {\log {\left (x^{2} + 1 \right )} \operatorname {atan}{\left (x \right )}}{3 x^{3}} \]

[In]

integrate(atan(x)*ln(x**2+1)/x**4,x)

[Out]

2*log(x)/3 + log(2*x**2)/6 + log(x**2 + 1)**2/12 - log(x**2 + 1)/3 - log(2*x**2 + 2)/6 - atan(x)**2/3 + polylo
g(2, x**2*exp_polar(I*pi))/6 - 2*atan(x)/(3*x) - log(x**2 + 1)/(6*x**2) - log(x**2 + 1)*atan(x)/(3*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.17 \[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^4} \, dx=-\frac {1}{3} \, {\left (\frac {2}{x} + \frac {\log \left (x^{2} + 1\right )}{x^{3}} + 2 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) + \frac {4 \, x^{2} \arctan \left (x\right )^{2} + x^{2} \log \left (x^{2} + 1\right )^{2} - 2 \, x^{2} {\rm Li}_2\left (x^{2} + 1\right ) + 12 \, x^{2} \log \left (x\right ) - 2 \, {\left (x^{2} \log \left (-x^{2}\right ) + 3 \, x^{2} + 1\right )} \log \left (x^{2} + 1\right )}{12 \, x^{2}} \]

[In]

integrate(arctan(x)*log(x^2+1)/x^4,x, algorithm="maxima")

[Out]

-1/3*(2/x + log(x^2 + 1)/x^3 + 2*arctan(x))*arctan(x) + 1/12*(4*x^2*arctan(x)^2 + x^2*log(x^2 + 1)^2 - 2*x^2*d
ilog(x^2 + 1) + 12*x^2*log(x) - 2*(x^2*log(-x^2) + 3*x^2 + 1)*log(x^2 + 1))/x^2

Giac [F]

\[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^4} \, dx=\int { \frac {\arctan \left (x\right ) \log \left (x^{2} + 1\right )}{x^{4}} \,d x } \]

[In]

integrate(arctan(x)*log(x^2+1)/x^4,x, algorithm="giac")

[Out]

integrate(arctan(x)*log(x^2 + 1)/x^4, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^4} \, dx=\int \frac {\ln \left (x^2+1\right )\,\mathrm {atan}\left (x\right )}{x^4} \,d x \]

[In]

int((log(x^2 + 1)*atan(x))/x^4,x)

[Out]

int((log(x^2 + 1)*atan(x))/x^4, x)