\(\int \frac {\arctan (x) \log (1+x^2)}{x^6} \, dx\) [1285]

Optimal result

Integrand size = 12, antiderivative size = 114 \[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^6} \, dx=-\frac {7}{60 x^2}-\frac {2 \arctan (x)}{15 x^3}+\frac {2 \arctan (x)}{5 x}+\frac {\arctan (x)^2}{5}-\frac {5 \log (x)}{6}+\frac {5}{12} \log \left (1+x^2\right )-\frac {\log \left (1+x^2\right )}{20 x^4}+\frac {\log \left (1+x^2\right )}{10 x^2}-\frac {\arctan (x) \log \left (1+x^2\right )}{5 x^5}-\frac {1}{20} \log ^2\left (1+x^2\right )-\frac {\operatorname {PolyLog}\left (2,-x^2\right )}{10} \]

[Out]

-7/60/x^2-2/15*arctan(x)/x^3+2/5*arctan(x)/x+1/5*arctan(x)^2-5/6*ln(x)+5/12*ln(x^2+1)-1/20*ln(x^2+1)/x^4+1/10*
ln(x^2+1)/x^2-1/5*arctan(x)*ln(x^2+1)/x^5-1/20*ln(x^2+1)^2-1/10*polylog(2,-x^2)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.250, Rules used = {4946, 272, 46, 5137, 2525, 2457, 2437, 2338, 2442, 36, 29, 31, 2438, 5038, 5004} \[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^6} \, dx=-\frac {2 \arctan (x)}{15 x^3}-\frac {\arctan (x) \log \left (x^2+1\right )}{5 x^5}+\frac {\arctan (x)^2}{5}+\frac {2 \arctan (x)}{5 x}-\frac {\operatorname {PolyLog}\left (2,-x^2\right )}{10}-\frac {7}{60 x^2}-\frac {1}{20} \log ^2\left (x^2+1\right )+\frac {\log \left (x^2+1\right )}{10 x^2}+\frac {5}{12} \log \left (x^2+1\right )-\frac {\log \left (x^2+1\right )}{20 x^4}-\frac {5 \log (x)}{6} \]

[In]

Int[(ArcTan[x]*Log[1 + x^2])/x^6,x]

[Out]

-7/(60*x^2) - (2*ArcTan[x])/(15*x^3) + (2*ArcTan[x])/(5*x) + ArcTan[x]^2/5 - (5*Log[x])/6 + (5*Log[1 + x^2])/1
2 - Log[1 + x^2]/(20*x^4) + Log[1 + x^2]/(10*x^2) - (ArcTan[x]*Log[1 + x^2])/(5*x^5) - Log[1 + x^2]^2/20 - Pol
yLog[2, -x^2]/10

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 )}{5 x^5}+\frac {1}{5} \int \frac {\log \left (1+x^2\right )}{x^5 \left (1+x^2\right )} \, dx+\frac {2}{5} \int \frac {\arctan (x)}{x^4 \left (1+x^2\right )} \, dx \\ & = -\frac {\arctan (x) \log \left (1+x^2\right )}{5 x^5}+\frac {1}{10} \text {Subst}\left (\int \frac {\log (1+x)}{x^3 (1+x)} \, dx,x,x^2\right )+\frac {2}{5} \int \frac {\arctan (x)}{x^4} \, dx-\frac {2}{5} \int \frac {\arctan (x)}{x^2 \left (1+x^2\right )} \, dx \\ & = -\frac {2 \arctan (x)}{15 x^3}-\frac {\arctan (x) \log \left (1+x^2\right )}{5 x^5}+\frac {1}{10} \text {Subst}\left (\int \left (\frac {\log (1+x)}{-1-x}+\frac {\log (1+x)}{x^3}-\frac {\log (1+x)}{x^2}+\frac {\log (1+x)}{x}\right ) \, dx,x,x^2\right )+\frac {2}{15} \int \frac {1}{x^3 \left (1+x^2\right )} \, dx-\frac {2}{5} \int \frac {\arctan (x)}{x^2} \, dx+\frac {2}{5} \int \frac {\arctan (x)}{1+x^2} \, dx \\ & = -\frac {2 \arctan (x)}{15 x^3}+\frac {2 \arctan (x)}{5 x}+\frac {\arctan (x)^2}{5}-\frac {\arctan (x) \log \left (1+x^2\right )}{5 x^5}+\frac {1}{15} \text {Subst}\left (\int \frac {1}{x^2 (1+x)} \, dx,x,x^2\right )+\frac {1}{10} \text {Subst}\left (\int \frac {\log (1+x)}{-1-x} \, dx,x,x^2\right )+\frac {1}{10} \text {Subst}\left (\int \frac {\log (1+x)}{x^3} \, dx,x,x^2\right )-\frac {1}{10} \text {Subst}\left (\int \frac {\log (1+x)}{x^2} \, dx,x,x^2\right )+\frac {1}{10} \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,x^2\right )-\frac {2}{5} \int \frac {1}{x \left (1+x^2\right )} \, dx \\ & = -\frac {2 \arctan (x)}{15 x^3}+\frac {2 \arctan (x)}{5 x}+\frac {\arctan (x)^2}{5}-\frac {\log \left (1+x^2\right )}{20 x^4}+\frac {\log \left (1+x^2\right )}{10 x^2}-\frac {\arctan (x) \log \left (1+x^2\right )}{5 x^5}-\frac {\operatorname {PolyLog}\left (2,-x^2\right )}{10}+\frac {1}{20} \text {Subst}\left (\int \frac {1}{x^2 (1+x)} \, dx,x,x^2\right )+\frac {1}{15} \text {Subst}\left (\int \left (\frac {1}{x^2}-\frac {1}{x}+\frac {1}{1+x}\right ) \, dx,x,x^2\right )-\frac {1}{10} \text {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,x^2\right )-\frac {1}{10} \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1+x^2\right )-\frac {1}{5} \text {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,x^2\right ) \\ & = -\frac {1}{15 x^2}-\frac {2 \arctan (x)}{15 x^3}+\frac {2 \arctan (x)}{5 x}+\frac {\arctan (x)^2}{5}-\frac {2 \log (x)}{15}+\frac {1}{15} \log \left (1+x^2\right )-\frac {\log \left (1+x^2\right )}{20 x^4}+\frac {\log \left (1+x^2\right )}{10 x^2}-\frac {\arctan (x) \log \left (1+x^2\right )}{5 x^5}-\frac {1}{20} \log ^2\left (1+x^2\right )-\frac {\operatorname {PolyLog}\left (2,-x^2\right )}{10}+\frac {1}{20} \text {Subst}\left (\int \left (\frac {1}{x^2}-\frac {1}{x}+\frac {1}{1+x}\right ) \, dx,x,x^2\right )-\frac {1}{10} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{10} \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^2\right )-\frac {1}{5} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{5} \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^2\right ) \\ & = -\frac {7}{60 x^2}-\frac {2 \arctan (x)}{15 x^3}+\frac {2 \arctan (x)}{5 x}+\frac {\arctan (x)^2}{5}-\frac {5 \log (x)}{6}+\frac {5}{12} \log \left (1+x^2\right )-\frac {\log \left (1+x^2\right )}{20 x^4}+\frac {\log \left (1+x^2\right )}{10 x^2}-\frac {\arctan (x) \log \left (1+x^2\right )}{5 x^5}-\frac {1}{20} \log ^2\left (1+x^2\right )-\frac {\operatorname {PolyLog}\left (2,-x^2\right )}{10} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00 \[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^6} \, dx=-\frac {7}{60 x^2}-\frac {2 \arctan (x)}{15 x^3}+\frac {2 \arctan (x)}{5 x}+\frac {\arctan (x)^2}{5}-\frac {5 \log (x)}{6}+\frac {5}{12} \log \left (1+x^2\right )-\frac {\log \left (1+x^2\right )}{20 x^4}+\frac {\log \left (1+x^2\right )}{10 x^2}-\frac {\arctan (x) \log \left (1+x^2\right )}{5 x^5}-\frac {1}{20} \log ^2\left (1+x^2\right )-\frac {\operatorname {PolyLog}\left (2,-x^2\right )}{10} \]

[In]

Integrate[(ArcTan[x]*Log[1 + x^2])/x^6,x]

[Out]

-7/(60*x^2) - (2*ArcTan[x])/(15*x^3) + (2*ArcTan[x])/(5*x) + ArcTan[x]^2/5 - (5*Log[x])/6 + (5*Log[1 + x^2])/1
2 - Log[1 + x^2]/(20*x^4) + Log[1 + x^2]/(10*x^2) - (ArcTan[x]*Log[1 + x^2])/(5*x^5) - Log[1 + x^2]^2/20 - Pol
yLog[2, -x^2]/10

Maple [F]

\[\int \frac {\arctan \left (x \right ) \ln \left (x^{2}+1\right )}{x^{6}}d x\]

[In]

int(arctan(x)*ln(x^2+1)/x^6,x)

[Out]

int(arctan(x)*ln(x^2+1)/x^6,x)

Fricas [F]

\[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^6} \, dx=\int { \frac {\arctan \left (x\right ) \log \left (x^{2} + 1\right )}{x^{6}} \,d x } \]

[In]

integrate(arctan(x)*log(x^2+1)/x^6,x, algorithm="fricas")

[Out]

integral(arctan(x)*log(x^2 + 1)/x^6, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 17.97 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.18 \[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^6} \, dx=- \frac {8 \log {\left (x \right )}}{15} - \frac {\log {\left (x^{2} \right )}}{20} - \frac {\log {\left (2 x^{2} \right )}}{10} - \frac {\log {\left (x^{2} + 1 \right )}^{2}}{20} + \frac {19 \log {\left (x^{2} + 1 \right )}}{60} + \frac {\log {\left (2 x^{2} + 2 \right )}}{10} + \frac {\operatorname {atan}^{2}{\left (x \right )}}{5} - \frac {\operatorname {Li}_{2}\left (x^{2} e^{i \pi }\right )}{10} + \frac {2 \operatorname {atan}{\left (x \right )}}{5 x} + \frac {\log {\left (x^{2} + 1 \right )}}{10 x^{2}} - \frac {7}{60 x^{2}} - \frac {2 \operatorname {atan}{\left (x \right )}}{15 x^{3}} - \frac {\log {\left (x^{2} + 1 \right )}}{20 x^{4}} - \frac {\log {\left (x^{2} + 1 \right )} \operatorname {atan}{\left (x \right )}}{5 x^{5}} \]

[In]

integrate(atan(x)*ln(x**2+1)/x**6,x)

[Out]

-8*log(x)/15 - log(x**2)/20 - log(2*x**2)/10 - log(x**2 + 1)**2/20 + 19*log(x**2 + 1)/60 + log(2*x**2 + 2)/10
+ atan(x)**2/5 - polylog(2, x**2*exp_polar(I*pi))/10 + 2*atan(x)/(5*x) + log(x**2 + 1)/(10*x**2) - 7/(60*x**2)
- 2*atan(x)/(15*x**3) - log(x**2 + 1)/(20*x**4) - log(x**2 + 1)*atan(x)/(5*x**5)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.01 \[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^6} \, dx=\frac {1}{15} \, {\left (\frac {2 \, {\left (3 \, x^{2} - 1\right )}}{x^{3}} - \frac {3 \, \log \left (x^{2} + 1\right )}{x^{5}} + 6 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) - \frac {12 \, x^{4} \arctan \left (x\right )^{2} + 3 \, x^{4} \log \left (x^{2} + 1\right )^{2} - 6 \, x^{4} {\rm Li}_2\left (x^{2} + 1\right ) + 50 \, x^{4} \log \left (x\right ) + 7 \, x^{2} - {\left (6 \, x^{4} \log \left (-x^{2}\right ) + 25 \, x^{4} + 6 \, x^{2} - 3\right )} \log \left (x^{2} + 1\right )}{60 \, x^{4}} \]

[In]

integrate(arctan(x)*log(x^2+1)/x^6,x, algorithm="maxima")

[Out]

1/15*(2*(3*x^2 - 1)/x^3 - 3*log(x^2 + 1)/x^5 + 6*arctan(x))*arctan(x) - 1/60*(12*x^4*arctan(x)^2 + 3*x^4*log(x
^2 + 1)^2 - 6*x^4*dilog(x^2 + 1) + 50*x^4*log(x) + 7*x^2 - (6*x^4*log(-x^2) + 25*x^4 + 6*x^2 - 3)*log(x^2 + 1)
)/x^4

Giac [F]

\[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^6} \, dx=\int { \frac {\arctan \left (x\right ) \log \left (x^{2} + 1\right )}{x^{6}} \,d x } \]

[In]

integrate(arctan(x)*log(x^2+1)/x^6,x, algorithm="giac")

[Out]

integrate(arctan(x)*log(x^2 + 1)/x^6, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\arctan (x) \log \left (1+x^2\right )}{x^6} \, dx=\int \frac {\ln \left (x^2+1\right )\,\mathrm {atan}\left (x\right )}{x^6} \,d x \]

[In]

int((log(x^2 + 1)*atan(x))/x^6,x)

[Out]

int((log(x^2 + 1)*atan(x))/x^6, x)