3.1275 \(\int x^4 \tan ^{-1}(x) \log (1+x^2) \, dx\)
Optimal. Leaf size=111 \[ -\frac {2}{25} x^5 \tan ^{-1}(x)+\frac {9 x^4}{200}+\frac {2}{15} x^3 \tan ^{-1}(x)-\frac {77 x^2}{300}-\frac {1}{20} \log ^2\left (x^2+1\right )+\frac {1}{10} x^2 \log \left (x^2+1\right )+\frac {137}{300} \log \left (x^2+1\right )+\frac {1}{5} x^5 \log \left (x^2+1\right ) \tan ^{-1}(x)-\frac {1}{20} x^4 \log \left (x^2+1\right )-\frac {2}{5} x \tan ^{-1}(x)+\frac {1}{5} \tan ^{-1}(x)^2 \]
[Out]
-77/300*x^2+9/200*x^4-2/5*x*arctan(x)+2/15*x^3*arctan(x)-2/25*x^5*arctan(x)+1/5*arctan(x)^2+137/300*ln(x^2+1)+
1/10*x^2*ln(x^2+1)-1/20*x^4*ln(x^2+1)+1/5*x^5*arctan(x)*ln(x^2+1)-1/20*ln(x^2+1)^2
________________________________________________________________________________________
Rubi [A] time = 0.44, antiderivative size = 111, normalized size of antiderivative = 1.00,
number of steps used = 24, number of rules used = 14, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.167, Rules
used = {4852, 266, 43, 5021, 6725, 446, 77, 4916, 4846, 260, 4884, 2475, 2390, 2301}
\[ \frac {9 x^4}{200}-\frac {77 x^2}{300}-\frac {1}{20} \log ^2\left (x^2+1\right )-\frac {1}{20} x^4 \log \left (x^2+1\right )+\frac {1}{10} x^2 \log \left (x^2+1\right )+\frac {137}{300} \log \left (x^2+1\right )-\frac {2}{25} x^5 \tan ^{-1}(x)+\frac {2}{15} x^3 \tan ^{-1}(x)+\frac {1}{5} x^5 \log \left (x^2+1\right ) \tan ^{-1}(x)-\frac {2}{5} x \tan ^{-1}(x)+\frac {1}{5} \tan ^{-1}(x)^2 \]
Antiderivative was successfully verified.
[In]
Int[x^4*ArcTan[x]*Log[1 + x^2],x]
[Out]
(-77*x^2)/300 + (9*x^4)/200 - (2*x*ArcTan[x])/5 + (2*x^3*ArcTan[x])/15 - (2*x^5*ArcTan[x])/25 + ArcTan[x]^2/5
+ (137*Log[1 + x^2])/300 + (x^2*Log[1 + x^2])/10 - (x^4*Log[1 + x^2])/20 + (x^5*ArcTan[x]*Log[1 + x^2])/5 - Lo
g[1 + x^2]^2/20
Rule 43
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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Rule 77
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Rule 260
Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]
Rule 266
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 446
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Rule 2301
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 2390
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 2475
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 4846
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x])^p, x] - Dist[b*c*p, Int[
(x*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]
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 4884
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 4916
Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2/
e, Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x])^p, x], x] - Dist[(d*f^2)/e, 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] && GtQ[m, 1]
Rule 5021
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.))*(x_)^(m_.), x_Symbol] :> With
[{u = IntHide[x^m*(a + b*ArcTan[c*x]), x]}, Dist[d + e*Log[f + g*x^2], u, x] - Dist[2*e*g, Int[ExpandIntegrand
[(x*u)/(f + g*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && IntegerQ[m] && NeQ[m, -1]
Rule 6725
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]
Rubi steps
\begin {align*} \int x^4 \tan ^{-1}(x) \log \left (1+x^2\right ) \, dx &=\frac {1}{10} x^2 \log \left (1+x^2\right )-\frac {1}{20} x^4 \log \left (1+x^2\right )+\frac {1}{5} x^5 \tan ^{-1}(x) \log \left (1+x^2\right )-\frac {1}{10} \log ^2\left (1+x^2\right )-2 \int \left (\frac {x^3 \left (2-x^2+4 x^3 \tan ^{-1}(x)\right )}{20 \left (1+x^2\right )}-\frac {x \log \left (1+x^2\right )}{10 \left (1+x^2\right )}\right ) \, dx\\ &=\frac {1}{10} x^2 \log \left (1+x^2\right )-\frac {1}{20} x^4 \log \left (1+x^2\right )+\frac {1}{5} x^5 \tan ^{-1}(x) \log \left (1+x^2\right )-\frac {1}{10} \log ^2\left (1+x^2\right )-\frac {1}{10} \int \frac {x^3 \left (2-x^2+4 x^3 \tan ^{-1}(x)\right )}{1+x^2} \, dx+\frac {1}{5} \int \frac {x \log \left (1+x^2\right )}{1+x^2} \, dx\\ &=\frac {1}{10} x^2 \log \left (1+x^2\right )-\frac {1}{20} x^4 \log \left (1+x^2\right )+\frac {1}{5} x^5 \tan ^{-1}(x) \log \left (1+x^2\right )-\frac {1}{10} \log ^2\left (1+x^2\right )-\frac {1}{10} \int \left (-\frac {x^3 \left (-2+x^2\right )}{1+x^2}+\frac {4 x^6 \tan ^{-1}(x)}{1+x^2}\right ) \, dx+\frac {1}{10} \operatorname {Subst}\left (\int \frac {\log (1+x)}{1+x} \, dx,x,x^2\right )\\ &=\frac {1}{10} x^2 \log \left (1+x^2\right )-\frac {1}{20} x^4 \log \left (1+x^2\right )+\frac {1}{5} x^5 \tan ^{-1}(x) \log \left (1+x^2\right )-\frac {1}{10} \log ^2\left (1+x^2\right )+\frac {1}{10} \int \frac {x^3 \left (-2+x^2\right )}{1+x^2} \, dx+\frac {1}{10} \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1+x^2\right )-\frac {2}{5} \int \frac {x^6 \tan ^{-1}(x)}{1+x^2} \, dx\\ &=\frac {1}{10} x^2 \log \left (1+x^2\right )-\frac {1}{20} x^4 \log \left (1+x^2\right )+\frac {1}{5} x^5 \tan ^{-1}(x) \log \left (1+x^2\right )-\frac {1}{20} \log ^2\left (1+x^2\right )+\frac {1}{20} \operatorname {Subst}\left (\int \frac {(-2+x) x}{1+x} \, dx,x,x^2\right )-\frac {2}{5} \int x^4 \tan ^{-1}(x) \, dx+\frac {2}{5} \int \frac {x^4 \tan ^{-1}(x)}{1+x^2} \, dx\\ &=-\frac {2}{25} x^5 \tan ^{-1}(x)+\frac {1}{10} x^2 \log \left (1+x^2\right )-\frac {1}{20} x^4 \log \left (1+x^2\right )+\frac {1}{5} x^5 \tan ^{-1}(x) \log \left (1+x^2\right )-\frac {1}{20} \log ^2\left (1+x^2\right )+\frac {1}{20} \operatorname {Subst}\left (\int \left (-3+x+\frac {3}{1+x}\right ) \, dx,x,x^2\right )+\frac {2}{25} \int \frac {x^5}{1+x^2} \, dx+\frac {2}{5} \int x^2 \tan ^{-1}(x) \, dx-\frac {2}{5} \int \frac {x^2 \tan ^{-1}(x)}{1+x^2} \, dx\\ &=-\frac {3 x^2}{20}+\frac {x^4}{40}+\frac {2}{15} x^3 \tan ^{-1}(x)-\frac {2}{25} x^5 \tan ^{-1}(x)+\frac {3}{20} \log \left (1+x^2\right )+\frac {1}{10} x^2 \log \left (1+x^2\right )-\frac {1}{20} x^4 \log \left (1+x^2\right )+\frac {1}{5} x^5 \tan ^{-1}(x) \log \left (1+x^2\right )-\frac {1}{20} \log ^2\left (1+x^2\right )+\frac {1}{25} \operatorname {Subst}\left (\int \frac {x^2}{1+x} \, dx,x,x^2\right )-\frac {2}{15} \int \frac {x^3}{1+x^2} \, dx-\frac {2}{5} \int \tan ^{-1}(x) \, dx+\frac {2}{5} \int \frac {\tan ^{-1}(x)}{1+x^2} \, dx\\ &=-\frac {3 x^2}{20}+\frac {x^4}{40}-\frac {2}{5} x \tan ^{-1}(x)+\frac {2}{15} x^3 \tan ^{-1}(x)-\frac {2}{25} x^5 \tan ^{-1}(x)+\frac {1}{5} \tan ^{-1}(x)^2+\frac {3}{20} \log \left (1+x^2\right )+\frac {1}{10} x^2 \log \left (1+x^2\right )-\frac {1}{20} x^4 \log \left (1+x^2\right )+\frac {1}{5} x^5 \tan ^{-1}(x) \log \left (1+x^2\right )-\frac {1}{20} \log ^2\left (1+x^2\right )+\frac {1}{25} \operatorname {Subst}\left (\int \left (-1+x+\frac {1}{1+x}\right ) \, dx,x,x^2\right )-\frac {1}{15} \operatorname {Subst}\left (\int \frac {x}{1+x} \, dx,x,x^2\right )+\frac {2}{5} \int \frac {x}{1+x^2} \, dx\\ &=-\frac {19 x^2}{100}+\frac {9 x^4}{200}-\frac {2}{5} x \tan ^{-1}(x)+\frac {2}{15} x^3 \tan ^{-1}(x)-\frac {2}{25} x^5 \tan ^{-1}(x)+\frac {1}{5} \tan ^{-1}(x)^2+\frac {39}{100} \log \left (1+x^2\right )+\frac {1}{10} x^2 \log \left (1+x^2\right )-\frac {1}{20} x^4 \log \left (1+x^2\right )+\frac {1}{5} x^5 \tan ^{-1}(x) \log \left (1+x^2\right )-\frac {1}{20} \log ^2\left (1+x^2\right )-\frac {1}{15} \operatorname {Subst}\left (\int \left (1+\frac {1}{-1-x}\right ) \, dx,x,x^2\right )\\ &=-\frac {77 x^2}{300}+\frac {9 x^4}{200}-\frac {2}{5} x \tan ^{-1}(x)+\frac {2}{15} x^3 \tan ^{-1}(x)-\frac {2}{25} x^5 \tan ^{-1}(x)+\frac {1}{5} \tan ^{-1}(x)^2+\frac {137}{300} \log \left (1+x^2\right )+\frac {1}{10} x^2 \log \left (1+x^2\right )-\frac {1}{20} x^4 \log \left (1+x^2\right )+\frac {1}{5} x^5 \tan ^{-1}(x) \log \left (1+x^2\right )-\frac {1}{20} \log ^2\left (1+x^2\right )\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.03, size = 79, normalized size = 0.71 \[ \frac {1}{600} \left (\left (27 x^2-154\right ) x^2-30 \log ^2\left (x^2+1\right )+\left (-30 x^4+60 x^2+274\right ) \log \left (x^2+1\right )+8 x \left (-6 x^4+10 x^2+15 x^4 \log \left (x^2+1\right )-30\right ) \tan ^{-1}(x)+120 \tan ^{-1}(x)^2\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[x^4*ArcTan[x]*Log[1 + x^2],x]
[Out]
(x^2*(-154 + 27*x^2) + 120*ArcTan[x]^2 + (274 + 60*x^2 - 30*x^4)*Log[1 + x^2] - 30*Log[1 + x^2]^2 + 8*x*ArcTan
[x]*(-30 + 10*x^2 - 6*x^4 + 15*x^4*Log[1 + x^2]))/600
________________________________________________________________________________________
fricas [A] time = 0.39, size = 72, normalized size = 0.65 \[ \frac {9}{200} \, x^{4} - \frac {77}{300} \, x^{2} - \frac {2}{75} \, {\left (3 \, x^{5} - 5 \, x^{3} + 15 \, x\right )} \arctan \relax (x) + \frac {1}{5} \, \arctan \relax (x)^{2} + \frac {1}{300} \, {\left (60 \, x^{5} \arctan \relax (x) - 15 \, x^{4} + 30 \, x^{2} + 137\right )} \log \left (x^{2} + 1\right ) - \frac {1}{20} \, \log \left (x^{2} + 1\right )^{2} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^4*arctan(x)*log(x^2+1),x, algorithm="fricas")
[Out]
9/200*x^4 - 77/300*x^2 - 2/75*(3*x^5 - 5*x^3 + 15*x)*arctan(x) + 1/5*arctan(x)^2 + 1/300*(60*x^5*arctan(x) - 1
5*x^4 + 30*x^2 + 137)*log(x^2 + 1) - 1/20*log(x^2 + 1)^2
________________________________________________________________________________________
giac [A] time = 4.70, size = 168, normalized size = 1.51 \[ \frac {1}{10} \, \pi x^{5} \log \left (x^{2} + 1\right ) \mathrm {sgn}\relax (x) - \frac {1}{5} \, x^{5} \arctan \left (\frac {1}{x}\right ) \log \left (x^{2} + 1\right ) - \frac {1}{25} \, \pi x^{5} \mathrm {sgn}\relax (x) + \frac {2}{25} \, x^{5} \arctan \left (\frac {1}{x}\right ) - \frac {1}{20} \, x^{4} \log \left (x^{2} + 1\right ) + \frac {1}{15} \, \pi x^{3} \mathrm {sgn}\relax (x) + \frac {9}{200} \, x^{4} - \frac {2}{15} \, x^{3} \arctan \left (\frac {1}{x}\right ) + \frac {1}{10} \, x^{2} \log \left (x^{2} + 1\right ) - \frac {3}{10} \, \pi ^{2} \mathrm {sgn}\relax (x) - \frac {1}{5} \, \pi x \mathrm {sgn}\relax (x) - \frac {1}{5} \, \pi \arctan \left (\frac {1}{x}\right ) \mathrm {sgn}\relax (x) + \frac {1}{10} \, \pi ^{2} - \frac {77}{300} \, x^{2} + \frac {1}{5} \, \pi \arctan \relax (x) + \frac {1}{5} \, \pi \arctan \left (\frac {1}{x}\right ) + \frac {2}{5} \, x \arctan \left (\frac {1}{x}\right ) + \frac {1}{5} \, \arctan \left (\frac {1}{x}\right )^{2} - \frac {1}{20} \, \log \left (x^{2} + 1\right )^{2} + \frac {137}{300} \, \log \left (x^{2} + 1\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^4*arctan(x)*log(x^2+1),x, algorithm="giac")
[Out]
1/10*pi*x^5*log(x^2 + 1)*sgn(x) - 1/5*x^5*arctan(1/x)*log(x^2 + 1) - 1/25*pi*x^5*sgn(x) + 2/25*x^5*arctan(1/x)
- 1/20*x^4*log(x^2 + 1) + 1/15*pi*x^3*sgn(x) + 9/200*x^4 - 2/15*x^3*arctan(1/x) + 1/10*x^2*log(x^2 + 1) - 3/1
0*pi^2*sgn(x) - 1/5*pi*x*sgn(x) - 1/5*pi*arctan(1/x)*sgn(x) + 1/10*pi^2 - 77/300*x^2 + 1/5*pi*arctan(x) + 1/5*
pi*arctan(1/x) + 2/5*x*arctan(1/x) + 1/5*arctan(1/x)^2 - 1/20*log(x^2 + 1)^2 + 137/300*log(x^2 + 1)
________________________________________________________________________________________
maple [C] time = 3.78, size = 3626, normalized size = 32.67 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^4*arctan(x)*ln(x^2+1),x)
[Out]
-2/5*x*arctan(x)+2/15*x^3*arctan(x)-2/25*x^5*arctan(x)+1/10*I*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)^2/(x^2+
1)/((1+I*x)^2/(x^2+1)+1)^2)^2*arctan(x)*Pi*x^5-1/10*I*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)/(x^2+1)^(1/2))^
2*arctan(x)*Pi*x^5+1/10*I*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^2*csgn(I/((1+I*x)^2/(x^2+1)+1)^2)*
arctan(x)*Pi*x^5+1/10*I*csgn(I*((1+I*x)^2/(x^2+1)+1))^2*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)*arctan(x)*Pi*x^5-1/5*I
*csgn(I*((1+I*x)^2/(x^2+1)+1))*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)^2*arctan(x)*Pi*x^5+1/40*I*csgn(I*(1+I*x)^2/(x^2
+1))*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)*csgn(I/((1+I*x)^2/(x^2+1)+1)^2)*Pi*x^4-1/20*I*csgn(I*(1
+I*x)^2/(x^2+1))*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)*csgn(I/((1+I*x)^2/(x^2+1)+1)^2)*Pi*x^2+1/5*
I*csgn(I*(1+I*x)^2/(x^2+1))^2*csgn(I*(1+I*x)/(x^2+1)^(1/2))*arctan(x)*Pi*x^5-1/10*I*ln((1+I*x)^2/(x^2+1)+1)*Pi
*csgn(I/((1+I*x)^2/(x^2+1)+1)^2)*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)+1
/5*x^2*ln(2)-1/5*ln((1+I*x)^2/(x^2+1)+1)*x^2+1/10*ln((1+I*x)^2/(x^2+1)+1)*x^4+46/75*I*arctan(x)+1/10*(-4*I*arc
tan(x)+4*x^5*arctan(x)+4*ln((1+I*x)^2/(x^2+1)+1)+3+2*x^2-x^4)*ln((1+I*x)/(x^2+1)^(1/2))-181/600-1/10*csgn(I*(1
+I*x)^2/(x^2+1))^3*arctan(x)*Pi-1/10*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^3*arctan(x)*Pi+1/10*csg
n(I*((1+I*x)^2/(x^2+1)+1)^2)^3*arctan(x)*Pi+2/5*ln(2)*arctan(x)*x^5-3/40*I*Pi*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x
)^2/(x^2+1)+1)^2)^3+3/40*I*Pi*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)^3-3/40*I*Pi*csgn(I*(1+I*x)^2/(x^2+1))^3-2/5*I*ln
(2)*arctan(x)-2/5*ln((1+I*x)^2/(x^2+1)+1)*arctan(x)*x^5+1/5*csgn(I*(1+I*x)^2/(x^2+1))^2*csgn(I*(1+I*x)/(x^2+1)
^(1/2))*arctan(x)*Pi+1/10*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^2*arctan
(x)*Pi-1/10*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)/(x^2+1)^(1/2))^2*arctan(x)*Pi+1/10*csgn(I*(1+I*x)^2/(x^2+
1)/((1+I*x)^2/(x^2+1)+1)^2)^2*csgn(I/((1+I*x)^2/(x^2+1)+1)^2)*arctan(x)*Pi+1/10*csgn(I*((1+I*x)^2/(x^2+1)+1))^
2*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)*arctan(x)*Pi-1/5*csgn(I*((1+I*x)^2/(x^2+1)+1))*csgn(I*((1+I*x)^2/(x^2+1)+1)^
2)^2*arctan(x)*Pi+3/20*I*Pi*csgn(I*(1+I*x)/(x^2+1)^(1/2))*csgn(I*(1+I*x)^2/(x^2+1))^2-3/20*I*Pi*csgn(I*((1+I*x
)^2/(x^2+1)+1))*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)^2+1/40*I*csgn(I*(1+I*x)^2/(x^2+1))^3*Pi*x^4+1/40*I*csgn(I*(1+I
*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^3*Pi*x^4-1/40*I*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)^3*Pi*x^4-1/20*I*csgn(I*
(1+I*x)^2/(x^2+1))^3*Pi*x^2-1/20*I*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^3*Pi*x^2+1/20*I*csgn(I*((
1+I*x)^2/(x^2+1)+1)^2)^3*Pi*x^2+2/5*ln(2)*ln((1+I*x)^2/(x^2+1)+1)-1/10*ln(2)*x^4-1/10*I*csgn(I*(1+I*x)^2/(x^2+
1))*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)*csgn(I/((1+I*x)^2/(x^2+1)+1)^2)*arctan(x)*Pi*x^5+3/40*I*
Pi*csgn(I*((1+I*x)^2/(x^2+1)+1))^2*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)+3/40*I*Pi*csgn(I/((1+I*x)^2/(x^2+1)+1)^2)*c
sgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^2-3/40*I*Pi*csgn(I*(1+I*x)/(x^2+1)^(1/2))^2*csgn(I*(1+I*x)^2/
(x^2+1))+3/40*I*Pi*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^2-1/10*I*ln((1+
I*x)^2/(x^2+1)+1)*Pi*csgn(I*(1+I*x)^2/(x^2+1))^3+1/10*I*ln((1+I*x)^2/(x^2+1)+1)*Pi*csgn(I*((1+I*x)^2/(x^2+1)+1
)^2)^3-1/10*I*ln((1+I*x)^2/(x^2+1)+1)*Pi*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^3+1/10*I*ln((1+I*x)
^2/(x^2+1)+1)*Pi*csgn(I/((1+I*x)^2/(x^2+1)+1)^2)*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^2+1/10*I*ln
((1+I*x)^2/(x^2+1)+1)*Pi*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^2-1/5*I*l
n((1+I*x)^2/(x^2+1)+1)*Pi*csgn(I*((1+I*x)^2/(x^2+1)+1))*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)^2-1/10*I*ln((1+I*x)^2/
(x^2+1)+1)*Pi*csgn(I*(1+I*x)/(x^2+1)^(1/2))^2*csgn(I*(1+I*x)^2/(x^2+1))-1/40*I*csgn(I*(1+I*x)^2/(x^2+1))*csgn(
I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^2*Pi*x^4+1/40*I*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)/(x^2+1)^
(1/2))^2*Pi*x^4-1/40*I*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^2*csgn(I/((1+I*x)^2/(x^2+1)+1)^2)*Pi*
x^4-1/40*I*csgn(I*((1+I*x)^2/(x^2+1)+1))^2*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)*Pi*x^4+1/20*I*csgn(I*(1+I*x)^2/(x^2
+1))*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^2*Pi*x^2-1/10*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)^
2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)*csgn(I/((1+I*x)^2/(x^2+1)+1)^2)*arctan(x)*Pi+1/20*I*csgn(I*(1+I*x)^2/(x^2+1
)/((1+I*x)^2/(x^2+1)+1)^2)^2*csgn(I/((1+I*x)^2/(x^2+1)+1)^2)*Pi*x^2+1/20*I*csgn(I*((1+I*x)^2/(x^2+1)+1))^2*csg
n(I*((1+I*x)^2/(x^2+1)+1)^2)*Pi*x^2-1/10*I*csgn(I*((1+I*x)^2/(x^2+1)+1))*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)^2*Pi*
x^2-3/40*I*Pi*csgn(I/((1+I*x)^2/(x^2+1)+1)^2)*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x
^2+1)+1)^2)-1/10*I*csgn(I*(1+I*x)^2/(x^2+1))^3*arctan(x)*Pi*x^5-1/10*I*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^
2+1)+1)^2)^3*arctan(x)*Pi*x^5+1/10*I*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)^3*arctan(x)*Pi*x^5-1/20*I*csgn(I*(1+I*x)^
2/(x^2+1))^2*csgn(I*(1+I*x)/(x^2+1)^(1/2))*Pi*x^4+1/5*I*ln((1+I*x)^2/(x^2+1)+1)*Pi*csgn(I*(1+I*x)/(x^2+1)^(1/2
))*csgn(I*(1+I*x)^2/(x^2+1))^2+1/10*I*ln((1+I*x)^2/(x^2+1)+1)*Pi*csgn(I*((1+I*x)^2/(x^2+1)+1))^2*csgn(I*((1+I*
x)^2/(x^2+1)+1)^2)+1/20*I*csgn(I*((1+I*x)^2/(x^2+1)+1))*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)^2*Pi*x^4+1/10*I*csgn(I
*(1+I*x)^2/(x^2+1))^2*csgn(I*(1+I*x)/(x^2+1)^(1/2))*Pi*x^2-1/20*I*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)/(x^
2+1)^(1/2))^2*Pi*x^2-77/300*x^2+3/10*ln(2)-137/150*ln((1+I*x)^2/(x^2+1)+1)-1/5*ln((1+I*x)^2/(x^2+1)+1)^2+9/200
*x^4
________________________________________________________________________________________
maxima [A] time = 0.42, size = 80, normalized size = 0.72 \[ \frac {9}{200} \, x^{4} - \frac {77}{300} \, x^{2} + \frac {1}{75} \, {\left (15 \, x^{5} \log \left (x^{2} + 1\right ) - 6 \, x^{5} + 10 \, x^{3} - 30 \, x + 30 \, \arctan \relax (x)\right )} \arctan \relax (x) - \frac {1}{5} \, \arctan \relax (x)^{2} - \frac {1}{300} \, {\left (15 \, x^{4} - 30 \, x^{2} - 137\right )} \log \left (x^{2} + 1\right ) - \frac {1}{20} \, \log \left (x^{2} + 1\right )^{2} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^4*arctan(x)*log(x^2+1),x, algorithm="maxima")
[Out]
9/200*x^4 - 77/300*x^2 + 1/75*(15*x^5*log(x^2 + 1) - 6*x^5 + 10*x^3 - 30*x + 30*arctan(x))*arctan(x) - 1/5*arc
tan(x)^2 - 1/300*(15*x^4 - 30*x^2 - 137)*log(x^2 + 1) - 1/20*log(x^2 + 1)^2
________________________________________________________________________________________
mupad [B] time = 0.48, size = 82, normalized size = 0.74 \[ \frac {137\,\ln \left (x^2+1\right )}{300}-\frac {{\ln \left (x^2+1\right )}^2}{20}+\frac {{\mathrm {atan}\relax (x)}^2}{5}-\mathrm {atan}\relax (x)\,\left (\frac {2\,x}{5}-\frac {2\,x^3}{15}+\frac {2\,x^5}{25}-\frac {x^5\,\ln \left (x^2+1\right )}{5}\right )+\ln \left (x^2+1\right )\,\left (\frac {x^2}{10}-\frac {x^4}{20}\right )-\frac {77\,x^2}{300}+\frac {9\,x^4}{200} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^4*log(x^2 + 1)*atan(x),x)
[Out]
(137*log(x^2 + 1))/300 - log(x^2 + 1)^2/20 + atan(x)^2/5 - atan(x)*((2*x)/5 - (2*x^3)/15 + (2*x^5)/25 - (x^5*l
og(x^2 + 1))/5) + log(x^2 + 1)*(x^2/10 - x^4/20) - (77*x^2)/300 + (9*x^4)/200
________________________________________________________________________________________
sympy [A] time = 4.64, size = 107, normalized size = 0.96 \[ \frac {x^{5} \log {\left (x^{2} + 1 \right )} \operatorname {atan}{\relax (x )}}{5} - \frac {2 x^{5} \operatorname {atan}{\relax (x )}}{25} - \frac {x^{4} \log {\left (x^{2} + 1 \right )}}{20} + \frac {9 x^{4}}{200} + \frac {2 x^{3} \operatorname {atan}{\relax (x )}}{15} + \frac {x^{2} \log {\left (x^{2} + 1 \right )}}{10} - \frac {77 x^{2}}{300} - \frac {2 x \operatorname {atan}{\relax (x )}}{5} - \frac {\log {\left (x^{2} + 1 \right )}^{2}}{20} + \frac {137 \log {\left (x^{2} + 1 \right )}}{300} + \frac {\operatorname {atan}^{2}{\relax (x )}}{5} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**4*atan(x)*ln(x**2+1),x)
[Out]
x**5*log(x**2 + 1)*atan(x)/5 - 2*x**5*atan(x)/25 - x**4*log(x**2 + 1)/20 + 9*x**4/200 + 2*x**3*atan(x)/15 + x*
*2*log(x**2 + 1)/10 - 77*x**2/300 - 2*x*atan(x)/5 - log(x**2 + 1)**2/20 + 137*log(x**2 + 1)/300 + atan(x)**2/5
________________________________________________________________________________________