\(\int x^4 \arctan (x) \log (1+x^2) \, dx\) [1275]

Optimal result

Integrand size = 12, antiderivative size = 111 \[ \int x^4 \arctan (x) \log \left (1+x^2\right ) \, dx=-\frac {77 x^2}{300}+\frac {9 x^4}{200}-\frac {2}{5} x \arctan (x)+\frac {2}{15} x^3 \arctan (x)-\frac {2}{25} x^5 \arctan (x)+\frac {\arctan (x)^2}{5}+\frac {137}{300} \log \left (1+x^2\right )+\frac {1}{10} x^2 \log \left (1+x^2\right )-\frac {1}{20} x^4 \log \left (1+x^2\right )+\frac {1}{5} x^5 \arctan (x) \log \left (1+x^2\right )-\frac {1}{20} \log ^2\left (1+x^2\right ) \]

[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] (verified)

Time = 0.30 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.167, Rules used = {4946, 272, 45, 5141, 6857, 457, 78, 5036, 4930, 266, 5004, 2525, 2437, 2338} \[ \int x^4 \arctan (x) \log \left (1+x^2\right ) \, dx=-\frac {2}{25} x^5 \arctan (x)+\frac {2}{15} x^3 \arctan (x)+\frac {1}{5} x^5 \arctan (x) \log \left (x^2+1\right )-\frac {2}{5} x \arctan (x)+\frac {\arctan (x)^2}{5}+\frac {9 x^4}{200}-\frac {77 x^2}{300}-\frac {1}{20} \log ^2\left (x^2+1\right )+\frac {1}{10} x^2 \log \left (x^2+1\right )+\frac {137}{300} \log \left (x^2+1\right )-\frac {1}{20} x^4 \log \left (x^2+1\right ) \]

[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 45

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 78

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 266

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 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 457

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 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 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 4930

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

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 5036

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 5141

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 6857

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*} \text {integral}& = \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 \arctan (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 \arctan (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 \arctan (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 \arctan (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 \arctan (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 \arctan (x)}{1+x^2}\right ) \, dx+\frac {1}{10} \text {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 \arctan (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} \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1+x^2\right )-\frac {2}{5} \int \frac {x^6 \arctan (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 \arctan (x) \log \left (1+x^2\right )-\frac {1}{20} \log ^2\left (1+x^2\right )+\frac {1}{20} \text {Subst}\left (\int \frac {(-2+x) x}{1+x} \, dx,x,x^2\right )-\frac {2}{5} \int x^4 \arctan (x) \, dx+\frac {2}{5} \int \frac {x^4 \arctan (x)}{1+x^2} \, dx \\ & = -\frac {2}{25} x^5 \arctan (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 \arctan (x) \log \left (1+x^2\right )-\frac {1}{20} \log ^2\left (1+x^2\right )+\frac {1}{20} \text {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 \arctan (x) \, dx-\frac {2}{5} \int \frac {x^2 \arctan (x)}{1+x^2} \, dx \\ & = -\frac {3 x^2}{20}+\frac {x^4}{40}+\frac {2}{15} x^3 \arctan (x)-\frac {2}{25} x^5 \arctan (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 \arctan (x) \log \left (1+x^2\right )-\frac {1}{20} \log ^2\left (1+x^2\right )+\frac {1}{25} \text {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 \arctan (x) \, dx+\frac {2}{5} \int \frac {\arctan (x)}{1+x^2} \, dx \\ & = -\frac {3 x^2}{20}+\frac {x^4}{40}-\frac {2}{5} x \arctan (x)+\frac {2}{15} x^3 \arctan (x)-\frac {2}{25} x^5 \arctan (x)+\frac {\arctan (x)^2}{5}+\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 \arctan (x) \log \left (1+x^2\right )-\frac {1}{20} \log ^2\left (1+x^2\right )+\frac {1}{25} \text {Subst}\left (\int \left (-1+x+\frac {1}{1+x}\right ) \, dx,x,x^2\right )-\frac {1}{15} \text {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 \arctan (x)+\frac {2}{15} x^3 \arctan (x)-\frac {2}{25} x^5 \arctan (x)+\frac {\arctan (x)^2}{5}+\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 \arctan (x) \log \left (1+x^2\right )-\frac {1}{20} \log ^2\left (1+x^2\right )-\frac {1}{15} \text {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 \arctan (x)+\frac {2}{15} x^3 \arctan (x)-\frac {2}{25} x^5 \arctan (x)+\frac {\arctan (x)^2}{5}+\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 \arctan (x) \log \left (1+x^2\right )-\frac {1}{20} \log ^2\left (1+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.71 \[ \int x^4 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {1}{600} \left (x^2 \left (-154+27 x^2\right )+120 \arctan (x)^2+\left (274+60 x^2-30 x^4\right ) \log \left (1+x^2\right )-30 \log ^2\left (1+x^2\right )+8 x \arctan (x) \left (-30+10 x^2-6 x^4+15 x^4 \log \left (1+x^2\right )\right )\right ) \]

[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

Maple [A] (verified)

Time = 2.10 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.82

method	result	size
parallelrisch	\(\frac {x^{5} \arctan \left (x \right ) \ln \left (x^{2}+1\right )}{5}-\frac {2 x^{5} \arctan \left (x \right )}{25}-\frac {x^{4} \ln \left (x^{2}+1\right )}{20}+\frac {9 x^{4}}{200}+\frac {2 x^{3} \arctan \left (x \right )}{15}+\frac {x^{2} \ln \left (x^{2}+1\right )}{10}-\frac {77 x^{2}}{300}-\frac {2 x \arctan \left (x \right )}{5}+\frac {\arctan \left (x \right )^{2}}{5}-\frac {\ln \left (x^{2}+1\right )^{2}}{20}+\frac {137 \ln \left (x^{2}+1\right )}{300}+\frac {77}{300}\)	\(91\)
default	\(\text {Expression too large to display}\)	\(3626\)
risch	\(\text {Expression too large to display}\)	\(5733\)

[In]

int(x^4*arctan(x)*ln(x^2+1),x,method=_RETURNVERBOSE)

[Out]

1/5*x^5*arctan(x)*ln(x^2+1)-2/25*x^5*arctan(x)-1/20*x^4*ln(x^2+1)+9/200*x^4+2/15*x^3*arctan(x)+1/10*x^2*ln(x^2
+1)-77/300*x^2-2/5*x*arctan(x)+1/5*arctan(x)^2-1/20*ln(x^2+1)^2+137/300*ln(x^2+1)+77/300

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.65 \[ \int x^4 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {9}{200} \, x^{4} - \frac {77}{300} \, x^{2} - \frac {2}{75} \, {\left (3 \, x^{5} - 5 \, x^{3} + 15 \, x\right )} \arctan \left (x\right ) + \frac {1}{5} \, \arctan \left (x\right )^{2} + \frac {1}{300} \, {\left (60 \, x^{5} \arctan \left (x\right ) - 15 \, x^{4} + 30 \, x^{2} + 137\right )} \log \left (x^{2} + 1\right ) - \frac {1}{20} \, \log \left (x^{2} + 1\right )^{2} \]

[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

Sympy [A] (verification not implemented)

Time = 0.87 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.96 \[ \int x^4 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {x^{5} \log {\left (x^{2} + 1 \right )} \operatorname {atan}{\left (x \right )}}{5} - \frac {2 x^{5} \operatorname {atan}{\left (x \right )}}{25} - \frac {x^{4} \log {\left (x^{2} + 1 \right )}}{20} + \frac {9 x^{4}}{200} + \frac {2 x^{3} \operatorname {atan}{\left (x \right )}}{15} + \frac {x^{2} \log {\left (x^{2} + 1 \right )}}{10} - \frac {77 x^{2}}{300} - \frac {2 x \operatorname {atan}{\left (x \right )}}{5} - \frac {\log {\left (x^{2} + 1 \right )}^{2}}{20} + \frac {137 \log {\left (x^{2} + 1 \right )}}{300} + \frac {\operatorname {atan}^{2}{\left (x \right )}}{5} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.72 \[ \int x^4 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {9}{200} \, x^{4} - \frac {77}{300} \, x^{2} + \frac {1}{75} \, {\left (15 \, x^{5} \log \left (x^{2} + 1\right ) - 6 \, x^{5} + 10 \, x^{3} - 30 \, x + 30 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) - \frac {1}{5} \, \arctan \left (x\right )^{2} - \frac {1}{300} \, {\left (15 \, x^{4} - 30 \, x^{2} - 137\right )} \log \left (x^{2} + 1\right ) - \frac {1}{20} \, \log \left (x^{2} + 1\right )^{2} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.51 \[ \int x^4 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {1}{10} \, \pi x^{5} \log \left (x^{2} + 1\right ) \mathrm {sgn}\left (x\right ) - \frac {1}{5} \, x^{5} \arctan \left (\frac {1}{x}\right ) \log \left (x^{2} + 1\right ) - \frac {1}{25} \, \pi x^{5} \mathrm {sgn}\left (x\right ) + \frac {2}{25} \, x^{5} \arctan \left (\frac {1}{x}\right ) - \frac {1}{20} \, x^{4} \log \left (x^{2} + 1\right ) + \frac {1}{15} \, \pi x^{3} \mathrm {sgn}\left (x\right ) + \frac {9}{200} \, x^{4} - \frac {2}{15} \, x^{3} \arctan \left (\frac {1}{x}\right ) + \frac {1}{10} \, x^{2} \log \left (x^{2} + 1\right ) - \frac {3}{10} \, \pi ^{2} \mathrm {sgn}\left (x\right ) - \frac {1}{5} \, \pi x \mathrm {sgn}\left (x\right ) - \frac {1}{5} \, \pi \arctan \left (\frac {1}{x}\right ) \mathrm {sgn}\left (x\right ) + \frac {1}{10} \, \pi ^{2} - \frac {77}{300} \, x^{2} + \frac {1}{5} \, \pi \arctan \left (x\right ) + \frac {1}{5} \, \pi \arctan \left (\frac {1}{x}\right ) + \frac {2}{5} \, x \arctan \left (\frac {1}{x}\right ) + \frac {1}{5} \, \arctan \left (\frac {1}{x}\right )^{2} - \frac {1}{20} \, \log \left (x^{2} + 1\right )^{2} + \frac {137}{300} \, \log \left (x^{2} + 1\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.74 \[ \int x^4 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {137\,\ln \left (x^2+1\right )}{300}-\frac {{\ln \left (x^2+1\right )}^2}{20}+\frac {{\mathrm {atan}\left (x\right )}^2}{5}-\mathrm {atan}\left (x\right )\,\left (\frac {2\,x}{5}-\frac {2\,x^3}{15}+\frac {2\,x^5}{25}-\frac {x^5\,\ln \left (x^2+1\right )}{5}\right )+\ln \left (x^2+1\right )\,\left (\frac {x^2}{10}-\frac {x^4}{20}\right )-\frac {77\,x^2}{300}+\frac {9\,x^4}{200} \]

[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