3.1.46 \(\int x \arctan (x)^2 \log (1+x^2) \, dx\) [46]

3.1.46.1 Optimal result
3.1.46.2 Mathematica [A] (verified)
3.1.46.3 Rubi [A] (verified)
3.1.46.4 Maple [A] (verified)
3.1.46.5 Fricas [A] (verification not implemented)
3.1.46.6 Sympy [A] (verification not implemented)
3.1.46.7 Maxima [A] (verification not implemented)
3.1.46.8 Giac [F]
3.1.46.9 Mupad [B] (verification not implemented)

3.1.46.1 Optimal result

Integrand size = 12, antiderivative size = 77 \[ \int x \arctan (x)^2 \log \left (1+x^2\right ) \, dx=3 x \arctan (x)-\frac {3 \arctan (x)^2}{2}-\frac {1}{2} x^2 \arctan (x)^2-\frac {3}{2} \log \left (1+x^2\right )-x \arctan (x) \log \left (1+x^2\right )+\frac {1}{2} \left (1+x^2\right ) \arctan (x)^2 \log \left (1+x^2\right )+\frac {1}{4} \log ^2\left (1+x^2\right ) \]

output
3*x*arctan(x)-3/2*arctan(x)^2-1/2*x^2*arctan(x)^2-3/2*ln(x^2+1)-x*arctan(x 
)*ln(x^2+1)+1/2*(x^2+1)*arctan(x)^2*ln(x^2+1)+1/4*ln(x^2+1)^2
 
3.1.46.2 Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.75 \[ \int x \arctan (x)^2 \log \left (1+x^2\right ) \, dx=\frac {1}{4} \left (-4 x \arctan (x) \left (-3+\log \left (1+x^2\right )\right )+\left (-6+\log \left (1+x^2\right )\right ) \log \left (1+x^2\right )+2 \arctan (x)^2 \left (-3-x^2+\left (1+x^2\right ) \log \left (1+x^2\right )\right )\right ) \]

input
Integrate[x*ArcTan[x]^2*Log[1 + x^2],x]
 
output
(-4*x*ArcTan[x]*(-3 + Log[1 + x^2]) + (-6 + Log[1 + x^2])*Log[1 + x^2] + 2 
*ArcTan[x]^2*(-3 - x^2 + (1 + x^2)*Log[1 + x^2]))/4
 
3.1.46.3 Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.31, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.083, Rules used = {5558, 5451, 5345, 240, 5419, 5544, 2925, 2837, 2738, 5451, 5345, 240, 5419}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x \arctan (x)^2 \log \left (x^2+1\right ) \, dx\)

\(\Big \downarrow \) 5558

\(\displaystyle \int \frac {x^2 \arctan (x)}{x^2+1}dx-\int \arctan (x) \log \left (x^2+1\right )dx-\frac {1}{2} x^2 \arctan (x)^2+\frac {1}{2} \left (x^2+1\right ) \arctan (x)^2 \log \left (x^2+1\right )\)

\(\Big \downarrow \) 5451

\(\displaystyle -\int \frac {\arctan (x)}{x^2+1}dx-\int \arctan (x) \log \left (x^2+1\right )dx+\int \arctan (x)dx-\frac {1}{2} x^2 \arctan (x)^2+\frac {1}{2} \left (x^2+1\right ) \arctan (x)^2 \log \left (x^2+1\right )\)

\(\Big \downarrow \) 5345

\(\displaystyle -\int \frac {\arctan (x)}{x^2+1}dx-\int \arctan (x) \log \left (x^2+1\right )dx-\int \frac {x}{x^2+1}dx-\frac {1}{2} x^2 \arctan (x)^2+\frac {1}{2} \left (x^2+1\right ) \arctan (x)^2 \log \left (x^2+1\right )+x \arctan (x)\)

\(\Big \downarrow \) 240

\(\displaystyle -\int \frac {\arctan (x)}{x^2+1}dx-\int \arctan (x) \log \left (x^2+1\right )dx-\frac {1}{2} x^2 \arctan (x)^2+\frac {1}{2} \left (x^2+1\right ) \arctan (x)^2 \log \left (x^2+1\right )+x \arctan (x)-\frac {1}{2} \log \left (x^2+1\right )\)

\(\Big \downarrow \) 5419

\(\displaystyle -\int \arctan (x) \log \left (x^2+1\right )dx-\frac {1}{2} x^2 \arctan (x)^2+\frac {1}{2} \left (x^2+1\right ) \arctan (x)^2 \log \left (x^2+1\right )-\frac {\arctan (x)^2}{2}+x \arctan (x)-\frac {1}{2} \log \left (x^2+1\right )\)

\(\Big \downarrow \) 5544

\(\displaystyle 2 \int \frac {x^2 \arctan (x)}{x^2+1}dx+\int \frac {x \log \left (x^2+1\right )}{x^2+1}dx-\frac {1}{2} x^2 \arctan (x)^2+\frac {1}{2} \left (x^2+1\right ) \arctan (x)^2 \log \left (x^2+1\right )-x \arctan (x) \log \left (x^2+1\right )-\frac {\arctan (x)^2}{2}+x \arctan (x)-\frac {1}{2} \log \left (x^2+1\right )\)

\(\Big \downarrow \) 2925

\(\displaystyle 2 \int \frac {x^2 \arctan (x)}{x^2+1}dx+\frac {1}{2} \int \frac {\log \left (x^2+1\right )}{x^2+1}dx^2-\frac {1}{2} x^2 \arctan (x)^2+\frac {1}{2} \left (x^2+1\right ) \arctan (x)^2 \log \left (x^2+1\right )-x \arctan (x) \log \left (x^2+1\right )-\frac {\arctan (x)^2}{2}+x \arctan (x)-\frac {1}{2} \log \left (x^2+1\right )\)

\(\Big \downarrow \) 2837

\(\displaystyle 2 \int \frac {x^2 \arctan (x)}{x^2+1}dx+\frac {1}{2} \int \frac {\log \left (x^2+1\right )}{x^2}d\left (x^2+1\right )-\frac {1}{2} x^2 \arctan (x)^2+\frac {1}{2} \left (x^2+1\right ) \arctan (x)^2 \log \left (x^2+1\right )-x \arctan (x) \log \left (x^2+1\right )-\frac {\arctan (x)^2}{2}+x \arctan (x)-\frac {1}{2} \log \left (x^2+1\right )\)

\(\Big \downarrow \) 2738

\(\displaystyle 2 \int \frac {x^2 \arctan (x)}{x^2+1}dx-\frac {1}{2} x^2 \arctan (x)^2+\frac {1}{2} \left (x^2+1\right ) \arctan (x)^2 \log \left (x^2+1\right )-x \arctan (x) \log \left (x^2+1\right )-\frac {\arctan (x)^2}{2}+x \arctan (x)+\frac {1}{4} \log ^2\left (x^2+1\right )-\frac {1}{2} \log \left (x^2+1\right )\)

\(\Big \downarrow \) 5451

\(\displaystyle 2 \left (\int \arctan (x)dx-\int \frac {\arctan (x)}{x^2+1}dx\right )-\frac {1}{2} x^2 \arctan (x)^2+\frac {1}{2} \left (x^2+1\right ) \arctan (x)^2 \log \left (x^2+1\right )-x \arctan (x) \log \left (x^2+1\right )-\frac {\arctan (x)^2}{2}+x \arctan (x)+\frac {1}{4} \log ^2\left (x^2+1\right )-\frac {1}{2} \log \left (x^2+1\right )\)

\(\Big \downarrow \) 5345

\(\displaystyle 2 \left (-\int \frac {\arctan (x)}{x^2+1}dx-\int \frac {x}{x^2+1}dx+x \arctan (x)\right )-\frac {1}{2} x^2 \arctan (x)^2+\frac {1}{2} \left (x^2+1\right ) \arctan (x)^2 \log \left (x^2+1\right )-x \arctan (x) \log \left (x^2+1\right )-\frac {\arctan (x)^2}{2}+x \arctan (x)+\frac {1}{4} \log ^2\left (x^2+1\right )-\frac {1}{2} \log \left (x^2+1\right )\)

\(\Big \downarrow \) 240

\(\displaystyle 2 \left (-\int \frac {\arctan (x)}{x^2+1}dx+x \arctan (x)-\frac {1}{2} \log \left (x^2+1\right )\right )-\frac {1}{2} x^2 \arctan (x)^2+\frac {1}{2} \left (x^2+1\right ) \arctan (x)^2 \log \left (x^2+1\right )-x \arctan (x) \log \left (x^2+1\right )-\frac {\arctan (x)^2}{2}+x \arctan (x)+\frac {1}{4} \log ^2\left (x^2+1\right )-\frac {1}{2} \log \left (x^2+1\right )\)

\(\Big \downarrow \) 5419

\(\displaystyle -\frac {1}{2} x^2 \arctan (x)^2+\frac {1}{2} \left (x^2+1\right ) \arctan (x)^2 \log \left (x^2+1\right )-x \arctan (x) \log \left (x^2+1\right )+2 \left (-\frac {1}{2} \arctan (x)^2+x \arctan (x)-\frac {1}{2} \log \left (x^2+1\right )\right )-\frac {\arctan (x)^2}{2}+x \arctan (x)+\frac {1}{4} \log ^2\left (x^2+1\right )-\frac {1}{2} \log \left (x^2+1\right )\)

input
Int[x*ArcTan[x]^2*Log[1 + x^2],x]
 
output
x*ArcTan[x] - ArcTan[x]^2/2 - (x^2*ArcTan[x]^2)/2 + 2*(x*ArcTan[x] - ArcTa 
n[x]^2/2 - Log[1 + x^2]/2) - Log[1 + x^2]/2 - x*ArcTan[x]*Log[1 + x^2] + ( 
(1 + x^2)*ArcTan[x]^2*Log[1 + x^2])/2 + Log[1 + x^2]^2/4
 

3.1.46.3.1 Defintions of rubi rules used

rule 240
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x 
^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
 

rule 2738
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Lo 
g[c*x^n])^2/(2*b*n), x] /; FreeQ[{a, b, c, n}, x]
 

rule 2837
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_. 
)*(x_))^(q_.), x_Symbol] :> Simp[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 2925
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m 
_.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Si 
mplify[(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] && Integer 
Q[r] && IntegerQ[s/n] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0 
] || IGtQ[q, 0])
 

rule 5345
Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a 
+ b*ArcTan[c*x^n])^p, x] - Simp[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 5419
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbo 
l] :> 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 5451
Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e 
_.)*(x_)^2), x_Symbol] :> Simp[f^2/e   Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x] 
)^p, x], x] - Simp[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 5544
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*( 
e_.)), x_Symbol] :> Simp[x*(d + e*Log[f + g*x^2])*(a + b*ArcTan[c*x]), x] + 
 (-Simp[b*c   Int[x*((d + e*Log[f + g*x^2])/(1 + c^2*x^2)), x], x] - Simp[2 
*e*g   Int[x^2*((a + b*ArcTan[c*x])/(f + g*x^2)), x], x]) /; FreeQ[{a, b, c 
, d, e, f, g}, x]
 

rule 5558
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^2*((d_.) + Log[(f_) + (g_.)*(x_)^2]* 
(e_.))*(x_), x_Symbol] :> Simp[(f + g*x^2)*(d + e*Log[f + g*x^2])*((a + b*A 
rcTan[c*x])^2/(2*g)), x] + (-Simp[e*x^2*((a + b*ArcTan[c*x])^2/2), x] - Sim 
p[b/c   Int[(d + e*Log[f + g*x^2])*(a + b*ArcTan[c*x]), x], x] + Simp[b*c*e 
   Int[x^2*((a + b*ArcTan[c*x])/(1 + c^2*x^2)), x], x]) /; FreeQ[{a, b, c, 
d, e, f, g}, x] && EqQ[g, c^2*f]
 
3.1.46.4 Maple [A] (verified)

Time = 4.50 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01

method result size
parallelrisch \(\frac {\ln \left (x^{2}+1\right ) \arctan \left (x \right )^{2} x^{2}}{2}-\frac {x^{2} \arctan \left (x \right )^{2}}{2}-x \arctan \left (x \right ) \ln \left (x^{2}+1\right )+\frac {\arctan \left (x \right )^{2} \ln \left (x^{2}+1\right )}{2}+3 x \arctan \left (x \right )-\frac {3 \arctan \left (x \right )^{2}}{2}+\frac {\ln \left (x^{2}+1\right )^{2}}{4}-\frac {3 \ln \left (x^{2}+1\right )}{2}\) \(78\)
default \(\text {Expression too large to display}\) \(3134\)
risch \(\text {Expression too large to display}\) \(113915\)

input
int(x*arctan(x)^2*ln(x^2+1),x,method=_RETURNVERBOSE)
 
output
1/2*ln(x^2+1)*arctan(x)^2*x^2-1/2*x^2*arctan(x)^2-x*arctan(x)*ln(x^2+1)+1/ 
2*arctan(x)^2*ln(x^2+1)+3*x*arctan(x)-3/2*arctan(x)^2+1/4*ln(x^2+1)^2-3/2* 
ln(x^2+1)
 
3.1.46.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.68 \[ \int x \arctan (x)^2 \log \left (1+x^2\right ) \, dx=-\frac {1}{2} \, {\left (x^{2} + 3\right )} \arctan \left (x\right )^{2} + 3 \, x \arctan \left (x\right ) + \frac {1}{2} \, {\left ({\left (x^{2} + 1\right )} \arctan \left (x\right )^{2} - 2 \, x \arctan \left (x\right ) - 3\right )} \log \left (x^{2} + 1\right ) + \frac {1}{4} \, \log \left (x^{2} + 1\right )^{2} \]

input
integrate(x*arctan(x)^2*log(x^2+1),x, algorithm="fricas")
 
output
-1/2*(x^2 + 3)*arctan(x)^2 + 3*x*arctan(x) + 1/2*((x^2 + 1)*arctan(x)^2 - 
2*x*arctan(x) - 3)*log(x^2 + 1) + 1/4*log(x^2 + 1)^2
 
3.1.46.6 Sympy [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.13 \[ \int x \arctan (x)^2 \log \left (1+x^2\right ) \, dx=\frac {x^{2} \log {\left (x^{2} + 1 \right )} \operatorname {atan}^{2}{\left (x \right )}}{2} - \frac {x^{2} \operatorname {atan}^{2}{\left (x \right )}}{2} - x \log {\left (x^{2} + 1 \right )} \operatorname {atan}{\left (x \right )} + 3 x \operatorname {atan}{\left (x \right )} + \frac {\log {\left (x^{2} + 1 \right )}^{2}}{4} + \frac {\log {\left (x^{2} + 1 \right )} \operatorname {atan}^{2}{\left (x \right )}}{2} - \frac {3 \log {\left (x^{2} + 1 \right )}}{2} - \frac {3 \operatorname {atan}^{2}{\left (x \right )}}{2} \]

input
integrate(x*atan(x)**2*ln(x**2+1),x)
 
output
x**2*log(x**2 + 1)*atan(x)**2/2 - x**2*atan(x)**2/2 - x*log(x**2 + 1)*atan 
(x) + 3*x*atan(x) + log(x**2 + 1)**2/4 + log(x**2 + 1)*atan(x)**2/2 - 3*lo 
g(x**2 + 1)/2 - 3*atan(x)**2/2
 
3.1.46.7 Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.87 \[ \int x \arctan (x)^2 \log \left (1+x^2\right ) \, dx=-\frac {1}{2} \, {\left (x^{2} - {\left (x^{2} + 1\right )} \log \left (x^{2} + 1\right ) + 1\right )} \arctan \left (x\right )^{2} - {\left (x \log \left (x^{2} + 1\right ) - 3 \, x + 2 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) + \arctan \left (x\right )^{2} + \frac {1}{4} \, \log \left (x^{2} + 1\right )^{2} - \frac {3}{2} \, \log \left (x^{2} + 1\right ) \]

input
integrate(x*arctan(x)^2*log(x^2+1),x, algorithm="maxima")
 
output
-1/2*(x^2 - (x^2 + 1)*log(x^2 + 1) + 1)*arctan(x)^2 - (x*log(x^2 + 1) - 3* 
x + 2*arctan(x))*arctan(x) + arctan(x)^2 + 1/4*log(x^2 + 1)^2 - 3/2*log(x^ 
2 + 1)
 
3.1.46.8 Giac [F]

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

input
integrate(x*arctan(x)^2*log(x^2+1),x, algorithm="giac")
 
output
integrate(x*arctan(x)^2*log(x^2 + 1), x)
 
3.1.46.9 Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01 \[ \int x \arctan (x)^2 \log \left (1+x^2\right ) \, dx=\frac {{\ln \left (x^2+1\right )}^2}{4}-\frac {3\,\ln \left (x^2+1\right )}{2}-\frac {3\,{\mathrm {atan}\left (x\right )}^2}{2}+\frac {\ln \left (x^2+1\right )\,{\mathrm {atan}\left (x\right )}^2}{2}+x\,\left (3\,\mathrm {atan}\left (x\right )-\ln \left (x^2+1\right )\,\mathrm {atan}\left (x\right )\right )-x^2\,\left (\frac {{\mathrm {atan}\left (x\right )}^2}{2}-\frac {\ln \left (x^2+1\right )\,{\mathrm {atan}\left (x\right )}^2}{2}\right ) \]

input
int(x*log(x^2 + 1)*atan(x)^2,x)
 
output
log(x^2 + 1)^2/4 - (3*log(x^2 + 1))/2 - (3*atan(x)^2)/2 + (log(x^2 + 1)*at 
an(x)^2)/2 + x*(3*atan(x) - log(x^2 + 1)*atan(x)) - x^2*(atan(x)^2/2 - (lo 
g(x^2 + 1)*atan(x)^2)/2)