3.1276 \(\int x^3 \tan ^{-1}(x) \log (1+x^2) \, dx\)
Optimal. Leaf size=88 \[ -\frac {1}{8} x^4 \tan ^{-1}(x)+\frac {7 x^3}{72}+\frac {1}{4} x \log \left (x^2+1\right )+\frac {1}{4} x^2 \tan ^{-1}(x)-\frac {1}{4} \log \left (x^2+1\right ) \tan ^{-1}(x)+\frac {1}{4} x^4 \log \left (x^2+1\right ) \tan ^{-1}(x)-\frac {1}{12} x^3 \log \left (x^2+1\right )-\frac {25 x}{24}+\frac {25}{24} \tan ^{-1}(x) \]
[Out]
-25/24*x+7/72*x^3+25/24*arctan(x)+1/4*x^2*arctan(x)-1/8*x^4*arctan(x)+1/4*x*ln(x^2+1)-1/12*x^3*ln(x^2+1)-1/4*a
rctan(x)*ln(x^2+1)+1/4*x^4*arctan(x)*ln(x^2+1)
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Rubi [A] time = 0.12, antiderivative size = 88, normalized size of antiderivative =
1.00, number of steps used = 14, number of rules used = 12, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) =
1.000, Rules used = {4852, 302, 203, 2454, 2395, 43, 5019, 459, 321, 2471, 2448, 2455}
\[ \frac {7 x^3}{72}-\frac {1}{12} x^3 \log \left (x^2+1\right )+\frac {1}{4} x \log \left (x^2+1\right )-\frac {1}{8} x^4 \tan ^{-1}(x)+\frac {1}{4} x^2 \tan ^{-1}(x)+\frac {1}{4} x^4 \log \left (x^2+1\right ) \tan ^{-1}(x)-\frac {1}{4} \log \left (x^2+1\right ) \tan ^{-1}(x)-\frac {25 x}{24}+\frac {25}{24} \tan ^{-1}(x) \]
Antiderivative was successfully verified.
[In]
Int[x^3*ArcTan[x]*Log[1 + x^2],x]
[Out]
(-25*x)/24 + (7*x^3)/72 + (25*ArcTan[x])/24 + (x^2*ArcTan[x])/4 - (x^4*ArcTan[x])/8 + (x*Log[1 + x^2])/4 - (x^
3*Log[1 + x^2])/12 - (ArcTan[x]*Log[1 + x^2])/4 + (x^4*ArcTan[x]*Log[1 + x^2])/4
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 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 302
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]
Rule 321
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]
Rule 459
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]
Rule 2395
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 2448
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]
Rule 2454
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) && !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])
Rule 2455
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^(m_.), x_Symbol] :> Simp[((f*x)^(m
+ 1)*(a + b*Log[c*(d + e*x^n)^p]))/(f*(m + 1)), x] - Dist[(b*e*n*p)/(f*(m + 1)), Int[(x^(n - 1)*(f*x)^(m + 1))
/(d + e*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]
Rule 2471
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol]
:> With[{t = ExpandIntegrand[(a + b*Log[c*(d + e*x^n)^p])^q, (f + g*x^s)^r, x]}, Int[t, x] /; SumQ[t]] /; Free
Q[{a, b, c, d, e, f, g, n, p, q, r, s}, x] && IntegerQ[n] && IGtQ[q, 0] && IntegerQ[r] && IntegerQ[s] && (EqQ[
q, 1] || (GtQ[r, 0] && GtQ[s, 1]) || (LtQ[s, 0] && LtQ[r, 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 5019
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.))*(x_)^(m_.), x_Symbol] :> With
[{u = IntHide[x^m*(d + e*Log[f + g*x^2]), x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[ExpandIntegrand[u
/(1 + c^2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[(m + 1)/2, 0]
Rubi steps
\begin {align*} \int x^3 \tan ^{-1}(x) \log \left (1+x^2\right ) \, dx &=\frac {1}{4} x^2 \tan ^{-1}(x)-\frac {1}{8} x^4 \tan ^{-1}(x)-\frac {1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{4} x^4 \tan ^{-1}(x) \log \left (1+x^2\right )-\int \left (-\frac {x^2 \left (-2+x^2\right )}{8 \left (1+x^2\right )}+\frac {1}{4} \left (-1+x^2\right ) \log \left (1+x^2\right )\right ) \, dx\\ &=\frac {1}{4} x^2 \tan ^{-1}(x)-\frac {1}{8} x^4 \tan ^{-1}(x)-\frac {1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{4} x^4 \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{8} \int \frac {x^2 \left (-2+x^2\right )}{1+x^2} \, dx-\frac {1}{4} \int \left (-1+x^2\right ) \log \left (1+x^2\right ) \, dx\\ &=\frac {x^3}{24}+\frac {1}{4} x^2 \tan ^{-1}(x)-\frac {1}{8} x^4 \tan ^{-1}(x)-\frac {1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{4} x^4 \tan ^{-1}(x) \log \left (1+x^2\right )-\frac {1}{4} \int \left (-\log \left (1+x^2\right )+x^2 \log \left (1+x^2\right )\right ) \, dx-\frac {3}{8} \int \frac {x^2}{1+x^2} \, dx\\ &=-\frac {3 x}{8}+\frac {x^3}{24}+\frac {1}{4} x^2 \tan ^{-1}(x)-\frac {1}{8} x^4 \tan ^{-1}(x)-\frac {1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{4} x^4 \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{4} \int \log \left (1+x^2\right ) \, dx-\frac {1}{4} \int x^2 \log \left (1+x^2\right ) \, dx+\frac {3}{8} \int \frac {1}{1+x^2} \, dx\\ &=-\frac {3 x}{8}+\frac {x^3}{24}+\frac {3}{8} \tan ^{-1}(x)+\frac {1}{4} x^2 \tan ^{-1}(x)-\frac {1}{8} x^4 \tan ^{-1}(x)+\frac {1}{4} x \log \left (1+x^2\right )-\frac {1}{12} x^3 \log \left (1+x^2\right )-\frac {1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{4} x^4 \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{6} \int \frac {x^4}{1+x^2} \, dx-\frac {1}{2} \int \frac {x^2}{1+x^2} \, dx\\ &=-\frac {7 x}{8}+\frac {x^3}{24}+\frac {3}{8} \tan ^{-1}(x)+\frac {1}{4} x^2 \tan ^{-1}(x)-\frac {1}{8} x^4 \tan ^{-1}(x)+\frac {1}{4} x \log \left (1+x^2\right )-\frac {1}{12} x^3 \log \left (1+x^2\right )-\frac {1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{4} x^4 \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{6} \int \left (-1+x^2+\frac {1}{1+x^2}\right ) \, dx+\frac {1}{2} \int \frac {1}{1+x^2} \, dx\\ &=-\frac {25 x}{24}+\frac {7 x^3}{72}+\frac {7}{8} \tan ^{-1}(x)+\frac {1}{4} x^2 \tan ^{-1}(x)-\frac {1}{8} x^4 \tan ^{-1}(x)+\frac {1}{4} x \log \left (1+x^2\right )-\frac {1}{12} x^3 \log \left (1+x^2\right )-\frac {1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{4} x^4 \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{6} \int \frac {1}{1+x^2} \, dx\\ &=-\frac {25 x}{24}+\frac {7 x^3}{72}+\frac {25}{24} \tan ^{-1}(x)+\frac {1}{4} x^2 \tan ^{-1}(x)-\frac {1}{8} x^4 \tan ^{-1}(x)+\frac {1}{4} x \log \left (1+x^2\right )-\frac {1}{12} x^3 \log \left (1+x^2\right )-\frac {1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{4} x^4 \tan ^{-1}(x) \log \left (1+x^2\right )\\ \end {align*}
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Mathematica [A] time = 0.03, size = 56, normalized size = 0.64 \[ \frac {1}{72} \left (x \left (7 x^2-6 \left (x^2-3\right ) \log \left (x^2+1\right )-75\right )+3 \left (-3 x^4+6 x^2+6 \left (x^4-1\right ) \log \left (x^2+1\right )+25\right ) \tan ^{-1}(x)\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[x^3*ArcTan[x]*Log[1 + x^2],x]
[Out]
(x*(-75 + 7*x^2 - 6*(-3 + x^2)*Log[1 + x^2]) + 3*ArcTan[x]*(25 + 6*x^2 - 3*x^4 + 6*(-1 + x^4)*Log[1 + x^2]))/7
2
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fricas [A] time = 0.39, size = 49, normalized size = 0.56 \[ \frac {7}{72} \, x^{3} - \frac {1}{24} \, {\left (3 \, x^{4} - 6 \, x^{2} - 25\right )} \arctan \relax (x) - \frac {1}{12} \, {\left (x^{3} - 3 \, {\left (x^{4} - 1\right )} \arctan \relax (x) - 3 \, x\right )} \log \left (x^{2} + 1\right ) - \frac {25}{24} \, x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*arctan(x)*log(x^2+1),x, algorithm="fricas")
[Out]
7/72*x^3 - 1/24*(3*x^4 - 6*x^2 - 25)*arctan(x) - 1/12*(x^3 - 3*(x^4 - 1)*arctan(x) - 3*x)*log(x^2 + 1) - 25/24
*x
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giac [A] time = 5.98, size = 124, normalized size = 1.41 \[ \frac {1}{8} \, \pi x^{4} \log \left (x^{2} + 1\right ) \mathrm {sgn}\relax (x) - \frac {1}{4} \, x^{4} \arctan \left (\frac {1}{x}\right ) \log \left (x^{2} + 1\right ) - \frac {1}{16} \, \pi x^{4} \mathrm {sgn}\relax (x) + \frac {1}{8} \, x^{4} \arctan \left (\frac {1}{x}\right ) - \frac {1}{12} \, x^{3} \log \left (x^{2} + 1\right ) + \frac {1}{8} \, \pi x^{2} \mathrm {sgn}\relax (x) + \frac {7}{72} \, x^{3} - \frac {1}{4} \, x^{2} \arctan \left (\frac {1}{x}\right ) - \frac {1}{8} \, \pi \log \left (x^{2} + 1\right ) \mathrm {sgn}\relax (x) + \frac {1}{4} \, x \log \left (x^{2} + 1\right ) + \frac {1}{4} \, \arctan \left (\frac {1}{x}\right ) \log \left (x^{2} + 1\right ) - \frac {25}{24} \, \pi \mathrm {sgn}\relax (x) - \frac {25}{24} \, x + \frac {25}{24} \, \arctan \relax (x) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*arctan(x)*log(x^2+1),x, algorithm="giac")
[Out]
1/8*pi*x^4*log(x^2 + 1)*sgn(x) - 1/4*x^4*arctan(1/x)*log(x^2 + 1) - 1/16*pi*x^4*sgn(x) + 1/8*x^4*arctan(1/x) -
1/12*x^3*log(x^2 + 1) + 1/8*pi*x^2*sgn(x) + 7/72*x^3 - 1/4*x^2*arctan(1/x) - 1/8*pi*log(x^2 + 1)*sgn(x) + 1/4
*x*log(x^2 + 1) + 1/4*arctan(1/x)*log(x^2 + 1) - 25/24*pi*sgn(x) - 25/24*x + 25/24*arctan(x)
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maple [C] time = 3.22, size = 2849, normalized size = 32.38 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^3*arctan(x)*ln(x^2+1),x)
[Out]
1/6*(4+3*I*arctan(x)+I*x-3*x*arctan(x)-x^2-3*I*arctan(x)*x^2+3*x^3*arctan(x))*(x+I)*ln((1+I*x)/(x^2+1)^(1/2))+
1/4*x^2*arctan(x)-1/8*x^4*arctan(x)+1/2*ln((1+I*x)^2/(x^2+1)+1)*arctan(x)-1/2*ln(2)*arctan(x)-1/2*ln((1+I*x)^2
/(x^2+1)+1)*x+1/6*ln((1+I*x)^2/(x^2+1)+1)*x^3-1/6*x^3*ln(2)+1/6*Pi*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)
+1)^2)^3-1/6*Pi*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)^3+1/6*Pi*csgn(I*(1+I*x)^2/(x^2+1))^3+1/24*I*csgn(I*(1+I*x)^2/(
x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^3*Pi*x^3+1/24*I*csgn(I*(1+I*x)^2/(x^2+1))^3*Pi*x^3-1/24*I*csgn(I*((1+I*x)^2/(x
^2+1)+1)^2)^3*Pi*x^3+1/8*I*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^3*arctan(x)*Pi-1/8*I*csgn(I*(1+I*
x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^3*Pi*x+1/8*I*csgn(I*(1+I*x)^2/(x^2+1))^3*arctan(x)*Pi-1/8*I*csgn(I*(1+I*
x)^2/(x^2+1))^3*Pi*x-1/8*I*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)^3*arctan(x)*Pi+1/8*I*csgn(I*((1+I*x)^2/(x^2+1)+1)^2
)^3*Pi*x+1/24*I*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)/(x^2+1)^(1/2))^2*Pi*x^3-1/24*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^3+1/12*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^3-1/8*I*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*arc
tan(x)*Pi+1/8*I*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*Pi*x-1/8*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))*arctan(x)*Pi+1/8*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))*Pi*x-1/4*I*csgn(I*(1+I*x)^2/(x^2+1))^2*csgn(
I*(1+I*x)/(x^2+1)^(1/2))*arctan(x)*Pi+1/4*I*csgn(I*(1+I*x)^2/(x^2+1))^2*csgn(I*(1+I*x)/(x^2+1)^(1/2))*Pi*x+1/8
*I*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)/(x^2+1)^(1/2))^2*arctan(x)*Pi-1/8*I*csgn(I*(1+I*x)^2/(x^2+1))*csgn
(I*(1+I*x)/(x^2+1)^(1/2))^2*Pi*x-1/8*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+1/8*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+1/4*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-1/4*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-1/8*I*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)*csg
n(I*(1+I*x)^2/(x^2+1))*arctan(x)*Pi*x^4+1/2*ln(2)*x+1/6*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)+1/8*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^4-1/4*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^4+1/24*I*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)*csgn(I*(1+
I*x)^2/(x^2+1))*Pi*x^3+1/8*I*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)
*csgn(I*(1+I*x)^2/(x^2+1))*arctan(x)*Pi-1/8*I*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)*csgn(I*(1+I*x)^2/(x^2+1))*Pi*x+2/3*I*ln(2)-41/36*I-1/8*I*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2
/(x^2+1)+1)^2)^3*arctan(x)*Pi*x^4-1/8*I*csgn(I*(1+I*x)^2/(x^2+1))^3*arctan(x)*Pi*x^4+1/8*I*csgn(I*((1+I*x)^2/(
x^2+1)+1)^2)^3*arctan(x)*Pi*x^4-1/24*I*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*Pi*x^3-1/24*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))*Pi*x^
3-1/12*I*csgn(I*(1+I*x)^2/(x^2+1))^2*csgn(I*(1+I*x)/(x^2+1)^(1/2))*Pi*x^3-25/24*x+1/6*Pi*csgn(I*(1+I*x)/(x^2+1
)^(1/2))^2*csgn(I*(1+I*x)^2/(x^2+1))-1/3*Pi*csgn(I*(1+I*x)/(x^2+1)^(1/2))*csgn(I*(1+I*x)^2/(x^2+1))^2-1/6*Pi*c
sgn(I*((1+I*x)^2/(x^2+1)+1))^2*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)+1/3*Pi*csgn(I*((1+I*x)^2/(x^2+1)+1))*csgn(I*((1
+I*x)^2/(x^2+1)+1)^2)^2-1/6*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)^2-1/6*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)^2-1/2*ln((1+I*x)^2/(x^
2+1)+1)*arctan(x)*x^4+1/2*ln(2)*arctan(x)*x^4+7/72*x^3+41/24*arctan(x)+1/8*I*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*arctan(x)*Pi*x^4+1/8*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))*arctan(x)*Pi*x^4+1/4*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^4-1/8*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^4
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maxima [A] time = 0.41, size = 62, normalized size = 0.70 \[ \frac {7}{72} \, x^{3} + \frac {1}{8} \, {\left (2 \, x^{4} \log \left (x^{2} + 1\right ) - x^{4} + 2 \, x^{2} - 2 \, \log \left (x^{2} + 1\right )\right )} \arctan \relax (x) - \frac {1}{12} \, {\left (x^{3} - 3 \, x\right )} \log \left (x^{2} + 1\right ) - \frac {25}{24} \, x + \frac {25}{24} \, \arctan \relax (x) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*arctan(x)*log(x^2+1),x, algorithm="maxima")
[Out]
7/72*x^3 + 1/8*(2*x^4*log(x^2 + 1) - x^4 + 2*x^2 - 2*log(x^2 + 1))*arctan(x) - 1/12*(x^3 - 3*x)*log(x^2 + 1) -
25/24*x + 25/24*arctan(x)
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mupad [B] time = 0.53, size = 69, normalized size = 0.78 \[ \frac {25\,\mathrm {atan}\relax (x)}{24}+\frac {x^2\,\mathrm {atan}\relax (x)}{4}+x\,\left (\frac {\ln \left (x^2+1\right )}{4}-\frac {25}{24}\right )-x^3\,\left (\frac {\ln \left (x^2+1\right )}{12}-\frac {7}{72}\right )-x^4\,\left (\frac {\mathrm {atan}\relax (x)}{8}-\frac {\ln \left (x^2+1\right )\,\mathrm {atan}\relax (x)}{4}\right )-\frac {\ln \left (x^2+1\right )\,\mathrm {atan}\relax (x)}{4} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^3*log(x^2 + 1)*atan(x),x)
[Out]
(25*atan(x))/24 + (x^2*atan(x))/4 + x*(log(x^2 + 1)/4 - 25/24) - x^3*(log(x^2 + 1)/12 - 7/72) - x^4*(atan(x)/8
- (log(x^2 + 1)*atan(x))/4) - (log(x^2 + 1)*atan(x))/4
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sympy [A] time = 2.85, size = 83, normalized size = 0.94 \[ \frac {x^{4} \log {\left (x^{2} + 1 \right )} \operatorname {atan}{\relax (x )}}{4} - \frac {x^{4} \operatorname {atan}{\relax (x )}}{8} - \frac {x^{3} \log {\left (x^{2} + 1 \right )}}{12} + \frac {7 x^{3}}{72} + \frac {x^{2} \operatorname {atan}{\relax (x )}}{4} + \frac {x \log {\left (x^{2} + 1 \right )}}{4} - \frac {25 x}{24} - \frac {\log {\left (x^{2} + 1 \right )} \operatorname {atan}{\relax (x )}}{4} + \frac {25 \operatorname {atan}{\relax (x )}}{24} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**3*atan(x)*ln(x**2+1),x)
[Out]
x**4*log(x**2 + 1)*atan(x)/4 - x**4*atan(x)/8 - x**3*log(x**2 + 1)/12 + 7*x**3/72 + x**2*atan(x)/4 + x*log(x**
2 + 1)/4 - 25*x/24 - log(x**2 + 1)*atan(x)/4 + 25*atan(x)/24
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