\(\int x^3 \arctan (x) \log (1+x^2) \, dx\) [1276]

Optimal result

Integrand size = 12, antiderivative size = 88 \[ \int x^3 \arctan (x) \log \left (1+x^2\right ) \, dx=-\frac {25 x}{24}+\frac {7 x^3}{72}+\frac {25 \arctan (x)}{24}+\frac {1}{4} x^2 \arctan (x)-\frac {1}{8} x^4 \arctan (x)+\frac {1}{4} x \log \left (1+x^2\right )-\frac {1}{12} x^3 \log \left (1+x^2\right )-\frac {1}{4} \arctan (x) \log \left (1+x^2\right )+\frac {1}{4} x^4 \arctan (x) \log \left (1+x^2\right ) \]

[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)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {4946, 308, 209, 2504, 2442, 45, 5139, 470, 327, 2521, 2498, 2505} \[ \int x^3 \arctan (x) \log \left (1+x^2\right ) \, dx=-\frac {1}{8} x^4 \arctan (x)+\frac {1}{4} x^2 \arctan (x)-\frac {1}{4} \arctan (x) \log \left (x^2+1\right )+\frac {1}{4} x^4 \arctan (x) \log \left (x^2+1\right )+\frac {25 \arctan (x)}{24}+\frac {7 x^3}{72}+\frac {1}{4} x \log \left (x^2+1\right )-\frac {1}{12} x^3 \log \left (x^2+1\right )-\frac {25 x}{24} \]

[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 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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 308

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 327

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 470

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 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 2498

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 2504

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 2505

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 2521

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 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 5139

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*} \text {integral}& = \frac {1}{4} x^2 \arctan (x)-\frac {1}{8} x^4 \arctan (x)-\frac {1}{4} \arctan (x) \log \left (1+x^2\right )+\frac {1}{4} x^4 \arctan (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 \arctan (x)-\frac {1}{8} x^4 \arctan (x)-\frac {1}{4} \arctan (x) \log \left (1+x^2\right )+\frac {1}{4} x^4 \arctan (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 \arctan (x)-\frac {1}{8} x^4 \arctan (x)-\frac {1}{4} \arctan (x) \log \left (1+x^2\right )+\frac {1}{4} x^4 \arctan (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 \arctan (x)-\frac {1}{8} x^4 \arctan (x)-\frac {1}{4} \arctan (x) \log \left (1+x^2\right )+\frac {1}{4} x^4 \arctan (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 \arctan (x)}{8}+\frac {1}{4} x^2 \arctan (x)-\frac {1}{8} x^4 \arctan (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} \arctan (x) \log \left (1+x^2\right )+\frac {1}{4} x^4 \arctan (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 \arctan (x)}{8}+\frac {1}{4} x^2 \arctan (x)-\frac {1}{8} x^4 \arctan (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} \arctan (x) \log \left (1+x^2\right )+\frac {1}{4} x^4 \arctan (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 \arctan (x)}{8}+\frac {1}{4} x^2 \arctan (x)-\frac {1}{8} x^4 \arctan (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} \arctan (x) \log \left (1+x^2\right )+\frac {1}{4} x^4 \arctan (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 \arctan (x)}{24}+\frac {1}{4} x^2 \arctan (x)-\frac {1}{8} x^4 \arctan (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} \arctan (x) \log \left (1+x^2\right )+\frac {1}{4} x^4 \arctan (x) \log \left (1+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.64 \[ \int x^3 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {1}{72} \left (x \left (-75+7 x^2-6 \left (-3+x^2\right ) \log \left (1+x^2\right )\right )+3 \arctan (x) \left (25+6 x^2-3 x^4+6 \left (-1+x^4\right ) \log \left (1+x^2\right )\right )\right ) \]

[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

Maple [A] (verified)

Time = 1.63 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.81

method	result	size
parallelrisch	\(-\frac {25 x}{24}+\frac {7 x^{3}}{72}+\frac {25 \arctan \left (x \right )}{24}+\frac {x^{2} \arctan \left (x \right )}{4}-\frac {x^{4} \arctan \left (x \right )}{8}+\frac {x \ln \left (x^{2}+1\right )}{4}-\frac {x^{3} \ln \left (x^{2}+1\right )}{12}-\frac {\arctan \left (x \right ) \ln \left (x^{2}+1\right )}{4}+\frac {x^{4} \arctan \left (x \right ) \ln \left (x^{2}+1\right )}{4}\)	\(71\)
default	\(\text {Expression too large to display}\)	\(2834\)
risch	\(\text {Expression too large to display}\)	\(16521\)

[In]

int(x^3*arctan(x)*ln(x^2+1),x,method=_RETURNVERBOSE)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.56 \[ \int x^3 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {7}{72} \, x^{3} - \frac {1}{24} \, {\left (3 \, x^{4} - 6 \, x^{2} - 25\right )} \arctan \left (x\right ) - \frac {1}{12} \, {\left (x^{3} - 3 \, {\left (x^{4} - 1\right )} \arctan \left (x\right ) - 3 \, x\right )} \log \left (x^{2} + 1\right ) - \frac {25}{24} \, x \]

[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

Sympy [A] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.94 \[ \int x^3 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {x^{4} \log {\left (x^{2} + 1 \right )} \operatorname {atan}{\left (x \right )}}{4} - \frac {x^{4} \operatorname {atan}{\left (x \right )}}{8} - \frac {x^{3} \log {\left (x^{2} + 1 \right )}}{12} + \frac {7 x^{3}}{72} + \frac {x^{2} \operatorname {atan}{\left (x \right )}}{4} + \frac {x \log {\left (x^{2} + 1 \right )}}{4} - \frac {25 x}{24} - \frac {\log {\left (x^{2} + 1 \right )} \operatorname {atan}{\left (x \right )}}{4} + \frac {25 \operatorname {atan}{\left (x \right )}}{24} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.70 \[ \int x^3 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {7}{72} \, x^{3} + \frac {1}{8} \, {\left (2 \, x^{4} \log \left (x^{2} + 1\right ) - x^{4} + 2 \, x^{2} - 2 \, \log \left (x^{2} + 1\right )\right )} \arctan \left (x\right ) - \frac {1}{12} \, {\left (x^{3} - 3 \, x\right )} \log \left (x^{2} + 1\right ) - \frac {25}{24} \, x + \frac {25}{24} \, \arctan \left (x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.41 \[ \int x^3 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {1}{8} \, \pi x^{4} \log \left (x^{2} + 1\right ) \mathrm {sgn}\left (x\right ) - \frac {1}{4} \, x^{4} \arctan \left (\frac {1}{x}\right ) \log \left (x^{2} + 1\right ) - \frac {1}{16} \, \pi x^{4} \mathrm {sgn}\left (x\right ) + \frac {1}{8} \, x^{4} \arctan \left (\frac {1}{x}\right ) - \frac {1}{12} \, x^{3} \log \left (x^{2} + 1\right ) + \frac {1}{8} \, \pi x^{2} \mathrm {sgn}\left (x\right ) + \frac {7}{72} \, x^{3} - \frac {1}{4} \, x^{2} \arctan \left (\frac {1}{x}\right ) - \frac {1}{8} \, \pi \log \left (x^{2} + 1\right ) \mathrm {sgn}\left (x\right ) + \frac {1}{4} \, x \log \left (x^{2} + 1\right ) + \frac {1}{4} \, \arctan \left (\frac {1}{x}\right ) \log \left (x^{2} + 1\right ) - \frac {25}{24} \, \pi \mathrm {sgn}\left (x\right ) - \frac {25}{24} \, x + \frac {25}{24} \, \arctan \left (x\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78 \[ \int x^3 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {25\,\mathrm {atan}\left (x\right )}{24}+\frac {x^2\,\mathrm {atan}\left (x\right )}{4}+x\,\left (\frac {\ln \left (x^2+1\right )}{4}-\frac {25}{24}\right )-x^3\,\left (\frac {\ln \left (x^2+1\right )}{12}-\frac {7}{72}\right )-x^4\,\left (\frac {\mathrm {atan}\left (x\right )}{8}-\frac {\ln \left (x^2+1\right )\,\mathrm {atan}\left (x\right )}{4}\right )-\frac {\ln \left (x^2+1\right )\,\mathrm {atan}\left (x\right )}{4} \]

[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