Integrand size = 9, antiderivative size = 38 \[ \int \arctan (x) \log \left (1+x^2\right ) \, dx=-2 x \arctan (x)+\arctan (x)^2+\log \left (1+x^2\right )+x \arctan (x) \log \left (1+x^2\right )-\frac {1}{4} \log ^2\left (1+x^2\right ) \] Output:
-2*x*arctan(x)+arctan(x)^2+ln(x^2+1)+x*arctan(x)*ln(x^2+1)-1/4*ln(x^2+1)^2
Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \arctan (x) \log \left (1+x^2\right ) \, dx=-2 x \arctan (x)+\arctan (x)^2+\log \left (1+x^2\right )+x \arctan (x) \log \left (1+x^2\right )-\frac {1}{4} \log ^2\left (1+x^2\right ) \] Input:
Integrate[ArcTan[x]*Log[1 + x^2],x]
Output:
-2*x*ArcTan[x] + ArcTan[x]^2 + Log[1 + x^2] + x*ArcTan[x]*Log[1 + x^2] - L og[1 + x^2]^2/4
Time = 0.60 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.889, Rules used = {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.
\(\displaystyle \int \arctan (x) \log \left (x^2+1\right ) \, dx\) |
\(\Big \downarrow \) 5544 |
\(\displaystyle -2 \int \frac {x^2 \arctan (x)}{x^2+1}dx-\int \frac {x \log \left (x^2+1\right )}{x^2+1}dx+x \arctan (x) \log \left (x^2+1\right )\) |
\(\Big \downarrow \) 2925 |
\(\displaystyle -2 \int \frac {x^2 \arctan (x)}{x^2+1}dx-\frac {1}{2} \int \frac {\log \left (x^2+1\right )}{x^2+1}dx^2+x \arctan (x) \log \left (x^2+1\right )\) |
\(\Big \downarrow \) 2837 |
\(\displaystyle -2 \int \frac {x^2 \arctan (x)}{x^2+1}dx-\frac {1}{2} \int \frac {\log \left (x^2+1\right )}{x^2}d\left (x^2+1\right )+x \arctan (x) \log \left (x^2+1\right )\) |
\(\Big \downarrow \) 2738 |
\(\displaystyle -2 \int \frac {x^2 \arctan (x)}{x^2+1}dx+x \arctan (x) \log \left (x^2+1\right )-\frac {1}{4} \log ^2\left (x^2+1\right )\) |
\(\Big \downarrow \) 5451 |
\(\displaystyle -2 \left (\int \arctan (x)dx-\int \frac {\arctan (x)}{x^2+1}dx\right )+x \arctan (x) \log \left (x^2+1\right )-\frac {1}{4} \log ^2\left (x^2+1\right )\) |
\(\Big \downarrow \) 5345 |
\(\displaystyle -2 \left (-\int \frac {\arctan (x)}{x^2+1}dx-\int \frac {x}{x^2+1}dx+x \arctan (x)\right )+x \arctan (x) \log \left (x^2+1\right )-\frac {1}{4} \log ^2\left (x^2+1\right )\) |
\(\Big \downarrow \) 240 |
\(\displaystyle -2 \left (-\int \frac {\arctan (x)}{x^2+1}dx+x \arctan (x)-\frac {1}{2} \log \left (x^2+1\right )\right )+x \arctan (x) \log \left (x^2+1\right )-\frac {1}{4} \log ^2\left (x^2+1\right )\) |
\(\Big \downarrow \) 5419 |
\(\displaystyle x \arctan (x) \log \left (x^2+1\right )-2 \left (-\frac {1}{2} \arctan (x)^2+x \arctan (x)-\frac {1}{2} \log \left (x^2+1\right )\right )-\frac {1}{4} \log ^2\left (x^2+1\right )\) |
Input:
Int[ArcTan[x]*Log[1 + x^2],x]
Output:
-2*(x*ArcTan[x] - ArcTan[x]^2/2 - Log[1 + x^2]/2) + x*ArcTan[x]*Log[1 + x^ 2] - Log[1 + x^2]^2/4
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
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]
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]
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])
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])
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]
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]
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]
Time = 1.84 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97
method | result | size |
parallelrisch | \(-2 x \arctan \left (x \right )+\arctan \left (x \right )^{2}+\ln \left (x^{2}+1\right )+x \arctan \left (x \right ) \ln \left (x^{2}+1\right )-\frac {\ln \left (x^{2}+1\right )^{2}}{4}\) | \(37\) |
default | \(\text {Expression too large to display}\) | \(1913\) |
risch | \(\text {Expression too large to display}\) | \(4618\) |
Input:
int(arctan(x)*ln(x^2+1),x,method=_RETURNVERBOSE)
Output:
-2*x*arctan(x)+arctan(x)^2+ln(x^2+1)+x*arctan(x)*ln(x^2+1)-1/4*ln(x^2+1)^2
Time = 0.10 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87 \[ \int \arctan (x) \log \left (1+x^2\right ) \, dx=-2 \, x \arctan \left (x\right ) + \arctan \left (x\right )^{2} + {\left (x \arctan \left (x\right ) + 1\right )} \log \left (x^{2} + 1\right ) - \frac {1}{4} \, \log \left (x^{2} + 1\right )^{2} \] Input:
integrate(arctan(x)*log(x^2+1),x, algorithm="fricas")
Output:
-2*x*arctan(x) + arctan(x)^2 + (x*arctan(x) + 1)*log(x^2 + 1) - 1/4*log(x^ 2 + 1)^2
Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.03 \[ \int \arctan (x) \log \left (1+x^2\right ) \, dx=x \log {\left (x^{2} + 1 \right )} \operatorname {atan}{\left (x \right )} - 2 x \operatorname {atan}{\left (x \right )} - \frac {\log {\left (x^{2} + 1 \right )}^{2}}{4} + \log {\left (x^{2} + 1 \right )} + \operatorname {atan}^{2}{\left (x \right )} \] Input:
integrate(atan(x)*ln(x**2+1),x)
Output:
x*log(x**2 + 1)*atan(x) - 2*x*atan(x) - log(x**2 + 1)**2/4 + log(x**2 + 1) + atan(x)**2
Time = 0.14 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.11 \[ \int \arctan (x) \log \left (1+x^2\right ) \, dx={\left (x \log \left (x^{2} + 1\right ) - 2 \, x + 2 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) - \arctan \left (x\right )^{2} - \frac {1}{4} \, \log \left (x^{2} + 1\right )^{2} + \log \left (x^{2} + 1\right ) \] Input:
integrate(arctan(x)*log(x^2+1),x, algorithm="maxima")
Output:
(x*log(x^2 + 1) - 2*x + 2*arctan(x))*arctan(x) - arctan(x)^2 - 1/4*log(x^2 + 1)^2 + log(x^2 + 1)
Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (36) = 72\).
Time = 0.13 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.42 \[ \int \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {1}{2} \, \pi x \log \left (x^{2} + 1\right ) \mathrm {sgn}\left (x\right ) - x \arctan \left (\frac {1}{x}\right ) \log \left (x^{2} + 1\right ) - \frac {3}{2} \, \pi ^{2} \mathrm {sgn}\left (x\right ) - \pi x \mathrm {sgn}\left (x\right ) - \pi \arctan \left (\frac {1}{x}\right ) \mathrm {sgn}\left (x\right ) + \frac {1}{2} \, \pi ^{2} + \pi \arctan \left (x\right ) + \pi \arctan \left (\frac {1}{x}\right ) + 2 \, x \arctan \left (\frac {1}{x}\right ) + \arctan \left (\frac {1}{x}\right )^{2} - \frac {1}{4} \, \log \left (x^{2} + 1\right )^{2} + \log \left (x^{2} + 1\right ) \] Input:
integrate(arctan(x)*log(x^2+1),x, algorithm="giac")
Output:
1/2*pi*x*log(x^2 + 1)*sgn(x) - x*arctan(1/x)*log(x^2 + 1) - 3/2*pi^2*sgn(x ) - pi*x*sgn(x) - pi*arctan(1/x)*sgn(x) + 1/2*pi^2 + pi*arctan(x) + pi*arc tan(1/x) + 2*x*arctan(1/x) + arctan(1/x)^2 - 1/4*log(x^2 + 1)^2 + log(x^2 + 1)
Time = 0.77 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.03 \[ \int \arctan (x) \log \left (1+x^2\right ) \, dx=\ln \left (x^2+1\right )-\frac {{\ln \left (x^2+1\right )}^2}{4}+{\mathrm {atan}\left (x\right )}^2-x\,\left (2\,\mathrm {atan}\left (x\right )-\ln \left (x^2+1\right )\,\mathrm {atan}\left (x\right )\right ) \] Input:
int(log(x^2 + 1)*atan(x),x)
Output:
log(x^2 + 1) - log(x^2 + 1)^2/4 + atan(x)^2 - x*(2*atan(x) - log(x^2 + 1)* atan(x))
Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.95 \[ \int \arctan (x) \log \left (1+x^2\right ) \, dx=\mathit {atan} \left (x \right )^{2}+\mathit {atan} \left (x \right ) \mathrm {log}\left (x^{2}+1\right ) x -2 \mathit {atan} \left (x \right ) x -\frac {\mathrm {log}\left (x^{2}+1\right )^{2}}{4}+\mathrm {log}\left (x^{2}+1\right ) \] Input:
int(atan(x)*log(x^2+1),x)
Output:
(4*atan(x)**2 + 4*atan(x)*log(x**2 + 1)*x - 8*atan(x)*x - log(x**2 + 1)**2 + 4*log(x**2 + 1))/4