Integrand size = 8, antiderivative size = 53 \[ \int x^3 \arctan (x)^2 \, dx=\frac {x^2}{12}+\frac {1}{2} x \arctan (x)-\frac {1}{6} x^3 \arctan (x)-\frac {\arctan (x)^2}{4}+\frac {1}{4} x^4 \arctan (x)^2-\frac {1}{3} \log \left (1+x^2\right ) \]
1/12*x^2+1/2*x*arctan(x)-1/6*x^3*arctan(x)-1/4*arctan(x)^2+1/4*x^4*arctan( x)^2-1/3*ln(x^2+1)
Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.70 \[ \int x^3 \arctan (x)^2 \, dx=\frac {1}{12} \left (x^2-2 x \left (-3+x^2\right ) \arctan (x)+3 \left (-1+x^4\right ) \arctan (x)^2-4 \log \left (1+x^2\right )\right ) \]
Time = 0.52 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.21, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.250, Rules used = {5361, 5451, 5361, 243, 49, 2009, 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.
\(\displaystyle \int x^3 \arctan (x)^2 \, dx\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle \frac {1}{4} x^4 \arctan (x)^2-\frac {1}{2} \int \frac {x^4 \arctan (x)}{x^2+1}dx\) |
\(\Big \downarrow \) 5451 |
\(\displaystyle \frac {1}{2} \left (\int \frac {x^2 \arctan (x)}{x^2+1}dx-\int x^2 \arctan (x)dx\right )+\frac {1}{4} x^4 \arctan (x)^2\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle \frac {1}{2} \left (\int \frac {x^2 \arctan (x)}{x^2+1}dx+\frac {1}{3} \int \frac {x^3}{x^2+1}dx-\frac {1}{3} x^3 \arctan (x)\right )+\frac {1}{4} x^4 \arctan (x)^2\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \left (\int \frac {x^2 \arctan (x)}{x^2+1}dx+\frac {1}{6} \int \frac {x^2}{x^2+1}dx^2-\frac {1}{3} x^3 \arctan (x)\right )+\frac {1}{4} x^4 \arctan (x)^2\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{2} \left (\int \frac {x^2 \arctan (x)}{x^2+1}dx+\frac {1}{6} \int \left (1+\frac {1}{-x^2-1}\right )dx^2-\frac {1}{3} x^3 \arctan (x)\right )+\frac {1}{4} x^4 \arctan (x)^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\int \frac {x^2 \arctan (x)}{x^2+1}dx-\frac {1}{3} x^3 \arctan (x)+\frac {1}{6} \left (x^2-\log \left (x^2+1\right )\right )\right )+\frac {1}{4} x^4 \arctan (x)^2\) |
\(\Big \downarrow \) 5451 |
\(\displaystyle \frac {1}{2} \left (-\int \frac {\arctan (x)}{x^2+1}dx+\int \arctan (x)dx-\frac {1}{3} x^3 \arctan (x)+\frac {1}{6} \left (x^2-\log \left (x^2+1\right )\right )\right )+\frac {1}{4} x^4 \arctan (x)^2\) |
\(\Big \downarrow \) 5345 |
\(\displaystyle \frac {1}{2} \left (-\int \frac {\arctan (x)}{x^2+1}dx-\int \frac {x}{x^2+1}dx-\frac {1}{3} x^3 \arctan (x)+x \arctan (x)+\frac {1}{6} \left (x^2-\log \left (x^2+1\right )\right )\right )+\frac {1}{4} x^4 \arctan (x)^2\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \frac {1}{2} \left (-\int \frac {\arctan (x)}{x^2+1}dx-\frac {1}{3} x^3 \arctan (x)+x \arctan (x)+\frac {1}{6} \left (x^2-\log \left (x^2+1\right )\right )-\frac {1}{2} \log \left (x^2+1\right )\right )+\frac {1}{4} x^4 \arctan (x)^2\) |
\(\Big \downarrow \) 5419 |
\(\displaystyle \frac {1}{4} x^4 \arctan (x)^2+\frac {1}{2} \left (-\frac {1}{3} x^3 \arctan (x)+x \arctan (x)-\frac {\arctan (x)^2}{2}+\frac {1}{6} \left (x^2-\log \left (x^2+1\right )\right )-\frac {1}{2} \log \left (x^2+1\right )\right )\) |
(x^4*ArcTan[x]^2)/4 + (x*ArcTan[x] - (x^3*ArcTan[x])/3 - ArcTan[x]^2/2 + ( x^2 - Log[1 + x^2])/6 - Log[1 + x^2]/2)/2
3.7.47.3.1 Defintions of rubi rules used
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}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
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_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^n])^p/(m + 1)), x] - Simp[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]
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]
Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.79
method | result | size |
default | \(\frac {x^{2}}{12}+\frac {x \arctan \left (x \right )}{2}-\frac {x^{3} \arctan \left (x \right )}{6}-\frac {\arctan \left (x \right )^{2}}{4}+\frac {x^{4} \arctan \left (x \right )^{2}}{4}-\frac {\ln \left (x^{2}+1\right )}{3}\) | \(42\) |
parts | \(\frac {x^{2}}{12}+\frac {x \arctan \left (x \right )}{2}-\frac {x^{3} \arctan \left (x \right )}{6}-\frac {\arctan \left (x \right )^{2}}{4}+\frac {x^{4} \arctan \left (x \right )^{2}}{4}-\frac {\ln \left (x^{2}+1\right )}{3}\) | \(42\) |
parallelrisch | \(\frac {x^{4} \arctan \left (x \right )^{2}}{4}-\frac {x^{3} \arctan \left (x \right )}{6}+\frac {x^{2}}{12}+\frac {x \arctan \left (x \right )}{2}-\frac {\arctan \left (x \right )^{2}}{4}-\frac {\ln \left (x^{2}+1\right )}{3}-\frac {1}{12}\) | \(43\) |
risch | \(-\frac {\left (\frac {x^{4}}{4}-\frac {1}{4}\right ) \ln \left (i x +1\right )^{2}}{4}-\frac {\left (-\frac {x^{4} \ln \left (-i x +1\right )}{2}-\frac {i x^{3}}{3}+i x +\frac {\ln \left (-i x +1\right )}{2}\right ) \ln \left (i x +1\right )}{4}-\frac {x^{4} \ln \left (-i x +1\right )^{2}}{16}+\frac {\ln \left (-i x +1\right )^{2}}{16}-\frac {i x^{3} \ln \left (-i x +1\right )}{12}+\frac {i x \ln \left (-i x +1\right )}{4}+\frac {x^{2}}{12}-\frac {\ln \left (x^{2}+1\right )}{3}\) | \(123\) |
1/12*x^2+1/2*x*arctan(x)-1/6*x^3*arctan(x)-1/4*arctan(x)^2+1/4*x^4*arctan( x)^2-1/3*ln(x^2+1)
Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.68 \[ \int x^3 \arctan (x)^2 \, dx=\frac {1}{4} \, {\left (x^{4} - 1\right )} \arctan \left (x\right )^{2} + \frac {1}{12} \, x^{2} - \frac {1}{6} \, {\left (x^{3} - 3 \, x\right )} \arctan \left (x\right ) - \frac {1}{3} \, \log \left (x^{2} + 1\right ) \]
Time = 0.14 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83 \[ \int x^3 \arctan (x)^2 \, dx=\frac {x^{4} \operatorname {atan}^{2}{\left (x \right )}}{4} - \frac {x^{3} \operatorname {atan}{\left (x \right )}}{6} + \frac {x^{2}}{12} + \frac {x \operatorname {atan}{\left (x \right )}}{2} - \frac {\log {\left (x^{2} + 1 \right )}}{3} - \frac {\operatorname {atan}^{2}{\left (x \right )}}{4} \]
Time = 0.28 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83 \[ \int x^3 \arctan (x)^2 \, dx=\frac {1}{4} \, x^{4} \arctan \left (x\right )^{2} + \frac {1}{12} \, x^{2} - \frac {1}{6} \, {\left (x^{3} - 3 \, x + 3 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) + \frac {1}{4} \, \arctan \left (x\right )^{2} - \frac {1}{3} \, \log \left (x^{2} + 1\right ) \]
1/4*x^4*arctan(x)^2 + 1/12*x^2 - 1/6*(x^3 - 3*x + 3*arctan(x))*arctan(x) + 1/4*arctan(x)^2 - 1/3*log(x^2 + 1)
Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.77 \[ \int x^3 \arctan (x)^2 \, dx=\frac {1}{4} \, x^{4} \arctan \left (x\right )^{2} - \frac {1}{6} \, x^{3} \arctan \left (x\right ) + \frac {1}{12} \, x^{2} + \frac {1}{2} \, x \arctan \left (x\right ) - \frac {1}{4} \, \arctan \left (x\right )^{2} - \frac {1}{3} \, \log \left (x^{2} + 1\right ) \]
1/4*x^4*arctan(x)^2 - 1/6*x^3*arctan(x) + 1/12*x^2 + 1/2*x*arctan(x) - 1/4 *arctan(x)^2 - 1/3*log(x^2 + 1)
Time = 0.41 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.77 \[ \int x^3 \arctan (x)^2 \, dx=\frac {x^4\,{\mathrm {atan}\left (x\right )}^2}{4}-\frac {x^3\,\mathrm {atan}\left (x\right )}{6}-\frac {{\mathrm {atan}\left (x\right )}^2}{4}-\frac {\ln \left (x^2+1\right )}{3}+\frac {x\,\mathrm {atan}\left (x\right )}{2}+\frac {x^2}{12} \]
Time = 0.00 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.77 \[ \int x^3 \arctan (x)^2 \, dx=\frac {\mathit {atan} \left (x \right )^{2} x^{4}}{4}-\frac {\mathit {atan} \left (x \right )^{2}}{4}-\frac {\mathit {atan} \left (x \right ) x^{3}}{6}+\frac {\mathit {atan} \left (x \right ) x}{2}-\frac {\mathrm {log}\left (x^{2}+1\right )}{3}+\frac {x^{2}}{12} \]