Integrand size = 13, antiderivative size = 67 \[ \int \frac {x^3 \arctan (x)}{1+x^2} \, dx=-\frac {x}{2}+\frac {\arctan (x)}{2}+\frac {1}{2} x^2 \arctan (x)+\frac {1}{2} i \arctan (x)^2+\arctan (x) \log \left (\frac {2}{1+i x}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{1+i x}\right ) \]
-1/2*x+1/2*arctan(x)+1/2*x^2*arctan(x)+1/2*I*arctan(x)^2+arctan(x)*ln(2/(1 +I*x))+1/2*I*polylog(2,1-2/(1+I*x))
Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.85 \[ \int \frac {x^3 \arctan (x)}{1+x^2} \, dx=\frac {1}{2} \left (-x+i \arctan (x)^2+\arctan (x) \left (1+x^2+2 \log \left (-\frac {2 i}{-i+x}\right )\right )+i \operatorname {PolyLog}\left (2,\frac {i+x}{-i+x}\right )\right ) \]
(-x + I*ArcTan[x]^2 + ArcTan[x]*(1 + x^2 + 2*Log[(-2*I)/(-I + x)]) + I*Pol yLog[2, (I + x)/(-I + x)])/2
Time = 0.41 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {5451, 5361, 262, 216, 5455, 5379, 2849, 2752}
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.
\(\displaystyle \int \frac {x^3 \arctan (x)}{x^2+1} \, dx\) |
\(\Big \downarrow \) 5451 |
\(\displaystyle \int x \arctan (x)dx-\int \frac {x \arctan (x)}{x^2+1}dx\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle -\int \frac {x \arctan (x)}{x^2+1}dx-\frac {1}{2} \int \frac {x^2}{x^2+1}dx+\frac {1}{2} x^2 \arctan (x)\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\int \frac {x \arctan (x)}{x^2+1}dx+\frac {1}{2} \left (\int \frac {1}{x^2+1}dx-x\right )+\frac {1}{2} x^2 \arctan (x)\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\int \frac {x \arctan (x)}{x^2+1}dx+\frac {1}{2} x^2 \arctan (x)+\frac {1}{2} (\arctan (x)-x)\) |
\(\Big \downarrow \) 5455 |
\(\displaystyle \int \frac {\arctan (x)}{i-x}dx+\frac {1}{2} x^2 \arctan (x)+\frac {1}{2} i \arctan (x)^2+\frac {1}{2} (\arctan (x)-x)\) |
\(\Big \downarrow \) 5379 |
\(\displaystyle -\int \frac {\log \left (\frac {2}{i x+1}\right )}{x^2+1}dx+\frac {1}{2} x^2 \arctan (x)+\frac {1}{2} i \arctan (x)^2+\frac {1}{2} (\arctan (x)-x)+\arctan (x) \log \left (\frac {2}{1+i x}\right )\) |
\(\Big \downarrow \) 2849 |
\(\displaystyle i \int \frac {\log \left (\frac {2}{i x+1}\right )}{1-\frac {2}{i x+1}}d\frac {1}{i x+1}+\frac {1}{2} x^2 \arctan (x)+\frac {1}{2} i \arctan (x)^2+\frac {1}{2} (\arctan (x)-x)+\arctan (x) \log \left (\frac {2}{1+i x}\right )\) |
\(\Big \downarrow \) 2752 |
\(\displaystyle \frac {1}{2} x^2 \arctan (x)+\frac {1}{2} i \arctan (x)^2+\frac {1}{2} (\arctan (x)-x)+\arctan (x) \log \left (\frac {2}{1+i x}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{i x+1}\right )\) |
(x^2*ArcTan[x])/2 + (I/2)*ArcTan[x]^2 + (-x + ArcTan[x])/2 + ArcTan[x]*Log [2/(1 + I*x)] + (I/2)*PolyLog[2, 1 - 2/(1 + I*x)]
3.7.72.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Simp [-e/g Subst[Int[Log[2*d*x]/(1 - 2*d*x), x], x, 1/(d + e*x)], x] /; FreeQ[ {c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]
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_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(Log[2/(1 + e*(x/d))]/e), x] + Simp[b*c*( p/e) Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2/(1 + e*(x/d))]/(1 + c^2*x^2)) , x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0 ]
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_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*((a + b*ArcTan[c*x])^(p + 1)/(b*e*(p + 1))), x] - Si mp[1/(c*d) Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (53 ) = 106\).
Time = 0.44 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.88
method | result | size |
default | \(\frac {x^{2} \arctan \left (x \right )}{2}-\frac {\arctan \left (x \right ) \ln \left (x^{2}+1\right )}{2}-\frac {x}{2}+\frac {\arctan \left (x \right )}{2}-\frac {i \left (\ln \left (x -i\right ) \ln \left (x^{2}+1\right )-\frac {\ln \left (x -i\right )^{2}}{2}-\operatorname {dilog}\left (-\frac {i \left (x +i\right )}{2}\right )-\ln \left (x -i\right ) \ln \left (-\frac {i \left (x +i\right )}{2}\right )\right )}{4}+\frac {i \left (\ln \left (x +i\right ) \ln \left (x^{2}+1\right )-\frac {\ln \left (x +i\right )^{2}}{2}-\operatorname {dilog}\left (\frac {i \left (x -i\right )}{2}\right )-\ln \left (x +i\right ) \ln \left (\frac {i \left (x -i\right )}{2}\right )\right )}{4}\) | \(126\) |
parts | \(\frac {x^{2} \arctan \left (x \right )}{2}-\frac {\arctan \left (x \right ) \ln \left (x^{2}+1\right )}{2}-\frac {x}{2}+\frac {\arctan \left (x \right )}{2}-\frac {i \left (\ln \left (x -i\right ) \ln \left (x^{2}+1\right )-\frac {\ln \left (x -i\right )^{2}}{2}-\operatorname {dilog}\left (-\frac {i \left (x +i\right )}{2}\right )-\ln \left (x -i\right ) \ln \left (-\frac {i \left (x +i\right )}{2}\right )\right )}{4}+\frac {i \left (\ln \left (x +i\right ) \ln \left (x^{2}+1\right )-\frac {\ln \left (x +i\right )^{2}}{2}-\operatorname {dilog}\left (\frac {i \left (x -i\right )}{2}\right )-\ln \left (x +i\right ) \ln \left (\frac {i \left (x -i\right )}{2}\right )\right )}{4}\) | \(126\) |
risch | \(-\frac {i \ln \left (i x +1\right )}{4}-\frac {i \ln \left (-i x +1\right )^{2}}{8}+\frac {i \operatorname {dilog}\left (\frac {1}{2}-\frac {i x}{2}\right )}{4}-\frac {x}{2}+\frac {i \ln \left (\frac {1}{2}-\frac {i x}{2}\right ) \ln \left (i x +1\right )}{4}-\frac {i \ln \left (\frac {1}{2}+\frac {i x}{2}\right ) \ln \left (-i x +1\right )}{4}+\frac {i \ln \left (i x +1\right )^{2}}{8}+\frac {i x^{2} \ln \left (-i x +1\right )}{4}-\frac {i x^{2} \ln \left (i x +1\right )}{4}-\frac {i \operatorname {dilog}\left (\frac {1}{2}+\frac {i x}{2}\right )}{4}+\frac {i \ln \left (-i x +1\right )}{4}\) | \(129\) |
1/2*x^2*arctan(x)-1/2*arctan(x)*ln(x^2+1)-1/2*x+1/2*arctan(x)-1/4*I*(ln(x- I)*ln(x^2+1)-1/2*ln(x-I)^2-dilog(-1/2*I*(x+I))-ln(x-I)*ln(-1/2*I*(x+I)))+1 /4*I*(ln(x+I)*ln(x^2+1)-1/2*ln(x+I)^2-dilog(1/2*I*(x-I))-ln(x+I)*ln(1/2*I* (x-I)))
\[ \int \frac {x^3 \arctan (x)}{1+x^2} \, dx=\int { \frac {x^{3} \arctan \left (x\right )}{x^{2} + 1} \,d x } \]
\[ \int \frac {x^3 \arctan (x)}{1+x^2} \, dx=\int \frac {x^{3} \operatorname {atan}{\left (x \right )}}{x^{2} + 1}\, dx \]
\[ \int \frac {x^3 \arctan (x)}{1+x^2} \, dx=\int { \frac {x^{3} \arctan \left (x\right )}{x^{2} + 1} \,d x } \]
\[ \int \frac {x^3 \arctan (x)}{1+x^2} \, dx=\int { \frac {x^{3} \arctan \left (x\right )}{x^{2} + 1} \,d x } \]
Timed out. \[ \int \frac {x^3 \arctan (x)}{1+x^2} \, dx=\int \frac {x^3\,\mathrm {atan}\left (x\right )}{x^2+1} \,d x \]
\[ \int \frac {x^3 \arctan (x)}{1+x^2} \, dx=\int \frac {\mathit {atan} \left (x \right ) x^{3}}{x^{2}+1}d x \]