Integrand size = 15, antiderivative size = 79 \[ \int \frac {x^3 \arctan (x)^2}{\left (1+x^2\right )^3} \, dx=-\frac {1}{32 \left (1+x^2\right )^2}+\frac {5}{32 \left (1+x^2\right )}+\frac {x^3 \arctan (x)}{8 \left (1+x^2\right )^2}+\frac {3 x \arctan (x)}{16 \left (1+x^2\right )}-\frac {3 \arctan (x)^2}{32}+\frac {x^4 \arctan (x)^2}{4 \left (1+x^2\right )^2} \]
-1/32/(x^2+1)^2+5/32/(x^2+1)+1/8*x^3*arctan(x)/(x^2+1)^2+3/16*x*arctan(x)/ (x^2+1)-3/32*arctan(x)^2+1/4*x^4*arctan(x)^2/(x^2+1)^2
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.59 \[ \int \frac {x^3 \arctan (x)^2}{\left (1+x^2\right )^3} \, dx=\frac {4+5 x^2+2 x \left (3+5 x^2\right ) \arctan (x)+\left (-3-6 x^2+5 x^4\right ) \arctan (x)^2}{32 \left (1+x^2\right )^2} \]
Time = 0.43 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.16, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {5479, 5473, 5469, 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.
\(\displaystyle \int \frac {x^3 \arctan (x)^2}{\left (x^2+1\right )^3} \, dx\) |
\(\Big \downarrow \) 5479 |
\(\displaystyle \frac {x^4 \arctan (x)^2}{4 \left (x^2+1\right )^2}-\frac {1}{2} \int \frac {x^4 \arctan (x)}{\left (x^2+1\right )^3}dx\) |
\(\Big \downarrow \) 5473 |
\(\displaystyle \frac {1}{2} \left (-\frac {3}{4} \int \frac {x^2 \arctan (x)}{\left (x^2+1\right )^2}dx+\frac {x^3 \arctan (x)}{4 \left (x^2+1\right )^2}-\frac {x^4}{16 \left (x^2+1\right )^2}\right )+\frac {x^4 \arctan (x)^2}{4 \left (x^2+1\right )^2}\) |
\(\Big \downarrow \) 5469 |
\(\displaystyle \frac {1}{2} \left (-\frac {3}{4} \left (\frac {1}{2} \int \frac {\arctan (x)}{x^2+1}dx-\frac {x \arctan (x)}{2 \left (x^2+1\right )}-\frac {1}{4 \left (x^2+1\right )}\right )+\frac {x^3 \arctan (x)}{4 \left (x^2+1\right )^2}-\frac {x^4}{16 \left (x^2+1\right )^2}\right )+\frac {x^4 \arctan (x)^2}{4 \left (x^2+1\right )^2}\) |
\(\Big \downarrow \) 5419 |
\(\displaystyle \frac {x^4 \arctan (x)^2}{4 \left (x^2+1\right )^2}+\frac {1}{2} \left (-\frac {3}{4} \left (-\frac {x \arctan (x)}{2 \left (x^2+1\right )}+\frac {\arctan (x)^2}{4}-\frac {1}{4 \left (x^2+1\right )}\right )+\frac {x^3 \arctan (x)}{4 \left (x^2+1\right )^2}-\frac {x^4}{16 \left (x^2+1\right )^2}\right )\) |
(x^4*ArcTan[x]^2)/(4*(1 + x^2)^2) + (-1/16*x^4/(1 + x^2)^2 + (x^3*ArcTan[x ])/(4*(1 + x^2)^2) - (3*(-1/4*1/(1 + x^2) - (x*ArcTan[x])/(2*(1 + x^2)) + ArcTan[x]^2/4))/4)/2
3.7.82.3.1 Defintions of rubi rules used
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_.))*(x_)^2*((d_) + (e_.)*(x_)^2)^(q_), x _Symbol] :> Simp[(-b)*((d + e*x^2)^(q + 1)/(4*c^3*d*(q + 1)^2)), x] + (Simp [x*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])/(2*c^2*d*(q + 1))), x] - Simp[1 /(2*c^2*d*(q + 1)) Int[(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]), x], x]) / ; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && LtQ[q, -1] && NeQ[q, -5/2]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_) ^2)^(q_), x_Symbol] :> Simp[b*(f*x)^m*((d + e*x^2)^(q + 1)/(c*d*m^2)), x] + (-Simp[f*(f*x)^(m - 1)*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])/(c^2*d*m)) , x] + Simp[f^2*((m - 1)/(c^2*d*m)) Int[(f*x)^(m - 2)*(d + e*x^2)^(q + 1) *(a + b*ArcTan[c*x]), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2 *d] && EqQ[m + 2*q + 2, 0] && LtQ[q, -1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_)^2)^(q_.), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^p/(d*f*(m + 1))), x] - Simp[b*c*(p/(f*(m + 1))) Int[(f*x) ^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[e, c^2*d] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]
Time = 0.56 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.99
method | result | size |
default | \(\frac {\arctan \left (x \right )^{2}}{4 \left (x^{2}+1\right )^{2}}-\frac {\arctan \left (x \right )^{2}}{2 \left (x^{2}+1\right )}+\frac {5 x^{3} \arctan \left (x \right )}{16 \left (x^{2}+1\right )^{2}}+\frac {3 x \arctan \left (x \right )}{16 \left (x^{2}+1\right )^{2}}+\frac {5 \arctan \left (x \right )^{2}}{32}-\frac {1}{32 \left (x^{2}+1\right )^{2}}+\frac {5}{32 \left (x^{2}+1\right )}\) | \(78\) |
parts | \(\frac {\arctan \left (x \right )^{2}}{4 \left (x^{2}+1\right )^{2}}-\frac {\arctan \left (x \right )^{2}}{2 \left (x^{2}+1\right )}+\frac {5 x^{3} \arctan \left (x \right )}{16 \left (x^{2}+1\right )^{2}}+\frac {3 x \arctan \left (x \right )}{16 \left (x^{2}+1\right )^{2}}+\frac {5 \arctan \left (x \right )^{2}}{32}-\frac {1}{32 \left (x^{2}+1\right )^{2}}+\frac {5}{32 \left (x^{2}+1\right )}\) | \(78\) |
risch | \(-\frac {\left (5 x^{4}-6 x^{2}-3\right ) \ln \left (i x +1\right )^{2}}{128 \left (x^{2}+1\right )^{2}}+\frac {\left (-6 x^{2} \ln \left (-i x +1\right )-3 \ln \left (-i x +1\right )+5 x^{4} \ln \left (-i x +1\right )-10 i x^{3}-6 i x \right ) \ln \left (i x +1\right )}{64 \left (x +i\right )^{2} \left (x -i\right )^{2}}-\frac {5 x^{4} \ln \left (-i x +1\right )^{2}-6 x^{2} \ln \left (-i x +1\right )^{2}-3 \ln \left (-i x +1\right )^{2}-20 i x^{3} \ln \left (-i x +1\right )-12 i x \ln \left (-i x +1\right )-20 x^{2}-16}{128 \left (x +i\right )^{2} \left (x -i\right )^{2}}\) | \(181\) |
1/4*arctan(x)^2/(x^2+1)^2-1/2*arctan(x)^2/(x^2+1)+5/16*x^3*arctan(x)/(x^2+ 1)^2+3/16*x*arctan(x)/(x^2+1)^2+5/32*arctan(x)^2-1/32/(x^2+1)^2+5/32/(x^2+ 1)
Time = 0.24 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.65 \[ \int \frac {x^3 \arctan (x)^2}{\left (1+x^2\right )^3} \, dx=\frac {{\left (5 \, x^{4} - 6 \, x^{2} - 3\right )} \arctan \left (x\right )^{2} + 5 \, x^{2} + 2 \, {\left (5 \, x^{3} + 3 \, x\right )} \arctan \left (x\right ) + 4}{32 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \]
1/32*((5*x^4 - 6*x^2 - 3)*arctan(x)^2 + 5*x^2 + 2*(5*x^3 + 3*x)*arctan(x) + 4)/(x^4 + 2*x^2 + 1)
\[ \int \frac {x^3 \arctan (x)^2}{\left (1+x^2\right )^3} \, dx=\int \frac {x^{3} \operatorname {atan}^{2}{\left (x \right )}}{\left (x^{2} + 1\right )^{3}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.19 \[ \int \frac {x^3 \arctan (x)^2}{\left (1+x^2\right )^3} \, dx=\frac {1}{16} \, {\left (\frac {5 \, x^{3} + 3 \, x}{x^{4} + 2 \, x^{2} + 1} + 5 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) - \frac {{\left (2 \, x^{2} + 1\right )} \arctan \left (x\right )^{2}}{4 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} - \frac {5 \, {\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \left (x\right )^{2} - 5 \, x^{2} - 4}{32 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \]
1/16*((5*x^3 + 3*x)/(x^4 + 2*x^2 + 1) + 5*arctan(x))*arctan(x) - 1/4*(2*x^ 2 + 1)*arctan(x)^2/(x^4 + 2*x^2 + 1) - 1/32*(5*(x^4 + 2*x^2 + 1)*arctan(x) ^2 - 5*x^2 - 4)/(x^4 + 2*x^2 + 1)
\[ \int \frac {x^3 \arctan (x)^2}{\left (1+x^2\right )^3} \, dx=\int { \frac {x^{3} \arctan \left (x\right )^{2}}{{\left (x^{2} + 1\right )}^{3}} \,d x } \]
Time = 0.40 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.71 \[ \int \frac {x^3 \arctan (x)^2}{\left (1+x^2\right )^3} \, dx=-\frac {-5\,x^4\,{\mathrm {atan}\left (x\right )}^2+4\,x^4-10\,x^3\,\mathrm {atan}\left (x\right )+6\,x^2\,{\mathrm {atan}\left (x\right )}^2+3\,x^2-6\,x\,\mathrm {atan}\left (x\right )+3\,{\mathrm {atan}\left (x\right )}^2}{32\,{\left (x^2+1\right )}^2} \]
-(3*atan(x)^2 - 10*x^3*atan(x) + 6*x^2*atan(x)^2 - 5*x^4*atan(x)^2 - 6*x*a tan(x) + 3*x^2 + 4*x^4)/(32*(x^2 + 1)^2)
Time = 0.00 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.73 \[ \int \frac {x^3 \arctan (x)^2}{\left (1+x^2\right )^3} \, dx=\frac {10 \mathit {atan} \left (x \right )^{2} x^{4}-12 \mathit {atan} \left (x \right )^{2} x^{2}-6 \mathit {atan} \left (x \right )^{2}+20 \mathit {atan} \left (x \right ) x^{3}+12 \mathit {atan} \left (x \right ) x -5 x^{4}+3}{64 x^{4}+128 x^{2}+64} \]