Integrand size = 12, antiderivative size = 111 \[ \int x^4 \arctan (x) \log \left (1+x^2\right ) \, dx=-\frac {77 x^2}{300}+\frac {9 x^4}{200}-\frac {2}{5} x \arctan (x)+\frac {2}{15} x^3 \arctan (x)-\frac {2}{25} x^5 \arctan (x)+\frac {\arctan (x)^2}{5}+\frac {137}{300} \log \left (1+x^2\right )+\frac {1}{10} x^2 \log \left (1+x^2\right )-\frac {1}{20} x^4 \log \left (1+x^2\right )+\frac {1}{5} x^5 \arctan (x) \log \left (1+x^2\right )-\frac {1}{20} \log ^2\left (1+x^2\right ) \]
-77/300*x^2+9/200*x^4-2/5*x*arctan(x)+2/15*x^3*arctan(x)-2/25*x^5*arctan(x )+1/5*arctan(x)^2+137/300*ln(x^2+1)+1/10*x^2*ln(x^2+1)-1/20*x^4*ln(x^2+1)+ 1/5*x^5*arctan(x)*ln(x^2+1)-1/20*ln(x^2+1)^2
Time = 0.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.71 \[ \int x^4 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {1}{600} \left (x^2 \left (-154+27 x^2\right )+120 \arctan (x)^2+\left (274+60 x^2-30 x^4\right ) \log \left (1+x^2\right )-30 \log ^2\left (1+x^2\right )+8 x \arctan (x) \left (-30+10 x^2-6 x^4+15 x^4 \log \left (1+x^2\right )\right )\right ) \]
(x^2*(-154 + 27*x^2) + 120*ArcTan[x]^2 + (274 + 60*x^2 - 30*x^4)*Log[1 + x ^2] - 30*Log[1 + x^2]^2 + 8*x*ArcTan[x]*(-30 + 10*x^2 - 6*x^4 + 15*x^4*Log [1 + x^2]))/600
Time = 0.57 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.14, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5556, 2009}
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.
\(\displaystyle \int x^4 \arctan (x) \log \left (x^2+1\right ) \, dx\) |
\(\Big \downarrow \) 5556 |
\(\displaystyle -2 \int \left (\frac {x^3 \left (4 \arctan (x) x^3-x^2+2\right )}{20 \left (x^2+1\right )}-\frac {x \log \left (x^2+1\right )}{10 \left (x^2+1\right )}\right )dx+\frac {1}{5} x^5 \arctan (x) \log \left (x^2+1\right )-\frac {1}{10} \log ^2\left (x^2+1\right )+\frac {1}{10} x^2 \log \left (x^2+1\right )-\frac {1}{20} x^4 \log \left (x^2+1\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} x^5 \arctan (x) \log \left (x^2+1\right )-2 \left (\frac {1}{25} x^5 \arctan (x)-\frac {1}{15} x^3 \arctan (x)+\frac {1}{5} x \arctan (x)-\frac {\arctan (x)^2}{10}-\frac {9 x^4}{400}+\frac {77 x^2}{600}-\frac {1}{40} \log ^2\left (x^2+1\right )-\frac {137}{600} \log \left (x^2+1\right )\right )-\frac {1}{10} \log ^2\left (x^2+1\right )+\frac {1}{10} x^2 \log \left (x^2+1\right )-\frac {1}{20} x^4 \log \left (x^2+1\right )\) |
(x^2*Log[1 + x^2])/10 - (x^4*Log[1 + x^2])/20 + (x^5*ArcTan[x]*Log[1 + x^2 ])/5 - Log[1 + x^2]^2/10 - 2*((77*x^2)/600 - (9*x^4)/400 + (x*ArcTan[x])/5 - (x^3*ArcTan[x])/15 + (x^5*ArcTan[x])/25 - ArcTan[x]^2/10 - (137*Log[1 + x^2])/600 - Log[1 + x^2]^2/40)
3.13.75.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*( e_.))*(x_)^(m_.), x_Symbol] :> With[{u = IntHide[x^m*(a + b*ArcTan[c*x]), x ]}, Simp[(d + e*Log[f + g*x^2]) u, x] - Simp[2*e*g Int[ExpandIntegrand[ x*(u/(f + g*x^2)), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && Intege rQ[m] && NeQ[m, -1]
Time = 2.10 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.82
method | result | size |
parallelrisch | \(\frac {x^{5} \arctan \left (x \right ) \ln \left (x^{2}+1\right )}{5}-\frac {2 x^{5} \arctan \left (x \right )}{25}-\frac {x^{4} \ln \left (x^{2}+1\right )}{20}+\frac {9 x^{4}}{200}+\frac {2 x^{3} \arctan \left (x \right )}{15}+\frac {x^{2} \ln \left (x^{2}+1\right )}{10}-\frac {77 x^{2}}{300}-\frac {2 x \arctan \left (x \right )}{5}+\frac {\arctan \left (x \right )^{2}}{5}-\frac {\ln \left (x^{2}+1\right )^{2}}{20}+\frac {137 \ln \left (x^{2}+1\right )}{300}+\frac {77}{300}\) | \(91\) |
default | \(\text {Expression too large to display}\) | \(3626\) |
risch | \(\text {Expression too large to display}\) | \(5733\) |
1/5*x^5*arctan(x)*ln(x^2+1)-2/25*x^5*arctan(x)-1/20*x^4*ln(x^2+1)+9/200*x^ 4+2/15*x^3*arctan(x)+1/10*x^2*ln(x^2+1)-77/300*x^2-2/5*x*arctan(x)+1/5*arc tan(x)^2-1/20*ln(x^2+1)^2+137/300*ln(x^2+1)+77/300
Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.65 \[ \int x^4 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {9}{200} \, x^{4} - \frac {77}{300} \, x^{2} - \frac {2}{75} \, {\left (3 \, x^{5} - 5 \, x^{3} + 15 \, x\right )} \arctan \left (x\right ) + \frac {1}{5} \, \arctan \left (x\right )^{2} + \frac {1}{300} \, {\left (60 \, x^{5} \arctan \left (x\right ) - 15 \, x^{4} + 30 \, x^{2} + 137\right )} \log \left (x^{2} + 1\right ) - \frac {1}{20} \, \log \left (x^{2} + 1\right )^{2} \]
9/200*x^4 - 77/300*x^2 - 2/75*(3*x^5 - 5*x^3 + 15*x)*arctan(x) + 1/5*arcta n(x)^2 + 1/300*(60*x^5*arctan(x) - 15*x^4 + 30*x^2 + 137)*log(x^2 + 1) - 1 /20*log(x^2 + 1)^2
Time = 0.87 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.96 \[ \int x^4 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {x^{5} \log {\left (x^{2} + 1 \right )} \operatorname {atan}{\left (x \right )}}{5} - \frac {2 x^{5} \operatorname {atan}{\left (x \right )}}{25} - \frac {x^{4} \log {\left (x^{2} + 1 \right )}}{20} + \frac {9 x^{4}}{200} + \frac {2 x^{3} \operatorname {atan}{\left (x \right )}}{15} + \frac {x^{2} \log {\left (x^{2} + 1 \right )}}{10} - \frac {77 x^{2}}{300} - \frac {2 x \operatorname {atan}{\left (x \right )}}{5} - \frac {\log {\left (x^{2} + 1 \right )}^{2}}{20} + \frac {137 \log {\left (x^{2} + 1 \right )}}{300} + \frac {\operatorname {atan}^{2}{\left (x \right )}}{5} \]
x**5*log(x**2 + 1)*atan(x)/5 - 2*x**5*atan(x)/25 - x**4*log(x**2 + 1)/20 + 9*x**4/200 + 2*x**3*atan(x)/15 + x**2*log(x**2 + 1)/10 - 77*x**2/300 - 2* x*atan(x)/5 - log(x**2 + 1)**2/20 + 137*log(x**2 + 1)/300 + atan(x)**2/5
Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.72 \[ \int x^4 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {9}{200} \, x^{4} - \frac {77}{300} \, x^{2} + \frac {1}{75} \, {\left (15 \, x^{5} \log \left (x^{2} + 1\right ) - 6 \, x^{5} + 10 \, x^{3} - 30 \, x + 30 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) - \frac {1}{5} \, \arctan \left (x\right )^{2} - \frac {1}{300} \, {\left (15 \, x^{4} - 30 \, x^{2} - 137\right )} \log \left (x^{2} + 1\right ) - \frac {1}{20} \, \log \left (x^{2} + 1\right )^{2} \]
9/200*x^4 - 77/300*x^2 + 1/75*(15*x^5*log(x^2 + 1) - 6*x^5 + 10*x^3 - 30*x + 30*arctan(x))*arctan(x) - 1/5*arctan(x)^2 - 1/300*(15*x^4 - 30*x^2 - 13 7)*log(x^2 + 1) - 1/20*log(x^2 + 1)^2
Time = 0.27 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.51 \[ \int x^4 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {1}{10} \, \pi x^{5} \log \left (x^{2} + 1\right ) \mathrm {sgn}\left (x\right ) - \frac {1}{5} \, x^{5} \arctan \left (\frac {1}{x}\right ) \log \left (x^{2} + 1\right ) - \frac {1}{25} \, \pi x^{5} \mathrm {sgn}\left (x\right ) + \frac {2}{25} \, x^{5} \arctan \left (\frac {1}{x}\right ) - \frac {1}{20} \, x^{4} \log \left (x^{2} + 1\right ) + \frac {1}{15} \, \pi x^{3} \mathrm {sgn}\left (x\right ) + \frac {9}{200} \, x^{4} - \frac {2}{15} \, x^{3} \arctan \left (\frac {1}{x}\right ) + \frac {1}{10} \, x^{2} \log \left (x^{2} + 1\right ) - \frac {3}{10} \, \pi ^{2} \mathrm {sgn}\left (x\right ) - \frac {1}{5} \, \pi x \mathrm {sgn}\left (x\right ) - \frac {1}{5} \, \pi \arctan \left (\frac {1}{x}\right ) \mathrm {sgn}\left (x\right ) + \frac {1}{10} \, \pi ^{2} - \frac {77}{300} \, x^{2} + \frac {1}{5} \, \pi \arctan \left (x\right ) + \frac {1}{5} \, \pi \arctan \left (\frac {1}{x}\right ) + \frac {2}{5} \, x \arctan \left (\frac {1}{x}\right ) + \frac {1}{5} \, \arctan \left (\frac {1}{x}\right )^{2} - \frac {1}{20} \, \log \left (x^{2} + 1\right )^{2} + \frac {137}{300} \, \log \left (x^{2} + 1\right ) \]
1/10*pi*x^5*log(x^2 + 1)*sgn(x) - 1/5*x^5*arctan(1/x)*log(x^2 + 1) - 1/25* pi*x^5*sgn(x) + 2/25*x^5*arctan(1/x) - 1/20*x^4*log(x^2 + 1) + 1/15*pi*x^3 *sgn(x) + 9/200*x^4 - 2/15*x^3*arctan(1/x) + 1/10*x^2*log(x^2 + 1) - 3/10* pi^2*sgn(x) - 1/5*pi*x*sgn(x) - 1/5*pi*arctan(1/x)*sgn(x) + 1/10*pi^2 - 77 /300*x^2 + 1/5*pi*arctan(x) + 1/5*pi*arctan(1/x) + 2/5*x*arctan(1/x) + 1/5 *arctan(1/x)^2 - 1/20*log(x^2 + 1)^2 + 137/300*log(x^2 + 1)
Time = 0.53 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.74 \[ \int x^4 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {137\,\ln \left (x^2+1\right )}{300}-\frac {{\ln \left (x^2+1\right )}^2}{20}+\frac {{\mathrm {atan}\left (x\right )}^2}{5}-\mathrm {atan}\left (x\right )\,\left (\frac {2\,x}{5}-\frac {2\,x^3}{15}+\frac {2\,x^5}{25}-\frac {x^5\,\ln \left (x^2+1\right )}{5}\right )+\ln \left (x^2+1\right )\,\left (\frac {x^2}{10}-\frac {x^4}{20}\right )-\frac {77\,x^2}{300}+\frac {9\,x^4}{200} \]