Integrand size = 13, antiderivative size = 40 \[ \int \frac {x^4 \cot ^{-1}(x)}{1+x^2} \, dx=\frac {x^2}{6}-x \cot ^{-1}(x)+\frac {1}{3} x^3 \cot ^{-1}(x)-\frac {1}{2} \cot ^{-1}(x)^2-\frac {2}{3} \log \left (1+x^2\right ) \]
Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80 \[ \int \frac {x^4 \cot ^{-1}(x)}{1+x^2} \, dx=\frac {1}{6} \left (x^2+2 x \left (-3+x^2\right ) \cot ^{-1}(x)-3 \cot ^{-1}(x)^2-4 \log \left (1+x^2\right )\right ) \]
Time = 0.51 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.22, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {5452, 5362, 243, 49, 2009, 5452, 5346, 240, 5420}
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.
\(\displaystyle \int \frac {x^4 \cot ^{-1}(x)}{x^2+1} \, dx\) |
\(\Big \downarrow \) 5452 |
\(\displaystyle \int x^2 \cot ^{-1}(x)dx-\int \frac {x^2 \cot ^{-1}(x)}{x^2+1}dx\) |
\(\Big \downarrow \) 5362 |
\(\displaystyle -\int \frac {x^2 \cot ^{-1}(x)}{x^2+1}dx+\frac {1}{3} \int \frac {x^3}{x^2+1}dx+\frac {1}{3} x^3 \cot ^{-1}(x)\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{6} \int \frac {x^2}{x^2+1}dx^2-\int \frac {x^2 \cot ^{-1}(x)}{x^2+1}dx+\frac {1}{3} x^3 \cot ^{-1}(x)\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{6} \int \left (1+\frac {1}{-x^2-1}\right )dx^2-\int \frac {x^2 \cot ^{-1}(x)}{x^2+1}dx+\frac {1}{3} x^3 \cot ^{-1}(x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {x^2 \cot ^{-1}(x)}{x^2+1}dx+\frac {1}{3} x^3 \cot ^{-1}(x)+\frac {1}{6} \left (x^2-\log \left (x^2+1\right )\right )\) |
\(\Big \downarrow \) 5452 |
\(\displaystyle \int \frac {\cot ^{-1}(x)}{x^2+1}dx-\int \cot ^{-1}(x)dx+\frac {1}{3} x^3 \cot ^{-1}(x)+\frac {1}{6} \left (x^2-\log \left (x^2+1\right )\right )\) |
\(\Big \downarrow \) 5346 |
\(\displaystyle -\int \frac {x}{x^2+1}dx+\int \frac {\cot ^{-1}(x)}{x^2+1}dx+\frac {1}{3} x^3 \cot ^{-1}(x)+\frac {1}{6} \left (x^2-\log \left (x^2+1\right )\right )-x \cot ^{-1}(x)\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \int \frac {\cot ^{-1}(x)}{x^2+1}dx+\frac {1}{3} x^3 \cot ^{-1}(x)+\frac {1}{6} \left (x^2-\log \left (x^2+1\right )\right )-\frac {1}{2} \log \left (x^2+1\right )-x \cot ^{-1}(x)\) |
\(\Big \downarrow \) 5420 |
\(\displaystyle \frac {1}{3} x^3 \cot ^{-1}(x)+\frac {1}{6} \left (x^2-\log \left (x^2+1\right )\right )-\frac {1}{2} \log \left (x^2+1\right )-x \cot ^{-1}(x)-\frac {1}{2} \cot ^{-1}(x)^2\) |
-(x*ArcCot[x]) + (x^3*ArcCot[x])/3 - ArcCot[x]^2/2 + (x^2 - Log[1 + x^2])/ 6 - Log[1 + x^2]/2
3.1.37.3.1 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_.) + ArcCot[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcCot[c*x^n])^p, x] + Simp[b*c*n*p Int[x^n*((a + b*ArcCot[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_.) + ArcCot[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcCot[c*x^n])^p/(m + 1)), x] + Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcCot[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_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbo l] :> Simp[-(a + b*ArcCot[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_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e _.)*(x_)^2), x_Symbol] :> Simp[f^2/e Int[(f*x)^(m - 2)*(a + b*ArcCot[c*x] )^p, x], x] - Simp[d*(f^2/e) Int[(f*x)^(m - 2)*((a + b*ArcCot[c*x])^p/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]
Time = 0.48 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85
method | result | size |
parallelrisch | \(\frac {x^{3} \operatorname {arccot}\left (x \right )}{3}+\frac {x^{2}}{6}-x \,\operatorname {arccot}\left (x \right )-\frac {\operatorname {arccot}\left (x \right )^{2}}{2}-\frac {2 \ln \left (x^{2}+1\right )}{3}-\frac {1}{3}\) | \(34\) |
default | \(\frac {x^{3} \operatorname {arccot}\left (x \right )}{3}-x \,\operatorname {arccot}\left (x \right )+\operatorname {arccot}\left (x \right ) \arctan \left (x \right )+\frac {x^{2}}{6}-\frac {2 \ln \left (x^{2}+1\right )}{3}+\frac {\arctan \left (x \right )^{2}}{2}\) | \(38\) |
parts | \(\frac {x^{3} \operatorname {arccot}\left (x \right )}{3}-x \,\operatorname {arccot}\left (x \right )+\operatorname {arccot}\left (x \right ) \arctan \left (x \right )+\frac {x^{2}}{6}-\frac {2 \ln \left (x^{2}+1\right )}{3}+\frac {\arctan \left (x \right )^{2}}{2}\) | \(38\) |
risch | \(\frac {\ln \left (i x +1\right )^{2}}{8}+\left (\frac {i x^{3}}{6}-\frac {i x}{2}-\frac {\ln \left (-i x +1\right )}{4}\right ) \ln \left (i x +1\right )+\frac {\ln \left (-i x +1\right )^{2}}{8}-\frac {i x^{3} \ln \left (-i x +1\right )}{6}+\frac {i \ln \left (-i x +1\right ) x}{2}+\frac {\pi \,x^{3}}{6}-\frac {\pi x}{2}+\frac {x^{2}}{6}-\frac {2 \ln \left (x^{2}+1\right )}{3}+\frac {\pi \arctan \left (x \right )}{2}\) | \(104\) |
Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.78 \[ \int \frac {x^4 \cot ^{-1}(x)}{1+x^2} \, dx=\frac {1}{6} \, x^{2} + \frac {1}{3} \, {\left (x^{3} - 3 \, x\right )} \operatorname {arccot}\left (x\right ) - \frac {1}{2} \, \operatorname {arccot}\left (x\right )^{2} - \frac {2}{3} \, \log \left (x^{2} + 1\right ) \]
Time = 0.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85 \[ \int \frac {x^4 \cot ^{-1}(x)}{1+x^2} \, dx=\frac {x^{3} \operatorname {acot}{\left (x \right )}}{3} + \frac {x^{2}}{6} - x \operatorname {acot}{\left (x \right )} - \frac {2 \log {\left (x^{2} + 1 \right )}}{3} - \frac {\operatorname {acot}^{2}{\left (x \right )}}{2} \]
Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.88 \[ \int \frac {x^4 \cot ^{-1}(x)}{1+x^2} \, dx=\frac {1}{6} \, x^{2} + \frac {1}{3} \, {\left (x^{3} - 3 \, x + 3 \, \arctan \left (x\right )\right )} \operatorname {arccot}\left (x\right ) + \frac {1}{2} \, \arctan \left (x\right )^{2} - \frac {2}{3} \, \log \left (x^{2} + 1\right ) \]
\[ \int \frac {x^4 \cot ^{-1}(x)}{1+x^2} \, dx=\int { \frac {x^{4} \operatorname {arccot}\left (x\right )}{x^{2} + 1} \,d x } \]
Time = 0.74 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80 \[ \int \frac {x^4 \cot ^{-1}(x)}{1+x^2} \, dx=\frac {x^3\,\mathrm {acot}\left (x\right )}{3}-\frac {2\,\ln \left (x^2+1\right )}{3}-\frac {{\mathrm {acot}\left (x\right )}^2}{2}-x\,\mathrm {acot}\left (x\right )+\frac {x^2}{6} \]