Integrand size = 13, antiderivative size = 60 \[ \int \frac {\left (1+x^2\right ) \arctan (x)^2}{x^5} \, dx=-\frac {1}{12 x^2}-\frac {\arctan (x)}{6 x^3}-\frac {\arctan (x)}{2 x}-\frac {\left (1+x^2\right )^2 \arctan (x)^2}{4 x^4}+\frac {\log (x)}{3}-\frac {1}{6} \log \left (1+x^2\right ) \]
-1/12/x^2-1/6*arctan(x)/x^3-1/2*arctan(x)/x-1/4*(x^2+1)^2*arctan(x)^2/x^4+ 1/3*ln(x)-1/6*ln(x^2+1)
Time = 0.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.93 \[ \int \frac {\left (1+x^2\right ) \arctan (x)^2}{x^5} \, dx=\frac {-2 \left (x+3 x^3\right ) \arctan (x)-3 \left (1+x^2\right )^2 \arctan (x)^2+x^2 \left (-1+4 x^2 \log (x)-2 x^2 \log \left (1+x^2\right )\right )}{12 x^4} \]
(-2*(x + 3*x^3)*ArcTan[x] - 3*(1 + x^2)^2*ArcTan[x]^2 + x^2*(-1 + 4*x^2*Lo g[x] - 2*x^2*Log[1 + x^2]))/(12*x^4)
Time = 0.33 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.32, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {5479, 5485, 5361, 243, 47, 14, 16, 54, 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.
| \(\displaystyle \int \frac {\left (x^2+1\right ) \arctan (x)^2}{x^5} \, dx\) |
| \(\Big \downarrow \) 5479 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (x^2+1\right ) \arctan (x)}{x^4}dx-\frac {\left (x^2+1\right )^2 \arctan (x)^2}{4 x^4}\) |
| \(\Big \downarrow \) 5485 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {\arctan (x)}{x^4}dx+\int \frac {\arctan (x)}{x^2}dx\right )-\frac {\left (x^2+1\right )^2 \arctan (x)^2}{4 x^4}\) |
| \(\Big \downarrow \) 5361 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {1}{x \left (x^2+1\right )}dx+\frac {1}{3} \int \frac {1}{x^3 \left (x^2+1\right )}dx-\frac {\arctan (x)}{3 x^3}-\frac {\arctan (x)}{x}\right )-\frac {\left (x^2+1\right )^2 \arctan (x)^2}{4 x^4}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2 \left (x^2+1\right )}dx^2+\frac {1}{6} \int \frac {1}{x^4 \left (x^2+1\right )}dx^2-\frac {\arctan (x)}{3 x^3}-\frac {\arctan (x)}{x}\right )-\frac {\left (x^2+1\right )^2 \arctan (x)^2}{4 x^4}\) |
| \(\Big \downarrow \) 47 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\int \frac {1}{x^2}dx^2-\int \frac {1}{x^2+1}dx^2\right )+\frac {1}{6} \int \frac {1}{x^4 \left (x^2+1\right )}dx^2-\frac {\arctan (x)}{3 x^3}-\frac {\arctan (x)}{x}\right )-\frac {\left (x^2+1\right )^2 \arctan (x)^2}{4 x^4}\) |
| \(\Big \downarrow \) 14 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\log \left (x^2\right )-\int \frac {1}{x^2+1}dx^2\right )+\frac {1}{6} \int \frac {1}{x^4 \left (x^2+1\right )}dx^2-\frac {\arctan (x)}{3 x^3}-\frac {\arctan (x)}{x}\right )-\frac {\left (x^2+1\right )^2 \arctan (x)^2}{4 x^4}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{6} \int \frac {1}{x^4 \left (x^2+1\right )}dx^2-\frac {\arctan (x)}{3 x^3}-\frac {\arctan (x)}{x}+\frac {1}{2} \left (\log \left (x^2\right )-\log \left (x^2+1\right )\right )\right )-\frac {\left (x^2+1\right )^2 \arctan (x)^2}{4 x^4}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{6} \int \left (-\frac {1}{x^2}+\frac {1}{x^4}+\frac {1}{x^2+1}\right )dx^2-\frac {\arctan (x)}{3 x^3}-\frac {\arctan (x)}{x}+\frac {1}{2} \left (\log \left (x^2\right )-\log \left (x^2+1\right )\right )\right )-\frac {\left (x^2+1\right )^2 \arctan (x)^2}{4 x^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\arctan (x)}{3 x^3}-\frac {\arctan (x)}{x}+\frac {1}{2} \left (\log \left (x^2\right )-\log \left (x^2+1\right )\right )+\frac {1}{6} \left (-\frac {1}{x^2}-\log \left (x^2\right )+\log \left (x^2+1\right )\right )\right )-\frac {\left (x^2+1\right )^2 \arctan (x)^2}{4 x^4}\) |
-1/4*((1 + x^2)^2*ArcTan[x]^2)/x^4 + (-1/3*ArcTan[x]/x^3 - ArcTan[x]/x + ( Log[x^2] - Log[1 + x^2])/2 + (-x^(-2) - Log[x^2] + Log[1 + x^2])/6)/2
3.7.81.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x), x ], x] /; FreeQ[{a, b, c, d}, x]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
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_)^(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_.)*((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]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_. )*(x_)^2)^(q_.), x_Symbol] :> Simp[d Int[(f*x)^m*(d + e*x^2)^(q - 1)*(a + b*ArcTan[c*x])^p, x], x] + Simp[c^2*(d/f^2) Int[(f*x)^(m + 2)*(d + e*x^2 )^(q - 1)*(a + b*ArcTan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[q, 0] && IGtQ[p, 0] && (RationalQ[m] || (EqQ[p, 1] && IntegerQ[q]))
Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.95
| method | result | size |
| default | \(-\frac {\arctan \left (x \right )^{2}}{2 x^{2}}-\frac {\arctan \left (x \right )^{2}}{4 x^{4}}-\frac {\arctan \left (x \right )^{2}}{4}-\frac {\arctan \left (x \right )}{6 x^{3}}-\frac {\arctan \left (x \right )}{2 x}-\frac {1}{12 x^{2}}+\frac {\ln \left (x \right )}{3}-\frac {\ln \left (x^{2}+1\right )}{6}\) | \(57\) |
| parts | \(-\frac {\arctan \left (x \right )^{2}}{2 x^{2}}-\frac {\arctan \left (x \right )^{2}}{4 x^{4}}-\frac {\arctan \left (x \right )^{2}}{4}-\frac {\arctan \left (x \right )}{6 x^{3}}-\frac {\arctan \left (x \right )}{2 x}-\frac {1}{12 x^{2}}+\frac {\ln \left (x \right )}{3}-\frac {\ln \left (x^{2}+1\right )}{6}\) | \(57\) |
| parallelrisch | \(\frac {-3 x^{4} \arctan \left (x \right )^{2}+4 x^{4} \ln \left (x \right )-2 \ln \left (x^{2}+1\right ) x^{4}-6 x^{3} \arctan \left (x \right )-6 x^{2} \arctan \left (x \right )^{2}-x^{2}-2 x \arctan \left (x \right )-3 \arctan \left (x \right )^{2}}{12 x^{4}}\) | \(66\) |
| risch | \(\frac {\left (x^{4}+2 x^{2}+1\right ) \ln \left (i x +1\right )^{2}}{16 x^{4}}-\frac {\left (3 x^{4} \ln \left (-i x +1\right )-6 i x^{3}+6 x^{2} \ln \left (-i x +1\right )-2 i x +3 \ln \left (-i x +1\right )\right ) \ln \left (i x +1\right )}{24 x^{4}}+\frac {3 x^{4} \ln \left (-i x +1\right )^{2}-12 i x^{3} \ln \left (-i x +1\right )+16 x^{4} \ln \left (x \right )-8 \ln \left (x^{2}+1\right ) x^{4}+6 x^{2} \ln \left (-i x +1\right )^{2}-4 i x \ln \left (-i x +1\right )-4 x^{2}+3 \ln \left (-i x +1\right )^{2}}{48 x^{4}}\) | \(174\) |
-1/2*arctan(x)^2/x^2-1/4*arctan(x)^2/x^4-1/4*arctan(x)^2-1/6/x^3*arctan(x) -1/2/x*arctan(x)-1/12/x^2+1/3*ln(x)-1/6*ln(x^2+1)
Time = 0.26 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.90 \[ \int \frac {\left (1+x^2\right ) \arctan (x)^2}{x^5} \, dx=-\frac {2 \, x^{4} \log \left (x^{2} + 1\right ) - 4 \, x^{4} \log \left (x\right ) + 3 \, {\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \left (x\right )^{2} + x^{2} + 2 \, {\left (3 \, x^{3} + x\right )} \arctan \left (x\right )}{12 \, x^{4}} \]
-1/12*(2*x^4*log(x^2 + 1) - 4*x^4*log(x) + 3*(x^4 + 2*x^2 + 1)*arctan(x)^2 + x^2 + 2*(3*x^3 + x)*arctan(x))/x^4
Time = 0.24 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02 \[ \int \frac {\left (1+x^2\right ) \arctan (x)^2}{x^5} \, dx=\frac {\log {\left (x \right )}}{3} - \frac {\log {\left (x^{2} + 1 \right )}}{6} - \frac {\operatorname {atan}^{2}{\left (x \right )}}{4} - \frac {\operatorname {atan}{\left (x \right )}}{2 x} - \frac {\operatorname {atan}^{2}{\left (x \right )}}{2 x^{2}} - \frac {1}{12 x^{2}} - \frac {\operatorname {atan}{\left (x \right )}}{6 x^{3}} - \frac {\operatorname {atan}^{2}{\left (x \right )}}{4 x^{4}} \]
log(x)/3 - log(x**2 + 1)/6 - atan(x)**2/4 - atan(x)/(2*x) - atan(x)**2/(2* x**2) - 1/(12*x**2) - atan(x)/(6*x**3) - atan(x)**2/(4*x**4)
Time = 0.29 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.18 \[ \int \frac {\left (1+x^2\right ) \arctan (x)^2}{x^5} \, dx=-\frac {1}{6} \, {\left (\frac {3 \, x^{2} + 1}{x^{3}} + 3 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) + \frac {3 \, x^{2} \arctan \left (x\right )^{2} - 2 \, x^{2} \log \left (x^{2} + 1\right ) + 4 \, x^{2} \log \left (x\right ) - 1}{12 \, x^{2}} - \frac {{\left (2 \, x^{2} + 1\right )} \arctan \left (x\right )^{2}}{4 \, x^{4}} \]
-1/6*((3*x^2 + 1)/x^3 + 3*arctan(x))*arctan(x) + 1/12*(3*x^2*arctan(x)^2 - 2*x^2*log(x^2 + 1) + 4*x^2*log(x) - 1)/x^2 - 1/4*(2*x^2 + 1)*arctan(x)^2/ x^4
\[ \int \frac {\left (1+x^2\right ) \arctan (x)^2}{x^5} \, dx=\int { \frac {{\left (x^{2} + 1\right )} \arctan \left (x\right )^{2}}{x^{5}} \,d x } \]
Time = 0.12 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.85 \[ \int \frac {\left (1+x^2\right ) \arctan (x)^2}{x^5} \, dx=\frac {\ln \left (x\right )}{3}-\frac {\ln \left (x^2+1\right )}{6}-{\mathrm {atan}\left (x\right )}^2\,\left (\frac {\frac {x^2}{2}+\frac {1}{4}}{x^4}+\frac {1}{4}\right )-\frac {1}{12\,x^2}-\frac {\mathrm {atan}\left (x\right )\,\left (\frac {x^2}{2}+\frac {1}{6}\right )}{x^3} \]
log(x)/3 - log(x^2 + 1)/6 - atan(x)^2*((x^2/2 + 1/4)/x^4 + 1/4) - 1/(12*x^ 2) - (atan(x)*(x^2/2 + 1/6))/x^3
Time = 0.00 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.08 \[ \int \frac {\left (1+x^2\right ) \arctan (x)^2}{x^5} \, dx=\frac {-3 \mathit {atan} \left (x \right )^{2} x^{4}-6 \mathit {atan} \left (x \right )^{2} x^{2}-3 \mathit {atan} \left (x \right )^{2}-6 \mathit {atan} \left (x \right ) x^{3}-2 \mathit {atan} \left (x \right ) x -2 \,\mathrm {log}\left (x^{2}+1\right ) x^{4}+4 \,\mathrm {log}\left (x \right ) x^{4}-x^{2}}{12 x^{4}} \]