Integrand size = 8, antiderivative size = 39 \[ \int \frac {\arctan (x)^2}{x^3} \, dx=-\frac {\arctan (x)}{x}-\frac {\arctan (x)^2}{2}-\frac {\arctan (x)^2}{2 x^2}+\log (x)-\frac {1}{2} \log \left (1+x^2\right ) \] Output:
-arctan(x)/x-1/2*arctan(x)^2-1/2*arctan(x)^2/x^2+ln(x)-1/2*ln(x^2+1)
Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.97 \[ \int \frac {\arctan (x)^2}{x^3} \, dx=-\frac {\arctan (x)}{x}+\frac {\left (-1-x^2\right ) \arctan (x)^2}{2 x^2}+\log (x)-\frac {1}{2} \log \left (1+x^2\right ) \] Input:
Integrate[ArcTan[x]^2/x^3,x]
Output:
-(ArcTan[x]/x) + ((-1 - x^2)*ArcTan[x]^2)/(2*x^2) + Log[x] - Log[1 + x^2]/ 2
Time = 0.34 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5361, 5453, 5361, 243, 47, 14, 16, 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.
\(\displaystyle \int \frac {\arctan (x)^2}{x^3} \, dx\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle \int \frac {\arctan (x)}{x^2 \left (x^2+1\right )}dx-\frac {\arctan (x)^2}{2 x^2}\) |
\(\Big \downarrow \) 5453 |
\(\displaystyle \int \frac {\arctan (x)}{x^2}dx-\int \frac {\arctan (x)}{x^2+1}dx-\frac {\arctan (x)^2}{2 x^2}\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle -\int \frac {\arctan (x)}{x^2+1}dx+\int \frac {1}{x \left (x^2+1\right )}dx-\frac {\arctan (x)^2}{2 x^2}-\frac {\arctan (x)}{x}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\int \frac {\arctan (x)}{x^2+1}dx+\frac {1}{2} \int \frac {1}{x^2 \left (x^2+1\right )}dx^2-\frac {\arctan (x)^2}{2 x^2}-\frac {\arctan (x)}{x}\) |
\(\Big \downarrow \) 47 |
\(\displaystyle -\int \frac {\arctan (x)}{x^2+1}dx+\frac {1}{2} \left (\int \frac {1}{x^2}dx^2-\int \frac {1}{x^2+1}dx^2\right )-\frac {\arctan (x)^2}{2 x^2}-\frac {\arctan (x)}{x}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle -\int \frac {\arctan (x)}{x^2+1}dx+\frac {1}{2} \left (\log \left (x^2\right )-\int \frac {1}{x^2+1}dx^2\right )-\frac {\arctan (x)^2}{2 x^2}-\frac {\arctan (x)}{x}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle -\int \frac {\arctan (x)}{x^2+1}dx-\frac {\arctan (x)^2}{2 x^2}-\frac {\arctan (x)}{x}+\frac {1}{2} \left (\log \left (x^2\right )-\log \left (x^2+1\right )\right )\) |
\(\Big \downarrow \) 5419 |
\(\displaystyle -\frac {\arctan (x)^2}{2 x^2}-\frac {\arctan (x)^2}{2}-\frac {\arctan (x)}{x}+\frac {1}{2} \left (\log \left (x^2\right )-\log \left (x^2+1\right )\right )\) |
Input:
Int[ArcTan[x]^2/x^3,x]
Output:
-(ArcTan[x]/x) - ArcTan[x]^2/2 - ArcTan[x]^2/(2*x^2) + (Log[x^2] - Log[1 + x^2])/2
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[(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_.)/((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_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e _.)*(x_)^2), x_Symbol] :> Simp[1/d Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Simp[e/(d*f^2) 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] && LtQ[m, -1]
Time = 0.16 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.87
method | result | size |
default | \(-\frac {\arctan \left (x \right )}{x}-\frac {\arctan \left (x \right )^{2}}{2}-\frac {\arctan \left (x \right )^{2}}{2 x^{2}}+\ln \left (x \right )-\frac {\ln \left (x^{2}+1\right )}{2}\) | \(34\) |
parts | \(-\frac {\arctan \left (x \right )}{x}-\frac {\arctan \left (x \right )^{2}}{2}-\frac {\arctan \left (x \right )^{2}}{2 x^{2}}+\ln \left (x \right )-\frac {\ln \left (x^{2}+1\right )}{2}\) | \(34\) |
parallelrisch | \(\frac {-x^{2} \arctan \left (x \right )^{2}+2 x^{2} \ln \left (x \right )-\ln \left (x^{2}+1\right ) x^{2}-2 x \arctan \left (x \right )-\arctan \left (x \right )^{2}}{2 x^{2}}\) | \(45\) |
risch | \(\frac {\left (x^{2}+1\right ) \ln \left (i x +1\right )^{2}}{8 x^{2}}-\frac {\left (x^{2} \ln \left (-i x +1\right )-2 i x +\ln \left (-i x +1\right )\right ) \ln \left (i x +1\right )}{4 x^{2}}+\frac {x^{2} \ln \left (-i x +1\right )^{2}-4 i x \ln \left (-i x +1\right )+8 x^{2} \ln \left (x \right )-4 \ln \left (x^{2}+1\right ) x^{2}+\ln \left (-i x +1\right )^{2}}{8 x^{2}}\) | \(113\) |
Input:
int(arctan(x)^2/x^3,x,method=_RETURNVERBOSE)
Output:
-1/x*arctan(x)-1/2*arctan(x)^2-1/2*arctan(x)^2/x^2+ln(x)-1/2*ln(x^2+1)
Time = 0.06 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.97 \[ \int \frac {\arctan (x)^2}{x^3} \, dx=-\frac {{\left (x^{2} + 1\right )} \arctan \left (x\right )^{2} + x^{2} \log \left (x^{2} + 1\right ) - 2 \, x^{2} \log \left (x\right ) + 2 \, x \arctan \left (x\right )}{2 \, x^{2}} \] Input:
integrate(arctan(x)^2/x^3,x, algorithm="fricas")
Output:
-1/2*((x^2 + 1)*arctan(x)^2 + x^2*log(x^2 + 1) - 2*x^2*log(x) + 2*x*arctan (x))/x^2
Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.82 \[ \int \frac {\arctan (x)^2}{x^3} \, dx=\log {\left (x \right )} - \frac {\log {\left (x^{2} + 1 \right )}}{2} - \frac {\operatorname {atan}^{2}{\left (x \right )}}{2} - \frac {\operatorname {atan}{\left (x \right )}}{x} - \frac {\operatorname {atan}^{2}{\left (x \right )}}{2 x^{2}} \] Input:
integrate(atan(x)**2/x**3,x)
Output:
log(x) - log(x**2 + 1)/2 - atan(x)**2/2 - atan(x)/x - atan(x)**2/(2*x**2)
Time = 0.11 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.92 \[ \int \frac {\arctan (x)^2}{x^3} \, dx=-{\left (\frac {1}{x} + \arctan \left (x\right )\right )} \arctan \left (x\right ) + \frac {1}{2} \, \arctan \left (x\right )^{2} - \frac {\arctan \left (x\right )^{2}}{2 \, x^{2}} - \frac {1}{2} \, \log \left (x^{2} + 1\right ) + \log \left (x\right ) \] Input:
integrate(arctan(x)^2/x^3,x, algorithm="maxima")
Output:
-(1/x + arctan(x))*arctan(x) + 1/2*arctan(x)^2 - 1/2*arctan(x)^2/x^2 - 1/2 *log(x^2 + 1) + log(x)
\[ \int \frac {\arctan (x)^2}{x^3} \, dx=\int { \frac {\arctan \left (x\right )^{2}}{x^{3}} \,d x } \] Input:
integrate(arctan(x)^2/x^3,x, algorithm="giac")
Output:
integrate(arctan(x)^2/x^3, x)
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.79 \[ \int \frac {\arctan (x)^2}{x^3} \, dx=\ln \left (x\right )-\frac {\ln \left (x^2+1\right )}{2}-\frac {\mathrm {atan}\left (x\right )}{x}-{\mathrm {atan}\left (x\right )}^2\,\left (\frac {1}{2\,x^2}+\frac {1}{2}\right ) \] Input:
int(atan(x)^2/x^3,x)
Output:
log(x) - log(x^2 + 1)/2 - atan(x)/x - atan(x)^2*(1/(2*x^2) + 1/2)
Time = 0.15 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.13 \[ \int \frac {\arctan (x)^2}{x^3} \, dx=\frac {-\mathit {atan} \left (x \right )^{2} x^{2}-\mathit {atan} \left (x \right )^{2}-2 \mathit {atan} \left (x \right ) x -\mathrm {log}\left (x^{2}+1\right ) x^{2}+2 \,\mathrm {log}\left (x \right ) x^{2}}{2 x^{2}} \] Input:
int(atan(x)^2/x^3,x)
Output:
( - atan(x)**2*x**2 - atan(x)**2 - 2*atan(x)*x - log(x**2 + 1)*x**2 + 2*lo g(x)*x**2)/(2*x**2)