Integrand size = 17, antiderivative size = 57 \[ \int \frac {\arctan (x)}{x^2 \sqrt {1-x^2}} \, dx=-\frac {\sqrt {1-x^2} \arctan (x)}{x}-\text {arctanh}\left (\sqrt {1-x^2}\right )+\sqrt {2} \text {arctanh}\left (\frac {\sqrt {1-x^2}}{\sqrt {2}}\right ) \] Output:
-arctanh((-x^2+1)^(1/2))+arctanh(1/2*2^(1/2)*(-x^2+1)^(1/2))*2^(1/2)-arcta n(x)*(-x^2+1)^(1/2)/x
Time = 0.05 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.35 \[ \int \frac {\arctan (x)}{x^2 \sqrt {1-x^2}} \, dx=-\frac {\sqrt {1-x^2} \arctan (x)}{x}+\log (x)-\frac {\log \left (1+x^2\right )}{\sqrt {2}}+\frac {\log \left (3-x^2+2 \sqrt {2-2 x^2}\right )}{\sqrt {2}}-\log \left (1+\sqrt {1-x^2}\right ) \] Input:
Integrate[ArcTan[x]/(x^2*Sqrt[1 - x^2]),x]
Output:
-((Sqrt[1 - x^2]*ArcTan[x])/x) + Log[x] - Log[1 + x^2]/Sqrt[2] + Log[3 - x ^2 + 2*Sqrt[2 - 2*x^2]]/Sqrt[2] - Log[1 + Sqrt[1 - x^2]]
Time = 0.24 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {5511, 25, 354, 94, 73, 219}
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.
\(\displaystyle \int \frac {\arctan (x)}{x^2 \sqrt {1-x^2}} \, dx\) |
\(\Big \downarrow \) 5511 |
\(\displaystyle -\int -\frac {\sqrt {1-x^2}}{x \left (x^2+1\right )}dx-\frac {\sqrt {1-x^2} \arctan (x)}{x}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {\sqrt {1-x^2}}{x \left (x^2+1\right )}dx-\frac {\sqrt {1-x^2} \arctan (x)}{x}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\sqrt {1-x^2}}{x^2 \left (x^2+1\right )}dx^2-\frac {\sqrt {1-x^2} \arctan (x)}{x}\) |
\(\Big \downarrow \) 94 |
\(\displaystyle \frac {1}{2} \left (\int \frac {1}{x^2 \sqrt {1-x^2}}dx^2-2 \int \frac {1}{\sqrt {1-x^2} \left (x^2+1\right )}dx^2\right )-\frac {\sqrt {1-x^2} \arctan (x)}{x}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (4 \int \frac {1}{2-x^4}d\sqrt {1-x^2}-2 \int \frac {1}{1-x^4}d\sqrt {1-x^2}\right )-\frac {\sqrt {1-x^2} \arctan (x)}{x}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1-x^2}}{\sqrt {2}}\right )-2 \text {arctanh}\left (\sqrt {1-x^2}\right )\right )-\frac {\sqrt {1-x^2} \arctan (x)}{x}\) |
Input:
Int[ArcTan[x]/(x^2*Sqrt[1 - x^2]),x]
Output:
-((Sqrt[1 - x^2]*ArcTan[x])/x) + (-2*ArcTanh[Sqrt[1 - x^2]] + 2*Sqrt[2]*Ar cTanh[Sqrt[1 - x^2]/Sqrt[2]])/2
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_] :> Simp[(b*e - a*f)/(b*c - a*d) Int[(e + f*x)^(p - 1)/(a + b*x), x], x] - Simp[(d*e - c*f)/(b*c - a*d) Int[(e + f*x)^(p - 1)/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[0, p, 1]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x _)^2)^(q_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^q, x]}, Sim p[(a + b*ArcTan[c*x]) u, x] - Simp[b*c Int[SimplifyIntegrand[u/(1 + c^2 *x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && ((IGtQ[q, 0] && !(ILtQ[(m - 1)/2, 0] && GtQ[m + 2*q + 3, 0])) || (IGtQ[(m + 1)/2, 0] && !(ILtQ[q, 0] && GtQ[m + 2*q + 3, 0])) || (ILtQ[(m + 2*q + 1)/2, 0] && !ILt Q[(m - 1)/2, 0]))
\[\int \frac {\arctan \left (x \right )}{x^{2} \sqrt {-x^{2}+1}}d x\]
Input:
int(arctan(x)/x^2/(-x^2+1)^(1/2),x)
Output:
int(arctan(x)/x^2/(-x^2+1)^(1/2),x)
Time = 0.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.42 \[ \int \frac {\arctan (x)}{x^2 \sqrt {1-x^2}} \, dx=\frac {\sqrt {2} x \log \left (\frac {x^{2} - 2 \, \sqrt {2} \sqrt {-x^{2} + 1} - 3}{x^{2} + 1}\right ) - x \log \left (\sqrt {-x^{2} + 1} + 1\right ) + x \log \left (\sqrt {-x^{2} + 1} - 1\right ) - 2 \, \sqrt {-x^{2} + 1} \arctan \left (x\right )}{2 \, x} \] Input:
integrate(arctan(x)/x^2/(-x^2+1)^(1/2),x, algorithm="fricas")
Output:
1/2*(sqrt(2)*x*log((x^2 - 2*sqrt(2)*sqrt(-x^2 + 1) - 3)/(x^2 + 1)) - x*log (sqrt(-x^2 + 1) + 1) + x*log(sqrt(-x^2 + 1) - 1) - 2*sqrt(-x^2 + 1)*arctan (x))/x
\[ \int \frac {\arctan (x)}{x^2 \sqrt {1-x^2}} \, dx=\int \frac {\operatorname {atan}{\left (x \right )}}{x^{2} \sqrt {- \left (x - 1\right ) \left (x + 1\right )}}\, dx \] Input:
integrate(atan(x)/x**2/(-x**2+1)**(1/2),x)
Output:
Integral(atan(x)/(x**2*sqrt(-(x - 1)*(x + 1))), x)
\[ \int \frac {\arctan (x)}{x^2 \sqrt {1-x^2}} \, dx=\int { \frac {\arctan \left (x\right )}{\sqrt {-x^{2} + 1} x^{2}} \,d x } \] Input:
integrate(arctan(x)/x^2/(-x^2+1)^(1/2),x, algorithm="maxima")
Output:
integrate(arctan(x)/(sqrt(-x^2 + 1)*x^2), x)
Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (48) = 96\).
Time = 0.14 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.82 \[ \int \frac {\arctan (x)}{x^2 \sqrt {1-x^2}} \, dx=\frac {1}{2} \, {\left (\frac {x}{\sqrt {-x^{2} + 1} - 1} - \frac {\sqrt {-x^{2} + 1} - 1}{x}\right )} \arctan \left (x\right ) - \frac {1}{2} \, \sqrt {2} \log \left (\frac {\sqrt {2} - \sqrt {-x^{2} + 1}}{\sqrt {2} + \sqrt {-x^{2} + 1}}\right ) - \frac {1}{2} \, \log \left (\sqrt {-x^{2} + 1} + 1\right ) + \frac {1}{2} \, \log \left (-\sqrt {-x^{2} + 1} + 1\right ) \] Input:
integrate(arctan(x)/x^2/(-x^2+1)^(1/2),x, algorithm="giac")
Output:
1/2*(x/(sqrt(-x^2 + 1) - 1) - (sqrt(-x^2 + 1) - 1)/x)*arctan(x) - 1/2*sqrt (2)*log((sqrt(2) - sqrt(-x^2 + 1))/(sqrt(2) + sqrt(-x^2 + 1))) - 1/2*log(s qrt(-x^2 + 1) + 1) + 1/2*log(-sqrt(-x^2 + 1) + 1)
Timed out. \[ \int \frac {\arctan (x)}{x^2 \sqrt {1-x^2}} \, dx=\int \frac {\mathrm {atan}\left (x\right )}{x^2\,\sqrt {1-x^2}} \,d x \] Input:
int(atan(x)/(x^2*(1 - x^2)^(1/2)),x)
Output:
int(atan(x)/(x^2*(1 - x^2)^(1/2)), x)
Time = 0.17 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.74 \[ \int \frac {\arctan (x)}{x^2 \sqrt {1-x^2}} \, dx=\frac {-2 \sqrt {-x^{2}+1}\, \mathit {atan} \left (\frac {2 \tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )}{\tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )^{2}+1}\right )-\sqrt {2}\, \mathrm {log}\left (-\sqrt {2}\, i +\tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )+i \right ) x -\sqrt {2}\, \mathrm {log}\left (\sqrt {2}\, i +\tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )-i \right ) x +\sqrt {2}\, \mathrm {log}\left (2 \sqrt {2}+\tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )^{2}+3\right ) x +2 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )\right ) x}{2 x} \] Input:
int(atan(x)/x^2/(-x^2+1)^(1/2),x)
Output:
( - 2*sqrt( - x**2 + 1)*atan((2*tan(asin(x)/2))/(tan(asin(x)/2)**2 + 1)) - sqrt(2)*log( - sqrt(2)*i + tan(asin(x)/2) + i)*x - sqrt(2)*log(sqrt(2)*i + tan(asin(x)/2) - i)*x + sqrt(2)*log(2*sqrt(2) + tan(asin(x)/2)**2 + 3)*x + 2*log(tan(asin(x)/2))*x)/(2*x)