Integrand size = 27, antiderivative size = 152 \[ \int \frac {x \arctan \left (x+\sqrt {1-x^2}\right )}{\sqrt {1-x^2}} \, dx=-\frac {\arcsin (x)}{2}+\frac {1}{4} \sqrt {3} \arctan \left (\frac {-1+\sqrt {3} x}{\sqrt {1-x^2}}\right )+\frac {1}{4} \sqrt {3} \arctan \left (\frac {1+\sqrt {3} x}{\sqrt {1-x^2}}\right )-\frac {1}{4} \sqrt {3} \arctan \left (\frac {-1+2 x^2}{\sqrt {3}}\right )-\sqrt {1-x^2} \arctan \left (x+\sqrt {1-x^2}\right )+\frac {1}{4} \text {arctanh}\left (x \sqrt {1-x^2}\right )+\frac {1}{8} \log \left (1-x^2+x^4\right ) \]
-1/2*arcsin(x)+1/4*arctanh(x*(-x^2+1)^(1/2))+1/8*ln(x^4-x^2+1)-1/4*arctan( 1/3*(2*x^2-1)*3^(1/2))*3^(1/2)+1/4*arctan((-1+x*3^(1/2))/(-x^2+1)^(1/2))*3 ^(1/2)+1/4*arctan((1+x*3^(1/2))/(-x^2+1)^(1/2))*3^(1/2)-arctan(x+(-x^2+1)^ (1/2))*(-x^2+1)^(1/2)
Result contains complex when optimal does not.
Time = 3.09 (sec) , antiderivative size = 2175, normalized size of antiderivative = 14.31 \[ \int \frac {x \arctan \left (x+\sqrt {1-x^2}\right )}{\sqrt {1-x^2}} \, dx=\text {Result too large to show} \]
(-24*ArcSin[x] - 48*Sqrt[1 - x^2]*ArcTan[x + Sqrt[1 - x^2]] + (2*(-3*I + S qrt[3])*ArcTan[(3 - I*Sqrt[3] + (-3 - I*Sqrt[3])*x^4 + 2*x*(-6*I + 2*Sqrt[ 3] - I*Sqrt[2 - (2*I)*Sqrt[3]]*Sqrt[1 - x^2]) - 2*x^3*(6*I + 2*Sqrt[3] + I *Sqrt[2 - (2*I)*Sqrt[3]]*Sqrt[1 - x^2]) - (2*I)*Sqrt[3]*x^2*(6 + Sqrt[2 - (2*I)*Sqrt[3]]*Sqrt[1 - x^2]))/(I - Sqrt[3] + (6*I)*(I + Sqrt[3])*x - 2*(- 15*I + Sqrt[3])*x^2 + 6*(1 + (3*I)*Sqrt[3])*x^3 + (11*I + 3*Sqrt[3])*x^4)] )/Sqrt[(1 - I*Sqrt[3])/6] - (2*(-3*I + Sqrt[3])*ArcTan[(3 - I*Sqrt[3] + (- 3 - I*Sqrt[3])*x^4 + 2*x^3*(6*I + 2*Sqrt[3] + I*Sqrt[2 - (2*I)*Sqrt[3]]*Sq rt[1 - x^2]) + x*(12*I - 4*Sqrt[3] + (2*I)*Sqrt[2 - (2*I)*Sqrt[3]]*Sqrt[1 - x^2]) - (2*I)*Sqrt[3]*x^2*(6 + Sqrt[2 - (2*I)*Sqrt[3]]*Sqrt[1 - x^2]))/( I - Sqrt[3] + (6 - (6*I)*Sqrt[3])*x - 2*(-15*I + Sqrt[3])*x^2 + (-6 - (18* I)*Sqrt[3])*x^3 + (11*I + 3*Sqrt[3])*x^4)])/Sqrt[(1 - I*Sqrt[3])/6] - (2*( 3*I + Sqrt[3])*ArcTan[(-3 - I*Sqrt[3] + (3 - I*Sqrt[3])*x^4 + 2*x^3*(-6*I + 2*Sqrt[3] - I*Sqrt[2 + (2*I)*Sqrt[3]]*Sqrt[1 - x^2]) - 2*x*(6*I + 2*Sqrt [3] + I*Sqrt[2 + (2*I)*Sqrt[3]]*Sqrt[1 - x^2]) - (2*I)*Sqrt[3]*x^2*(6 + Sq rt[2 + (2*I)*Sqrt[3]]*Sqrt[1 - x^2]))/(-I - Sqrt[3] + (-6 - (6*I)*Sqrt[3]) *x - 2*(15*I + Sqrt[3])*x^2 + 6*(1 - (3*I)*Sqrt[3])*x^3 + (-11*I + 3*Sqrt[ 3])*x^4)])/Sqrt[(1 + I*Sqrt[3])/6] + (2*(3*I + Sqrt[3])*ArcTan[(-3 - I*Sqr t[3] + (3 - I*Sqrt[3])*x^4 + 2*x*(6*I + 2*Sqrt[3] + I*Sqrt[2 + (2*I)*Sqrt[ 3]]*Sqrt[1 - x^2]) + x^3*(12*I - 4*Sqrt[3] + (2*I)*Sqrt[2 + (2*I)*Sqrt[...
Result contains complex when optimal does not.
Time = 0.69 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.86, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {5730, 27, 7293, 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.
| \(\displaystyle \int \frac {x \arctan \left (\sqrt {1-x^2}+x\right )}{\sqrt {1-x^2}} \, dx\) |
| \(\Big \downarrow \) 5730 |
| \(\displaystyle -\int \frac {x-\sqrt {1-x^2}}{2 \left (\sqrt {1-x^2} x+1\right )}dx-\sqrt {1-x^2} \arctan \left (\sqrt {1-x^2}+x\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {1}{2} \int \frac {x-\sqrt {1-x^2}}{\sqrt {1-x^2} x+1}dx-\sqrt {1-x^2} \arctan \left (\sqrt {1-x^2}+x\right )\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {1}{2} \int \left (\frac {x}{\sqrt {1-x^2} x+1}-\frac {\sqrt {1-x^2}}{\sqrt {1-x^2} x+1}\right )dx-\sqrt {1-x^2} \arctan \left (\sqrt {1-x^2}+x\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\sqrt {1-x^2} \arctan \left (\sqrt {1-x^2}+x\right )+\frac {1}{2} \left (-\arcsin (x)+\frac {1}{2} \sqrt {3} \arctan \left (\frac {1-2 x^2}{\sqrt {3}}\right )-\frac {1}{6} \left (-\sqrt {3}+3 i\right ) \arctan \left (\frac {x}{\sqrt {-\frac {-\sqrt {3}+i}{\sqrt {3}+i}} \sqrt {1-x^2}}\right )+\frac {\arctan \left (\frac {x}{\sqrt {-\frac {-\sqrt {3}+i}{\sqrt {3}+i}} \sqrt {1-x^2}}\right )}{\sqrt {3}}+\frac {1}{6} \left (\sqrt {3}+3 i\right ) \arctan \left (\frac {\sqrt {-\frac {-\sqrt {3}+i}{\sqrt {3}+i}} x}{\sqrt {1-x^2}}\right )+\frac {\arctan \left (\frac {\sqrt {-\frac {-\sqrt {3}+i}{\sqrt {3}+i}} x}{\sqrt {1-x^2}}\right )}{\sqrt {3}}+\frac {1}{4} \log \left (x^4-x^2+1\right )\right )\) |
-(Sqrt[1 - x^2]*ArcTan[x + Sqrt[1 - x^2]]) + (-ArcSin[x] + (Sqrt[3]*ArcTan [(1 - 2*x^2)/Sqrt[3]])/2 + ArcTan[x/(Sqrt[-((I - Sqrt[3])/(I + Sqrt[3]))]* Sqrt[1 - x^2])]/Sqrt[3] - ((3*I - Sqrt[3])*ArcTan[x/(Sqrt[-((I - Sqrt[3])/ (I + Sqrt[3]))]*Sqrt[1 - x^2])])/6 + ArcTan[(Sqrt[-((I - Sqrt[3])/(I + Sqr t[3]))]*x)/Sqrt[1 - x^2]]/Sqrt[3] + ((3*I + Sqrt[3])*ArcTan[(Sqrt[-((I - S qrt[3])/(I + Sqrt[3]))]*x)/Sqrt[1 - x^2]])/6 + Log[1 - x^2 + x^4]/4)/2
3.1.13.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + ArcTan[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Simp[(a + b*ArcTan[u]) w, x] - Simp[b Int[SimplifyIntegrand[w*(D[u, x] /(1 + u^2)), x], x], x] /; InverseFunctionFreeQ[w, x]] /; FreeQ[{a, b}, x] && InverseFunctionFreeQ[u, x] && !MatchQ[v, ((c_.) + (d_.)*x)^(m_.) /; Fre eQ[{c, d, m}, x]] && FalseQ[FunctionOfLinear[v*(a + b*ArcTan[u]), x]]
\[\int \frac {x \arctan \left (x +\sqrt {-x^{2}+1}\right )}{\sqrt {-x^{2}+1}}d x\]
Time = 0.27 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.32 \[ \int \frac {x \arctan \left (x+\sqrt {1-x^2}\right )}{\sqrt {1-x^2}} \, dx=-\frac {1}{4} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \sqrt {-x^{2} + 1} \arctan \left (x + \sqrt {-x^{2} + 1}\right ) - \frac {1}{8} \, \sqrt {3} \arctan \left (\frac {4 \, \sqrt {3} \sqrt {-x^{2} + 1} x + \sqrt {3}}{3 \, {\left (2 \, x^{2} - 1\right )}}\right ) - \frac {1}{8} \, \sqrt {3} \arctan \left (\frac {4 \, \sqrt {3} \sqrt {-x^{2} + 1} x - \sqrt {3}}{3 \, {\left (2 \, x^{2} - 1\right )}}\right ) + \frac {1}{2} \, \arctan \left (\frac {\sqrt {-x^{2} + 1} x}{x^{2} - 1}\right ) + \frac {1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) + \frac {1}{16} \, \log \left (-x^{4} + x^{2} + 2 \, \sqrt {-x^{2} + 1} x + 1\right ) - \frac {1}{16} \, \log \left (-x^{4} + x^{2} - 2 \, \sqrt {-x^{2} + 1} x + 1\right ) \]
-1/4*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) - sqrt(-x^2 + 1)*arctan(x + s qrt(-x^2 + 1)) - 1/8*sqrt(3)*arctan(1/3*(4*sqrt(3)*sqrt(-x^2 + 1)*x + sqrt (3))/(2*x^2 - 1)) - 1/8*sqrt(3)*arctan(1/3*(4*sqrt(3)*sqrt(-x^2 + 1)*x - s qrt(3))/(2*x^2 - 1)) + 1/2*arctan(sqrt(-x^2 + 1)*x/(x^2 - 1)) + 1/8*log(x^ 4 - x^2 + 1) + 1/16*log(-x^4 + x^2 + 2*sqrt(-x^2 + 1)*x + 1) - 1/16*log(-x ^4 + x^2 - 2*sqrt(-x^2 + 1)*x + 1)
Timed out. \[ \int \frac {x \arctan \left (x+\sqrt {1-x^2}\right )}{\sqrt {1-x^2}} \, dx=\text {Timed out} \]
\[ \int \frac {x \arctan \left (x+\sqrt {1-x^2}\right )}{\sqrt {1-x^2}} \, dx=\int { \frac {x \arctan \left (x + \sqrt {-x^{2} + 1}\right )}{\sqrt {-x^{2} + 1}} \,d x } \]
-sqrt(x + 1)*sqrt(-x + 1)*arctan(x + sqrt(x + 1)*sqrt(-x + 1)) - integrate (x/(x^2 + 2*x*e^(1/2*log(x + 1) + 1/2*log(-x + 1)) + e^(log(x + 1) + log(- x + 1)) + 1), x)
Leaf count of result is larger than twice the leaf count of optimal. 373 vs. \(2 (119) = 238\).
Time = 0.34 (sec) , antiderivative size = 373, normalized size of antiderivative = 2.45 \[ \int \frac {x \arctan \left (x+\sqrt {1-x^2}\right )}{\sqrt {1-x^2}} \, dx=-\frac {1}{4} \, \pi \mathrm {sgn}\left (x\right ) + \frac {1}{8} \, \sqrt {3} {\left (\pi \mathrm {sgn}\left (x\right ) + 2 \, \arctan \left (-\frac {\sqrt {3} x {\left (\frac {\sqrt {-x^{2} + 1} - 1}{x} + \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{3 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right )\right )} + \frac {1}{8} \, \sqrt {3} {\left (\pi \mathrm {sgn}\left (x\right ) + 2 \, \arctan \left (\frac {\sqrt {3} x {\left (\frac {\sqrt {-x^{2} + 1} - 1}{x} - \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} + 1\right )}}{3 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right )\right )} - \frac {1}{4} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \sqrt {-x^{2} + 1} \arctan \left (x + \sqrt {-x^{2} + 1}\right ) - \frac {1}{2} \, \arctan \left (-\frac {x {\left (\frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{2 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right ) + \frac {1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) - \frac {1}{8} \, \log \left ({\left (\frac {x}{\sqrt {-x^{2} + 1} - 1} - \frac {\sqrt {-x^{2} + 1} - 1}{x}\right )}^{2} + \frac {2 \, x}{\sqrt {-x^{2} + 1} - 1} - \frac {2 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}{x} + 4\right ) + \frac {1}{8} \, \log \left ({\left (\frac {x}{\sqrt {-x^{2} + 1} - 1} - \frac {\sqrt {-x^{2} + 1} - 1}{x}\right )}^{2} - \frac {2 \, x}{\sqrt {-x^{2} + 1} - 1} + \frac {2 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}{x} + 4\right ) \]
-1/4*pi*sgn(x) + 1/8*sqrt(3)*(pi*sgn(x) + 2*arctan(-1/3*sqrt(3)*x*((sqrt(- x^2 + 1) - 1)/x + (sqrt(-x^2 + 1) - 1)^2/x^2 - 1)/(sqrt(-x^2 + 1) - 1))) + 1/8*sqrt(3)*(pi*sgn(x) + 2*arctan(1/3*sqrt(3)*x*((sqrt(-x^2 + 1) - 1)/x - (sqrt(-x^2 + 1) - 1)^2/x^2 + 1)/(sqrt(-x^2 + 1) - 1))) - 1/4*sqrt(3)*arct an(1/3*sqrt(3)*(2*x^2 - 1)) - sqrt(-x^2 + 1)*arctan(x + sqrt(-x^2 + 1)) - 1/2*arctan(-1/2*x*((sqrt(-x^2 + 1) - 1)^2/x^2 - 1)/(sqrt(-x^2 + 1) - 1)) + 1/8*log(x^4 - x^2 + 1) - 1/8*log((x/(sqrt(-x^2 + 1) - 1) - (sqrt(-x^2 + 1 ) - 1)/x)^2 + 2*x/(sqrt(-x^2 + 1) - 1) - 2*(sqrt(-x^2 + 1) - 1)/x + 4) + 1 /8*log((x/(sqrt(-x^2 + 1) - 1) - (sqrt(-x^2 + 1) - 1)/x)^2 - 2*x/(sqrt(-x^ 2 + 1) - 1) + 2*(sqrt(-x^2 + 1) - 1)/x + 4)
Timed out. \[ \int \frac {x \arctan \left (x+\sqrt {1-x^2}\right )}{\sqrt {1-x^2}} \, dx=\int \frac {x\,\mathrm {atan}\left (x+\sqrt {1-x^2}\right )}{\sqrt {1-x^2}} \,d x \]