Integrand size = 19, antiderivative size = 66 \[ \int \frac {2 \sqrt {x}}{1+2 x-x^2} \, dx=-\sqrt {2 \left (-1+\sqrt {2}\right )} \arctan \left (\frac {\sqrt {x}}{\sqrt {-1+\sqrt {2}}}\right )+\sqrt {2 \left (1+\sqrt {2}\right )} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt {1+\sqrt {2}}}\right ) \] Output:
-(-2+2*2^(1/2))^(1/2)*arctan(x^(1/2)/(2^(1/2)-1)^(1/2))+(2+2*2^(1/2))^(1/2 )*arctanh(x^(1/2)/(1+2^(1/2))^(1/2))
Time = 0.14 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.09 \[ \int \frac {2 \sqrt {x}}{1+2 x-x^2} \, dx=2 \left (-\sqrt {\frac {1}{2} \left (-1+\sqrt {2}\right )} \arctan \left (\sqrt {1+\sqrt {2}} \sqrt {x}\right )+\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \text {arctanh}\left (\sqrt {-1+\sqrt {2}} \sqrt {x}\right )\right ) \] Input:
Integrate[(2*Sqrt[x])/(1 + 2*x - x^2),x]
Output:
2*(-(Sqrt[(-1 + Sqrt[2])/2]*ArcTan[Sqrt[1 + Sqrt[2]]*Sqrt[x]]) + Sqrt[(1 + Sqrt[2])/2]*ArcTanh[Sqrt[-1 + Sqrt[2]]*Sqrt[x]])
Time = 0.36 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.29, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {27, 1148, 1450, 217, 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.
\(\displaystyle \int \frac {2 \sqrt {x}}{-x^2+2 x+1} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {\sqrt {x}}{-x^2+2 x+1}dx\) |
\(\Big \downarrow \) 1148 |
\(\displaystyle 4 \int \frac {x}{-x^2+2 x+1}d\sqrt {x}\) |
\(\Big \downarrow \) 1450 |
\(\displaystyle 4 \left (\frac {1}{4} \left (2-\sqrt {2}\right ) \int \frac {1}{-x-\sqrt {2}+1}d\sqrt {x}+\frac {1}{4} \left (2+\sqrt {2}\right ) \int \frac {1}{-x+\sqrt {2}+1}d\sqrt {x}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle 4 \left (\frac {1}{4} \left (2+\sqrt {2}\right ) \int \frac {1}{-x+\sqrt {2}+1}d\sqrt {x}-\frac {\left (2-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {x}}{\sqrt {\sqrt {2}-1}}\right )}{4 \sqrt {\sqrt {2}-1}}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 4 \left (\frac {\left (2+\sqrt {2}\right ) \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt {1+\sqrt {2}}}\right )}{4 \sqrt {1+\sqrt {2}}}-\frac {\left (2-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {x}}{\sqrt {\sqrt {2}-1}}\right )}{4 \sqrt {\sqrt {2}-1}}\right )\) |
Input:
Int[(2*Sqrt[x])/(1 + 2*x - x^2),x]
Output:
4*(-1/4*((2 - Sqrt[2])*ArcTan[Sqrt[x]/Sqrt[-1 + Sqrt[2]]])/Sqrt[-1 + Sqrt[ 2]] + ((2 + Sqrt[2])*ArcTanh[Sqrt[x]/Sqrt[1 + Sqrt[2]]])/(4*Sqrt[1 + Sqrt[ 2]]))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
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[Sqrt[(d_.) + (e_.)*(x_)]/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[2*e Subst[Int[x^2/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c *x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x]
Int[((d_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Wi th[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(d^2/2)*(b/q + 1) Int[(d*x)^(m - 2)/(b/ 2 + q/2 + c*x^2), x], x] - Simp[(d^2/2)*(b/q - 1) Int[(d*x)^(m - 2)/(b/2 - q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - 4*a*c, 0] && GeQ[m, 2]
Time = 2.06 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.74
method | result | size |
derivativedivides | \(-\sqrt {\sqrt {2}-1}\, \sqrt {2}\, \arctan \left (\frac {\sqrt {x}}{\sqrt {\sqrt {2}-1}}\right )+\sqrt {1+\sqrt {2}}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {x}}{\sqrt {1+\sqrt {2}}}\right )\) | \(49\) |
default | \(-\sqrt {\sqrt {2}-1}\, \sqrt {2}\, \arctan \left (\frac {\sqrt {x}}{\sqrt {\sqrt {2}-1}}\right )+\sqrt {1+\sqrt {2}}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {x}}{\sqrt {1+\sqrt {2}}}\right )\) | \(49\) |
trager | \(2 \operatorname {RootOf}\left (64 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}-1\right ) \ln \left (\frac {-8 \operatorname {RootOf}\left (64 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}-1\right )^{3} x +8 \operatorname {RootOf}\left (64 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}-1\right )^{3}+3 \operatorname {RootOf}\left (64 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}-1\right ) x +\sqrt {x}-\operatorname {RootOf}\left (64 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}-1\right )}{8 \operatorname {RootOf}\left (64 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}-1\right )^{2} x -8 \operatorname {RootOf}\left (64 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}-1\right )^{2}-x -1}\right )+\operatorname {RootOf}\left (\textit {\_Z}^{2}+4 \operatorname {RootOf}\left (64 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}-1\right )^{2}-1\right ) \ln \left (\frac {8 \operatorname {RootOf}\left (\textit {\_Z}^{2}+4 \operatorname {RootOf}\left (64 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}-1\right )^{2}-1\right ) x \operatorname {RootOf}\left (64 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}-1\right )^{2}-8 \operatorname {RootOf}\left (64 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}-1\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+4 \operatorname {RootOf}\left (64 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}-1\right )^{2}-1\right )+\operatorname {RootOf}\left (\textit {\_Z}^{2}+4 \operatorname {RootOf}\left (64 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}-1\right )^{2}-1\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{2}+4 \operatorname {RootOf}\left (64 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}-1\right )^{2}-1\right )+2 \sqrt {x}}{8 \operatorname {RootOf}\left (64 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}-1\right )^{2} x -8 \operatorname {RootOf}\left (64 \textit {\_Z}^{4}-16 \textit {\_Z}^{2}-1\right )^{2}-x +3}\right )\) | \(334\) |
Input:
int(2*x^(1/2)/(-x^2+2*x+1),x,method=_RETURNVERBOSE)
Output:
-(2^(1/2)-1)^(1/2)*2^(1/2)*arctan(x^(1/2)/(2^(1/2)-1)^(1/2))+(1+2^(1/2))^( 1/2)*2^(1/2)*arctanh(x^(1/2)/(1+2^(1/2))^(1/2))
Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.35 \[ \int \frac {2 \sqrt {x}}{1+2 x-x^2} \, dx=-2 \, \sqrt {\frac {1}{2} \, \sqrt {2} - \frac {1}{2}} \arctan \left (\sqrt {x} {\left (\sqrt {2} + 2\right )} \sqrt {\frac {1}{2} \, \sqrt {2} - \frac {1}{2}}\right ) + \sqrt {\frac {1}{2} \, \sqrt {2} + \frac {1}{2}} \log \left (\sqrt {2} \sqrt {\frac {1}{2} \, \sqrt {2} + \frac {1}{2}} + \sqrt {x}\right ) - \sqrt {\frac {1}{2} \, \sqrt {2} + \frac {1}{2}} \log \left (-\sqrt {2} \sqrt {\frac {1}{2} \, \sqrt {2} + \frac {1}{2}} + \sqrt {x}\right ) \] Input:
integrate(2*x^(1/2)/(-x^2+2*x+1),x, algorithm="fricas")
Output:
-2*sqrt(1/2*sqrt(2) - 1/2)*arctan(sqrt(x)*(sqrt(2) + 2)*sqrt(1/2*sqrt(2) - 1/2)) + sqrt(1/2*sqrt(2) + 1/2)*log(sqrt(2)*sqrt(1/2*sqrt(2) + 1/2) + sqr t(x)) - sqrt(1/2*sqrt(2) + 1/2)*log(-sqrt(2)*sqrt(1/2*sqrt(2) + 1/2) + sqr t(x))
Leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (56) = 112\).
Time = 1.95 (sec) , antiderivative size = 340, normalized size of antiderivative = 5.15 \[ \int \frac {2 \sqrt {x}}{1+2 x-x^2} \, dx=\frac {4 \log {\left (\sqrt {x} - \sqrt {1 + \sqrt {2}} \right )}}{- 12 \sqrt {1 + \sqrt {2}} + 8 \sqrt {2} \sqrt {1 + \sqrt {2}}} - \frac {2 \sqrt {2} \log {\left (\sqrt {x} - \sqrt {1 + \sqrt {2}} \right )}}{- 12 \sqrt {1 + \sqrt {2}} + 8 \sqrt {2} \sqrt {1 + \sqrt {2}}} + \frac {2 \sqrt {2} \log {\left (\sqrt {x} + \sqrt {1 + \sqrt {2}} \right )}}{- 12 \sqrt {1 + \sqrt {2}} + 8 \sqrt {2} \sqrt {1 + \sqrt {2}}} - \frac {4 \log {\left (\sqrt {x} + \sqrt {1 + \sqrt {2}} \right )}}{- 12 \sqrt {1 + \sqrt {2}} + 8 \sqrt {2} \sqrt {1 + \sqrt {2}}} + \frac {12 \sqrt {2} \sqrt {-1 + \sqrt {2}} \sqrt {1 + \sqrt {2}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt {-1 + \sqrt {2}}} \right )}}{- 12 \sqrt {1 + \sqrt {2}} + 8 \sqrt {2} \sqrt {1 + \sqrt {2}}} - \frac {16 \sqrt {-1 + \sqrt {2}} \sqrt {1 + \sqrt {2}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt {-1 + \sqrt {2}}} \right )}}{- 12 \sqrt {1 + \sqrt {2}} + 8 \sqrt {2} \sqrt {1 + \sqrt {2}}} \] Input:
integrate(2*x**(1/2)/(-x**2+2*x+1),x)
Output:
4*log(sqrt(x) - sqrt(1 + sqrt(2)))/(-12*sqrt(1 + sqrt(2)) + 8*sqrt(2)*sqrt (1 + sqrt(2))) - 2*sqrt(2)*log(sqrt(x) - sqrt(1 + sqrt(2)))/(-12*sqrt(1 + sqrt(2)) + 8*sqrt(2)*sqrt(1 + sqrt(2))) + 2*sqrt(2)*log(sqrt(x) + sqrt(1 + sqrt(2)))/(-12*sqrt(1 + sqrt(2)) + 8*sqrt(2)*sqrt(1 + sqrt(2))) - 4*log(s qrt(x) + sqrt(1 + sqrt(2)))/(-12*sqrt(1 + sqrt(2)) + 8*sqrt(2)*sqrt(1 + sq rt(2))) + 12*sqrt(2)*sqrt(-1 + sqrt(2))*sqrt(1 + sqrt(2))*atan(sqrt(x)/sqr t(-1 + sqrt(2)))/(-12*sqrt(1 + sqrt(2)) + 8*sqrt(2)*sqrt(1 + sqrt(2))) - 1 6*sqrt(-1 + sqrt(2))*sqrt(1 + sqrt(2))*atan(sqrt(x)/sqrt(-1 + sqrt(2)))/(- 12*sqrt(1 + sqrt(2)) + 8*sqrt(2)*sqrt(1 + sqrt(2)))
\[ \int \frac {2 \sqrt {x}}{1+2 x-x^2} \, dx=\int { -\frac {2 \, \sqrt {x}}{x^{2} - 2 \, x - 1} \,d x } \] Input:
integrate(2*x^(1/2)/(-x^2+2*x+1),x, algorithm="maxima")
Output:
-2*integrate(sqrt(x)/(x^2 - 2*x - 1), x)
Time = 0.21 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.11 \[ \int \frac {2 \sqrt {x}}{1+2 x-x^2} \, dx=-\sqrt {2 \, \sqrt {2} - 2} \arctan \left (\frac {\sqrt {x}}{\sqrt {\sqrt {2} - 1}}\right ) + \frac {1}{2} \, \sqrt {2 \, \sqrt {2} + 2} \log \left (\sqrt {x} + \sqrt {\sqrt {2} + 1}\right ) - \frac {1}{2} \, \sqrt {2 \, \sqrt {2} + 2} \log \left ({\left | \sqrt {x} - \sqrt {\sqrt {2} + 1} \right |}\right ) \] Input:
integrate(2*x^(1/2)/(-x^2+2*x+1),x, algorithm="giac")
Output:
-sqrt(2*sqrt(2) - 2)*arctan(sqrt(x)/sqrt(sqrt(2) - 1)) + 1/2*sqrt(2*sqrt(2 ) + 2)*log(sqrt(x) + sqrt(sqrt(2) + 1)) - 1/2*sqrt(2*sqrt(2) + 2)*log(abs( sqrt(x) - sqrt(sqrt(2) + 1)))
Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.30 \[ \int \frac {2 \sqrt {x}}{1+2 x-x^2} \, dx=\sqrt {2}\,\mathrm {atanh}\left (\sqrt {2}\,\sqrt {1-\sqrt {2}}\,\left (\sqrt {x}\,\left (\frac {\sqrt {2}}{2}-\frac {1}{2}\right )+\frac {3\,\sqrt {x}}{2}\right )\right )\,\sqrt {1-\sqrt {2}}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {-2\,\sqrt {2}-2}\,\left (\sqrt {x}\,\left (\frac {\sqrt {2}}{2}+\frac {1}{2}\right )-\frac {3\,\sqrt {x}}{2}\right )\right )\,\sqrt {\sqrt {2}+1}\,1{}\mathrm {i} \] Input:
int((2*x^(1/2))/(2*x - x^2 + 1),x)
Output:
2^(1/2)*atan((- 2*2^(1/2) - 2)^(1/2)*(x^(1/2)*(2^(1/2)/2 + 1/2) - (3*x^(1/ 2))/2))*(2^(1/2) + 1)^(1/2)*1i + 2^(1/2)*atanh(2^(1/2)*(1 - 2^(1/2))^(1/2) *(x^(1/2)*(2^(1/2)/2 - 1/2) + (3*x^(1/2))/2))*(1 - 2^(1/2))^(1/2)
Time = 0.19 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.85 \[ \int \frac {2 \sqrt {x}}{1+2 x-x^2} \, dx=\frac {\sqrt {2}\, \left (-2 \sqrt {\sqrt {2}-1}\, \mathit {atan} \left (\frac {\sqrt {x}}{\sqrt {\sqrt {2}-1}}\right )-\sqrt {\sqrt {2}+1}\, \mathrm {log}\left (-\sqrt {\sqrt {2}+1}+\sqrt {x}\right )+\sqrt {\sqrt {2}+1}\, \mathrm {log}\left (\sqrt {\sqrt {2}+1}+\sqrt {x}\right )\right )}{2} \] Input:
int(2*x^(1/2)/(-x^2+2*x+1),x)
Output:
(sqrt(2)*( - 2*sqrt(sqrt(2) - 1)*atan(sqrt(x)/sqrt(sqrt(2) - 1)) - sqrt(sq rt(2) + 1)*log( - sqrt(sqrt(2) + 1) + sqrt(x)) + sqrt(sqrt(2) + 1)*log(sqr t(sqrt(2) + 1) + sqrt(x))))/2