Integrand size = 16, antiderivative size = 65 \[ \int \frac {1}{x+\sqrt {-3-2 x+x^2}} \, dx=-\frac {2}{1-x-\sqrt {-3-2 x+x^2}}+2 \log \left (1-x-\sqrt {-3-2 x+x^2}\right )-\frac {3}{2} \log \left (x+\sqrt {-3-2 x+x^2}\right ) \] Output:
-2/(1-x-(x^2-2*x-3)^(1/2))+2*ln(1-x-(x^2-2*x-3)^(1/2))-3/2*ln(x+(x^2-2*x-3 )^(1/2))
Time = 0.32 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.20 \[ \int \frac {1}{x+\sqrt {-3-2 x+x^2}} \, dx=\frac {1}{2} \left (x-\sqrt {-3-2 x+x^2}-\log \left (-1-x+\sqrt {-3-2 x+x^2}\right )+4 \log \left (1+x+\sqrt {-3-2 x+x^2}\right )-3 \log \left (3+3 x+\sqrt {-3-2 x+x^2}\right )\right ) \] Input:
Integrate[(x + Sqrt[-3 - 2*x + x^2])^(-1),x]
Output:
(x - Sqrt[-3 - 2*x + x^2] - Log[-1 - x + Sqrt[-3 - 2*x + x^2]] + 4*Log[1 + x + Sqrt[-3 - 2*x + x^2]] - 3*Log[3 + 3*x + Sqrt[-3 - 2*x + x^2]])/2
Time = 0.22 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2541, 27, 1195, 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.
\(\displaystyle \int \frac {1}{\sqrt {x^2-2 x-3}+x} \, dx\) |
\(\Big \downarrow \) 2541 |
\(\displaystyle 2 \int -\frac {-\left (x+\sqrt {x^2-2 x-3}\right )^2+2 \left (x+\sqrt {x^2-2 x-3}\right )+3}{4 \left (-x-\sqrt {x^2-2 x-3}+1\right )^2 \left (x+\sqrt {x^2-2 x-3}\right )}d\left (x+\sqrt {x^2-2 x-3}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{2} \int \frac {-\left (x+\sqrt {x^2-2 x-3}\right )^2+2 \left (x+\sqrt {x^2-2 x-3}\right )+3}{\left (-x-\sqrt {x^2-2 x-3}+1\right )^2 \left (x+\sqrt {x^2-2 x-3}\right )}d\left (x+\sqrt {x^2-2 x-3}\right )\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {3}{x+\sqrt {x^2-2 x-3}}-\frac {4}{x+\sqrt {x^2-2 x-3}-1}+\frac {4}{\left (x+\sqrt {x^2-2 x-3}-1\right )^2}\right )d\left (x+\sqrt {x^2-2 x-3}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {4}{-\sqrt {x^2-2 x-3}-x+1}+4 \log \left (-\sqrt {x^2-2 x-3}-x+1\right )-3 \log \left (\sqrt {x^2-2 x-3}+x\right )\right )\) |
Input:
Int[(x + Sqrt[-3 - 2*x + x^2])^(-1),x]
Output:
(-4/(1 - x - Sqrt[-3 - 2*x + x^2]) + 4*Log[1 - x - Sqrt[-3 - 2*x + x^2]] - 3*Log[x + Sqrt[-3 - 2*x + x^2]])/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c _.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Simp[2 Subst[Int[(g + h*x^n)^p*((d ^2*e - (b*d - a*e)*f^2 - (2*d*e - b*f^2)*x + e*x^2)/(-2*d*e + b*f^2 + 2*e*x )^2), x], x, d + e*x + f*Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e , f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]
Time = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.09
method | result | size |
default | \(-\frac {\sqrt {4 \left (x +\frac {3}{2}\right )^{2}-20 x -21}}{4}+\frac {5 \ln \left (-1+x +\sqrt {\left (x +\frac {3}{2}\right )^{2}-5 x -\frac {21}{4}}\right )}{4}+\frac {3 \,\operatorname {arctanh}\left (\frac {-2-\frac {10 x}{3}}{\sqrt {4 \left (x +\frac {3}{2}\right )^{2}-20 x -21}}\right )}{4}+\frac {x}{2}-\frac {3 \ln \left (2 x +3\right )}{4}\) | \(71\) |
trager | \(\frac {x}{2}-\frac {\sqrt {x^{2}-2 x -3}}{2}-\frac {\ln \left (\sqrt {x^{2}-2 x -3}\, x^{3}-x^{4}+3 x^{2} \sqrt {x^{2}-2 x -3}-2 x^{3}+\sqrt {x^{2}-2 x -3}\, x +4 x^{2}-3 \sqrt {x^{2}-2 x -3}+12 x +9\right )}{2}\) | \(93\) |
Input:
int(1/(x+(x^2-2*x-3)^(1/2)),x,method=_RETURNVERBOSE)
Output:
-1/4*(4*(x+3/2)^2-20*x-21)^(1/2)+5/4*ln(-1+x+((x+3/2)^2-5*x-21/4)^(1/2))+3 /4*arctanh(2/3*(-3-5*x)/(4*(x+3/2)^2-20*x-21)^(1/2))+1/2*x-3/4*ln(2*x+3)
Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.18 \[ \int \frac {1}{x+\sqrt {-3-2 x+x^2}} \, dx=\frac {1}{2} \, x - \frac {1}{2} \, \sqrt {x^{2} - 2 \, x - 3} - \frac {3}{4} \, \log \left (2 \, x + 3\right ) - \frac {5}{4} \, \log \left (-x + \sqrt {x^{2} - 2 \, x - 3} + 1\right ) + \frac {3}{4} \, \log \left (-x + \sqrt {x^{2} - 2 \, x - 3}\right ) - \frac {3}{4} \, \log \left (-x + \sqrt {x^{2} - 2 \, x - 3} - 3\right ) \] Input:
integrate(1/(x+(x^2-2*x-3)^(1/2)),x, algorithm="fricas")
Output:
1/2*x - 1/2*sqrt(x^2 - 2*x - 3) - 3/4*log(2*x + 3) - 5/4*log(-x + sqrt(x^2 - 2*x - 3) + 1) + 3/4*log(-x + sqrt(x^2 - 2*x - 3)) - 3/4*log(-x + sqrt(x ^2 - 2*x - 3) - 3)
\[ \int \frac {1}{x+\sqrt {-3-2 x+x^2}} \, dx=\int \frac {1}{x + \sqrt {x^{2} - 2 x - 3}}\, dx \] Input:
integrate(1/(x+(x**2-2*x-3)**(1/2)),x)
Output:
Integral(1/(x + sqrt(x**2 - 2*x - 3)), x)
\[ \int \frac {1}{x+\sqrt {-3-2 x+x^2}} \, dx=\int { \frac {1}{x + \sqrt {x^{2} - 2 \, x - 3}} \,d x } \] Input:
integrate(1/(x+(x^2-2*x-3)^(1/2)),x, algorithm="maxima")
Output:
integrate(1/(x + sqrt(x^2 - 2*x - 3)), x)
Time = 0.13 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.25 \[ \int \frac {1}{x+\sqrt {-3-2 x+x^2}} \, dx=\frac {1}{2} \, x - \frac {1}{2} \, \sqrt {x^{2} - 2 \, x - 3} - \frac {3}{4} \, \log \left ({\left | 2 \, x + 3 \right |}\right ) - \frac {5}{4} \, \log \left ({\left | -x + \sqrt {x^{2} - 2 \, x - 3} + 1 \right |}\right ) + \frac {3}{4} \, \log \left ({\left | -x + \sqrt {x^{2} - 2 \, x - 3} \right |}\right ) - \frac {3}{4} \, \log \left ({\left | -x + \sqrt {x^{2} - 2 \, x - 3} - 3 \right |}\right ) \] Input:
integrate(1/(x+(x^2-2*x-3)^(1/2)),x, algorithm="giac")
Output:
1/2*x - 1/2*sqrt(x^2 - 2*x - 3) - 3/4*log(abs(2*x + 3)) - 5/4*log(abs(-x + sqrt(x^2 - 2*x - 3) + 1)) + 3/4*log(abs(-x + sqrt(x^2 - 2*x - 3))) - 3/4* log(abs(-x + sqrt(x^2 - 2*x - 3) - 3))
Timed out. \[ \int \frac {1}{x+\sqrt {-3-2 x+x^2}} \, dx=\frac {x}{2}-\frac {3\,\ln \left (x+\frac {3}{2}\right )}{4}-\int \frac {\sqrt {x^2-2\,x-3}}{2\,x+3} \,d x \] Input:
int(1/(x + (x^2 - 2*x - 3)^(1/2)),x)
Output:
x/2 - (3*log(x + 3/2))/4 - int((x^2 - 2*x - 3)^(1/2)/(2*x + 3), x)
Time = 0.16 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x+\sqrt {-3-2 x+x^2}} \, dx=-\frac {\sqrt {x^{2}-2 x -3}}{2}-\frac {3 \,\mathrm {log}\left (\sqrt {x^{2}-2 x -3}+x \right )}{2}+2 \,\mathrm {log}\left (\frac {\sqrt {x^{2}-2 x -3}}{2}+\frac {x}{2}-\frac {1}{2}\right )+\frac {x}{2}-\frac {1}{2} \] Input:
int(1/(x+(x^2-2*x-3)^(1/2)),x)
Output:
( - sqrt(x**2 - 2*x - 3) - 3*log(sqrt(x**2 - 2*x - 3) + x) + 4*log((sqrt(x **2 - 2*x - 3) + x - 1)/2) + x - 1)/2