Integrand size = 18, antiderivative size = 61 \[ \int \frac {\sqrt {-1-x+x^2}}{1+x} \, dx=\sqrt {-1-x+x^2}+\frac {3}{2} \text {arctanh}\left (\frac {1-2 x}{2 \sqrt {-1-x+x^2}}\right )+\text {arctanh}\left (\frac {1+3 x}{2 \sqrt {-1-x+x^2}}\right ) \] Output:
(x^2-x-1)^(1/2)+3/2*arctanh(1/2*(1-2*x)/(x^2-x-1)^(1/2))+arctanh(1/2*(1+3* x)/(x^2-x-1)^(1/2))
Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {-1-x+x^2}}{1+x} \, dx=\sqrt {-1-x+x^2}+2 \text {arctanh}\left (1+x-\sqrt {-1-x+x^2}\right )+\frac {3}{2} \log \left (1-2 x+2 \sqrt {-1-x+x^2}\right ) \] Input:
Integrate[Sqrt[-1 - x + x^2]/(1 + x),x]
Output:
Sqrt[-1 - x + x^2] + 2*ArcTanh[1 + x - Sqrt[-1 - x + x^2]] + (3*Log[1 - 2* x + 2*Sqrt[-1 - x + x^2]])/2
Time = 0.24 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1162, 1269, 1092, 219, 1154, 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.
\(\displaystyle \int \frac {\sqrt {x^2-x-1}}{x+1} \, dx\) |
\(\Big \downarrow \) 1162 |
\(\displaystyle \sqrt {x^2-x-1}-\frac {1}{2} \int \frac {3 x+1}{(x+1) \sqrt {x^2-x-1}}dx\) |
\(\Big \downarrow \) 1269 |
\(\displaystyle \frac {1}{2} \left (2 \int \frac {1}{(x+1) \sqrt {x^2-x-1}}dx-3 \int \frac {1}{\sqrt {x^2-x-1}}dx\right )+\sqrt {x^2-x-1}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {1}{2} \left (2 \int \frac {1}{(x+1) \sqrt {x^2-x-1}}dx-6 \int \frac {1}{4-\frac {(1-2 x)^2}{x^2-x-1}}d\left (-\frac {1-2 x}{\sqrt {x^2-x-1}}\right )\right )+\sqrt {x^2-x-1}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (2 \int \frac {1}{(x+1) \sqrt {x^2-x-1}}dx+3 \text {arctanh}\left (\frac {1-2 x}{2 \sqrt {x^2-x-1}}\right )\right )+\sqrt {x^2-x-1}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {1}{2} \left (3 \text {arctanh}\left (\frac {1-2 x}{2 \sqrt {x^2-x-1}}\right )-4 \int \frac {1}{4-\frac {(3 x+1)^2}{x^2-x-1}}d\left (-\frac {3 x+1}{\sqrt {x^2-x-1}}\right )\right )+\sqrt {x^2-x-1}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (3 \text {arctanh}\left (\frac {1-2 x}{2 \sqrt {x^2-x-1}}\right )+2 \text {arctanh}\left (\frac {3 x+1}{2 \sqrt {x^2-x-1}}\right )\right )+\sqrt {x^2-x-1}\) |
Input:
Int[Sqrt[-1 - x + x^2]/(1 + x),x]
Output:
Sqrt[-1 - x + x^2] + (3*ArcTanh[(1 - 2*x)/(2*Sqrt[-1 - x + x^2])] + 2*ArcT anh[(1 + 3*x)/(2*Sqrt[-1 - x + x^2])])/2
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[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x ] - Simp[p/(e*(m + 2*p + 1)) Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x ] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || LtQ[m, 1]) && !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + b*x + c*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
Time = 0.85 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.82
method | result | size |
risch | \(\sqrt {x^{2}-x -1}-\frac {3 \ln \left (x -\frac {1}{2}+\sqrt {x^{2}-x -1}\right )}{2}-\operatorname {arctanh}\left (\frac {-3 x -1}{2 \sqrt {\left (x +1\right )^{2}-3 x -2}}\right )\) | \(50\) |
default | \(\sqrt {\left (x +1\right )^{2}-3 x -2}-\frac {3 \ln \left (-\frac {1}{2}+x +\sqrt {\left (x +1\right )^{2}-3 x -2}\right )}{2}-\operatorname {arctanh}\left (\frac {-3 x -1}{2 \sqrt {\left (x +1\right )^{2}-3 x -2}}\right )\) | \(54\) |
trager | \(\sqrt {x^{2}-x -1}-\frac {\ln \left (\frac {32 \sqrt {x^{2}-x -1}\, x^{4}+32 x^{5}+96 \sqrt {x^{2}-x -1}\, x^{3}+80 x^{4}+78 \sqrt {x^{2}-x -1}\, x^{2}+10 x^{3}-16 \sqrt {x^{2}-x -1}\, x -125 x^{2}-38 \sqrt {x^{2}-x -1}-120 x -41}{\left (x +1\right )^{2}}\right )}{2}\) | \(116\) |
Input:
int((x^2-x-1)^(1/2)/(x+1),x,method=_RETURNVERBOSE)
Output:
(x^2-x-1)^(1/2)-3/2*ln(x-1/2+(x^2-x-1)^(1/2))-arctanh(1/2*(-3*x-1)/((x+1)^ 2-3*x-2)^(1/2))
Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.05 \[ \int \frac {\sqrt {-1-x+x^2}}{1+x} \, dx=\sqrt {x^{2} - x - 1} - \log \left (-x + \sqrt {x^{2} - x - 1}\right ) + \log \left (-x + \sqrt {x^{2} - x - 1} - 2\right ) + \frac {3}{2} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x - 1} + 1\right ) \] Input:
integrate((x^2-x-1)^(1/2)/(1+x),x, algorithm="fricas")
Output:
sqrt(x^2 - x - 1) - log(-x + sqrt(x^2 - x - 1)) + log(-x + sqrt(x^2 - x - 1) - 2) + 3/2*log(-2*x + 2*sqrt(x^2 - x - 1) + 1)
\[ \int \frac {\sqrt {-1-x+x^2}}{1+x} \, dx=\int \frac {\sqrt {x^{2} - x - 1}}{x + 1}\, dx \] Input:
integrate((x**2-x-1)**(1/2)/(1+x),x)
Output:
Integral(sqrt(x**2 - x - 1)/(x + 1), x)
Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {-1-x+x^2}}{1+x} \, dx=\sqrt {x^{2} - x - 1} - \frac {3}{2} \, \log \left (2 \, x + 2 \, \sqrt {x^{2} - x - 1} - 1\right ) - \log \left (\frac {2 \, \sqrt {x^{2} - x - 1}}{{\left | x + 1 \right |}} + \frac {2}{{\left | x + 1 \right |}} - 3\right ) \] Input:
integrate((x^2-x-1)^(1/2)/(1+x),x, algorithm="maxima")
Output:
sqrt(x^2 - x - 1) - 3/2*log(2*x + 2*sqrt(x^2 - x - 1) - 1) - log(2*sqrt(x^ 2 - x - 1)/abs(x + 1) + 2/abs(x + 1) - 3)
Time = 0.16 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.10 \[ \int \frac {\sqrt {-1-x+x^2}}{1+x} \, dx=\sqrt {x^{2} - x - 1} - \log \left ({\left | -x + \sqrt {x^{2} - x - 1} \right |}\right ) + \log \left ({\left | -x + \sqrt {x^{2} - x - 1} - 2 \right |}\right ) + \frac {3}{2} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x - 1} + 1 \right |}\right ) \] Input:
integrate((x^2-x-1)^(1/2)/(1+x),x, algorithm="giac")
Output:
sqrt(x^2 - x - 1) - log(abs(-x + sqrt(x^2 - x - 1))) + log(abs(-x + sqrt(x ^2 - x - 1) - 2)) + 3/2*log(abs(-2*x + 2*sqrt(x^2 - x - 1) + 1))
Timed out. \[ \int \frac {\sqrt {-1-x+x^2}}{1+x} \, dx=\int \frac {\sqrt {x^2-x-1}}{x+1} \,d x \] Input:
int((x^2 - x - 1)^(1/2)/(x + 1),x)
Output:
int((x^2 - x - 1)^(1/2)/(x + 1), x)
Time = 0.70 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.30 \[ \int \frac {\sqrt {-1-x+x^2}}{1+x} \, dx=\sqrt {x^{2}-x -1}+\mathrm {log}\left (\frac {10 \sqrt {x^{2}-x -1}+10 x}{\sqrt {5}}\right )-\frac {3 \,\mathrm {log}\left (\frac {2 \sqrt {x^{2}-x -1}+2 x -1}{\sqrt {5}}\right )}{2}-\mathrm {log}\left (\frac {2 \sqrt {x^{2}-x -1}+2 x +4}{\sqrt {5}}\right ) \] Input:
int((x^2-x-1)^(1/2)/(1+x),x)
Output:
(2*sqrt(x**2 - x - 1) + 2*log((10*sqrt(x**2 - x - 1) + 10*x)/sqrt(5)) - 3* log((2*sqrt(x**2 - x - 1) + 2*x - 1)/sqrt(5)) - 2*log((2*sqrt(x**2 - x - 1 ) + 2*x + 4)/sqrt(5)))/2