3.6.76 \(\int \frac {\sqrt {-1+x} \sqrt {1+x}}{1+x-x^2} \, dx\)

Optimal. Leaf size=91 \[ \sqrt {\frac {2}{5} \left (\sqrt {5}-1\right )} \tan ^{-1}\left (\frac {\sqrt {x+1}}{\sqrt {\sqrt {5}-2} \sqrt {x-1}}\right )-\cosh ^{-1}(x)+\sqrt {\frac {2}{5} \left (1+\sqrt {5}\right )} \tanh ^{-1}\left (\frac {\sqrt {x+1}}{\sqrt {2+\sqrt {5}} \sqrt {x-1}}\right ) \]

________________________________________________________________________________________

Rubi [B]  time = 0.14, antiderivative size = 191, normalized size of antiderivative = 2.10, number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {901, 991, 217, 206, 1034, 725, 204} \begin {gather*} \frac {\sqrt {\frac {1}{10} \left (\sqrt {5}-1\right )} \sqrt {x-1} \sqrt {x+1} \tan ^{-1}\left (\frac {2-\left (1-\sqrt {5}\right ) x}{\sqrt {2 \left (\sqrt {5}-1\right )} \sqrt {x^2-1}}\right )}{\sqrt {x^2-1}}-\frac {\sqrt {x-1} \sqrt {x+1} \tanh ^{-1}\left (\frac {x}{\sqrt {x^2-1}}\right )}{\sqrt {x^2-1}}-\frac {\sqrt {\frac {1}{10} \left (1+\sqrt {5}\right )} \sqrt {x-1} \sqrt {x+1} \tanh ^{-1}\left (\frac {2-\left (1+\sqrt {5}\right ) x}{\sqrt {2 \left (1+\sqrt {5}\right )} \sqrt {x^2-1}}\right )}{\sqrt {x^2-1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[-1 + x]*Sqrt[1 + x])/(1 + x - x^2),x]

[Out]

(Sqrt[(-1 + Sqrt[5])/10]*Sqrt[-1 + x]*Sqrt[1 + x]*ArcTan[(2 - (1 - Sqrt[5])*x)/(Sqrt[2*(-1 + Sqrt[5])]*Sqrt[-1
+ x^2])])/Sqrt[-1 + x^2] - (Sqrt[-1 + x]*Sqrt[1 + x]*ArcTanh[x/Sqrt[-1 + x^2]])/Sqrt[-1 + x^2] - (Sqrt[(1 + S
qrt[5])/10]*Sqrt[-1 + x]*Sqrt[1 + x]*ArcTanh[(2 - (1 + Sqrt[5])*x)/(Sqrt[2*(1 + Sqrt[5])]*Sqrt[-1 + x^2])])/Sq
rt[-1 + x^2]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 901

Int[((d_) + (e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :>
Dist[((d + e*x)^FracPart[m]*(f + g*x)^FracPart[m])/(d*f + e*g*x^2)^FracPart[m], Int[(d*f + e*g*x^2)^m*(a + b*x
+ c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[m - n, 0] && EqQ[e*f + d*g, 0]

Rule 991

Int[Sqrt[(a_) + (c_.)*(x_)^2]/((d_) + (e_.)*(x_) + (f_.)*(x_)^2), x_Symbol] :> Dist[c/f, Int[1/Sqrt[a + c*x^2]
, x], x] - Dist[1/f, Int[(c*d - a*f + c*e*x)/(Sqrt[a + c*x^2]*(d + e*x + f*x^2)), x], x] /; FreeQ[{a, c, d, e,
f}, x] && NeQ[e^2 - 4*d*f, 0]

Rule 1034

Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[b^2 - 4*a*c, 2]}, Dist[(2*c*g - h*(b - q))/q, Int[1/((b - q + 2*c*x)*Sqrt[d + f*x^2]), x], x] - Dist[(2*c
*g - h*(b + q))/q, Int[1/((b + q + 2*c*x)*Sqrt[d + f*x^2]), x], x]] /; FreeQ[{a, b, c, d, f, g, h}, x] && NeQ[
b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c]

Rubi steps

\begin {align*} \int \frac {\sqrt {-1+x} \sqrt {1+x}}{1+x-x^2} \, dx &=\frac {\left (\sqrt {-1+x} \sqrt {1+x}\right ) \int \frac {\sqrt {-1+x^2}}{1+x-x^2} \, dx}{\sqrt {-1+x^2}}\\ &=-\frac {\left (\sqrt {-1+x} \sqrt {1+x}\right ) \int \frac {1}{\sqrt {-1+x^2}} \, dx}{\sqrt {-1+x^2}}+\frac {\left (\sqrt {-1+x} \sqrt {1+x}\right ) \int \frac {x}{\left (1+x-x^2\right ) \sqrt {-1+x^2}} \, dx}{\sqrt {-1+x^2}}\\ &=-\frac {\left (\sqrt {-1+x} \sqrt {1+x}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-1+x^2}}\right )}{\sqrt {-1+x^2}}+\frac {\left (\left (5-\sqrt {5}\right ) \sqrt {-1+x} \sqrt {1+x}\right ) \int \frac {1}{\left (1-\sqrt {5}-2 x\right ) \sqrt {-1+x^2}} \, dx}{5 \sqrt {-1+x^2}}+\frac {\left (\left (5+\sqrt {5}\right ) \sqrt {-1+x} \sqrt {1+x}\right ) \int \frac {1}{\left (1+\sqrt {5}-2 x\right ) \sqrt {-1+x^2}} \, dx}{5 \sqrt {-1+x^2}}\\ &=-\frac {\sqrt {-1+x} \sqrt {1+x} \tanh ^{-1}\left (\frac {x}{\sqrt {-1+x^2}}\right )}{\sqrt {-1+x^2}}-\frac {\left (\left (5-\sqrt {5}\right ) \sqrt {-1+x} \sqrt {1+x}\right ) \operatorname {Subst}\left (\int \frac {1}{-4+\left (1-\sqrt {5}\right )^2-x^2} \, dx,x,\frac {2-\left (1-\sqrt {5}\right ) x}{\sqrt {-1+x^2}}\right )}{5 \sqrt {-1+x^2}}-\frac {\left (\left (5+\sqrt {5}\right ) \sqrt {-1+x} \sqrt {1+x}\right ) \operatorname {Subst}\left (\int \frac {1}{-4+\left (1+\sqrt {5}\right )^2-x^2} \, dx,x,\frac {2-\left (1+\sqrt {5}\right ) x}{\sqrt {-1+x^2}}\right )}{5 \sqrt {-1+x^2}}\\ &=\frac {\sqrt {\frac {1}{10} \left (-1+\sqrt {5}\right )} \sqrt {-1+x} \sqrt {1+x} \tan ^{-1}\left (\frac {2-\left (1-\sqrt {5}\right ) x}{\sqrt {2 \left (-1+\sqrt {5}\right )} \sqrt {-1+x^2}}\right )}{\sqrt {-1+x^2}}-\frac {\sqrt {-1+x} \sqrt {1+x} \tanh ^{-1}\left (\frac {x}{\sqrt {-1+x^2}}\right )}{\sqrt {-1+x^2}}-\frac {\sqrt {\frac {1}{10} \left (1+\sqrt {5}\right )} \sqrt {-1+x} \sqrt {1+x} \tanh ^{-1}\left (\frac {2-\left (1+\sqrt {5}\right ) x}{\sqrt {2 \left (1+\sqrt {5}\right )} \sqrt {-1+x^2}}\right )}{\sqrt {-1+x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.21, size = 113, normalized size = 1.24 \begin {gather*} -\frac {1}{5} \sqrt {\sqrt {5}-2} \left (5+\sqrt {5}\right ) \tan ^{-1}\left (\sqrt {\sqrt {5}-2} \sqrt {\frac {x-1}{x+1}}\right )-2 \tanh ^{-1}\left (\sqrt {\frac {x-1}{x+1}}\right )-\frac {1}{5} \left (\sqrt {5}-5\right ) \sqrt {2+\sqrt {5}} \tanh ^{-1}\left (\sqrt {2+\sqrt {5}} \sqrt {\frac {x-1}{x+1}}\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sqrt[-1 + x]*Sqrt[1 + x])/(1 + x - x^2),x]

[Out]

-1/5*(Sqrt[-2 + Sqrt[5]]*(5 + Sqrt[5])*ArcTan[Sqrt[-2 + Sqrt[5]]*Sqrt[(-1 + x)/(1 + x)]]) - 2*ArcTanh[Sqrt[(-1
+ x)/(1 + x)]] - ((-5 + Sqrt[5])*Sqrt[2 + Sqrt[5]]*ArcTanh[Sqrt[2 + Sqrt[5]]*Sqrt[(-1 + x)/(1 + x)]])/5

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.31, size = 110, normalized size = 1.21 \begin {gather*} -\sqrt {\frac {1}{5} \left (2 \sqrt {5}-2\right )} \tan ^{-1}\left (\frac {\sqrt {\sqrt {5}-2} \sqrt {x-1}}{\sqrt {x+1}}\right )-2 \tanh ^{-1}\left (\frac {\sqrt {x-1}}{\sqrt {x+1}}\right )+\sqrt {\frac {1}{5} \left (2+2 \sqrt {5}\right )} \tanh ^{-1}\left (\frac {\sqrt {2+\sqrt {5}} \sqrt {x-1}}{\sqrt {x+1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[-1 + x]*Sqrt[1 + x])/(1 + x - x^2),x]

[Out]

-(Sqrt[(-2 + 2*Sqrt[5])/5]*ArcTan[(Sqrt[-2 + Sqrt[5]]*Sqrt[-1 + x])/Sqrt[1 + x]]) - 2*ArcTanh[Sqrt[-1 + x]/Sqr
t[1 + x]] + Sqrt[(2 + 2*Sqrt[5])/5]*ArcTanh[(Sqrt[2 + Sqrt[5]]*Sqrt[-1 + x])/Sqrt[1 + x]]

________________________________________________________________________________________

fricas [B]  time = 0.43, size = 214, normalized size = 2.35 \begin {gather*} \frac {2}{5} \, \sqrt {5} \sqrt {2 \, \sqrt {5} - 2} \arctan \left (\frac {1}{8} \, \sqrt {-4 \, {\left (2 \, x + \sqrt {5} - 1\right )} \sqrt {x + 1} \sqrt {x - 1} + 8 \, x^{2} + 4 \, \sqrt {5} x - 4 \, x} \sqrt {2 \, \sqrt {5} - 2} {\left (\sqrt {5} + 1\right )} - \frac {1}{4} \, {\left (\sqrt {x + 1} \sqrt {x - 1} {\left (\sqrt {5} + 1\right )} - \sqrt {5} x - x - 2\right )} \sqrt {2 \, \sqrt {5} - 2}\right ) + \frac {1}{10} \, \sqrt {5} \sqrt {2 \, \sqrt {5} + 2} \log \left (2 \, \sqrt {x + 1} \sqrt {x - 1} - 2 \, x + \sqrt {5} + \sqrt {2 \, \sqrt {5} + 2} + 1\right ) - \frac {1}{10} \, \sqrt {5} \sqrt {2 \, \sqrt {5} + 2} \log \left (2 \, \sqrt {x + 1} \sqrt {x - 1} - 2 \, x + \sqrt {5} - \sqrt {2 \, \sqrt {5} + 2} + 1\right ) + \log \left (\sqrt {x + 1} \sqrt {x - 1} - x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x)^(1/2)*(1+x)^(1/2)/(-x^2+x+1),x, algorithm="fricas")

[Out]

2/5*sqrt(5)*sqrt(2*sqrt(5) - 2)*arctan(1/8*sqrt(-4*(2*x + sqrt(5) - 1)*sqrt(x + 1)*sqrt(x - 1) + 8*x^2 + 4*sqr
t(5)*x - 4*x)*sqrt(2*sqrt(5) - 2)*(sqrt(5) + 1) - 1/4*(sqrt(x + 1)*sqrt(x - 1)*(sqrt(5) + 1) - sqrt(5)*x - x -
2)*sqrt(2*sqrt(5) - 2)) + 1/10*sqrt(5)*sqrt(2*sqrt(5) + 2)*log(2*sqrt(x + 1)*sqrt(x - 1) - 2*x + sqrt(5) + sq
rt(2*sqrt(5) + 2) + 1) - 1/10*sqrt(5)*sqrt(2*sqrt(5) + 2)*log(2*sqrt(x + 1)*sqrt(x - 1) - 2*x + sqrt(5) - sqrt
(2*sqrt(5) + 2) + 1) + log(sqrt(x + 1)*sqrt(x - 1) - x)

________________________________________________________________________________________

giac [A]  time = 0.20, size = 16, normalized size = 0.18 \begin {gather*} \log \left ({\left (\sqrt {x + 1} - \sqrt {x - 1}\right )}^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x)^(1/2)*(1+x)^(1/2)/(-x^2+x+1),x, algorithm="giac")

[Out]

log((sqrt(x + 1) - sqrt(x - 1))^2)

________________________________________________________________________________________

maple [B]  time = 0.09, size = 231, normalized size = 2.54 \begin {gather*} -\frac {\sqrt {x -1}\, \sqrt {x +1}\, \sqrt {5}\, \left (-\sqrt {5}\, \sqrt {-2+2 \sqrt {5}}\, \arctanh \left (\frac {\sqrt {5}\, x +x -2}{\sqrt {2 \sqrt {5}+2}\, \sqrt {x^{2}-1}}\right )-\sqrt {-2+2 \sqrt {5}}\, \arctanh \left (\frac {\sqrt {5}\, x +x -2}{\sqrt {2 \sqrt {5}+2}\, \sqrt {x^{2}-1}}\right )-\sqrt {5}\, \sqrt {2 \sqrt {5}+2}\, \arctan \left (\frac {\sqrt {5}\, x -x +2}{\sqrt {-2+2 \sqrt {5}}\, \sqrt {x^{2}-1}}\right )+\sqrt {2 \sqrt {5}+2}\, \arctan \left (\frac {\sqrt {5}\, x -x +2}{\sqrt {-2+2 \sqrt {5}}\, \sqrt {x^{2}-1}}\right )+\sqrt {5}\, \sqrt {2 \sqrt {5}+2}\, \sqrt {-2+2 \sqrt {5}}\, \ln \left (x +\sqrt {x^{2}-1}\right )\right )}{5 \sqrt {x^{2}-1}\, \sqrt {2 \sqrt {5}+2}\, \sqrt {-2+2 \sqrt {5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x-1)^(1/2)*(x+1)^(1/2)/(-x^2+x+1),x)

[Out]

-1/5*(x-1)^(1/2)*(x+1)^(1/2)*5^(1/2)*(5^(1/2)*ln(x+(x^2-1)^(1/2))*(2*5^(1/2)+2)^(1/2)*(-2+2*5^(1/2))^(1/2)-5^(
1/2)*arctan((x*5^(1/2)-x+2)/(-2+2*5^(1/2))^(1/2)/(x^2-1)^(1/2))*(2*5^(1/2)+2)^(1/2)-5^(1/2)*arctanh((x*5^(1/2)
+x-2)/(2*5^(1/2)+2)^(1/2)/(x^2-1)^(1/2))*(-2+2*5^(1/2))^(1/2)+arctan((x*5^(1/2)-x+2)/(-2+2*5^(1/2))^(1/2)/(x^2
-1)^(1/2))*(2*5^(1/2)+2)^(1/2)-arctanh((x*5^(1/2)+x-2)/(2*5^(1/2)+2)^(1/2)/(x^2-1)^(1/2))*(-2+2*5^(1/2))^(1/2)
)/(x^2-1)^(1/2)/(2*5^(1/2)+2)^(1/2)/(-2+2*5^(1/2))^(1/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {\sqrt {x + 1} \sqrt {x - 1}}{x^{2} - x - 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x)^(1/2)*(1+x)^(1/2)/(-x^2+x+1),x, algorithm="maxima")

[Out]

-integrate(sqrt(x + 1)*sqrt(x - 1)/(x^2 - x - 1), x)

________________________________________________________________________________________

mupad [B]  time = 5.02, size = 916, normalized size = 10.07 \begin {gather*} -4\,\mathrm {atanh}\left (\frac {\sqrt {x-1}-\mathrm {i}}{\sqrt {x+1}-1}\right )-\frac {\sqrt {10}\,\mathrm {atan}\left (\frac {3408370\,\sqrt {10}\,\sqrt {\sqrt {5}+1}-\sqrt {10}\,\sqrt {\sqrt {5}+1}\,\sqrt {x-1}\,300730{}\mathrm {i}-3408370\,\sqrt {10}\,\sqrt {\sqrt {5}+1}\,\sqrt {x+1}-1771398\,\sqrt {5}\,\sqrt {10}\,\sqrt {\sqrt {5}+1}+7836865\,\sqrt {10}\,x\,\sqrt {\sqrt {5}+1}+3066340\,\sqrt {10}\,x^2\,\sqrt {\sqrt {5}+1}-1294942\,\sqrt {5}\,\sqrt {10}\,x^2\,\sqrt {\sqrt {5}+1}+\sqrt {10}\,\sqrt {\sqrt {5}+1}\,\sqrt {x-1}\,\sqrt {x+1}\,300730{}\mathrm {i}-\sqrt {5}\,\sqrt {10}\,\sqrt {\sqrt {5}+1}\,\sqrt {x-1}\,134482{}\mathrm {i}+1771398\,\sqrt {5}\,\sqrt {10}\,\sqrt {\sqrt {5}+1}\,\sqrt {x+1}-\sqrt {10}\,x\,\sqrt {\sqrt {5}+1}\,\sqrt {x-1}\,300730{}\mathrm {i}-6132680\,\sqrt {10}\,x\,\sqrt {\sqrt {5}+1}\,\sqrt {x+1}-3475583\,\sqrt {5}\,\sqrt {10}\,x\,\sqrt {\sqrt {5}+1}+\sqrt {5}\,\sqrt {10}\,\sqrt {\sqrt {5}+1}\,\sqrt {x-1}\,\sqrt {x+1}\,134482{}\mathrm {i}+\sqrt {10}\,x\,\sqrt {\sqrt {5}+1}\,\sqrt {x-1}\,\sqrt {x+1}\,150365{}\mathrm {i}-\sqrt {5}\,\sqrt {10}\,x\,\sqrt {\sqrt {5}+1}\,\sqrt {x-1}\,134482{}\mathrm {i}+2589884\,\sqrt {5}\,\sqrt {10}\,x\,\sqrt {\sqrt {5}+1}\,\sqrt {x+1}+\sqrt {5}\,\sqrt {10}\,x\,\sqrt {\sqrt {5}+1}\,\sqrt {x-1}\,\sqrt {x+1}\,67241{}\mathrm {i}}{29119280\,x-24066900\,x\,\sqrt {x+1}-11518800\,\sqrt {5}\,x-10104760\,\sqrt {x+1}-7067880\,\sqrt {5}-3992430\,\sqrt {5}\,x^2+12033450\,x^2+7067880\,\sqrt {5}\,\sqrt {x+1}+7984860\,\sqrt {5}\,x\,\sqrt {x+1}[phone]}\right )\,\sqrt {\sqrt {5}+1}\,1{}\mathrm {i}}{5}-\frac {\sqrt {10}\,\mathrm {atan}\left (\frac {3408370\,\sqrt {10}\,\sqrt {1-\sqrt {5}}+3066340\,\sqrt {10}\,x^2\,\sqrt {1-\sqrt {5}}-\sqrt {10}\,\sqrt {1-\sqrt {5}}\,\sqrt {x-1}\,300730{}\mathrm {i}-3408370\,\sqrt {10}\,\sqrt {1-\sqrt {5}}\,\sqrt {x+1}+1771398\,\sqrt {5}\,\sqrt {10}\,\sqrt {1-\sqrt {5}}+7836865\,\sqrt {10}\,x\,\sqrt {1-\sqrt {5}}+3475583\,\sqrt {5}\,\sqrt {10}\,x\,\sqrt {1-\sqrt {5}}+1294942\,\sqrt {5}\,\sqrt {10}\,x^2\,\sqrt {1-\sqrt {5}}+\sqrt {10}\,\sqrt {1-\sqrt {5}}\,\sqrt {x-1}\,\sqrt {x+1}\,300730{}\mathrm {i}+\sqrt {5}\,\sqrt {10}\,\sqrt {1-\sqrt {5}}\,\sqrt {x-1}\,134482{}\mathrm {i}-1771398\,\sqrt {5}\,\sqrt {10}\,\sqrt {1-\sqrt {5}}\,\sqrt {x+1}-\sqrt {10}\,x\,\sqrt {1-\sqrt {5}}\,\sqrt {x-1}\,300730{}\mathrm {i}-6132680\,\sqrt {10}\,x\,\sqrt {1-\sqrt {5}}\,\sqrt {x+1}-\sqrt {5}\,\sqrt {10}\,\sqrt {1-\sqrt {5}}\,\sqrt {x-1}\,\sqrt {x+1}\,134482{}\mathrm {i}+\sqrt {10}\,x\,\sqrt {1-\sqrt {5}}\,\sqrt {x-1}\,\sqrt {x+1}\,150365{}\mathrm {i}+\sqrt {5}\,\sqrt {10}\,x\,\sqrt {1-\sqrt {5}}\,\sqrt {x-1}\,134482{}\mathrm {i}-2589884\,\sqrt {5}\,\sqrt {10}\,x\,\sqrt {1-\sqrt {5}}\,\sqrt {x+1}-\sqrt {5}\,\sqrt {10}\,x\,\sqrt {1-\sqrt {5}}\,\sqrt {x-1}\,\sqrt {x+1}\,67241{}\mathrm {i}}{29119280\,x-24066900\,x\,\sqrt {x+1}[phone]\,\sqrt {5}\,x-10104760\,\sqrt {x+1}+7067880\,\sqrt {5}+3992430\,\sqrt {5}\,x^2+12033450\,x^2-7067880\,\sqrt {5}\,\sqrt {x+1}-7984860\,\sqrt {5}\,x\,\sqrt {x+1}[phone]}\right )\,\sqrt {1-\sqrt {5}}\,1{}\mathrm {i}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x - 1)^(1/2)*(x + 1)^(1/2))/(x - x^2 + 1),x)

[Out]

- 4*atanh(((x - 1)^(1/2) - 1i)/((x + 1)^(1/2) - 1)) - (10^(1/2)*atan((3408370*10^(1/2)*(5^(1/2) + 1)^(1/2) - 1
0^(1/2)*(5^(1/2) + 1)^(1/2)*(x - 1)^(1/2)*300730i - 3408370*10^(1/2)*(5^(1/2) + 1)^(1/2)*(x + 1)^(1/2) - 17713
98*5^(1/2)*10^(1/2)*(5^(1/2) + 1)^(1/2) + 7836865*10^(1/2)*x*(5^(1/2) + 1)^(1/2) + 3066340*10^(1/2)*x^2*(5^(1/
2) + 1)^(1/2) - 1294942*5^(1/2)*10^(1/2)*x^2*(5^(1/2) + 1)^(1/2) + 10^(1/2)*(5^(1/2) + 1)^(1/2)*(x - 1)^(1/2)*
(x + 1)^(1/2)*300730i - 5^(1/2)*10^(1/2)*(5^(1/2) + 1)^(1/2)*(x - 1)^(1/2)*134482i + 1771398*5^(1/2)*10^(1/2)*
(5^(1/2) + 1)^(1/2)*(x + 1)^(1/2) - 10^(1/2)*x*(5^(1/2) + 1)^(1/2)*(x - 1)^(1/2)*300730i - 6132680*10^(1/2)*x*
(5^(1/2) + 1)^(1/2)*(x + 1)^(1/2) - 3475583*5^(1/2)*10^(1/2)*x*(5^(1/2) + 1)^(1/2) + 5^(1/2)*10^(1/2)*(5^(1/2)
+ 1)^(1/2)*(x - 1)^(1/2)*(x + 1)^(1/2)*134482i + 10^(1/2)*x*(5^(1/2) + 1)^(1/2)*(x - 1)^(1/2)*(x + 1)^(1/2)*1
50365i - 5^(1/2)*10^(1/2)*x*(5^(1/2) + 1)^(1/2)*(x - 1)^(1/2)*134482i + 2589884*5^(1/2)*10^(1/2)*x*(5^(1/2) +
1)^(1/2)*(x + 1)^(1/2) + 5^(1/2)*10^(1/2)*x*(5^(1/2) + 1)^(1/2)*(x - 1)^(1/2)*(x + 1)^(1/2)*67241i)/(29119280*
x - 24066900*x*(x + 1)^(1/2) - 11518800*5^(1/2)*x - 10104760*(x + 1)^(1/2) - 7067880*5^(1/2) - 3992430*5^(1/2)
*x^2 + 12033450*x^2 + 7067880*5^(1/2)*(x + 1)^(1/2) + 7984860*5^(1/2)*x*(x + 1)^(1/2) + 10104760))*(5^(1/2) +
1)^(1/2)*1i)/5 - (10^(1/2)*atan((3408370*10^(1/2)*(1 - 5^(1/2))^(1/2) + 3066340*10^(1/2)*x^2*(1 - 5^(1/2))^(1/
2) - 10^(1/2)*(1 - 5^(1/2))^(1/2)*(x - 1)^(1/2)*300730i - 3408370*10^(1/2)*(1 - 5^(1/2))^(1/2)*(x + 1)^(1/2) +
1771398*5^(1/2)*10^(1/2)*(1 - 5^(1/2))^(1/2) + 7836865*10^(1/2)*x*(1 - 5^(1/2))^(1/2) + 3475583*5^(1/2)*10^(1
/2)*x*(1 - 5^(1/2))^(1/2) + 1294942*5^(1/2)*10^(1/2)*x^2*(1 - 5^(1/2))^(1/2) + 10^(1/2)*(1 - 5^(1/2))^(1/2)*(x
- 1)^(1/2)*(x + 1)^(1/2)*300730i + 5^(1/2)*10^(1/2)*(1 - 5^(1/2))^(1/2)*(x - 1)^(1/2)*134482i - 1771398*5^(1/
2)*10^(1/2)*(1 - 5^(1/2))^(1/2)*(x + 1)^(1/2) - 10^(1/2)*x*(1 - 5^(1/2))^(1/2)*(x - 1)^(1/2)*300730i - 6132680
*10^(1/2)*x*(1 - 5^(1/2))^(1/2)*(x + 1)^(1/2) - 5^(1/2)*10^(1/2)*(1 - 5^(1/2))^(1/2)*(x - 1)^(1/2)*(x + 1)^(1/
2)*134482i + 10^(1/2)*x*(1 - 5^(1/2))^(1/2)*(x - 1)^(1/2)*(x + 1)^(1/2)*150365i + 5^(1/2)*10^(1/2)*x*(1 - 5^(1
/2))^(1/2)*(x - 1)^(1/2)*134482i - 2589884*5^(1/2)*10^(1/2)*x*(1 - 5^(1/2))^(1/2)*(x + 1)^(1/2) - 5^(1/2)*10^(
1/2)*x*(1 - 5^(1/2))^(1/2)*(x - 1)^(1/2)*(x + 1)^(1/2)*67241i)/(29119280*x - 24066900*x*(x + 1)^(1/2) + 115188
00*5^(1/2)*x - 10104760*(x + 1)^(1/2) + 7067880*5^(1/2) + 3992430*5^(1/2)*x^2 + 12033450*x^2 - 7067880*5^(1/2)
*(x + 1)^(1/2) - 7984860*5^(1/2)*x*(x + 1)^(1/2) + 10104760))*(1 - 5^(1/2))^(1/2)*1i)/5

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {\sqrt {x - 1} \sqrt {x + 1}}{x^{2} - x - 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x)**(1/2)*(1+x)**(1/2)/(-x**2+x+1),x)

[Out]

-Integral(sqrt(x - 1)*sqrt(x + 1)/(x**2 - x - 1), x)

________________________________________________________________________________________