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

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

[Out]

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

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Rubi [B]  time = 0.140556, antiderivative size = 191, normalized size of antiderivative = 2.1, number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {901, 991, 217, 206, 1034, 725, 204} \[ \frac{\sqrt{\frac{1}{10} \left (\sqrt{5}-1\right )} \sqrt{x-1} \sqrt{x+1} \tan ^{-1}\left (\frac{2-\left (1-\sqrt{5}\right ) x}{\sqrt{2 \left (\sqrt{5}-1\right )} \sqrt{x^2-1}}\right )}{\sqrt{x^2-1}}-\frac{\sqrt{x-1} \sqrt{x+1} \tanh ^{-1}\left (\frac{x}{\sqrt{x^2-1}}\right )}{\sqrt{x^2-1}}-\frac{\sqrt{\frac{1}{10} \left (1+\sqrt{5}\right )} \sqrt{x-1} \sqrt{x+1} \tanh ^{-1}\left (\frac{2-\left (1+\sqrt{5}\right ) x}{\sqrt{2 \left (1+\sqrt{5}\right )} \sqrt{x^2-1}}\right )}{\sqrt{x^2-1}} \]

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 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 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 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 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]

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 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])

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 = 0.207948, size = 113, normalized size = 1.24 \[ -\frac{1}{5} \sqrt{\sqrt{5}-2} \left (5+\sqrt{5}\right ) \tan ^{-1}\left (\sqrt{\sqrt{5}-2} \sqrt{\frac{x-1}{x+1}}\right )-2 \tanh ^{-1}\left (\sqrt{\frac{x-1}{x+1}}\right )-\frac{1}{5} \left (\sqrt{5}-5\right ) \sqrt{2+\sqrt{5}} \tanh ^{-1}\left (\sqrt{2+\sqrt{5}} \sqrt{\frac{x-1}{x+1}}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.125, size = 231, normalized size = 2.5 \begin{align*} -{\frac{\sqrt{5}}{5\,\sqrt{2+2\,\sqrt{5}}\sqrt{-2+2\,\sqrt{5}}}\sqrt{-1+x}\sqrt{1+x} \left ( \ln \left ( x+\sqrt{{x}^{2}-1} \right ) \sqrt{2+2\,\sqrt{5}}\sqrt{-2+2\,\sqrt{5}}\sqrt{5}-{\it Artanh} \left ({\frac{x\sqrt{5}+x-2}{\sqrt{2+2\,\sqrt{5}}}{\frac{1}{\sqrt{{x}^{2}-1}}}} \right ) \sqrt{-2+2\,\sqrt{5}}\sqrt{5}-\arctan \left ({\frac{x\sqrt{5}-x+2}{\sqrt{-2+2\,\sqrt{5}}}{\frac{1}{\sqrt{{x}^{2}-1}}}} \right ) \sqrt{2+2\,\sqrt{5}}\sqrt{5}-{\it Artanh} \left ({\frac{x\sqrt{5}+x-2}{\sqrt{2+2\,\sqrt{5}}}{\frac{1}{\sqrt{{x}^{2}-1}}}} \right ) \sqrt{-2+2\,\sqrt{5}}+\arctan \left ({\frac{x\sqrt{5}-x+2}{\sqrt{-2+2\,\sqrt{5}}}{\frac{1}{\sqrt{{x}^{2}-1}}}} \right ) \sqrt{2+2\,\sqrt{5}} \right ){\frac{1}{\sqrt{{x}^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)

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Fricas [B]  time = 1.68578, size = 671, normalized size = 7.37 \begin{align*} \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{align*}

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)

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

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)

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Giac [A]  time = 1.16558, size = 22, normalized size = 0.24 \begin{align*} \log \left ({\left (\sqrt{x + 1} - \sqrt{x - 1}\right )}^{2}\right ) \end{align*}

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)