3.290 \(\int \frac{-3 x+\sqrt{1+x+x^2}}{-1+\sqrt{1+x+x^2}} \, dx\)

Optimal. Leaf size=80 \[ -3 \sqrt{x^2+x+1}+4 \tanh ^{-1}\left (\frac{1-x}{2 \sqrt{x^2+x+1}}\right )-\tanh ^{-1}\left (\frac{x+2}{2 \sqrt{x^2+x+1}}\right )+x+\log (x)-4 \log (x+1)+\frac{5}{2} \sinh ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right ) \]

[Out]

x - 3*Sqrt[1 + x + x^2] + (5*ArcSinh[(1 + 2*x)/Sqrt[3]])/2 + 4*ArcTanh[(1 - x)/(2*Sqrt[1 + x + x^2])] - ArcTan
h[(2 + x)/(2*Sqrt[1 + x + x^2])] + Log[x] - 4*Log[1 + x]

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Rubi [A]  time = 0.343471, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 26, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {6742, 734, 843, 619, 215, 724, 206, 6740} \[ -3 \sqrt{x^2+x+1}+4 \tanh ^{-1}\left (\frac{1-x}{2 \sqrt{x^2+x+1}}\right )-\tanh ^{-1}\left (\frac{x+2}{2 \sqrt{x^2+x+1}}\right )+x+\log (x)-4 \log (x+1)+\frac{5}{2} \sinh ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(-3*x + Sqrt[1 + x + x^2])/(-1 + Sqrt[1 + x + x^2]),x]

[Out]

x - 3*Sqrt[1 + x + x^2] + (5*ArcSinh[(1 + 2*x)/Sqrt[3]])/2 + 4*ArcTanh[(1 - x)/(2*Sqrt[1 + x + x^2])] - ArcTan
h[(2 + x)/(2*Sqrt[1 + x + x^2])] + Log[x] - 4*Log[1 + x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 734

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[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] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 215

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

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-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] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 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 6740

Int[(v_)/((a_) + (b_.)*(u_)^(n_.)), x_Symbol] :> Int[ExpandIntegrand[PolynomialInSubst[v, u, x]/(a + b*x^n), x
] /. x -> u, x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && PolynomialInQ[v, u, x]

Rubi steps

\begin{align*} \int \frac{-3 x+\sqrt{1+x+x^2}}{-1+\sqrt{1+x+x^2}} \, dx &=\int \left (-\frac{3 x}{-1+\sqrt{1+x+x^2}}+\frac{\sqrt{1+x+x^2}}{-1+\sqrt{1+x+x^2}}\right ) \, dx\\ &=-\left (3 \int \frac{x}{-1+\sqrt{1+x+x^2}} \, dx\right )+\int \frac{\sqrt{1+x+x^2}}{-1+\sqrt{1+x+x^2}} \, dx\\ &=-\left (3 \int \left (\frac{1}{1+x}+\frac{\sqrt{1+x+x^2}}{1+x}\right ) \, dx\right )+\int \left (1+\frac{1}{-1+\sqrt{1+x+x^2}}\right ) \, dx\\ &=x-3 \log (1+x)-3 \int \frac{\sqrt{1+x+x^2}}{1+x} \, dx+\int \frac{1}{-1+\sqrt{1+x+x^2}} \, dx\\ &=x-3 \sqrt{1+x+x^2}-3 \log (1+x)+\frac{3}{2} \int \frac{-1+x}{(1+x) \sqrt{1+x+x^2}} \, dx+\int \left (\frac{1}{-1-x}+\frac{1}{x}+\frac{\sqrt{1+x+x^2}}{x}-\frac{\sqrt{1+x+x^2}}{1+x}\right ) \, dx\\ &=x-3 \sqrt{1+x+x^2}+\log (x)-4 \log (1+x)+\frac{3}{2} \int \frac{1}{\sqrt{1+x+x^2}} \, dx-3 \int \frac{1}{(1+x) \sqrt{1+x+x^2}} \, dx+\int \frac{\sqrt{1+x+x^2}}{x} \, dx-\int \frac{\sqrt{1+x+x^2}}{1+x} \, dx\\ &=x-3 \sqrt{1+x+x^2}+\log (x)-4 \log (1+x)-\frac{1}{2} \int \frac{-2-x}{x \sqrt{1+x+x^2}} \, dx+\frac{1}{2} \int \frac{-1+x}{(1+x) \sqrt{1+x+x^2}} \, dx+6 \operatorname{Subst}\left (\int \frac{1}{4-x^2} \, dx,x,\frac{1-x}{\sqrt{1+x+x^2}}\right )+\frac{1}{2} \sqrt{3} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{3}}} \, dx,x,1+2 x\right )\\ &=x-3 \sqrt{1+x+x^2}+\frac{3}{2} \sinh ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )+3 \tanh ^{-1}\left (\frac{1-x}{2 \sqrt{1+x+x^2}}\right )+\log (x)-4 \log (1+x)+2 \left (\frac{1}{2} \int \frac{1}{\sqrt{1+x+x^2}} \, dx\right )+\int \frac{1}{x \sqrt{1+x+x^2}} \, dx-\int \frac{1}{(1+x) \sqrt{1+x+x^2}} \, dx\\ &=x-3 \sqrt{1+x+x^2}+\frac{3}{2} \sinh ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )+3 \tanh ^{-1}\left (\frac{1-x}{2 \sqrt{1+x+x^2}}\right )+\log (x)-4 \log (1+x)+2 \operatorname{Subst}\left (\int \frac{1}{4-x^2} \, dx,x,\frac{1-x}{\sqrt{1+x+x^2}}\right )-2 \operatorname{Subst}\left (\int \frac{1}{4-x^2} \, dx,x,\frac{2+x}{\sqrt{1+x+x^2}}\right )+2 \frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{3}}} \, dx,x,1+2 x\right )}{2 \sqrt{3}}\\ &=x-3 \sqrt{1+x+x^2}+\frac{5}{2} \sinh ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )+4 \tanh ^{-1}\left (\frac{1-x}{2 \sqrt{1+x+x^2}}\right )-\tanh ^{-1}\left (\frac{2+x}{2 \sqrt{1+x+x^2}}\right )+\log (x)-4 \log (1+x)\\ \end{align*}

Mathematica [A]  time = 0.110335, size = 80, normalized size = 1. \[ -3 \sqrt{x^2+x+1}+4 \tanh ^{-1}\left (\frac{1-x}{2 \sqrt{x^2+x+1}}\right )-\tanh ^{-1}\left (\frac{x+2}{2 \sqrt{x^2+x+1}}\right )+x+\log (x)-4 \log (x+1)+\frac{5}{2} \sinh ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(-3*x + Sqrt[1 + x + x^2])/(-1 + Sqrt[1 + x + x^2]),x]

[Out]

x - 3*Sqrt[1 + x + x^2] + (5*ArcSinh[(1 + 2*x)/Sqrt[3]])/2 + 4*ArcTanh[(1 - x)/(2*Sqrt[1 + x + x^2])] - ArcTan
h[(2 + x)/(2*Sqrt[1 + x + x^2])] + Log[x] - 4*Log[1 + x]

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Maple [A]  time = 0.01, size = 80, normalized size = 1. \begin{align*} x-4\,\ln \left ( 1+x \right ) +\ln \left ( x \right ) +\sqrt{{x}^{2}+x+1}+{\frac{5}{2}{\it Arcsinh} \left ({\frac{2\,\sqrt{3}}{3} \left ( x+{\frac{1}{2}} \right ) } \right ) }-{\it Artanh} \left ({\frac{2+x}{2}{\frac{1}{\sqrt{{x}^{2}+x+1}}}} \right ) -4\,\sqrt{ \left ( 1+x \right ) ^{2}-x}+4\,{\it Artanh} \left ( 1/2\,{\frac{1-x}{\sqrt{ \left ( 1+x \right ) ^{2}-x}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*x+(x^2+x+1)^(1/2))/(-1+(x^2+x+1)^(1/2)),x)

[Out]

x-4*ln(1+x)+ln(x)+(x^2+x+1)^(1/2)+5/2*arcsinh(2/3*3^(1/2)*(x+1/2))-arctanh(1/2*(2+x)/(x^2+x+1)^(1/2))-4*((1+x)
^2-x)^(1/2)+4*arctanh(1/2*(1-x)/((1+x)^2-x)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{3}{4} \, x^{2} + \frac{1}{2} \, x + \int -\frac{3 \, x^{3} + 2 \, x^{2} - x}{2 \,{\left (x^{2} + x - 2 \, \sqrt{x^{2} + x + 1} + 2\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x+(x^2+x+1)^(1/2))/(-1+(x^2+x+1)^(1/2)),x, algorithm="maxima")

[Out]

3/4*x^2 + 1/2*x + integrate(-1/2*(3*x^3 + 2*x^2 - x)/(x^2 + x - 2*sqrt(x^2 + x + 1) + 2), x)

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Fricas [A]  time = 1.93481, size = 306, normalized size = 3.82 \begin{align*} x - 3 \, \sqrt{x^{2} + x + 1} - 4 \, \log \left (x + 1\right ) + \log \left (x\right ) - \log \left (-x + \sqrt{x^{2} + x + 1} + 1\right ) + 4 \, \log \left (-x + \sqrt{x^{2} + x + 1}\right ) + \log \left (-x + \sqrt{x^{2} + x + 1} - 1\right ) - 4 \, \log \left (-x + \sqrt{x^{2} + x + 1} - 2\right ) - \frac{5}{2} \, \log \left (-2 \, x + 2 \, \sqrt{x^{2} + x + 1} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x+(x^2+x+1)^(1/2))/(-1+(x^2+x+1)^(1/2)),x, algorithm="fricas")

[Out]

x - 3*sqrt(x^2 + x + 1) - 4*log(x + 1) + log(x) - log(-x + sqrt(x^2 + x + 1) + 1) + 4*log(-x + sqrt(x^2 + x +
1)) + log(-x + sqrt(x^2 + x + 1) - 1) - 4*log(-x + sqrt(x^2 + x + 1) - 2) - 5/2*log(-2*x + 2*sqrt(x^2 + x + 1)
- 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{3 x}{\sqrt{x^{2} + x + 1} - 1}\, dx - \int - \frac{\sqrt{x^{2} + x + 1}}{\sqrt{x^{2} + x + 1} - 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x+(x**2+x+1)**(1/2))/(-1+(x**2+x+1)**(1/2)),x)

[Out]

-Integral(3*x/(sqrt(x**2 + x + 1) - 1), x) - Integral(-sqrt(x**2 + x + 1)/(sqrt(x**2 + x + 1) - 1), x)

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Giac [A]  time = 1.14201, size = 142, normalized size = 1.78 \begin{align*} x - 3 \, \sqrt{x^{2} + x + 1} - \frac{5}{2} \, \log \left (-2 \, x + 2 \, \sqrt{x^{2} + x + 1} - 1\right ) - 4 \, \log \left ({\left | x + 1 \right |}\right ) + \log \left ({\left | x \right |}\right ) - \log \left ({\left | -x + \sqrt{x^{2} + x + 1} + 1 \right |}\right ) + 4 \, \log \left ({\left | -x + \sqrt{x^{2} + x + 1} \right |}\right ) + \log \left ({\left | -x + \sqrt{x^{2} + x + 1} - 1 \right |}\right ) - 4 \, \log \left ({\left | -x + \sqrt{x^{2} + x + 1} - 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x+(x^2+x+1)^(1/2))/(-1+(x^2+x+1)^(1/2)),x, algorithm="giac")

[Out]

x - 3*sqrt(x^2 + x + 1) - 5/2*log(-2*x + 2*sqrt(x^2 + x + 1) - 1) - 4*log(abs(x + 1)) + log(abs(x)) - log(abs(
-x + sqrt(x^2 + x + 1) + 1)) + 4*log(abs(-x + sqrt(x^2 + x + 1))) + log(abs(-x + sqrt(x^2 + x + 1) - 1)) - 4*l
og(abs(-x + sqrt(x^2 + x + 1) - 2))