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

Optimal. Leaf size=53 \[ \frac{1}{12} \left (2 x^3+6 x^2+\left (-2 x^2-3 x+4\right ) \sqrt{x^2+1}-6 \log \left (\sqrt{x^2+1}+1\right )-3 \sinh ^{-1}(x)\right ) \]

[Out]

(6*x^2 + 2*x^3 + (4 - 3*x - 2*x^2)*Sqrt[1 + x^2] - 3*ArcSinh[x] - 6*Log[1 + Sqrt[1 + x^2]])/12

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Rubi [A]  time = 0.214881, antiderivative size = 101, normalized size of antiderivative = 1.91, number of steps used = 12, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {6742, 2117, 893, 195, 215, 261} \[ \frac{x^3}{6}+\frac{x^2}{2}-\frac{1}{4} \sqrt{x^2+1} x-\frac{1}{6} \left (x^2+1\right )^{3/2}+\frac{1}{2 \left (\sqrt{x^2+1}+x\right )}+\frac{1}{2} \log \left (\sqrt{x^2+1}+x\right )-\log \left (\sqrt{x^2+1}+x+1\right )+\frac{x}{2}-\frac{1}{4} \sinh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/2 + x^2/2 + x^3/6 - (x*Sqrt[1 + x^2])/4 - (1 + x^2)^(3/2)/6 + 1/(2*(x + Sqrt[1 + x^2])) - ArcSinh[x]/4 + Log
[x + Sqrt[1 + x^2]]/2 - Log[1 + x + Sqrt[1 + x^2]]

Rule 6742

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

Rule 2117

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Dist[1/(2*
e), Subst[Int[((g + h*x^n)^p*(d^2 + a*f^2 - 2*d*x + x^2))/(d - x)^2, x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /
; FreeQ[{a, c, d, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]

Rule 893

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}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

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

Mathematica [A]  time = 0.143002, size = 88, normalized size = 1.66 \[ \frac{1}{12} \left (2 x^3+6 x^2-2 \left (x^2+1\right )^{3/2}+6 \left (\frac{1}{\sqrt{x^2+1}+x}+\log \left (\sqrt{x^2+1}+x\right )-2 \log \left (\sqrt{x^2+1}+x+1\right )\right )-3 \left (\sqrt{x^2+1} x+\sinh ^{-1}(x)\right )+6 x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*x + 6*x^2 + 2*x^3 - 2*(1 + x^2)^(3/2) - 3*(x*Sqrt[1 + x^2] + ArcSinh[x]) + 6*((x + Sqrt[1 + x^2])^(-1) + Lo
g[x + Sqrt[1 + x^2]] - 2*Log[1 + x + Sqrt[1 + x^2]]))/12

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Maple [A]  time = 0.005, size = 58, normalized size = 1.1 \begin{align*}{\frac{{x}^{2}}{2}}-{\frac{\ln \left ( x \right ) }{2}}+{\frac{{x}^{3}}{6}}-{\frac{x}{4}\sqrt{{x}^{2}+1}}-{\frac{{\it Arcsinh} \left ( x \right ) }{4}}+{\frac{1}{2}\sqrt{{x}^{2}+1}}-{\frac{1}{2}{\it Artanh} \left ({\frac{1}{\sqrt{{x}^{2}+1}}} \right ) }-{\frac{1}{6} \left ({x}^{2}+1 \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2-1/2*ln(x)+1/6*x^3-1/4*x*(x^2+1)^(1/2)-1/4*arcsinh(x)+1/2*(x^2+1)^(1/2)-1/2*arctanh(1/(x^2+1)^(1/2))-1/
6*(x^2+1)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{4} \, x^{2} - \frac{3}{56} \, \sqrt{7} \arctan \left (\frac{1}{7} \, \sqrt{7}{\left (4 \, x + 3\right )}\right ) + \frac{1}{4} \, x + \int \frac{x^{4} + x^{3} - x^{2}}{4 \, x^{5} + 12 \, x^{4} + 19 \, x^{3} + 19 \, x^{2} +{\left (4 \, x^{4} + 12 \, x^{3} + 17 \, x^{2} + 12 \, x + 4\right )} \sqrt{x^{2} + 1} + 12 \, x + 4}\,{d x} - \frac{7}{16} \, \log \left (2 \, x^{2} + 3 \, x + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^2 - 3/56*sqrt(7)*arctan(1/7*sqrt(7)*(4*x + 3)) + 1/4*x + integrate((x^4 + x^3 - x^2)/(4*x^5 + 12*x^4 + 1
9*x^3 + 19*x^2 + (4*x^4 + 12*x^3 + 17*x^2 + 12*x + 4)*sqrt(x^2 + 1) + 12*x + 4), x) - 7/16*log(2*x^2 + 3*x + 2
)

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Fricas [A]  time = 1.71586, size = 228, normalized size = 4.3 \begin{align*} \frac{1}{6} \, x^{3} + \frac{1}{2} \, x^{2} - \frac{1}{12} \,{\left (2 \, x^{2} + 3 \, x - 4\right )} \sqrt{x^{2} + 1} - \frac{1}{2} \, \log \left (x\right ) - \frac{1}{2} \, \log \left (-x + \sqrt{x^{2} + 1} + 1\right ) + \frac{1}{4} \, \log \left (-x + \sqrt{x^{2} + 1}\right ) + \frac{1}{2} \, \log \left (-x + \sqrt{x^{2} + 1} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3 + 1/2*x^2 - 1/12*(2*x^2 + 3*x - 4)*sqrt(x^2 + 1) - 1/2*log(x) - 1/2*log(-x + sqrt(x^2 + 1) + 1) + 1/4*
log(-x + sqrt(x^2 + 1)) + 1/2*log(-x + sqrt(x^2 + 1) - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12376, size = 108, normalized size = 2.04 \begin{align*} \frac{1}{6} \, x^{3} + \frac{1}{2} \, x^{2} - \frac{1}{12} \,{\left ({\left (2 \, x + 3\right )} x - 4\right )} \sqrt{x^{2} + 1} + \frac{1}{4} \, \log \left (-x + \sqrt{x^{2} + 1}\right ) - \frac{1}{2} \, \log \left ({\left | x \right |}\right ) - \frac{1}{2} \, \log \left ({\left | -x + \sqrt{x^{2} + 1} + 1 \right |}\right ) + \frac{1}{2} \, \log \left ({\left | -x + \sqrt{x^{2} + 1} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3 + 1/2*x^2 - 1/12*((2*x + 3)*x - 4)*sqrt(x^2 + 1) + 1/4*log(-x + sqrt(x^2 + 1)) - 1/2*log(abs(x)) - 1/2
*log(abs(-x + sqrt(x^2 + 1) + 1)) + 1/2*log(abs(-x + sqrt(x^2 + 1) - 1))