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

Optimal. Leaf size=65 \[ \frac{1}{2} \sqrt{x^2+1} x+\sqrt{x^2+1}+\frac{\sqrt{x^2+1}}{x}-\log \left (\sqrt{x^2+1}+1\right )-x-\frac{1}{x}-\frac{1}{2} \sinh ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.158976, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {6742, 277, 215, 1591, 190, 43, 195} \[ \frac{1}{2} \sqrt{x^2+1} x+\sqrt{x^2+1}+\frac{\sqrt{x^2+1}}{x}-\log \left (\sqrt{x^2+1}+1\right )-x-\frac{1}{x}-\frac{1}{2} \sinh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6742

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

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

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 1591

Int[((a_.) + (b_.)*(Pq_)^(n_.))^(p_.)*(Qr_), x_Symbol] :> With[{q = Expon[Pq, x], r = Expon[Qr, x]}, Dist[Coef
f[Qr, x, r]/(q*Coeff[Pq, x, q]), Subst[Int[(a + b*x^n)^p, x], x, Pq], x] /; EqQ[r, q - 1] && EqQ[Coeff[Qr, x,
r]*D[Pq, x], q*Coeff[Pq, x, q]*Qr]] /; FreeQ[{a, b, n, p}, x] && PolyQ[Pq, x] && PolyQ[Qr, x]

Rule 190

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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]])

Rubi steps

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

Mathematica [A]  time = 0.0570924, size = 65, normalized size = 1. \[ \frac{1}{2} \sqrt{x^2+1} x+\sqrt{x^2+1}+\frac{\sqrt{x^2+1}}{x}-\log \left (\sqrt{x^2+1}+1\right )-x-\frac{1}{x}-\frac{1}{2} \sinh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 56, normalized size = 0.9 \begin{align*} -x-{x}^{-1}-{\frac{x}{2}\sqrt{{x}^{2}+1}}-{\frac{{\it Arcsinh} \left ( x \right ) }{2}}+\sqrt{{x}^{2}+1}-{\it Artanh} \left ({\frac{1}{\sqrt{{x}^{2}+1}}} \right ) -\ln \left ( x \right ) +{\frac{1}{x} \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^2+1)^(1/2)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 2 \, x - 5 \, \arctan \left (\frac{1}{2} \, x\right ) + \int \frac{x^{6} + x^{5} - x^{4}}{3 \, x^{4} + 16 \, x^{2} +{\left (x^{4} + 8 \, x^{2} + 16\right )} \sqrt{x^{2} + 1} + 16}\,{d x} + \log \left (x^{2} + 4\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - 5*arctan(1/2*x) + integrate((x^6 + x^5 - x^4)/(3*x^4 + 16*x^2 + (x^4 + 8*x^2 + 16)*sqrt(x^2 + 1) + 16),
x) + log(x^2 + 4)

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Fricas [A]  time = 1.73818, size = 225, normalized size = 3.46 \begin{align*} -\frac{2 \, x^{2} + 2 \, x \log \left (x\right ) + 2 \, x \log \left (-x + \sqrt{x^{2} + 1} + 1\right ) - x \log \left (-x + \sqrt{x^{2} + 1}\right ) - 2 \, x \log \left (-x + \sqrt{x^{2} + 1} - 1\right ) -{\left (x^{2} + 2 \, x + 2\right )} \sqrt{x^{2} + 1} - 2 \, x + 2}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.75104, size = 63, normalized size = 0.97 \begin{align*} \frac{x \sqrt{x^{2} + 1}}{2} - x + \frac{x}{\sqrt{x^{2} + 1}} + \sqrt{x^{2} + 1} - \log{\left (\sqrt{x^{2} + 1} + 1 \right )} - \frac{\operatorname{asinh}{\left (x \right )}}{2} - \frac{1}{x} + \frac{1}{x \sqrt{x^{2} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*sqrt(x**2 + 1)/2 - x + x/sqrt(x**2 + 1) + sqrt(x**2 + 1) - log(sqrt(x**2 + 1) + 1) - asinh(x)/2 - 1/x + 1/(x
*sqrt(x**2 + 1))

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Giac [A]  time = 1.15208, size = 120, normalized size = 1.85 \begin{align*} \frac{1}{2} \, \sqrt{x^{2} + 1}{\left (x + 2\right )} - x - \frac{2}{{\left (x - \sqrt{x^{2} + 1}\right )}^{2} - 1} - \frac{1}{x} + \frac{1}{2} \, \log \left (-x + \sqrt{x^{2} + 1}\right ) - \log \left ({\left | x \right |}\right ) - \log \left ({\left | -x + \sqrt{x^{2} + 1} + 1 \right |}\right ) + \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)/((x^2+1)^(1/2)+1),x, algorithm="giac")

[Out]

1/2*sqrt(x^2 + 1)*(x + 2) - x - 2/((x - sqrt(x^2 + 1))^2 - 1) - 1/x + 1/2*log(-x + sqrt(x^2 + 1)) - log(abs(x)
) - log(abs(-x + sqrt(x^2 + 1) + 1)) + log(abs(-x + sqrt(x^2 + 1) - 1))