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

Optimal. Leaf size=31 \[ -\sqrt{2 x^2+1}+\tan ^{-1}\left (\sqrt{2 x^2+1}\right )-x+\tan ^{-1}(x) \]

[Out]

-x - Sqrt[1 + 2*x^2] + ArcTan[x] + ArcTan[Sqrt[1 + 2*x^2]]

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Rubi [A]  time = 0.0423509, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {2107, 321, 203, 444, 50, 63} \[ -\sqrt{2 x^2+1}+\tan ^{-1}\left (\sqrt{2 x^2+1}\right )-x+\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[x/(x - Sqrt[1 + 2*x^2]),x]

[Out]

-x - Sqrt[1 + 2*x^2] + ArcTan[x] + ArcTan[Sqrt[1 + 2*x^2]]

Rule 2107

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

Rule 321

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

Rule 203

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

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rubi steps

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

Mathematica [A]  time = 0.032193, size = 31, normalized size = 1. \[ -\sqrt{2 x^2+1}+\tan ^{-1}\left (\sqrt{2 x^2+1}\right )-x+\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[x/(x - Sqrt[1 + 2*x^2]),x]

[Out]

-x - Sqrt[1 + 2*x^2] + ArcTan[x] + ArcTan[Sqrt[1 + 2*x^2]]

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Maple [A]  time = 0.007, size = 28, normalized size = 0.9 \begin{align*} -x+\arctan \left ( x \right ) +\arctan \left ( \sqrt{2\,{x}^{2}+1} \right ) -\sqrt{2\,{x}^{2}+1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x+arctan(x)+arctan((2*x^2+1)^(1/2))-(2*x^2+1)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(x - sqrt(2*x^2 + 1)), x)

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Fricas [A]  time = 1.44933, size = 104, normalized size = 3.35 \begin{align*} -x - \sqrt{2 \, x^{2} + 1} + \arctan \left (x\right ) - \arctan \left (-\frac{x^{2} - \sqrt{2 \, x^{2} + 1} + 1}{x^{2}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/(x - sqrt(2*x**2 + 1)), x)

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Giac [B]  time = 1.16244, size = 85, normalized size = 2.74 \begin{align*} -\frac{1}{2} \, \pi - x - \sqrt{2 \, x^{2} + 1} + \arctan \left (x\right ) + \arctan \left (-\frac{{\left (\sqrt{2} x - \sqrt{2 \, x^{2} + 1}\right )}^{2} + 1}{2 \,{\left (\sqrt{2} x - \sqrt{2 \, x^{2} + 1}\right )}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*pi - x - sqrt(2*x^2 + 1) + arctan(x) + arctan(-1/2*((sqrt(2)*x - sqrt(2*x^2 + 1))^2 + 1)/(sqrt(2)*x - sqr
t(2*x^2 + 1)))