3.9.30 \(\int \frac {\sqrt {1+2 x^2}}{1+\sqrt {1+2 x^2}} \, dx\) [830]

Optimal. Leaf size=42 \[ -\frac {1}{2 x}+x+\frac {\sqrt {1+2 x^2}}{2 x}-\frac {\sinh ^{-1}\left (\sqrt {2} x\right )}{\sqrt {2}} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6872, 6874, 283, 221} \begin {gather*} \frac {\sqrt {2 x^2+1}}{2 x}+x-\frac {1}{2 x}-\frac {\sinh ^{-1}\left (\sqrt {2} x\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 221

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 283

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 6872

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]

Rule 6874

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

Rubi steps

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

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Mathematica [A]
time = 0.11, size = 53, normalized size = 1.26 \begin {gather*} \frac {-1+2 x^2+\sqrt {1+2 x^2}+\sqrt {2} x \log \left (-\sqrt {2} x+\sqrt {1+2 x^2}\right )}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.49, size = 45, normalized size = 1.07

method	result	size
default	\(x -\frac {1}{2 x}+\frac {\left (2 x^{2}+1\right )^{\frac {3}{2}}}{2 x}-x \sqrt {2 x^{2}+1}-\frac {\arcsinh \left (\sqrt {2}\, x \right ) \sqrt {2}}{2}\)	\(45\)
trager	\(\frac {\left (-1+x \right ) \left (2 x +1\right )}{2 x}+\frac {\sqrt {2 x^{2}+1}}{2 x}+\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (-\RootOf \left (\textit {\_Z}^{2}-2\right ) x +\sqrt {2 x^{2}+1}\right )}{2}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2+1)^(1/2)/(1+(2*x^2+1)^(1/2)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.34, size = 44, normalized size = 1.05 \begin {gather*} \frac {\sqrt {2} x \log \left (\sqrt {2} x - \sqrt {2 \, x^{2} + 1}\right ) + 2 \, x^{2} + \sqrt {2 \, x^{2} + 1} - 1}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {2 x^{2} + 1}}{\sqrt {2 x^{2} + 1} + 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 2.35, size = 57, normalized size = 1.36 \begin {gather*} \frac {1}{2} \, \sqrt {2} \log \left (-\sqrt {2} x + \sqrt {2 \, x^{2} + 1}\right ) + x - \frac {\sqrt {2}}{{\left (\sqrt {2} x - \sqrt {2 \, x^{2} + 1}\right )}^{2} - 1} - \frac {1}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.39, size = 31, normalized size = 0.74 \begin {gather*} x-\frac {\sqrt {2}\,\mathrm {asinh}\left (\sqrt {2}\,x\right )}{2}+\frac {\frac {\sqrt {2}\,\sqrt {x^2+\frac {1}{2}}}{2}-\frac {1}{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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