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

Optimal. Leaf size=61 \[ \frac{\sqrt{\frac{2 x}{x^2+1}+1} \left (x^2+1\right )}{x+1}+\frac{\sqrt{\frac{2 x}{x^2+1}+1} \sqrt{x^2+1} \sinh ^{-1}(x)}{x+1} \]

[Out]

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

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Rubi [A]  time = 0.0302999, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6723, 970, 641, 215} \[ \frac{\sqrt{\frac{2 x}{x^2+1}+1} \left (x^2+1\right )}{x+1}+\frac{\sqrt{\frac{2 x}{x^2+1}+1} \sqrt{x^2+1} \sinh ^{-1}(x)}{x+1} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + (2*x)/(1 + x^2)],x]

[Out]

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

Rule 6723

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_)*(x_)^(m_.))^(p_), x_Symbol] :> Dist[(a + b*x^m*v^n)^FracPart[p]/(v^(n*FracP
art[p])*(b*x^m + a/v^n)^FracPart[p]), Int[u*v^(n*p)*(b*x^m + a/v^n)^p, x], x] /; FreeQ[{a, b, m, p}, x] &&  !I
ntegerQ[p] && ILtQ[n, 0] && BinomialQ[v, x]

Rule 970

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_.), x_Symbol] :> Dist[(a + b*x + c*x^2)^F
racPart[p]/((4*c)^IntPart[p]*(b + 2*c*x)^(2*FracPart[p])), Int[(b + 2*c*x)^(2*p)*(d + f*x^2)^q, x], x] /; Free
Q[{a, b, c, d, f, p, q}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

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]

Rubi steps

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

Mathematica [A]  time = 0.0162468, size = 40, normalized size = 0.66 \[ \frac{\sqrt{\frac{(x+1)^2}{x^2+1}} \left (x^2+\sqrt{x^2+1} \sinh ^{-1}(x)+1\right )}{x+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 42, normalized size = 0.7 \begin{align*}{\frac{1}{1+x}\sqrt{{\frac{{x}^{2}+2\,x+1}{{x}^{2}+1}}}\sqrt{{x}^{2}+1} \left ( \sqrt{{x}^{2}+1}+{\it Arcsinh} \left ( x \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(2*x/(x^2 + 1) + 1), x)

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Fricas [A]  time = 1.49756, size = 178, normalized size = 2.92 \begin{align*} -\frac{{\left (x + 1\right )} \log \left (-\frac{x^{2} -{\left (x^{2} + 1\right )} \sqrt{\frac{x^{2} + 2 \, x + 1}{x^{2} + 1}} + x}{x + 1}\right ) -{\left (x^{2} + 1\right )} \sqrt{\frac{x^{2} + 2 \, x + 1}{x^{2} + 1}}}{x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(2*x/(x**2 + 1) + 1), x)

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Giac [A]  time = 1.09261, size = 66, normalized size = 1.08 \begin{align*} -{\left (\sqrt{2} - \log \left (\sqrt{2} + 1\right )\right )} \mathrm{sgn}\left (x + 1\right ) - \log \left (-x + \sqrt{x^{2} + 1}\right ) \mathrm{sgn}\left (x + 1\right ) + \sqrt{x^{2} + 1} \mathrm{sgn}\left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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