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

Optimal. Leaf size=28 \[ -\frac{(1-x) \sqrt{\frac{2 x}{x^2+1}+1}}{x+1} \]

[Out]

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

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Rubi [A]  time = 0.11607, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {6723, 970, 637} \[ -\frac{(1-x) \sqrt{\frac{2 x}{x^2+1}+1}}{x+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((1 - x)*Sqrt[1 + (2*x)/(1 + x^2)])/(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 637

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-(a*e) + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

Rubi steps

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

Mathematica [A]  time = 0.0106533, size = 26, normalized size = 0.93 \[ \frac{(x-1) \sqrt{\frac{(x+1)^2}{x^2+1}}}{x+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 28, normalized size = 1. \begin{align*}{\frac{x-1}{1+x}\sqrt{{\frac{{x}^{2}+2\,x+1}{{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [A]  time = 1.44015, size = 80, normalized size = 2.86 \begin{align*} \frac{{\left (x - 1\right )} \sqrt{\frac{x^{2} + 2 \, x + 1}{x^{2} + 1}} + x + 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^2+1),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.10619, size = 41, normalized size = 1.46 \begin{align*} \sqrt{2} \mathrm{sgn}\left (x + 1\right ) + \frac{x \mathrm{sgn}\left (x + 1\right ) - \mathrm{sgn}\left (x + 1\right )}{\sqrt{x^{2} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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