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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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 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 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 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]

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 = 0.02, 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)

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 75, normalized size = 1.23 \[ -\frac {{\left (x + 1\right )} \log \left (-\frac {x^{2} - {\left (x^{2} + 1\right )} \sqrt {\frac {x^{2} + 2 \, x + 1}{x^{2} + 1}} + x}{x + 1}\right ) - {\left (x^{2} + 1\right )} \sqrt {\frac {x^{2} + 2 \, x + 1}{x^{2} + 1}}}{x + 1} \]

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)

________________________________________________________________________________________

giac [A]  time = 0.48, size = 49, normalized size = 0.80 \[ -{\left (\sqrt {2} - \log \left (\sqrt {2} + 1\right )\right )} \mathrm {sgn}\left (x + 1\right ) - \log \left (-x + \sqrt {x^{2} + 1}\right ) \mathrm {sgn}\left (x + 1\right ) + \sqrt {x^{2} + 1} \mathrm {sgn}\left (x + 1\right ) \]

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)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 42, normalized size = 0.69 \[ \frac {\sqrt {\frac {x^{2}+2 x +1}{x^{2}+1}}\, \sqrt {x^{2}+1}\, \left (\arcsinh \relax (x )+\sqrt {x^{2}+1}\right )}{x +1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {2 \, x}{x^{2} + 1} + 1}\,{d x} \]

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \sqrt {\frac {2\,x}{x^2+1}+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x)/(x^2 + 1) + 1)^(1/2),x)

[Out]

int(((2*x)/(x^2 + 1) + 1)^(1/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {2 x}{x^{2} + 1} + 1}\, dx \]

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)

________________________________________________________________________________________