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

Optimal. Leaf size=109 \[ \frac{x+1}{\sqrt{\frac{2 x}{x^2+1}+1}}-\frac{(x+1) \sinh ^{-1}(x)}{\sqrt{x^2+1} \sqrt{\frac{2 x}{x^2+1}+1}}-\frac{\sqrt{2} (x+1) \tanh ^{-1}\left (\frac{1-x}{\sqrt{2} \sqrt{x^2+1}}\right )}{\sqrt{x^2+1} \sqrt{\frac{2 x}{x^2+1}+1}} \]

[Out]

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

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Rubi [A]  time = 0.0648028, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {6723, 970, 735, 844, 215, 725, 206} \[ \frac{x+1}{\sqrt{\frac{2 x}{x^2+1}+1}}-\frac{(x+1) \sinh ^{-1}(x)}{\sqrt{x^2+1} \sqrt{\frac{2 x}{x^2+1}+1}}-\frac{\sqrt{2} (x+1) \tanh ^{-1}\left (\frac{1-x}{\sqrt{2} \sqrt{x^2+1}}\right )}{\sqrt{x^2+1} \sqrt{\frac{2 x}{x^2+1}+1}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[1 + (2*x)/(1 + x^2)],x]

[Out]

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

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 735

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 2*p + 1)), x] + Dist[(2*p)/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[a*e - c*d*x, x]*(a + c*x^2)^(p - 1),
x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !Ration
alQ[m] || LtQ[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

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 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{1+\frac{2 x}{1+x^2}}} \, dx &=\frac{\sqrt{1+2 x+x^2} \int \frac{\sqrt{1+x^2}}{\sqrt{1+2 x+x^2}} \, dx}{\sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}}\\ &=\frac{(2+2 x) \int \frac{\sqrt{1+x^2}}{2+2 x} \, dx}{\sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}}\\ &=\frac{1+x}{\sqrt{1+\frac{2 x}{1+x^2}}}+\frac{(2+2 x) \int \frac{2-2 x}{(2+2 x) \sqrt{1+x^2}} \, dx}{2 \sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}}\\ &=\frac{1+x}{\sqrt{1+\frac{2 x}{1+x^2}}}-\frac{(2+2 x) \int \frac{1}{\sqrt{1+x^2}} \, dx}{2 \sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}}+\frac{(2 (2+2 x)) \int \frac{1}{(2+2 x) \sqrt{1+x^2}} \, dx}{\sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}}\\ &=\frac{1+x}{\sqrt{1+\frac{2 x}{1+x^2}}}-\frac{(1+x) \sinh ^{-1}(x)}{\sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}}-\frac{(2 (2+2 x)) \operatorname{Subst}\left (\int \frac{1}{8-x^2} \, dx,x,\frac{2-2 x}{\sqrt{1+x^2}}\right )}{\sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}}\\ &=\frac{1+x}{\sqrt{1+\frac{2 x}{1+x^2}}}-\frac{(1+x) \sinh ^{-1}(x)}{\sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}}-\frac{\sqrt{2} (1+x) \tanh ^{-1}\left (\frac{1-x}{\sqrt{2} \sqrt{1+x^2}}\right )}{\sqrt{1+x^2} \sqrt{1+\frac{2 x}{1+x^2}}}\\ \end{align*}

Mathematica [A]  time = 0.0282981, size = 72, normalized size = 0.66 \[ \frac{(x+1) \left (\sqrt{x^2+1}-\sqrt{2} \tanh ^{-1}\left (\frac{1-x}{\sqrt{2} \sqrt{x^2+1}}\right )-\sinh ^{-1}(x)\right )}{\sqrt{\frac{(x+1)^2}{x^2+1}} \sqrt{x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.03, size = 79, normalized size = 0.7 \begin{align*}{(1+x){\frac{1}{\sqrt{{\frac{ \left ( 1+x \right ) ^{2}}{{x}^{2}+1}}}}}}+{(1+x) \left ( -{\it Arcsinh} \left ( x \right ) -\sqrt{2}{\it Artanh} \left ({\frac{ \left ( 2-2\,x \right ) \sqrt{2}}{4}{\frac{1}{\sqrt{ \left ( 1+x \right ) ^{2}-2\,x}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{ \left ( 1+x \right ) ^{2}}{{x}^{2}+1}}}}}{\frac{1}{\sqrt{{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/((1+x)^2/(x^2+1))^(1/2)*(1+x)+(-arcsinh(x)-2^(1/2)*arctanh(1/4*(2-2*x)*2^(1/2)/((1+x)^2-2*x)^(1/2)))/((1+x)^
2/(x^2+1))^(1/2)/(x^2+1)^(1/2)*(1+x)

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

[Out]

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

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Fricas [A]  time = 1.49127, size = 360, normalized size = 3.3 \begin{align*} \frac{\sqrt{2}{\left (x + 1\right )} \log \left (-\frac{x^{2} + \sqrt{2}{\left (x^{2} - 1\right )} +{\left (2 \, x^{2} + \sqrt{2}{\left (x^{2} + 1\right )} + 2\right )} \sqrt{\frac{x^{2} + 2 \, x + 1}{x^{2} + 1}} - 1}{x^{2} + 2 \, x + 1}\right ) +{\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/(1+2*x/(x^2+1))^(1/2),x, algorithm="fricas")

[Out]

(sqrt(2)*(x + 1)*log(-(x^2 + sqrt(2)*(x^2 - 1) + (2*x^2 + sqrt(2)*(x^2 + 1) + 2)*sqrt((x^2 + 2*x + 1)/(x^2 + 1
)) - 1)/(x^2 + 2*x + 1)) + (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 \frac{1}{\sqrt{\frac{2 x}{x^{2} + 1} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17987, size = 119, normalized size = 1.09 \begin{align*} \frac{\sqrt{2} \log \left (\frac{{\left | -2 \, x - 2 \, \sqrt{2} + 2 \, \sqrt{x^{2} + 1} - 2 \right |}}{{\left | -2 \, x + 2 \, \sqrt{2} + 2 \, \sqrt{x^{2} + 1} - 2 \right |}}\right )}{\mathrm{sgn}\left (x + 1\right )} + \frac{\log \left (-x + \sqrt{x^{2} + 1}\right )}{\mathrm{sgn}\left (x + 1\right )} + \frac{\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/(1+2*x/(x^2+1))^(1/2),x, algorithm="giac")

[Out]

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