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

Optimal. Leaf size=48 \[ \frac{\sqrt{x^2+1} \sqrt{x^2+2 x+1}}{x+1}+\frac{\sqrt{x^2+2 x+1} \sinh ^{-1}(x)}{x+1} \]

[Out]

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

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Rubi [A]  time = 0.0153268, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {970, 641, 215} \[ \frac{\sqrt{x^2+1} \sqrt{x^2+2 x+1}}{x+1}+\frac{\sqrt{x^2+2 x+1} \sinh ^{-1}(x)}{x+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0176484, size = 27, normalized size = 0.56 \[ \frac{\sqrt{(x+1)^2} \left (\sqrt{x^2+1}+\sinh ^{-1}(x)\right )}{x+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.168, size = 16, normalized size = 0.3 \begin{align*}{\it csgn} \left ( 1+x \right ) \left ({\it Arcsinh} \left ( x \right ) +\sqrt{{x}^{2}+1} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.866251, size = 55, normalized size = 1.15 \begin{align*} \sqrt{x^{2} + 1} - \log \left (-x + \sqrt{x^{2} + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18228, size = 66, normalized size = 1.38 \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+x)^2)^(1/2)/(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)