3.13 \(\int \sqrt{4 x+x^2} \, dx\)

Optimal. Leaf size=35 \[ \frac{1}{2} (x+2) \sqrt{x^2+4 x}-4 \tanh ^{-1}\left (\frac{x}{\sqrt{x^2+4 x}}\right ) \]

[Out]

((2 + x)*Sqrt[4*x + x^2])/2 - 4*ArcTanh[x/Sqrt[4*x + x^2]]

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Rubi [A]  time = 0.0071598, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {612, 620, 206} \[ \frac{1}{2} (x+2) \sqrt{x^2+4 x}-4 \tanh ^{-1}\left (\frac{x}{\sqrt{x^2+4 x}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[4*x + x^2],x]

[Out]

((2 + x)*Sqrt[4*x + x^2])/2 - 4*ArcTanh[x/Sqrt[4*x + x^2]]

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, 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 \sqrt{4 x+x^2} \, dx &=\frac{1}{2} (2+x) \sqrt{4 x+x^2}-2 \int \frac{1}{\sqrt{4 x+x^2}} \, dx\\ &=\frac{1}{2} (2+x) \sqrt{4 x+x^2}-4 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{x}{\sqrt{4 x+x^2}}\right )\\ &=\frac{1}{2} (2+x) \sqrt{4 x+x^2}-4 \tanh ^{-1}\left (\frac{x}{\sqrt{4 x+x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0298817, size = 40, normalized size = 1.14 \[ \frac{1}{2} \sqrt{x (x+4)} \left (x-\frac{8 \sinh ^{-1}\left (\frac{\sqrt{x}}{2}\right )}{\sqrt{x+4} \sqrt{x}}+2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[4*x + x^2],x]

[Out]

(Sqrt[x*(4 + x)]*(2 + x - (8*ArcSinh[Sqrt[x]/2])/(Sqrt[x]*Sqrt[4 + x])))/2

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Maple [A]  time = 0.052, size = 33, normalized size = 0.9 \begin{align*}{\frac{2\,x+4}{4}\sqrt{{x}^{2}+4\,x}}-2\,\ln \left ( x+2+\sqrt{{x}^{2}+4\,x} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+4*x)^(1/2),x)

[Out]

1/4*(2*x+4)*(x^2+4*x)^(1/2)-2*ln(x+2+(x^2+4*x)^(1/2))

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Maxima [A]  time = 1.12924, size = 55, normalized size = 1.57 \begin{align*} \frac{1}{2} \, \sqrt{x^{2} + 4 \, x} x + \sqrt{x^{2} + 4 \, x} - 2 \, \log \left (2 \, x + 2 \, \sqrt{x^{2} + 4 \, x} + 4\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+4*x)^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(x^2 + 4*x)*x + sqrt(x^2 + 4*x) - 2*log(2*x + 2*sqrt(x^2 + 4*x) + 4)

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Fricas [A]  time = 2.16231, size = 85, normalized size = 2.43 \begin{align*} \frac{1}{2} \, \sqrt{x^{2} + 4 \, x}{\left (x + 2\right )} + 2 \, \log \left (-x + \sqrt{x^{2} + 4 \, x} - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+4*x)^(1/2),x, algorithm="fricas")

[Out]

1/2*sqrt(x^2 + 4*x)*(x + 2) + 2*log(-x + sqrt(x^2 + 4*x) - 2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{x^{2} + 4 x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+4*x)**(1/2),x)

[Out]

Integral(sqrt(x**2 + 4*x), x)

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Giac [A]  time = 1.25703, size = 45, normalized size = 1.29 \begin{align*} \frac{1}{2} \, \sqrt{x^{2} + 4 \, x}{\left (x + 2\right )} + 2 \, \log \left ({\left | -x + \sqrt{x^{2} + 4 \, x} - 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+4*x)^(1/2),x, algorithm="giac")

[Out]

1/2*sqrt(x^2 + 4*x)*(x + 2) + 2*log(abs(-x + sqrt(x^2 + 4*x) - 2))