3.704 \(\int \sqrt{2 x+\sqrt{-1+2 x}} \, dx\)

Optimal. Leaf size=80 \[ \frac{1}{3} \left (2 x+\sqrt{2 x-1}\right )^{3/2}-\frac{1}{8} \left (2 \sqrt{2 x-1}+1\right ) \sqrt{2 x+\sqrt{2 x-1}}-\frac{3}{16} \sinh ^{-1}\left (\frac{2 \sqrt{2 x-1}+1}{\sqrt{3}}\right ) \]

[Out]

(2*x + Sqrt[-1 + 2*x])^(3/2)/3 - (Sqrt[2*x + Sqrt[-1 + 2*x]]*(1 + 2*Sqrt[-1 + 2*x]))/8 - (3*ArcSinh[(1 + 2*Sqr
t[-1 + 2*x])/Sqrt[3]])/16

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Rubi [A]  time = 0.0393168, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {640, 612, 619, 215} \[ \frac{1}{3} \left (2 x+\sqrt{2 x-1}\right )^{3/2}-\frac{1}{8} \left (2 \sqrt{2 x-1}+1\right ) \sqrt{2 x+\sqrt{2 x-1}}-\frac{3}{16} \sinh ^{-1}\left (\frac{2 \sqrt{2 x-1}+1}{\sqrt{3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[2*x + Sqrt[-1 + 2*x]],x]

[Out]

(2*x + Sqrt[-1 + 2*x])^(3/2)/3 - (Sqrt[2*x + Sqrt[-1 + 2*x]]*(1 + 2*Sqrt[-1 + 2*x]))/8 - (3*ArcSinh[(1 + 2*Sqr
t[-1 + 2*x])/Sqrt[3]])/16

Rule 640

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

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 619

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

Rubi steps

\begin{align*} \int \sqrt{2 x+\sqrt{-1+2 x}} \, dx &=\operatorname{Subst}\left (\int x \sqrt{1+x+x^2} \, dx,x,\sqrt{-1+2 x}\right )\\ &=\frac{1}{3} \left (2 x+\sqrt{-1+2 x}\right )^{3/2}-\frac{1}{2} \operatorname{Subst}\left (\int \sqrt{1+x+x^2} \, dx,x,\sqrt{-1+2 x}\right )\\ &=\frac{1}{3} \left (2 x+\sqrt{-1+2 x}\right )^{3/2}-\frac{1}{8} \sqrt{2 x+\sqrt{-1+2 x}} \left (1+2 \sqrt{-1+2 x}\right )-\frac{3}{16} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x+x^2}} \, dx,x,\sqrt{-1+2 x}\right )\\ &=\frac{1}{3} \left (2 x+\sqrt{-1+2 x}\right )^{3/2}-\frac{1}{8} \sqrt{2 x+\sqrt{-1+2 x}} \left (1+2 \sqrt{-1+2 x}\right )-\frac{1}{16} \sqrt{3} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{3}}} \, dx,x,1+2 \sqrt{-1+2 x}\right )\\ &=\frac{1}{3} \left (2 x+\sqrt{-1+2 x}\right )^{3/2}-\frac{1}{8} \sqrt{2 x+\sqrt{-1+2 x}} \left (1+2 \sqrt{-1+2 x}\right )-\frac{3}{16} \sinh ^{-1}\left (\frac{1+2 \sqrt{-1+2 x}}{\sqrt{3}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0298329, size = 62, normalized size = 0.78 \[ \frac{1}{48} \left (2 \sqrt{2 x+\sqrt{2 x-1}} \left (16 x+2 \sqrt{2 x-1}-3\right )-9 \sinh ^{-1}\left (\frac{2 \sqrt{2 x-1}+1}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[2*x + Sqrt[-1 + 2*x]],x]

[Out]

(2*Sqrt[2*x + Sqrt[-1 + 2*x]]*(-3 + 16*x + 2*Sqrt[-1 + 2*x]) - 9*ArcSinh[(1 + 2*Sqrt[-1 + 2*x])/Sqrt[3]])/48

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Maple [A]  time = 0.006, size = 60, normalized size = 0.8 \begin{align*}{\frac{1}{3} \left ( 2\,x+\sqrt{2\,x-1} \right ) ^{{\frac{3}{2}}}}-{\frac{1}{8} \left ( 1+2\,\sqrt{2\,x-1} \right ) \sqrt{2\,x+\sqrt{2\,x-1}}}-{\frac{3}{16}{\it Arcsinh} \left ({\frac{2\,\sqrt{3}}{3} \left ( \sqrt{2\,x-1}+{\frac{1}{2}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*x+(2*x-1)^(1/2))^(3/2)-1/8*(1+2*(2*x-1)^(1/2))*(2*x+(2*x-1)^(1/2))^(1/2)-3/16*arcsinh(2/3*3^(1/2)*((2*x
-1)^(1/2)+1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{2 \, x + \sqrt{2 \, x - 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(2*x + sqrt(2*x - 1)), x)

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Fricas [A]  time = 4.11253, size = 207, normalized size = 2.59 \begin{align*} \frac{1}{24} \,{\left (16 \, x + 2 \, \sqrt{2 \, x - 1} - 3\right )} \sqrt{2 \, x + \sqrt{2 \, x - 1}} + \frac{3}{32} \, \log \left (-4 \, \sqrt{2 \, x + \sqrt{2 \, x - 1}}{\left (2 \, \sqrt{2 \, x - 1} + 1\right )} + 16 \, x + 8 \, \sqrt{2 \, x - 1} - 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(16*x + 2*sqrt(2*x - 1) - 3)*sqrt(2*x + sqrt(2*x - 1)) + 3/32*log(-4*sqrt(2*x + sqrt(2*x - 1))*(2*sqrt(2*
x - 1) + 1) + 16*x + 8*sqrt(2*x - 1) - 3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{2 x + \sqrt{2 x - 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.23049, size = 92, normalized size = 1.15 \begin{align*} \frac{1}{24} \,{\left (2 \, \sqrt{2 \, x - 1}{\left (4 \, \sqrt{2 \, x - 1} + 1\right )} + 5\right )} \sqrt{2 \, x + \sqrt{2 \, x - 1}} + \frac{3}{16} \, \log \left (2 \, \sqrt{2 \, x + \sqrt{2 \, x - 1}} - 2 \, \sqrt{2 \, x - 1} - 1\right ) - \frac{5}{24} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(2*sqrt(2*x - 1)*(4*sqrt(2*x - 1) + 1) + 5)*sqrt(2*x + sqrt(2*x - 1)) + 3/16*log(2*sqrt(2*x + sqrt(2*x -
1)) - 2*sqrt(2*x - 1) - 1) - 5/24