3.8.4 \(\int \sqrt {2 x+\sqrt {-1+2 x}} \, dx\) [704]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 80, normalized size of antiderivative = 1.00, 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 = {654, 626, 633, 221} \begin {gather*} \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 ) \end {gather*}

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 221

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 626

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 633

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 654

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]

Rubi steps

\begin {align*} \int \sqrt {2 x+\sqrt {-1+2 x}} \, dx &=\text {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} \text {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} \text {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} \text {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*}

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Mathematica [A]
time = 0.08, size = 75, normalized size = 0.94 \begin {gather*} \frac {1}{48} \left (2 \sqrt {2 x+\sqrt {-1+2 x}} \left (-3+16 x+2 \sqrt {-1+2 x}\right )+9 \log \left (-1-2 \sqrt {-1+2 x}+2 \sqrt {2 x+\sqrt {-1+2 x}}\right )\right ) \end {gather*}

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*Log[-1 - 2*Sqrt[-1 + 2*x] + 2*Sqrt[2*x + Sqrt
[-1 + 2*x]]])/48

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Maple [A]
time = 0.02, size = 60, normalized size = 0.75

method	result	size
derivativedivides	\(\frac {\left (2 x +\sqrt {2 x -1}\right )^{\frac {3}{2}}}{3}-\frac {\left (1+2 \sqrt {2 x -1}\right ) \sqrt {2 x +\sqrt {2 x -1}}}{8}-\frac {3 \arcsinh \left (\frac {2 \sqrt {3}\, \left (\sqrt {2 x -1}+\frac {1}{2}\right )}{3}\right )}{16}\)	\(60\)
default	\(\frac {\left (2 x +\sqrt {2 x -1}\right )^{\frac {3}{2}}}{3}-\frac {\left (1+2 \sqrt {2 x -1}\right ) \sqrt {2 x +\sqrt {2 x -1}}}{8}-\frac {3 \arcsinh \left (\frac {2 \sqrt {3}\, \left (\sqrt {2 x -1}+\frac {1}{2}\right )}{3}\right )}{16}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.59, size = 73, normalized size = 0.91 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {2 x + \sqrt {2 x - 1}}\, dx \end {gather*}

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 = 5.28, size = 67, normalized size = 0.84 \begin {gather*} \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 ) \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {2\,x+\sqrt {2\,x-1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + (2*x - 1)^(1/2))^(1/2),x)

[Out]

int((2*x + (2*x - 1)^(1/2))^(1/2), x)

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