Integrand size = 17, antiderivative size = 80 \[ \int \sqrt {2 x+\sqrt {-1+2 x}} \, dx=\frac {1}{3} \left (2 x+\sqrt {-1+2 x}\right )^{3/2}-\frac {1}{8} \sqrt {2 x+\sqrt {-1+2 x}} \left (1+2 \sqrt {-1+2 x}\right )-\frac {3}{16} \text {arcsinh}\left (\frac {1+2 \sqrt {-1+2 x}}{\sqrt {3}}\right ) \] Output:
1/3*(2*x+(-1+2*x)^(1/2))^(3/2)-1/8*(2*x+(-1+2*x)^(1/2))^(1/2)*(1+2*(-1+2*x )^(1/2))-3/16*arcsinh(1/3*(1+2*(-1+2*x)^(1/2))*3^(1/2))
Time = 0.14 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94 \[ \int \sqrt {2 x+\sqrt {-1+2 x}} \, dx=\frac {1}{48} \left (2 \sqrt {2 x+\sqrt {-1+2 x}} \left (-3+16 x+2 \sqrt {-1+2 x}\right )+9 \log \left (-1-2 \sqrt {-1+2 x}+2 \sqrt {2 x+\sqrt {-1+2 x}}\right )\right ) \] Input:
Integrate[Sqrt[2*x + Sqrt[-1 + 2*x]],x]
Output:
(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
Time = 0.41 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {7267, 1160, 1087, 1090, 222}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sqrt {2 x+\sqrt {2 x-1}} \, dx\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle \int \sqrt {2 x-1} \sqrt {2 x+\sqrt {2 x-1}}d\sqrt {2 x-1}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{3} \left (2 x+\sqrt {2 x-1}\right )^{3/2}-\frac {1}{2} \int \sqrt {2 x+\sqrt {2 x-1}}d\sqrt {2 x-1}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {1}{2} \left (-\frac {3}{8} \int \frac {1}{\sqrt {2 x+\sqrt {2 x-1}}}d\sqrt {2 x-1}-\frac {1}{4} \sqrt {2 x+\sqrt {2 x-1}} \left (2 \sqrt {2 x-1}+1\right )\right )+\frac {1}{3} \left (2 x+\sqrt {2 x-1}\right )^{3/2}\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{8} \sqrt {3} \int \frac {1}{\sqrt {\frac {1}{3} (2 x-1)+1}}d\left (2 \sqrt {2 x-1}+1\right )-\frac {1}{4} \sqrt {2 x+\sqrt {2 x-1}} \left (2 \sqrt {2 x-1}+1\right )\right )+\frac {1}{3} \left (2 x+\sqrt {2 x-1}\right )^{3/2}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{2} \left (-\frac {3}{8} \text {arcsinh}\left (\frac {2 \sqrt {2 x-1}+1}{\sqrt {3}}\right )-\frac {1}{4} \sqrt {2 x+\sqrt {2 x-1}} \left (2 \sqrt {2 x-1}+1\right )\right )+\frac {1}{3} \left (2 x+\sqrt {2 x-1}\right )^{3/2}\) |
Input:
Int[Sqrt[2*x + Sqrt[-1 + 2*x]],x]
Output:
(2*x + Sqrt[-1 + 2*x])^(3/2)/3 + (-1/4*(Sqrt[2*x + Sqrt[-1 + 2*x]]*(1 + 2* Sqrt[-1 + 2*x])) - (3*ArcSinh[(1 + 2*Sqrt[-1 + 2*x])/Sqrt[3]])/8)/2
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]
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] - Simp[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] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[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]
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] + Simp[(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[p, -1]
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.75
method | result | size |
derivativedivides | \(\frac {\left (2 x +\sqrt {2 x -1}\right )^{\frac {3}{2}}}{3}-\frac {\sqrt {2 x +\sqrt {2 x -1}}\, \left (1+2 \sqrt {2 x -1}\right )}{8}-\frac {3 \,\operatorname {arcsinh}\left (\frac {2 \sqrt {3}\, \left (\sqrt {2 x -1}+\frac {1}{2}\right )}{3}\right )}{16}\) | \(60\) |
default | \(\frac {\left (2 x +\sqrt {2 x -1}\right )^{\frac {3}{2}}}{3}-\frac {\sqrt {2 x +\sqrt {2 x -1}}\, \left (1+2 \sqrt {2 x -1}\right )}{8}-\frac {3 \,\operatorname {arcsinh}\left (\frac {2 \sqrt {3}\, \left (\sqrt {2 x -1}+\frac {1}{2}\right )}{3}\right )}{16}\) | \(60\) |
Input:
int((2*x+(2*x-1)^(1/2))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/3*(2*x+(2*x-1)^(1/2))^(3/2)-1/8*(2*x+(2*x-1)^(1/2))^(1/2)*(1+2*(2*x-1)^( 1/2))-3/16*arcsinh(2/3*3^(1/2)*((2*x-1)^(1/2)+1/2))
Time = 0.30 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.91 \[ \int \sqrt {2 x+\sqrt {-1+2 x}} \, dx=\frac {1}{24} \, {\left (16 \, x + 2 \, \sqrt {2 \, x - 1} - 3\right )} \sqrt {2 \, x + \sqrt {2 \, x - 1}} + \frac {3}{32} \, \log \left (-4 \, \sqrt {2 \, x + \sqrt {2 \, x - 1}} {\left (2 \, \sqrt {2 \, x - 1} + 1\right )} + 16 \, x + 8 \, \sqrt {2 \, x - 1} - 3\right ) \] Input:
integrate((2*x+(-1+2*x)^(1/2))^(1/2),x, algorithm="fricas")
Output:
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)
Time = 0.42 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.72 \[ \int \sqrt {2 x+\sqrt {-1+2 x}} \, dx=\sqrt {2 x + \sqrt {2 x - 1}} \cdot \left (\frac {2 x}{3} + \frac {\sqrt {2 x - 1}}{12} - \frac {1}{8}\right ) - \frac {3 \operatorname {asinh}{\left (\frac {2 \sqrt {3} \left (\sqrt {2 x - 1} + \frac {1}{2}\right )}{3} \right )}}{16} \] Input:
integrate((2*x+(-1+2*x)**(1/2))**(1/2),x)
Output:
sqrt(2*x + sqrt(2*x - 1))*(2*x/3 + sqrt(2*x - 1)/12 - 1/8) - 3*asinh(2*sqr t(3)*(sqrt(2*x - 1) + 1/2)/3)/16
\[ \int \sqrt {2 x+\sqrt {-1+2 x}} \, dx=\int { \sqrt {2 \, x + \sqrt {2 \, x - 1}} \,d x } \] Input:
integrate((2*x+(-1+2*x)^(1/2))^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(2*x + sqrt(2*x - 1)), x)
Time = 0.14 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.84 \[ \int \sqrt {2 x+\sqrt {-1+2 x}} \, dx=\frac {1}{24} \, {\left (2 \, \sqrt {2 \, x - 1} {\left (4 \, \sqrt {2 \, x - 1} + 1\right )} + 5\right )} \sqrt {2 \, x + \sqrt {2 \, x - 1}} + \frac {3}{16} \, \log \left (2 \, \sqrt {2 \, x + \sqrt {2 \, x - 1}} - 2 \, \sqrt {2 \, x - 1} - 1\right ) \] Input:
integrate((2*x+(-1+2*x)^(1/2))^(1/2),x, algorithm="giac")
Output:
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)
Timed out. \[ \int \sqrt {2 x+\sqrt {-1+2 x}} \, dx=\int \sqrt {2\,x+\sqrt {2\,x-1}} \,d x \] Input:
int((2*x + (2*x - 1)^(1/2))^(1/2),x)
Output:
int((2*x + (2*x - 1)^(1/2))^(1/2), x)
Time = 0.20 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.98 \[ \int \sqrt {2 x+\sqrt {-1+2 x}} \, dx=\frac {\sqrt {2 x -1}\, \sqrt {\sqrt {2 x -1}+2 x}}{12}+\frac {2 \sqrt {\sqrt {2 x -1}+2 x}\, x}{3}-\frac {\sqrt {\sqrt {2 x -1}+2 x}}{8}-\frac {3 \,\mathrm {log}\left (\frac {2 \sqrt {\sqrt {2 x -1}+2 x}+2 \sqrt {2 x -1}+1}{\sqrt {3}}\right )}{16} \] Input:
int((2*x+(-1+2*x)^(1/2))^(1/2),x)
Output:
(4*sqrt(2*x - 1)*sqrt(sqrt(2*x - 1) + 2*x) + 32*sqrt(sqrt(2*x - 1) + 2*x)* x - 6*sqrt(sqrt(2*x - 1) + 2*x) - 9*log((2*sqrt(sqrt(2*x - 1) + 2*x) + 2*s qrt(2*x - 1) + 1)/sqrt(3)))/48