Integrand size = 23, antiderivative size = 158 \[ \int \sqrt {x} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {\sqrt {x}}{12}-\frac {2 x^{3/2}}{9}-\frac {11 \sqrt {-x+x^2}}{12 \sqrt {x}}-\frac {2 \left (-x+x^2\right )^{3/2}}{9 x^{3/2}}+\frac {\sqrt {-x+x^2} \arctan \left (\frac {2}{3} \sqrt {2} \sqrt {-1+x}\right )}{24 \sqrt {2} \sqrt {-1+x} \sqrt {x}}-\frac {\arctan \left (2 \sqrt {2} \sqrt {x}\right )}{24 \sqrt {2}}+\frac {2}{3} x^{3/2} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right ) \]
-2/9*x^(3/2)-2/9*(x^2-x)^(3/2)/x^(3/2)+2/3*x^(3/2)*ln(-1+4*x+4*(x^2-x)^(1/ 2))-1/48*arctan(2*2^(1/2)*x^(1/2))*2^(1/2)+1/12*x^(1/2)-11/12/x^(1/2)*(x^2 -x)^(1/2)+1/48*arctan(2/3*2^(1/2)*(-1+x)^(1/2))*(x^2-x)^(1/2)*2^(1/2)/(-1+ x)^(1/2)/x^(1/2)
Result contains complex when optimal does not.
Time = 0.68 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.25 \[ \int \sqrt {x} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {-3 \sqrt {2} \sqrt {(-1+x) x} \arctan \left (\frac {2 \sqrt {2}-i \sqrt {x}}{3 \sqrt {-1+x}}\right )-3 \sqrt {2} \sqrt {(-1+x) x} \arctan \left (\frac {2 \sqrt {2}+i \sqrt {x}}{3 \sqrt {-1+x}}\right )+2 \sqrt {-1+x} \left (-3 \sqrt {2} \sqrt {x} \arctan \left (2 \sqrt {2} \sqrt {x}\right )-4 \left (-3 x+8 x^2+25 \sqrt {(-1+x) x}+8 x \sqrt {(-1+x) x}-24 x^2 \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )\right )\right )}{288 \sqrt {-1+x} \sqrt {x}} \]
(-3*Sqrt[2]*Sqrt[(-1 + x)*x]*ArcTan[(2*Sqrt[2] - I*Sqrt[x])/(3*Sqrt[-1 + x ])] - 3*Sqrt[2]*Sqrt[(-1 + x)*x]*ArcTan[(2*Sqrt[2] + I*Sqrt[x])/(3*Sqrt[-1 + x])] + 2*Sqrt[-1 + x]*(-3*Sqrt[2]*Sqrt[x]*ArcTan[2*Sqrt[2]*Sqrt[x]] - 4 *(-3*x + 8*x^2 + 25*Sqrt[(-1 + x)*x] + 8*x*Sqrt[(-1 + x)*x] - 24*x^2*Log[- 1 + 4*x + 4*Sqrt[(-1 + x)*x]])))/(288*Sqrt[-1 + x]*Sqrt[x])
Time = 0.54 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3017, 3015, 27, 2035, 7293, 2009}
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.
\(\displaystyle \int \sqrt {x} \log \left (4 x+4 \sqrt {(x-1) x}-1\right ) \, dx\) |
\(\Big \downarrow \) 3017 |
\(\displaystyle \int \sqrt {x} \log \left (4 \sqrt {x^2-x}+4 x-1\right )dx\) |
\(\Big \downarrow \) 3015 |
\(\displaystyle \frac {16}{3} \int -\frac {x^{3/2}}{4 \left (\sqrt {x^2-x} (2 x+1)+2 \left (x-x^2\right )\right )}dx+\frac {2}{3} x^{3/2} \log \left (4 \sqrt {x^2-x}+4 x-1\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{3} x^{3/2} \log \left (4 \sqrt {x^2-x}+4 x-1\right )-\frac {4}{3} \int \frac {x^{3/2}}{\sqrt {x^2-x} (2 x+1)+2 \left (x-x^2\right )}dx\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {2}{3} x^{3/2} \log \left (4 \sqrt {x^2-x}+4 x-1\right )-\frac {8}{3} \int \frac {x^2}{\sqrt {x^2-x} (2 x+1)+2 \left (x-x^2\right )}d\sqrt {x}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {2}{3} x^{3/2} \log \left (4 \sqrt {x^2-x}+4 x-1\right )-\frac {8}{3} \int \left (\frac {x}{3 \sqrt {x^2-x}}+\frac {x}{4}+\frac {\sqrt {x^2-x}}{12 (8 x+1)}+\frac {\sqrt {x^2-x}}{4}+\frac {1}{32 (8 x+1)}-\frac {1}{32}\right )d\sqrt {x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{3} x^{3/2} \log \left (4 \sqrt {x^2-x}+4 x-1\right )-\frac {8}{3} \left (-\frac {\sqrt {x^2-x} \arctan \left (\frac {2}{3} \sqrt {2} \sqrt {x-1}\right )}{64 \sqrt {2} \sqrt {x-1} \sqrt {x}}+\frac {\arctan \left (2 \sqrt {2} \sqrt {x}\right )}{64 \sqrt {2}}+\frac {x^{3/2}}{12}+\frac {11 \sqrt {x^2-x}}{32 \sqrt {x}}+\frac {\left (x^2-x\right )^{3/2}}{12 x^{3/2}}-\frac {\sqrt {x}}{32}\right )\) |
(-8*(-1/32*Sqrt[x] + x^(3/2)/12 + (11*Sqrt[-x + x^2])/(32*Sqrt[x]) + (-x + x^2)^(3/2)/(12*x^(3/2)) - (Sqrt[-x + x^2]*ArcTan[(2*Sqrt[2]*Sqrt[-1 + x]) /3])/(64*Sqrt[2]*Sqrt[-1 + x]*Sqrt[x]) + ArcTan[2*Sqrt[2]*Sqrt[x]]/(64*Sqr t[2])))/3 + (2*x^(3/2)*Log[-1 + 4*x + 4*Sqrt[-x + x^2]])/3
3.2.9.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[Log[(d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]] *((g_.)*(x_))^(m_.), x_Symbol] :> Simp[(g*x)^(m + 1)*(Log[d + e*x + f*Sqrt[ a + b*x + c*x^2]]/(g*(m + 1))), x] + Simp[f^2*((b^2 - 4*a*c)/(2*g*(m + 1))) Int[(g*x)^(m + 1)/((2*d*e - b*f^2)*(a + b*x + c*x^2) - f*(b*d - 2*a*e + (2*c*d - b*e)*x)*Sqrt[a + b*x + c*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[e^2 - c*f^2, 0] && NeQ[m, -1] && IntegerQ[2*m]
Int[Log[(d_.) + (f_.)*Sqrt[u_] + (e_.)*(x_)]*(v_.), x_Symbol] :> Int[v*Log[ d + e*x + f*Sqrt[ExpandToSum[u, x]]], x] /; FreeQ[{d, e, f}, x] && Quadrati cQ[u, x] && !QuadraticMatchQ[u, x] && (EqQ[v, 1] || MatchQ[v, ((g_.)*x)^(m _.) /; FreeQ[{g, m}, x]])
\[\int \sqrt {x}\, \ln \left (-1+4 x +4 \sqrt {\left (-1+x \right ) x}\right )d x\]
Time = 0.33 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.63 \[ \int \sqrt {x} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {96 \, x^{\frac {5}{2}} \log \left (4 \, x + 4 \, \sqrt {x^{2} - x} - 1\right ) - 3 \, \sqrt {2} x \arctan \left (2 \, \sqrt {2} \sqrt {x}\right ) - 3 \, \sqrt {2} x \arctan \left (\frac {3 \, \sqrt {2} \sqrt {x}}{4 \, \sqrt {x^{2} - x}}\right ) - 4 \, \sqrt {x^{2} - x} {\left (8 \, x + 25\right )} \sqrt {x} - 4 \, {\left (8 \, x^{2} - 3 \, x\right )} \sqrt {x}}{144 \, x} \]
1/144*(96*x^(5/2)*log(4*x + 4*sqrt(x^2 - x) - 1) - 3*sqrt(2)*x*arctan(2*sq rt(2)*sqrt(x)) - 3*sqrt(2)*x*arctan(3/4*sqrt(2)*sqrt(x)/sqrt(x^2 - x)) - 4 *sqrt(x^2 - x)*(8*x + 25)*sqrt(x) - 4*(8*x^2 - 3*x)*sqrt(x))/x
Timed out. \[ \int \sqrt {x} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\text {Timed out} \]
\[ \int \sqrt {x} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\int { \sqrt {x} \log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right ) \,d x } \]
2/3*x^(3/2)*log(4*sqrt(x - 1)*sqrt(x) + 4*x - 1) - 4/9*x^(3/2) - 2/3*sqrt( x) + integrate(1/3*(2*x^2 + x)/(4*x^(5/2) + 4*(x^2 - x)*sqrt(x - 1) - 5*x^ (3/2) + sqrt(x)), x) + 1/3*log(sqrt(x) + 1) - 1/3*log(sqrt(x) - 1)
Result contains complex when optimal does not.
Time = 0.38 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.77 \[ \int \sqrt {x} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {2}{3} \, x^{\frac {3}{2}} \log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right ) - \frac {1}{96} \, \sqrt {2} {\left (\pi - 2 \, \arctan \left (\frac {\sqrt {2} {\left ({\left (\sqrt {x - 1} - \sqrt {x}\right )}^{2} - 1\right )}}{3 \, {\left (\sqrt {x - 1} - \sqrt {x}\right )}}\right )\right )} + \frac {1}{288} \, \sqrt {2} {\left (-3 i \, \pi + 100 i \, \sqrt {2} - 6 \, \arctan \left (\frac {2}{3} i \, \sqrt {2}\right )\right )} - \frac {1}{36} \, {\left (8 \, x + 25\right )} \sqrt {x - 1} - \frac {2}{9} \, x^{\frac {3}{2}} - \frac {1}{48} \, \sqrt {2} \arctan \left (2 \, \sqrt {2} \sqrt {x}\right ) + \frac {1}{12} \, \sqrt {x} \]
2/3*x^(3/2)*log(4*x + 4*sqrt((x - 1)*x) - 1) - 1/96*sqrt(2)*(pi - 2*arctan (1/3*sqrt(2)*((sqrt(x - 1) - sqrt(x))^2 - 1)/(sqrt(x - 1) - sqrt(x)))) + 1 /288*sqrt(2)*(-3*I*pi + 100*I*sqrt(2) - 6*arctan(2/3*I*sqrt(2))) - 1/36*(8 *x + 25)*sqrt(x - 1) - 2/9*x^(3/2) - 1/48*sqrt(2)*arctan(2*sqrt(2)*sqrt(x) ) + 1/12*sqrt(x)
Timed out. \[ \int \sqrt {x} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\int \sqrt {x}\,\ln \left (4\,x+4\,\sqrt {x\,\left (x-1\right )}-1\right ) \,d x \]