3.2.10 \(\int \frac {\log (-1+4 x+4 \sqrt {(-1+x) x})}{\sqrt {x}} \, dx\) [110]

3.2.10.1 Optimal result

Integrand size = 23, antiderivative size = 118 \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{\sqrt {x}} \, dx=-2 \sqrt {x}-\frac {2 \sqrt {-x+x^2}}{\sqrt {x}}-\frac {\sqrt {-x+x^2} \arctan \left (\frac {2}{3} \sqrt {2} \sqrt {-1+x}\right )}{\sqrt {2} \sqrt {-1+x} \sqrt {x}}+\frac {\arctan \left (2 \sqrt {2} \sqrt {x}\right )}{\sqrt {2}}+2 \sqrt {x} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right ) \]

output

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

3.2.10.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.55 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.46 \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{\sqrt {x}} \, dx=\frac {\sqrt {2} \sqrt {(-1+x) x} \arctan \left (\frac {2 \sqrt {2}-i \sqrt {x}}{3 \sqrt {-1+x}}\right )+\sqrt {2} \sqrt {(-1+x) x} \arctan \left (\frac {2 \sqrt {2}+i \sqrt {x}}{3 \sqrt {-1+x}}\right )+2 \sqrt {-1+x} \left (\sqrt {2} \sqrt {x} \arctan \left (2 \sqrt {2} \sqrt {x}\right )-4 \left (x+\sqrt {(-1+x) x}-x \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )\right )\right )}{4 \sqrt {-1+x} \sqrt {x}} \]

input

Integrate[Log[-1 + 4*x + 4*Sqrt[(-1 + x)*x]]/Sqrt[x],x]
 

output

(Sqrt[2]*Sqrt[(-1 + x)*x]*ArcTan[(2*Sqrt[2] - I*Sqrt[x])/(3*Sqrt[-1 + x])] 
 + Sqrt[2]*Sqrt[(-1 + x)*x]*ArcTan[(2*Sqrt[2] + I*Sqrt[x])/(3*Sqrt[-1 + x] 
)] + 2*Sqrt[-1 + x]*(Sqrt[2]*Sqrt[x]*ArcTan[2*Sqrt[2]*Sqrt[x]] - 4*(x + Sq 
rt[(-1 + x)*x] - x*Log[-1 + 4*x + 4*Sqrt[(-1 + x)*x]])))/(4*Sqrt[-1 + x]*S 
qrt[x])
 

3.2.10.3 Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.10, 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.

\(\displaystyle \int \frac {\log \left (4 x+4 \sqrt {(x-1) x}-1\right )}{\sqrt {x}} \, dx\)

\(\Big \downarrow \) 3017

\(\displaystyle \int \frac {\log \left (4 \sqrt {x^2-x}+4 x-1\right )}{\sqrt {x}}dx\)

\(\Big \downarrow \) 3015

\(\displaystyle 16 \int -\frac {\sqrt {x}}{4 \left (\sqrt {x^2-x} (2 x+1)+2 \left (x-x^2\right )\right )}dx+2 \sqrt {x} \log \left (4 \sqrt {x^2-x}+4 x-1\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \sqrt {x} \log \left (4 \sqrt {x^2-x}+4 x-1\right )-4 \int \frac {\sqrt {x}}{\sqrt {x^2-x} (2 x+1)+2 \left (x-x^2\right )}dx\)

\(\Big \downarrow \) 2035

\(\displaystyle 2 \sqrt {x} \log \left (4 \sqrt {x^2-x}+4 x-1\right )-8 \int \frac {x}{\sqrt {x^2-x} (2 x+1)+2 \left (x-x^2\right )}d\sqrt {x}\)

\(\Big \downarrow \) 7293

\(\displaystyle 2 \sqrt {x} \log \left (4 \sqrt {x^2-x}+4 x-1\right )-8 \int \left (\frac {x}{3 \sqrt {x^2-x}}+\frac {2 \sqrt {x^2-x}}{3 (-8 x-1)}-\frac {1}{4 (8 x+1)}+\frac {1}{4}\right )d\sqrt {x}\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \sqrt {x} \log \left (4 \sqrt {x^2-x}+4 x-1\right )-8 \left (\frac {\sqrt {x^2-x} \arctan \left (\frac {2}{3} \sqrt {2} \sqrt {x-1}\right )}{8 \sqrt {2} \sqrt {x-1} \sqrt {x}}-\frac {\arctan \left (2 \sqrt {2} \sqrt {x}\right )}{8 \sqrt {2}}+\frac {\sqrt {x^2-x}}{4 \sqrt {x}}+\frac {\sqrt {x}}{4}\right )\)

input

Int[Log[-1 + 4*x + 4*Sqrt[(-1 + x)*x]]/Sqrt[x],x]
 

output

-8*(Sqrt[x]/4 + Sqrt[-x + x^2]/(4*Sqrt[x]) + (Sqrt[-x + x^2]*ArcTan[(2*Sqr 
t[2]*Sqrt[-1 + x])/3])/(8*Sqrt[2]*Sqrt[-1 + x]*Sqrt[x]) - ArcTan[2*Sqrt[2] 
*Sqrt[x]]/(8*Sqrt[2])) + 2*Sqrt[x]*Log[-1 + 4*x + 4*Sqrt[-x + x^2]]
 

3.2.10.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

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]
 

rule 3015

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]
 

rule 3017

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]])
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.2.10.4 Maple [F]

\[\int \frac {\ln \left (-1+4 x +4 \sqrt {\left (-1+x \right ) x}\right )}{\sqrt {x}}d x\]

input

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

output

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

3.2.10.5 Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.71 \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{\sqrt {x}} \, dx=\frac {\sqrt {2} x \arctan \left (2 \, \sqrt {2} \sqrt {x}\right ) + \sqrt {2} x \arctan \left (\frac {3 \, \sqrt {2} \sqrt {x}}{4 \, \sqrt {x^{2} - x}}\right ) + 4 \, x^{\frac {3}{2}} \log \left (4 \, x + 4 \, \sqrt {x^{2} - x} - 1\right ) - 4 \, x^{\frac {3}{2}} - 4 \, \sqrt {x^{2} - x} \sqrt {x}}{2 \, x} \]

input

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

output

1/2*(sqrt(2)*x*arctan(2*sqrt(2)*sqrt(x)) + sqrt(2)*x*arctan(3/4*sqrt(2)*sq 
rt(x)/sqrt(x^2 - x)) + 4*x^(3/2)*log(4*x + 4*sqrt(x^2 - x) - 1) - 4*x^(3/2 
) - 4*sqrt(x^2 - x)*sqrt(x))/x
 

3.2.10.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{\sqrt {x}} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.2.10.7 Maxima [F]

\[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{\sqrt {x}} \, dx=\int { \frac {\log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right )}{\sqrt {x}} \,d x } \]

input

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

output

2*sqrt(x)*log(4*sqrt(x - 1)*sqrt(x) + 4*x - 1) - 4*sqrt(x) + integrate((2* 
x^2 + x)/(4*x^(7/2) - 5*x^(5/2) + 4*(x^3 - x^2)*sqrt(x - 1) + x^(3/2)), x) 
 + log(sqrt(x) + 1) - log(sqrt(x) - 1)
 

3.2.10.8 Giac [F]

\[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{\sqrt {x}} \, dx=\int { \frac {\log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right )}{\sqrt {x}} \,d x } \]

input

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

output

integrate(log(4*x + 4*sqrt((x - 1)*x) - 1)/sqrt(x), x)
 

3.2.10.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{\sqrt {x}} \, dx=\int \frac {\ln \left (4\,x+4\,\sqrt {x\,\left (x-1\right )}-1\right )}{\sqrt {x}} \,d x \]

input

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

output

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