\(\int x^{3/2} \log (-1+4 x+4 \sqrt {(-1+x) x}) \, dx\) [108]

Optimal result

Integrand size = 23, antiderivative size = 187 \[ \int x^{3/2} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=-\frac {\sqrt {x}}{160}+\frac {x^{3/2}}{60}-\frac {2 x^{5/2}}{25}-\frac {17 \sqrt {-x+x^2}}{32 \sqrt {x}}-\frac {71 \left (-x+x^2\right )^{3/2}}{300 x^{3/2}}-\frac {2 \left (-x+x^2\right )^{3/2}}{25 \sqrt {x}}-\frac {\sqrt {-x+x^2} \arctan \left (\frac {2}{3} \sqrt {2} \sqrt {-1+x}\right )}{320 \sqrt {2} \sqrt {-1+x} \sqrt {x}}+\frac {\arctan \left (2 \sqrt {2} \sqrt {x}\right )}{320 \sqrt {2}}+\frac {2}{5} x^{5/2} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right ) \] Output:

-1/160*x^(1/2)+1/60*x^(3/2)-2/25*x^(5/2)-17/32*(x^2-x)^(1/2)/x^(1/2)-71/30 
0*(x^2-x)^(3/2)/x^(3/2)-2/25*(x^2-x)^(3/2)/x^(1/2)-1/640*(x^2-x)^(1/2)*arc 
tan(2/3*2^(1/2)*(-1+x)^(1/2))*2^(1/2)/(-1+x)^(1/2)/x^(1/2)+1/640*arctan(2* 
2^(1/2)*x^(1/2))*2^(1/2)+2/5*x^(5/2)*ln(-1+4*x+4*(x^2-x)^(1/2))
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.56 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.16 \[ \int x^{3/2} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {15 \sqrt {2} \sqrt {(-1+x) x} \arctan \left (\frac {2 \sqrt {2}-i \sqrt {x}}{3 \sqrt {-1+x}}\right )+15 \sqrt {2} \sqrt {(-1+x) x} \arctan \left (\frac {2 \sqrt {2}+i \sqrt {x}}{3 \sqrt {-1+x}}\right )-2 \sqrt {-1+x} \left (-15 \sqrt {2} \sqrt {x} \arctan \left (2 \sqrt {2} \sqrt {x}\right )+4 \left (192 x^3+707 \sqrt {(-1+x) x}+8 x^2 \left (-5+24 \sqrt {(-1+x) x}\right )+x \left (15+376 \sqrt {(-1+x) x}\right )-960 x^3 \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )\right )\right )}{19200 \sqrt {-1+x} \sqrt {x}} \] Input:

Integrate[x^(3/2)*Log[-1 + 4*x + 4*Sqrt[(-1 + x)*x]],x]
 

Output:

(15*Sqrt[2]*Sqrt[(-1 + x)*x]*ArcTan[(2*Sqrt[2] - I*Sqrt[x])/(3*Sqrt[-1 + x 
])] + 15*Sqrt[2]*Sqrt[(-1 + x)*x]*ArcTan[(2*Sqrt[2] + I*Sqrt[x])/(3*Sqrt[- 
1 + x])] - 2*Sqrt[-1 + x]*(-15*Sqrt[2]*Sqrt[x]*ArcTan[2*Sqrt[2]*Sqrt[x]] + 
 4*(192*x^3 + 707*Sqrt[(-1 + x)*x] + 8*x^2*(-5 + 24*Sqrt[(-1 + x)*x]) + x* 
(15 + 376*Sqrt[(-1 + x)*x]) - 960*x^3*Log[-1 + 4*x + 4*Sqrt[(-1 + x)*x]])) 
)/(19200*Sqrt[-1 + x]*Sqrt[x])
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 192, 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.

\(\displaystyle \int x^{3/2} \log \left (4 x+4 \sqrt {(x-1) x}-1\right ) \, dx\)

\(\Big \downarrow \) 3017

\(\displaystyle \int x^{3/2} \log \left (4 \sqrt {x^2-x}+4 x-1\right )dx\)

\(\Big \downarrow \) 3015

\(\displaystyle \frac {16}{5} \int -\frac {x^{5/2}}{4 \left (\sqrt {x^2-x} (2 x+1)+2 \left (x-x^2\right )\right )}dx+\frac {2}{5} x^{5/2} \log \left (4 \sqrt {x^2-x}+4 x-1\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{5} x^{5/2} \log \left (4 \sqrt {x^2-x}+4 x-1\right )-\frac {4}{5} \int \frac {x^{5/2}}{\sqrt {x^2-x} (2 x+1)+2 \left (x-x^2\right )}dx\)

\(\Big \downarrow \) 2035

\(\displaystyle \frac {2}{5} x^{5/2} \log \left (4 \sqrt {x^2-x}+4 x-1\right )-\frac {8}{5} \int \frac {x^3}{\sqrt {x^2-x} (2 x+1)+2 \left (x-x^2\right )}d\sqrt {x}\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {2}{5} x^{5/2} \log \left (4 \sqrt {x^2-x}+4 x-1\right )-\frac {8}{5} \int \left (\frac {x^2}{4}+\frac {1}{4} \sqrt {x^2-x} x+\frac {x}{3 \sqrt {x^2-x}}-\frac {x}{32}+\frac {\sqrt {x^2-x}}{96 (-8 x-1)}+\frac {11 \sqrt {x^2-x}}{32}-\frac {1}{256 (8 x+1)}+\frac {1}{256}\right )d\sqrt {x}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2}{5} x^{5/2} \log \left (4 \sqrt {x^2-x}+4 x-1\right )-\frac {8}{5} \left (\frac {\sqrt {x^2-x} \arctan \left (\frac {2}{3} \sqrt {2} \sqrt {x-1}\right )}{512 \sqrt {2} \sqrt {x-1} \sqrt {x}}-\frac {\arctan \left (2 \sqrt {2} \sqrt {x}\right )}{512 \sqrt {2}}+\frac {x^{5/2}}{20}-\frac {x^{3/2}}{96}+\frac {\left (x^2-x\right )^{3/2}}{20 \sqrt {x}}+\frac {85 \sqrt {x^2-x}}{256 \sqrt {x}}+\frac {71 \left (x^2-x\right )^{3/2}}{480 x^{3/2}}+\frac {\sqrt {x}}{256}\right )\)

Input:

Int[x^(3/2)*Log[-1 + 4*x + 4*Sqrt[(-1 + x)*x]],x]
 

Output:

(-8*(Sqrt[x]/256 - x^(3/2)/96 + x^(5/2)/20 + (85*Sqrt[-x + x^2])/(256*Sqrt 
[x]) + (71*(-x + x^2)^(3/2))/(480*x^(3/2)) + (-x + x^2)^(3/2)/(20*Sqrt[x]) 
 + (Sqrt[-x + x^2]*ArcTan[(2*Sqrt[2]*Sqrt[-1 + x])/3])/(512*Sqrt[2]*Sqrt[- 
1 + x]*Sqrt[x]) - ArcTan[2*Sqrt[2]*Sqrt[x]]/(512*Sqrt[2])))/5 + (2*x^(5/2) 
*Log[-1 + 4*x + 4*Sqrt[-x + x^2]])/5
 

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] 
]
 

Maple [F]

\[\int x^{\frac {3}{2}} \ln \left (-1+4 x +4 \sqrt {x \left (x -1\right )}\right )d x\]

Input:

int(x^(3/2)*ln(-1+4*x+4*(x*(x-1))^(1/2)),x)
 

Output:

int(x^(3/2)*ln(-1+4*x+4*(x*(x-1))^(1/2)),x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.59 \[ \int x^{3/2} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {3840 \, x^{\frac {7}{2}} \log \left (4 \, x + 4 \, \sqrt {x^{2} - x} - 1\right ) + 15 \, \sqrt {2} x \arctan \left (2 \, \sqrt {2} \sqrt {x}\right ) - 15 \, \sqrt {2} x \arctan \left (\frac {2 \, \sqrt {2} \sqrt {x^{2} - x}}{3 \, \sqrt {x}}\right ) - 4 \, {\left (192 \, x^{2} + 376 \, x + 707\right )} \sqrt {x^{2} - x} \sqrt {x} - 4 \, {\left (192 \, x^{3} - 40 \, x^{2} + 15 \, x\right )} \sqrt {x}}{9600 \, x} \] Input:

integrate(x^(3/2)*log(-1+4*x+4*((x-1)*x)^(1/2)),x, algorithm="fricas")
 

Output:

1/9600*(3840*x^(7/2)*log(4*x + 4*sqrt(x^2 - x) - 1) + 15*sqrt(2)*x*arctan( 
2*sqrt(2)*sqrt(x)) - 15*sqrt(2)*x*arctan(2/3*sqrt(2)*sqrt(x^2 - x)/sqrt(x) 
) - 4*(192*x^2 + 376*x + 707)*sqrt(x^2 - x)*sqrt(x) - 4*(192*x^3 - 40*x^2 
+ 15*x)*sqrt(x))/x
 

Sympy [F(-1)]

Timed out. \[ \int x^{3/2} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\text {Timed out} \] Input:

integrate(x**(3/2)*ln(-1+4*x+4*((x-1)*x)**(1/2)),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int x^{3/2} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\int { x^{\frac {3}{2}} \log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right ) \,d x } \] Input:

integrate(x^(3/2)*log(-1+4*x+4*((x-1)*x)^(1/2)),x, algorithm="maxima")
 

Output:

2/5*x^(5/2)*log(4*sqrt(x - 1)*sqrt(x) + 4*x - 1) - 2/25*(2*x^2 + 5)*sqrt(x 
) - 2/15*x^(3/2) + integrate(1/5*(2*x^(5/2) + x^(3/2))/(4*x^2 + 4*(x^(3/2) 
 - sqrt(x))*sqrt(x - 1) - 5*x + 1), x) + 1/5*log(sqrt(x) + 1) - 1/5*log(sq 
rt(x) - 1)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.59 \[ \int x^{3/2} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {2}{5} \, x^{\frac {5}{2}} \log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right ) - \frac {2}{25} \, x^{\frac {5}{2}} + \frac {1}{1280} \, \sqrt {2} {\left (\pi - 2 \, \arctan \left (\frac {\sqrt {2} {\left ({\left (\sqrt {x - 1} - \sqrt {x}\right )}^{2} - 1\right )}}{3 \, {\left (\sqrt {x - 1} - \sqrt {x}\right )}}\right )\right )} - \frac {1}{2400} \, {\left (8 \, {\left (24 \, x + 47\right )} x + 707\right )} \sqrt {x - 1} + \frac {1}{60} \, x^{\frac {3}{2}} + \frac {1}{640} \, \sqrt {2} \arctan \left (2 \, \sqrt {2} \sqrt {x}\right ) - \frac {1}{160} \, \sqrt {x} \] Input:

integrate(x^(3/2)*log(-1+4*x+4*((x-1)*x)^(1/2)),x, algorithm="giac")
 

Output:

2/5*x^(5/2)*log(4*x + 4*sqrt((x - 1)*x) - 1) - 2/25*x^(5/2) + 1/1280*sqrt( 
2)*(pi - 2*arctan(1/3*sqrt(2)*((sqrt(x - 1) - sqrt(x))^2 - 1)/(sqrt(x - 1) 
 - sqrt(x)))) - 1/2400*(8*(24*x + 47)*x + 707)*sqrt(x - 1) + 1/60*x^(3/2) 
+ 1/640*sqrt(2)*arctan(2*sqrt(2)*sqrt(x)) - 1/160*sqrt(x)
 

Mupad [F(-1)]

Timed out. \[ \int x^{3/2} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\int x^{3/2}\,\ln \left (4\,x+4\,\sqrt {x\,\left (x-1\right )}-1\right ) \,d x \] Input:

int(x^(3/2)*log(4*x + 4*(x*(x - 1))^(1/2) - 1),x)
 

Output:

int(x^(3/2)*log(4*x + 4*(x*(x - 1))^(1/2) - 1), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.43 \[ \int x^{3/2} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=-\frac {\sqrt {2}\, \mathit {atan} \left (\frac {2 \sqrt {x -1}+2 \sqrt {x}}{\sqrt {2}}\right )}{320}-\frac {2 \sqrt {x -1}\, x^{2}}{25}-\frac {47 \sqrt {x -1}\, x}{300}-\frac {707 \sqrt {x -1}}{2400}+\frac {2 \sqrt {x}\, \mathrm {log}\left (4 \sqrt {x}\, \sqrt {x -1}+4 x -1\right ) x^{2}}{5}-\frac {2 \sqrt {x}\, x^{2}}{25}+\frac {\sqrt {x}\, x}{60}-\frac {\sqrt {x}}{160} \] Input:

int(x^(3/2)*log(-1+4*x+4*(x*(x-1))^(1/2)),x)
 

Output:

( - 15*sqrt(2)*atan((2*sqrt(x - 1) + 2*sqrt(x))/sqrt(2)) - 384*sqrt(x - 1) 
*x**2 - 752*sqrt(x - 1)*x - 1414*sqrt(x - 1) + 1920*sqrt(x)*log(4*sqrt(x)* 
sqrt(x - 1) + 4*x - 1)*x**2 - 384*sqrt(x)*x**2 + 80*sqrt(x)*x - 30*sqrt(x) 
)/4800