\(\int x \log (-1+4 x+4 \sqrt {(-1+x) x}) \, dx\) [103]

Optimal result

Integrand size = 19, antiderivative size = 127 \[ \int x \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {x}{32}-\frac {x^2}{8}-\frac {11}{32} \sqrt {-x+x^2}+\frac {1}{16} (1-2 x) \sqrt {-x+x^2}+\frac {1}{256} \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {-x+x^2}}\right )-\frac {33}{128} \text {arctanh}\left (\frac {x}{\sqrt {-x+x^2}}\right )-\frac {1}{256} \log (1+8 x)+\frac {1}{2} x^2 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right ) \]

[Out]

1/32*x-1/8*x^2+1/256*arctanh(1/6*(1-10*x)/(x^2-x)^(1/2))-33/128*arctanh(x/(x^2-x)^(1/2))-1/256*ln(1+8*x)+1/2*x
^2*ln(-1+4*x+4*(x^2-x)^(1/2))-11/32*(x^2-x)^(1/2)+1/16*(1-2*x)*(x^2-x)^(1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {2617, 2615, 6874, 654, 634, 212, 626, 748, 857, 738} \[ \int x \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {1}{256} \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {x^2-x}}\right )-\frac {33}{128} \text {arctanh}\left (\frac {x}{\sqrt {x^2-x}}\right )-\frac {x^2}{8}+\frac {1}{16} (1-2 x) \sqrt {x^2-x}-\frac {11 \sqrt {x^2-x}}{32}+\frac {1}{2} x^2 \log \left (4 \sqrt {x^2-x}+4 x-1\right )+\frac {x}{32}-\frac {1}{256} \log (8 x+1) \]

[In]

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

[Out]

x/32 - x^2/8 - (11*Sqrt[-x + x^2])/32 + ((1 - 2*x)*Sqrt[-x + x^2])/16 + ArcTanh[(1 - 10*x)/(6*Sqrt[-x + x^2])]
/256 - (33*ArcTanh[x/Sqrt[-x + x^2]])/128 - Log[1 + 8*x]/256 + (x^2*Log[-1 + 4*x + 4*Sqrt[-x + x^2]])/2

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 634

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

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]

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 748

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 857

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 2615

Int[Log[(d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]]*((g_.)*(x_))^(m_.), x_Symbol] :> S
imp[(g*x)^(m + 1)*(Log[d + e*x + f*Sqrt[a + b*x + c*x^2]]/(g*(m + 1))), x] + Dist[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 2617

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] && QuadraticQ[u, x] &&  !QuadraticMatchQ[u, x] && (EqQ[v, 1] || MatchQ[v, ((g_.)*
x)^(m_.) /; FreeQ[{g, m}, x]])

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int x \log \left (-1+4 x+4 \sqrt {-x+x^2}\right ) \, dx \\ & = \frac {1}{2} x^2 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+4 \int \frac {x^2}{-4 (1+2 x) \sqrt {-x+x^2}+8 \left (-x+x^2\right )} \, dx \\ & = \frac {1}{2} x^2 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+4 \int \left (\frac {1}{128}-\frac {x}{16}-\frac {1}{128 (1+8 x)}-\frac {x}{12 \sqrt {-x+x^2}}-\frac {1}{16} \sqrt {-x+x^2}+\frac {\sqrt {-x+x^2}}{48 (-1-8 x)}\right ) \, dx \\ & = \frac {x}{32}-\frac {x^2}{8}-\frac {1}{256} \log (1+8 x)+\frac {1}{2} x^2 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+\frac {1}{12} \int \frac {\sqrt {-x+x^2}}{-1-8 x} \, dx-\frac {1}{4} \int \sqrt {-x+x^2} \, dx-\frac {1}{3} \int \frac {x}{\sqrt {-x+x^2}} \, dx \\ & = \frac {x}{32}-\frac {x^2}{8}-\frac {11}{32} \sqrt {-x+x^2}+\frac {1}{16} (1-2 x) \sqrt {-x+x^2}-\frac {1}{256} \log (1+8 x)+\frac {1}{2} x^2 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+\frac {1}{192} \int \frac {1-10 x}{(-1-8 x) \sqrt {-x+x^2}} \, dx+\frac {1}{32} \int \frac {1}{\sqrt {-x+x^2}} \, dx-\frac {1}{6} \int \frac {1}{\sqrt {-x+x^2}} \, dx \\ & = \frac {x}{32}-\frac {x^2}{8}-\frac {11}{32} \sqrt {-x+x^2}+\frac {1}{16} (1-2 x) \sqrt {-x+x^2}-\frac {1}{256} \log (1+8 x)+\frac {1}{2} x^2 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+\frac {5}{768} \int \frac {1}{\sqrt {-x+x^2}} \, dx+\frac {3}{256} \int \frac {1}{(-1-8 x) \sqrt {-x+x^2}} \, dx+\frac {1}{16} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-x+x^2}}\right )-\frac {1}{3} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-x+x^2}}\right ) \\ & = \frac {x}{32}-\frac {x^2}{8}-\frac {11}{32} \sqrt {-x+x^2}+\frac {1}{16} (1-2 x) \sqrt {-x+x^2}-\frac {13}{48} \tanh ^{-1}\left (\frac {x}{\sqrt {-x+x^2}}\right )-\frac {1}{256} \log (1+8 x)+\frac {1}{2} x^2 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+\frac {5}{384} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-x+x^2}}\right )-\frac {3}{128} \text {Subst}\left (\int \frac {1}{36-x^2} \, dx,x,\frac {-1+10 x}{\sqrt {-x+x^2}}\right ) \\ & = \frac {x}{32}-\frac {x^2}{8}-\frac {11}{32} \sqrt {-x+x^2}+\frac {1}{16} (1-2 x) \sqrt {-x+x^2}+\frac {1}{256} \tanh ^{-1}\left (\frac {1-10 x}{6 \sqrt {-x+x^2}}\right )-\frac {33}{128} \tanh ^{-1}\left (\frac {x}{\sqrt {-x+x^2}}\right )-\frac {1}{256} \log (1+8 x)+\frac {1}{2} x^2 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.50 \[ \int x \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {8 \sqrt {1-x} x^{3/2}-32 \sqrt {1-x} x^{5/2}-32 \sqrt {1-x} x^{3/2} \sqrt {(-1+x) x}-72 \sqrt {-(-1+x)^2 x^2}-66 \sqrt {(-1+x) x} \arcsin \left (\sqrt {1-x}\right )+\sqrt {-((-1+x) x)} \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {(-1+x) x}}\right )-\sqrt {-((-1+x) x)} \log (1+8 x)+128 \sqrt {1-x} x^{5/2} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{256 \sqrt {-((-1+x) x)}} \]

[In]

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

[Out]

(8*Sqrt[1 - x]*x^(3/2) - 32*Sqrt[1 - x]*x^(5/2) - 32*Sqrt[1 - x]*x^(3/2)*Sqrt[(-1 + x)*x] - 72*Sqrt[-((-1 + x)
^2*x^2)] - 66*Sqrt[(-1 + x)*x]*ArcSin[Sqrt[1 - x]] + Sqrt[-((-1 + x)*x)]*ArcTanh[(1 - 10*x)/(6*Sqrt[(-1 + x)*x
])] - Sqrt[-((-1 + x)*x)]*Log[1 + 8*x] + 128*Sqrt[1 - x]*x^(5/2)*Log[-1 + 4*x + 4*Sqrt[(-1 + x)*x]])/(256*Sqrt
[-((-1 + x)*x)])

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.38

method	result	size
parts	\(\frac {x^{2} \ln \left (-1+4 x +4 \sqrt {\left (-1+x \right ) x}\right )}{2}-\frac {61 \ln \left (-\frac {1}{2}+x +\sqrt {x^{2}-x}\right )}{512}-\frac {13 \sqrt {x^{2}-x}}{64}+\frac {\operatorname {arctanh}\left (\frac {\frac {4}{3}-\frac {40 x}{3}}{\sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}}\right )}{256}+\frac {x \sqrt {x^{2}-x}}{48}-\frac {x^{2} \sqrt {x^{2}-x}}{3}-\frac {x^{2}}{8}+\frac {x}{32}-\frac {\ln \left (1+8 x \right )}{256}+\frac {3 \left (2 x -1\right ) \sqrt {x^{2}-x}}{32}+\frac {\left (x^{2}-x \right )^{\frac {3}{2}}}{3}+\frac {\sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}}{512}-\frac {5 \ln \left (-\frac {1}{2}+x +\sqrt {\left (x +\frac {1}{8}\right )^{2}-\frac {5 x}{4}-\frac {1}{64}}\right )}{512}\)	\(175\)

[In]

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

[Out]

1/2*x^2*ln(-1+4*x+4*((-1+x)*x)^(1/2))-61/512*ln(-1/2+x+(x^2-x)^(1/2))-13/64*(x^2-x)^(1/2)+1/256*arctanh(32/3*(
1/8-5/4*x)/(64*(x+1/8)^2-80*x-1)^(1/2))+1/48*x*(x^2-x)^(1/2)-1/3*x^2*(x^2-x)^(1/2)-1/8*x^2+1/32*x-1/256*ln(1+8
*x)+3/32*(2*x-1)*(x^2-x)^(1/2)+1/3*(x^2-x)^(3/2)+1/512*(64*(x+1/8)^2-80*x-1)^(1/2)-5/512*ln(-1/2+x+((x+1/8)^2-
5/4*x-1/64)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.90 \[ \int x \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=-\frac {1}{8} \, x^{2} + \frac {1}{2} \, {\left (x^{2} - 1\right )} \log \left (4 \, x + 4 \, \sqrt {x^{2} - x} - 1\right ) - \frac {1}{32} \, \sqrt {x^{2} - x} {\left (4 \, x + 9\right )} + \frac {1}{32} \, x + \frac {63}{256} \, \log \left (8 \, x + 1\right ) - \frac {31}{256} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} + 1\right ) + \frac {63}{256} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} - 1\right ) - \frac {63}{256} \, \log \left (-4 \, x + 4 \, \sqrt {x^{2} - x} + 1\right ) \]

[In]

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

[Out]

-1/8*x^2 + 1/2*(x^2 - 1)*log(4*x + 4*sqrt(x^2 - x) - 1) - 1/32*sqrt(x^2 - x)*(4*x + 9) + 1/32*x + 63/256*log(8
*x + 1) - 31/256*log(-2*x + 2*sqrt(x^2 - x) + 1) + 63/256*log(-2*x + 2*sqrt(x^2 - x) - 1) - 63/256*log(-4*x +
4*sqrt(x^2 - x) + 1)

Sympy [F]

\[ \int x \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\int x \log {\left (4 x + 4 \sqrt {x^{2} - x} - 1 \right )}\, dx \]

[In]

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

[Out]

Integral(x*log(4*x + 4*sqrt(x**2 - x) - 1), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.90 \[ \int x \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {1}{2} \, x^{2} \log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right ) - \frac {1}{8} \, x^{2} - \frac {1}{32} \, \sqrt {x^{2} - x} {\left (4 \, x + 9\right )} + \frac {1}{32} \, x - \frac {1}{256} \, \log \left ({\left | 8 \, x + 1 \right |}\right ) + \frac {33}{256} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x} + 1 \right |}\right ) - \frac {1}{256} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x} - 1 \right |}\right ) + \frac {1}{256} \, \log \left ({\left | -4 \, x + 4 \, \sqrt {x^{2} - x} + 1 \right |}\right ) \]

[In]

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

[Out]

1/2*x^2*log(4*x + 4*sqrt((x - 1)*x) - 1) - 1/8*x^2 - 1/32*sqrt(x^2 - x)*(4*x + 9) + 1/32*x - 1/256*log(abs(8*x
+ 1)) + 33/256*log(abs(-2*x + 2*sqrt(x^2 - x) + 1)) - 1/256*log(abs(-2*x + 2*sqrt(x^2 - x) - 1)) + 1/256*log(
abs(-4*x + 4*sqrt(x^2 - x) + 1))

Mupad [F(-1)]

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

[In]

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

[Out]

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