\(\int x^3 \log (-1+4 x+4 \sqrt {(-1+x) x}) \, dx\) [101]

Optimal result

Integrand size = 21, antiderivative size = 172 \[ \int x^3 \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {x}{4096}-\frac {x^2}{1024}+\frac {x^3}{192}-\frac {x^4}{32}-\frac {683 \sqrt {-x+x^2}}{4096}+\frac {149 (1-2 x) \sqrt {-x+x^2}}{2048}-\frac {1}{12} \left (-x+x^2\right )^{3/2}-\frac {1}{32} x \left (-x+x^2\right )^{3/2}+\frac {\text {arctanh}\left (\frac {1-10 x}{6 \sqrt {-x+x^2}}\right )}{32768}-\frac {1537 \text {arctanh}\left (\frac {x}{\sqrt {-x+x^2}}\right )}{16384}-\frac {\log (1+8 x)}{32768}+\frac {1}{4} x^4 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right ) \]

[Out]

1/4096*x-1/1024*x^2+1/192*x^3-1/32*x^4-1/12*(x^2-x)^(3/2)-1/32*x*(x^2-x)^(3/2)+1/32768*arctanh(1/6*(1-10*x)/(x
^2-x)^(1/2))-1537/16384*arctanh(x/(x^2-x)^(1/2))-1/32768*ln(1+8*x)+1/4*x^4*ln(-1+4*x+4*(x^2-x)^(1/2))-683/4096
*(x^2-x)^(1/2)+149/2048*(1-2*x)*(x^2-x)^(1/2)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {2617, 2615, 6874, 654, 634, 212, 626, 748, 857, 738, 684} \[ \int x^3 \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {\text {arctanh}\left (\frac {1-10 x}{6 \sqrt {x^2-x}}\right )}{32768}-\frac {1537 \text {arctanh}\left (\frac {x}{\sqrt {x^2-x}}\right )}{16384}-\frac {x^4}{32}+\frac {x^3}{192}-\frac {x^2}{1024}-\frac {1}{32} \left (x^2-x\right )^{3/2} x-\frac {1}{12} \left (x^2-x\right )^{3/2}+\frac {149 (1-2 x) \sqrt {x^2-x}}{2048}-\frac {683 \sqrt {x^2-x}}{4096}+\frac {1}{4} x^4 \log \left (4 \sqrt {x^2-x}+4 x-1\right )+\frac {x}{4096}-\frac {\log (8 x+1)}{32768} \]

[In]

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

[Out]

x/4096 - x^2/1024 + x^3/192 - x^4/32 - (683*Sqrt[-x + x^2])/4096 + (149*(1 - 2*x)*Sqrt[-x + x^2])/2048 - (-x +
x^2)^(3/2)/12 - (x*(-x + x^2)^(3/2))/32 + ArcTanh[(1 - 10*x)/(6*Sqrt[-x + x^2])]/32768 - (1537*ArcTanh[x/Sqrt
[-x + x^2]])/16384 - Log[1 + 8*x]/32768 + (x^4*Log[-1 + 4*x + 4*Sqrt[-x + x^2]])/4

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 684

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

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^3 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right ) \, dx \\ & = \frac {1}{4} x^4 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+2 \int \frac {x^4}{-4 (1+2 x) \sqrt {-x+x^2}+8 \left (-x+x^2\right )} \, dx \\ & = \frac {1}{4} x^4 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+2 \int \left (\frac {1}{8192}-\frac {x}{1024}+\frac {x^2}{128}-\frac {x^3}{16}-\frac {1}{8192 (1+8 x)}-\frac {x}{12 \sqrt {-x+x^2}}-\frac {85 \sqrt {-x+x^2}}{1024}+\frac {\sqrt {-x+x^2}}{3072 (-1-8 x)}-\frac {11}{128} x \sqrt {-x+x^2}-\frac {1}{16} x^2 \sqrt {-x+x^2}\right ) \, dx \\ & = \frac {x}{4096}-\frac {x^2}{1024}+\frac {x^3}{192}-\frac {x^4}{32}-\frac {\log (1+8 x)}{32768}+\frac {1}{4} x^4 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+\frac {\int \frac {\sqrt {-x+x^2}}{-1-8 x} \, dx}{1536}-\frac {1}{8} \int x^2 \sqrt {-x+x^2} \, dx-\frac {85}{512} \int \sqrt {-x+x^2} \, dx-\frac {1}{6} \int \frac {x}{\sqrt {-x+x^2}} \, dx-\frac {11}{64} \int x \sqrt {-x+x^2} \, dx \\ & = \frac {x}{4096}-\frac {x^2}{1024}+\frac {x^3}{192}-\frac {x^4}{32}-\frac {683 \sqrt {-x+x^2}}{4096}+\frac {85 (1-2 x) \sqrt {-x+x^2}}{2048}-\frac {11}{192} \left (-x+x^2\right )^{3/2}-\frac {1}{32} x \left (-x+x^2\right )^{3/2}-\frac {\log (1+8 x)}{32768}+\frac {1}{4} x^4 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+\frac {\int \frac {1-10 x}{(-1-8 x) \sqrt {-x+x^2}} \, dx}{24576}+\frac {85 \int \frac {1}{\sqrt {-x+x^2}} \, dx}{4096}-\frac {5}{64} \int x \sqrt {-x+x^2} \, dx-\frac {1}{12} \int \frac {1}{\sqrt {-x+x^2}} \, dx-\frac {11}{128} \int \sqrt {-x+x^2} \, dx \\ & = \frac {x}{4096}-\frac {x^2}{1024}+\frac {x^3}{192}-\frac {x^4}{32}-\frac {683 \sqrt {-x+x^2}}{4096}+\frac {129 (1-2 x) \sqrt {-x+x^2}}{2048}-\frac {1}{12} \left (-x+x^2\right )^{3/2}-\frac {1}{32} x \left (-x+x^2\right )^{3/2}-\frac {\log (1+8 x)}{32768}+\frac {1}{4} x^4 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+\frac {5 \int \frac {1}{\sqrt {-x+x^2}} \, dx}{98304}+\frac {3 \int \frac {1}{(-1-8 x) \sqrt {-x+x^2}} \, dx}{32768}+\frac {11 \int \frac {1}{\sqrt {-x+x^2}} \, dx}{1024}-\frac {5}{128} \int \sqrt {-x+x^2} \, dx+\frac {85 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-x+x^2}}\right )}{2048}-\frac {1}{6} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-x+x^2}}\right ) \\ & = \frac {x}{4096}-\frac {x^2}{1024}+\frac {x^3}{192}-\frac {x^4}{32}-\frac {683 \sqrt {-x+x^2}}{4096}+\frac {149 (1-2 x) \sqrt {-x+x^2}}{2048}-\frac {1}{12} \left (-x+x^2\right )^{3/2}-\frac {1}{32} x \left (-x+x^2\right )^{3/2}-\frac {769 \tanh ^{-1}\left (\frac {x}{\sqrt {-x+x^2}}\right )}{6144}-\frac {\log (1+8 x)}{32768}+\frac {1}{4} x^4 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+\frac {5 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-x+x^2}}\right )}{49152}-\frac {3 \text {Subst}\left (\int \frac {1}{36-x^2} \, dx,x,\frac {-1+10 x}{\sqrt {-x+x^2}}\right )}{16384}+\frac {5 \int \frac {1}{\sqrt {-x+x^2}} \, dx}{1024}+\frac {11}{512} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-x+x^2}}\right ) \\ & = \frac {x}{4096}-\frac {x^2}{1024}+\frac {x^3}{192}-\frac {x^4}{32}-\frac {683 \sqrt {-x+x^2}}{4096}+\frac {149 (1-2 x) \sqrt {-x+x^2}}{2048}-\frac {1}{12} \left (-x+x^2\right )^{3/2}-\frac {1}{32} x \left (-x+x^2\right )^{3/2}+\frac {\tanh ^{-1}\left (\frac {1-10 x}{6 \sqrt {-x+x^2}}\right )}{32768}-\frac {1697 \tanh ^{-1}\left (\frac {x}{\sqrt {-x+x^2}}\right )}{16384}-\frac {\log (1+8 x)}{32768}+\frac {1}{4} x^4 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+\frac {5}{512} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-x+x^2}}\right ) \\ & = \frac {x}{4096}-\frac {x^2}{1024}+\frac {x^3}{192}-\frac {x^4}{32}-\frac {683 \sqrt {-x+x^2}}{4096}+\frac {149 (1-2 x) \sqrt {-x+x^2}}{2048}-\frac {1}{12} \left (-x+x^2\right )^{3/2}-\frac {1}{32} x \left (-x+x^2\right )^{3/2}+\frac {\tanh ^{-1}\left (\frac {1-10 x}{6 \sqrt {-x+x^2}}\right )}{32768}-\frac {1537 \tanh ^{-1}\left (\frac {x}{\sqrt {-x+x^2}}\right )}{16384}-\frac {\log (1+8 x)}{32768}+\frac {1}{4} x^4 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.59 \[ \int x^3 \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {24 \sqrt {1-x} x^{3/2}-96 \sqrt {1-x} x^{5/2}+512 \sqrt {1-x} x^{7/2}-3072 \sqrt {1-x} x^{9/2}-6112 \sqrt {1-x} x^{3/2} \sqrt {(-1+x) x}-5120 \sqrt {1-x} x^{5/2} \sqrt {(-1+x) x}-3072 \sqrt {1-x} x^{7/2} \sqrt {(-1+x) x}-9240 \sqrt {-(-1+x)^2 x^2}-9222 \sqrt {(-1+x) x} \arcsin \left (\sqrt {1-x}\right )+3 \sqrt {-((-1+x) x)} \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {(-1+x) x}}\right )-3 \sqrt {-((-1+x) x)} \log (1+8 x)+24576 \sqrt {1-x} x^{9/2} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{98304 \sqrt {-((-1+x) x)}} \]

[In]

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

[Out]

(24*Sqrt[1 - x]*x^(3/2) - 96*Sqrt[1 - x]*x^(5/2) + 512*Sqrt[1 - x]*x^(7/2) - 3072*Sqrt[1 - x]*x^(9/2) - 6112*S
qrt[1 - x]*x^(3/2)*Sqrt[(-1 + x)*x] - 5120*Sqrt[1 - x]*x^(5/2)*Sqrt[(-1 + x)*x] - 3072*Sqrt[1 - x]*x^(7/2)*Sqr
t[(-1 + x)*x] - 9240*Sqrt[-((-1 + x)^2*x^2)] - 9222*Sqrt[(-1 + x)*x]*ArcSin[Sqrt[1 - x]] + 3*Sqrt[-((-1 + x)*x
)]*ArcTanh[(1 - 10*x)/(6*Sqrt[(-1 + x)*x])] - 3*Sqrt[-((-1 + x)*x)]*Log[1 + 8*x] + 24576*Sqrt[1 - x]*x^(9/2)*L
og[-1 + 4*x + 4*Sqrt[(-1 + x)*x]])/(98304*Sqrt[-((-1 + x)*x)])

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.39

method	result	size
parts	\(\frac {\ln \left (-1+4 x +4 \sqrt {\left (-1+x \right ) x}\right ) x^{4}}{4}+\frac {x}{4096}+\frac {x^{3}}{192}-\frac {x^{4}}{32}+\frac {\sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}}{65536}-\frac {5 \ln \left (-\frac {1}{2}+x +\sqrt {\left (x +\frac {1}{8}\right )^{2}-\frac {5 x}{4}-\frac {1}{64}}\right )}{65536}-\frac {41 x^{2} \sqrt {x^{2}-x}}{960}-\frac {283 x \sqrt {x^{2}-x}}{6144}-\frac {x^{2}}{1024}-\frac {3069 \ln \left (-\frac {1}{2}+x +\sqrt {x^{2}-x}\right )}{65536}+\frac {\operatorname {arctanh}\left (\frac {\frac {4}{3}-\frac {40 x}{3}}{\sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}}\right )}{32768}+\frac {\left (x^{2}-x \right )^{\frac {3}{2}}}{16}-\frac {581 \sqrt {x^{2}-x}}{8192}-\frac {\ln \left (1+8 x \right )}{32768}+\frac {95 \left (2 x -1\right ) \sqrt {x^{2}-x}}{4096}+\frac {x^{2} \left (x^{2}-x \right )^{\frac {3}{2}}}{10}-\frac {x^{4} \sqrt {x^{2}-x}}{10}-\frac {x^{3} \sqrt {x^{2}-x}}{320}+\frac {23 x \left (x^{2}-x \right )^{\frac {3}{2}}}{320}\)	\(239\)

[In]

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

[Out]

1/4*ln(-1+4*x+4*((-1+x)*x)^(1/2))*x^4+1/4096*x+1/192*x^3-1/32*x^4+1/65536*(64*(x+1/8)^2-80*x-1)^(1/2)-5/65536*
ln(-1/2+x+((x+1/8)^2-5/4*x-1/64)^(1/2))-41/960*x^2*(x^2-x)^(1/2)-283/6144*x*(x^2-x)^(1/2)-1/1024*x^2-3069/6553
6*ln(-1/2+x+(x^2-x)^(1/2))+1/32768*arctanh(32/3*(1/8-5/4*x)/(64*(x+1/8)^2-80*x-1)^(1/2))+1/16*(x^2-x)^(3/2)-58
1/8192*(x^2-x)^(1/2)-1/32768*ln(1+8*x)+95/4096*(2*x-1)*(x^2-x)^(1/2)+1/10*x^2*(x^2-x)^(3/2)-1/10*x^4*(x^2-x)^(
1/2)-1/320*x^3*(x^2-x)^(1/2)+23/320*x*(x^2-x)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.78 \[ \int x^3 \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=-\frac {1}{32} \, x^{4} + \frac {1}{192} \, x^{3} - \frac {1}{1024} \, x^{2} + \frac {1}{4} \, {\left (x^{4} - 1\right )} \log \left (4 \, x + 4 \, \sqrt {x^{2} - x} - 1\right ) - \frac {1}{12288} \, {\left (384 \, x^{3} + 640 \, x^{2} + 764 \, x + 1155\right )} \sqrt {x^{2} - x} + \frac {1}{4096} \, x + \frac {4095}{32768} \, \log \left (8 \, x + 1\right ) - \frac {2559}{32768} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} + 1\right ) + \frac {4095}{32768} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} - 1\right ) - \frac {4095}{32768} \, \log \left (-4 \, x + 4 \, \sqrt {x^{2} - x} + 1\right ) \]

[In]

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

[Out]

-1/32*x^4 + 1/192*x^3 - 1/1024*x^2 + 1/4*(x^4 - 1)*log(4*x + 4*sqrt(x^2 - x) - 1) - 1/12288*(384*x^3 + 640*x^2
+ 764*x + 1155)*sqrt(x^2 - x) + 1/4096*x + 4095/32768*log(8*x + 1) - 2559/32768*log(-2*x + 2*sqrt(x^2 - x) +
1) + 4095/32768*log(-2*x + 2*sqrt(x^2 - x) - 1) - 4095/32768*log(-4*x + 4*sqrt(x^2 - x) + 1)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

integrate(x^3*log(4*x + 4*sqrt((x - 1)*x) - 1), x)

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.78 \[ \int x^3 \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {1}{4} \, x^{4} \log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right ) - \frac {1}{32} \, x^{4} + \frac {1}{192} \, x^{3} - \frac {1}{1024} \, x^{2} - \frac {1}{12288} \, {\left (4 \, {\left (32 \, {\left (3 \, x + 5\right )} x + 191\right )} x + 1155\right )} \sqrt {x^{2} - x} + \frac {1}{4096} \, x - \frac {1}{32768} \, \log \left ({\left | 8 \, x + 1 \right |}\right ) + \frac {1537}{32768} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x} + 1 \right |}\right ) - \frac {1}{32768} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x} - 1 \right |}\right ) + \frac {1}{32768} \, \log \left ({\left | -4 \, x + 4 \, \sqrt {x^{2} - x} + 1 \right |}\right ) \]

[In]

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

[Out]

1/4*x^4*log(4*x + 4*sqrt((x - 1)*x) - 1) - 1/32*x^4 + 1/192*x^3 - 1/1024*x^2 - 1/12288*(4*(32*(3*x + 5)*x + 19
1)*x + 1155)*sqrt(x^2 - x) + 1/4096*x - 1/32768*log(abs(8*x + 1)) + 1537/32768*log(abs(-2*x + 2*sqrt(x^2 - x)
+ 1)) - 1/32768*log(abs(-2*x + 2*sqrt(x^2 - x) - 1)) + 1/32768*log(abs(-4*x + 4*sqrt(x^2 - x) + 1))

Mupad [F(-1)]

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

[In]

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

[Out]

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