Integrand size = 21, antiderivative size = 172 \[ \int x^3 \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {x}{4096}-\frac {x^2}{1024}+\frac {x^3}{192}-\frac {x^4}{32}-\frac {683 \sqrt {-x+x^2}}{4096}+\frac {149 (1-2 x) \sqrt {-x+x^2}}{2048}-\frac {1}{12} \left (-x+x^2\right )^{3/2}-\frac {1}{32} x \left (-x+x^2\right )^{3/2}+\frac {\text {arctanh}\left (\frac {1-10 x}{6 \sqrt {-x+x^2}}\right )}{32768}-\frac {1537 \text {arctanh}\left (\frac {x}{\sqrt {-x+x^2}}\right )}{16384}-\frac {\log (1+8 x)}{32768}+\frac {1}{4} x^4 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right ) \]
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)
Time = 0.55 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.59 \[ \int x^3 \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {24 \sqrt {1-x} x^{3/2}-96 \sqrt {1-x} x^{5/2}+512 \sqrt {1-x} x^{7/2}-3072 \sqrt {1-x} x^{9/2}-6112 \sqrt {1-x} x^{3/2} \sqrt {(-1+x) x}-5120 \sqrt {1-x} x^{5/2} \sqrt {(-1+x) x}-3072 \sqrt {1-x} x^{7/2} \sqrt {(-1+x) x}-9240 \sqrt {-(-1+x)^2 x^2}-9222 \sqrt {(-1+x) x} \arcsin \left (\sqrt {1-x}\right )+3 \sqrt {-((-1+x) x)} \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {(-1+x) x}}\right )-3 \sqrt {-((-1+x) x)} \log (1+8 x)+24576 \sqrt {1-x} x^{9/2} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{98304 \sqrt {-((-1+x) x)}} \]
(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[ 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)*Log[- 1 + 4*x + 4*Sqrt[(-1 + x)*x]])/(98304*Sqrt[-((-1 + x)*x)])
Time = 0.57 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3017, 3015, 27, 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.
| \(\displaystyle \int x^3 \log \left (4 x+4 \sqrt {(x-1) x}-1\right ) \, dx\) |
| \(\Big \downarrow \) 3017 |
| \(\displaystyle \int x^3 \log \left (4 \sqrt {x^2-x}+4 x-1\right )dx\) |
| \(\Big \downarrow \) 3015 |
| \(\displaystyle 2 \int -\frac {x^4}{4 \left (\sqrt {x^2-x} (2 x+1)+2 \left (x-x^2\right )\right )}dx+\frac {1}{4} x^4 \log \left (4 \sqrt {x^2-x}+4 x-1\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} x^4 \log \left (4 \sqrt {x^2-x}+4 x-1\right )-\frac {1}{2} \int \frac {x^4}{\sqrt {x^2-x} (2 x+1)+2 \left (x-x^2\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {1}{4} x^4 \log \left (4 \sqrt {x^2-x}+4 x-1\right )-\frac {1}{2} \int \left (\frac {x^3}{4}+\frac {1}{4} \sqrt {x^2-x} x^2-\frac {x^2}{32}+\frac {11}{32} \sqrt {x^2-x} x+\frac {x}{3 \sqrt {x^2-x}}+\frac {x}{256}+\frac {\sqrt {x^2-x}}{768 (8 x+1)}+\frac {85 \sqrt {x^2-x}}{256}+\frac {1}{2048 (8 x+1)}-\frac {1}{2048}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (\frac {\text {arctanh}\left (\frac {1-10 x}{6 \sqrt {x^2-x}}\right )}{16384}-\frac {1537 \text {arctanh}\left (\frac {x}{\sqrt {x^2-x}}\right )}{8192}-\frac {x^4}{16}+\frac {x^3}{96}-\frac {x^2}{512}-\frac {1}{16} \left (x^2-x\right )^{3/2} x-\frac {1}{6} \left (x^2-x\right )^{3/2}+\frac {149 (1-2 x) \sqrt {x^2-x}}{1024}-\frac {683 \sqrt {x^2-x}}{2048}+\frac {x}{2048}-\frac {\log (8 x+1)}{16384}\right )+\frac {1}{4} x^4 \log \left (4 \sqrt {x^2-x}+4 x-1\right )\) |
(x/2048 - x^2/512 + x^3/96 - x^4/16 - (683*Sqrt[-x + x^2])/2048 + (149*(1 - 2*x)*Sqrt[-x + x^2])/1024 - (-x + x^2)^(3/2)/6 - (x*(-x + x^2)^(3/2))/16 + ArcTanh[(1 - 10*x)/(6*Sqrt[-x + x^2])]/16384 - (1537*ArcTanh[x/Sqrt[-x + x^2]])/8192 - Log[1 + 8*x]/16384)/2 + (x^4*Log[-1 + 4*x + 4*Sqrt[-x + x^ 2]])/4
3.2.1.3.1 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[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]])
Time = 0.12 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.39
| method | result | size |
| parts | \(\frac {\ln \left (-1+4 x +4 \sqrt {\left (-1+x \right ) x}\right ) x^{4}}{4}+\frac {x}{4096}+\frac {x^{3}}{192}-\frac {x^{4}}{32}+\frac {\sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}}{65536}-\frac {5 \ln \left (-\frac {1}{2}+x +\sqrt {\left (x +\frac {1}{8}\right )^{2}-\frac {5 x}{4}-\frac {1}{64}}\right )}{65536}-\frac {41 x^{2} \sqrt {x^{2}-x}}{960}-\frac {283 x \sqrt {x^{2}-x}}{6144}-\frac {x^{2}}{1024}-\frac {3069 \ln \left (-\frac {1}{2}+x +\sqrt {x^{2}-x}\right )}{65536}+\frac {\operatorname {arctanh}\left (\frac {\frac {4}{3}-\frac {40 x}{3}}{\sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}}\right )}{32768}+\frac {\left (x^{2}-x \right )^{\frac {3}{2}}}{16}-\frac {581 \sqrt {x^{2}-x}}{8192}-\frac {\ln \left (1+8 x \right )}{32768}+\frac {95 \left (2 x -1\right ) \sqrt {x^{2}-x}}{4096}+\frac {x^{2} \left (x^{2}-x \right )^{\frac {3}{2}}}{10}-\frac {x^{4} \sqrt {x^{2}-x}}{10}-\frac {x^{3} \sqrt {x^{2}-x}}{320}+\frac {23 x \left (x^{2}-x \right )^{\frac {3}{2}}}{320}\) | \(239\) |
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^2-3069/65536* 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)-581/8192*(x^2-x)^(1/2)-1/32768*ln(1+8*x)+9 5/4096*(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)
Time = 0.32 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.78 \[ \int x^3 \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=-\frac {1}{32} \, x^{4} + \frac {1}{192} \, x^{3} - \frac {1}{1024} \, x^{2} + \frac {1}{4} \, {\left (x^{4} - 1\right )} \log \left (4 \, x + 4 \, \sqrt {x^{2} - x} - 1\right ) - \frac {1}{12288} \, {\left (384 \, x^{3} + 640 \, x^{2} + 764 \, x + 1155\right )} \sqrt {x^{2} - x} + \frac {1}{4096} \, x + \frac {4095}{32768} \, \log \left (8 \, x + 1\right ) - \frac {2559}{32768} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} + 1\right ) + \frac {4095}{32768} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} - 1\right ) - \frac {4095}{32768} \, \log \left (-4 \, x + 4 \, \sqrt {x^{2} - x} + 1\right ) \]
-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) + 1/409 6*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*sq rt(x^2 - x) + 1)
Timed out. \[ \int x^3 \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\text {Timed out} \]
\[ \int x^3 \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\int { x^{3} \log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right ) \,d x } \]
Time = 0.38 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.78 \[ \int x^3 \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {1}{4} \, x^{4} \log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right ) - \frac {1}{32} \, x^{4} + \frac {1}{192} \, x^{3} - \frac {1}{1024} \, x^{2} - \frac {1}{12288} \, {\left (4 \, {\left (32 \, {\left (3 \, x + 5\right )} x + 191\right )} x + 1155\right )} \sqrt {x^{2} - x} + \frac {1}{4096} \, x - \frac {1}{32768} \, \log \left ({\left | 8 \, x + 1 \right |}\right ) + \frac {1537}{32768} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x} + 1 \right |}\right ) - \frac {1}{32768} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x} - 1 \right |}\right ) + \frac {1}{32768} \, \log \left ({\left | -4 \, x + 4 \, \sqrt {x^{2} - x} + 1 \right |}\right ) \]
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 + 191)*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))
Timed out. \[ \int x^3 \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\int x^3\,\ln \left (4\,x+4\,\sqrt {x\,\left (x-1\right )}-1\right ) \,d x \]