Integrand size = 70, antiderivative size = 21 \[ \int \frac {-4+9 x-2 x^2+\left (-x^2+2 x^3\right ) \log (2)+\left (-12+x-x^2+\left (24 x-2 x^2+2 x^3\right ) \log (2)\right ) \log \left (-12+x-x^2\right )}{12-x+x^2} \, dx=\left (4-x+x^2 \log (2)\right ) \log \left (-12+x-x^2\right ) \] Output:
ln(-x^2+x-12)*(4-x+x^2*ln(2))
Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {-4+9 x-2 x^2+\left (-x^2+2 x^3\right ) \log (2)+\left (-12+x-x^2+\left (24 x-2 x^2+2 x^3\right ) \log (2)\right ) \log \left (-12+x-x^2\right )}{12-x+x^2} \, dx=x (-1+x \log (2)) \log \left (-12+x-x^2\right )+4 \log \left (12-x+x^2\right ) \] Input:
Integrate[(-4 + 9*x - 2*x^2 + (-x^2 + 2*x^3)*Log[2] + (-12 + x - x^2 + (24 *x - 2*x^2 + 2*x^3)*Log[2])*Log[-12 + x - x^2])/(12 - x + x^2),x]
Output:
x*(-1 + x*Log[2])*Log[-12 + x - x^2] + 4*Log[12 - x + x^2]
Leaf count is larger than twice the leaf count of optimal. \(162\) vs. \(2(21)=42\).
Time = 0.64 (sec) , antiderivative size = 162, normalized size of antiderivative = 7.71, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {7279, 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.
\(\displaystyle \int \frac {-2 x^2+\left (-x^2+\left (2 x^3-2 x^2+24 x\right ) \log (2)+x-12\right ) \log \left (-x^2+x-12\right )+\left (2 x^3-x^2\right ) \log (2)+9 x-4}{x^2-x+12} \, dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {(2 x-1) \left (x^2 \log (2)-x+4\right )}{x^2-x+12}+(x \log (4)-1) \log \left (-x^2+x-12\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \sqrt {47} (2-\log (4)) \arctan \left (\frac {1-2 x}{\sqrt {47}}\right )-\sqrt {47} (1-\log (2)) \arctan \left (\frac {1-2 x}{\sqrt {47}}\right )-\frac {\left (2-23 \log ^2(4)-\log (16)\right ) \log \left (x^2-x+12\right )}{4 \log (4)}-\frac {1}{2} x^2 \log (4)+x^2 \log (2)+\frac {(1-x \log (4))^2 \log \left (-x^2+x-12\right )}{2 \log (4)}+\frac {1}{2} (7-23 \log (2)) \log \left (x^2-x+12\right )+\frac {1}{2} x (4-\log (4))-x (2-\log (2))\) |
Input:
Int[(-4 + 9*x - 2*x^2 + (-x^2 + 2*x^3)*Log[2] + (-12 + x - x^2 + (24*x - 2 *x^2 + 2*x^3)*Log[2])*Log[-12 + x - x^2])/(12 - x + x^2),x]
Output:
-(Sqrt[47]*ArcTan[(1 - 2*x)/Sqrt[47]]*(1 - Log[2])) - x*(2 - Log[2]) + x^2 *Log[2] + (Sqrt[47]*ArcTan[(1 - 2*x)/Sqrt[47]]*(2 - Log[4]))/2 + (x*(4 - L og[4]))/2 - (x^2*Log[4])/2 + ((1 - x*Log[4])^2*Log[-12 + x - x^2])/(2*Log[ 4]) + ((7 - 23*Log[2])*Log[12 - x + x^2])/2 - ((2 - 23*Log[4]^2 - Log[16]) *Log[12 - x + x^2])/(4*Log[4])
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 1.38 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57
method | result | size |
risch | \(\left (x^{2} \ln \left (2\right )-x \right ) \ln \left (-x^{2}+x -12\right )+4 \ln \left (x^{2}-x +12\right )\) | \(33\) |
default | \(\ln \left (2\right ) \ln \left (-x^{2}+x -12\right ) x^{2}+4 \ln \left (x^{2}-x +12\right )-\ln \left (-x^{2}+x -12\right ) x\) | \(40\) |
norman | \(4 \ln \left (-x^{2}+x -12\right )+\ln \left (2\right ) \ln \left (-x^{2}+x -12\right ) x^{2}-\ln \left (-x^{2}+x -12\right ) x\) | \(40\) |
parallelrisch | \(4 \ln \left (-x^{2}+x -12\right )+\ln \left (2\right ) \ln \left (-x^{2}+x -12\right ) x^{2}-\ln \left (-x^{2}+x -12\right ) x\) | \(40\) |
parts | \(2 \ln \left (2\right ) \left (\frac {\ln \left (-x^{2}+x -12\right ) x^{2}}{2}-\frac {x^{2}}{2}-\frac {x}{2}+\frac {23 \ln \left (x^{2}-x +12\right )}{4}+\frac {\sqrt {47}\, \arctan \left (\frac {\left (-1+2 x \right ) \sqrt {47}}{47}\right )}{2}\right )-\ln \left (-x^{2}+x -12\right ) x +\frac {\ln \left (x^{2}-x +12\right )}{2}-\sqrt {47}\, \arctan \left (\frac {\left (-1+2 x \right ) \sqrt {47}}{47}\right )+x^{2} \ln \left (2\right )+x \ln \left (2\right )+\frac {\left (-23 \ln \left (2\right )+7\right ) \ln \left (x^{2}-x +12\right )}{2}+\frac {2 \left (-\frac {47 \ln \left (2\right )}{2}+\frac {47}{2}\right ) \sqrt {47}\, \arctan \left (\frac {\left (-1+2 x \right ) \sqrt {47}}{47}\right )}{47}\) | \(144\) |
Input:
int((((2*x^3-2*x^2+24*x)*ln(2)-x^2+x-12)*ln(-x^2+x-12)+(2*x^3-x^2)*ln(2)-2 *x^2+9*x-4)/(x^2-x+12),x,method=_RETURNVERBOSE)
Output:
(x^2*ln(2)-x)*ln(-x^2+x-12)+4*ln(x^2-x+12)
Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {-4+9 x-2 x^2+\left (-x^2+2 x^3\right ) \log (2)+\left (-12+x-x^2+\left (24 x-2 x^2+2 x^3\right ) \log (2)\right ) \log \left (-12+x-x^2\right )}{12-x+x^2} \, dx={\left (x^{2} \log \left (2\right ) - x + 4\right )} \log \left (-x^{2} + x - 12\right ) \] Input:
integrate((((2*x^3-2*x^2+24*x)*log(2)-x^2+x-12)*log(-x^2+x-12)+(2*x^3-x^2) *log(2)-2*x^2+9*x-4)/(x^2-x+12),x, algorithm="fricas")
Output:
(x^2*log(2) - x + 4)*log(-x^2 + x - 12)
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {-4+9 x-2 x^2+\left (-x^2+2 x^3\right ) \log (2)+\left (-12+x-x^2+\left (24 x-2 x^2+2 x^3\right ) \log (2)\right ) \log \left (-12+x-x^2\right )}{12-x+x^2} \, dx=\left (x^{2} \log {\left (2 \right )} - x\right ) \log {\left (- x^{2} + x - 12 \right )} + 4 \log {\left (x^{2} - x + 12 \right )} \] Input:
integrate((((2*x**3-2*x**2+24*x)*ln(2)-x**2+x-12)*ln(-x**2+x-12)+(2*x**3-x **2)*ln(2)-2*x**2+9*x-4)/(x**2-x+12),x)
Output:
(x**2*log(2) - x)*log(-x**2 + x - 12) + 4*log(x**2 - x + 12)
Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (21) = 42\).
Time = 0.31 (sec) , antiderivative size = 165, normalized size of antiderivative = 7.86 \[ \int \frac {-4+9 x-2 x^2+\left (-x^2+2 x^3\right ) \log (2)+\left (-12+x-x^2+\left (24 x-2 x^2+2 x^3\right ) \log (2)\right ) \log \left (-12+x-x^2\right )}{12-x+x^2} \, dx=-x^{2} \log \left (2\right ) + \sqrt {47} {\left (\log \left (2\right ) - 1\right )} \arctan \left (\frac {1}{47} \, \sqrt {47} {\left (2 \, x - 1\right )}\right ) - x {\left (\log \left (2\right ) - 2\right )} + \frac {1}{47} \, {\left (47 \, x^{2} - 70 \, \sqrt {47} \arctan \left (\frac {1}{47} \, \sqrt {47} {\left (2 \, x - 1\right )}\right ) + 94 \, x - 517 \, \log \left (x^{2} - x + 12\right )\right )} \log \left (2\right ) + \frac {1}{94} \, {\left (46 \, \sqrt {47} \arctan \left (\frac {1}{47} \, \sqrt {47} {\left (2 \, x - 1\right )}\right ) - 94 \, x - 47 \, \log \left (x^{2} - x + 12\right )\right )} \log \left (2\right ) + \frac {1}{2} \, {\left (2 \, x^{2} \log \left (2\right ) - 2 \, x + 23 \, \log \left (2\right ) + 1\right )} \log \left (-x^{2} + x - 12\right ) + \sqrt {47} \arctan \left (\frac {1}{47} \, \sqrt {47} {\left (2 \, x - 1\right )}\right ) - 2 \, x + \frac {7}{2} \, \log \left (x^{2} - x + 12\right ) \] Input:
integrate((((2*x^3-2*x^2+24*x)*log(2)-x^2+x-12)*log(-x^2+x-12)+(2*x^3-x^2) *log(2)-2*x^2+9*x-4)/(x^2-x+12),x, algorithm="maxima")
Output:
-x^2*log(2) + sqrt(47)*(log(2) - 1)*arctan(1/47*sqrt(47)*(2*x - 1)) - x*(l og(2) - 2) + 1/47*(47*x^2 - 70*sqrt(47)*arctan(1/47*sqrt(47)*(2*x - 1)) + 94*x - 517*log(x^2 - x + 12))*log(2) + 1/94*(46*sqrt(47)*arctan(1/47*sqrt( 47)*(2*x - 1)) - 94*x - 47*log(x^2 - x + 12))*log(2) + 1/2*(2*x^2*log(2) - 2*x + 23*log(2) + 1)*log(-x^2 + x - 12) + sqrt(47)*arctan(1/47*sqrt(47)*( 2*x - 1)) - 2*x + 7/2*log(x^2 - x + 12)
Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \frac {-4+9 x-2 x^2+\left (-x^2+2 x^3\right ) \log (2)+\left (-12+x-x^2+\left (24 x-2 x^2+2 x^3\right ) \log (2)\right ) \log \left (-12+x-x^2\right )}{12-x+x^2} \, dx={\left (x^{2} \log \left (2\right ) - x\right )} \log \left (-x^{2} + x - 12\right ) + 4 \, \log \left (x^{2} - x + 12\right ) \] Input:
integrate((((2*x^3-2*x^2+24*x)*log(2)-x^2+x-12)*log(-x^2+x-12)+(2*x^3-x^2) *log(2)-2*x^2+9*x-4)/(x^2-x+12),x, algorithm="giac")
Output:
(x^2*log(2) - x)*log(-x^2 + x - 12) + 4*log(x^2 - x + 12)
Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {-4+9 x-2 x^2+\left (-x^2+2 x^3\right ) \log (2)+\left (-12+x-x^2+\left (24 x-2 x^2+2 x^3\right ) \log (2)\right ) \log \left (-12+x-x^2\right )}{12-x+x^2} \, dx=\ln \left (-x^2+x-12\right )\,\left (\ln \left (2\right )\,x^2-x+4\right ) \] Input:
int(-(log(2)*(x^2 - 2*x^3) - 9*x - log(x - x^2 - 12)*(x + log(2)*(24*x - 2 *x^2 + 2*x^3) - x^2 - 12) + 2*x^2 + 4)/(x^2 - x + 12),x)
Output:
log(x - x^2 - 12)*(x^2*log(2) - x + 4)
Time = 0.17 (sec) , antiderivative size = 76, normalized size of antiderivative = 3.62 \[ \int \frac {-4+9 x-2 x^2+\left (-x^2+2 x^3\right ) \log (2)+\left (-12+x-x^2+\left (24 x-2 x^2+2 x^3\right ) \log (2)\right ) \log \left (-12+x-x^2\right )}{12-x+x^2} \, dx=-\frac {23 \,\mathrm {log}\left (x^{2}-x +12\right ) \mathrm {log}\left (2\right )}{2}+\frac {7 \,\mathrm {log}\left (x^{2}-x +12\right )}{2}+\mathrm {log}\left (-x^{2}+x -12\right ) \mathrm {log}\left (2\right ) x^{2}+\frac {23 \,\mathrm {log}\left (-x^{2}+x -12\right ) \mathrm {log}\left (2\right )}{2}-\mathrm {log}\left (-x^{2}+x -12\right ) x +\frac {\mathrm {log}\left (-x^{2}+x -12\right )}{2} \] Input:
int((((2*x^3-2*x^2+24*x)*log(2)-x^2+x-12)*log(-x^2+x-12)+(2*x^3-x^2)*log(2 )-2*x^2+9*x-4)/(x^2-x+12),x)
Output:
( - 23*log(x**2 - x + 12)*log(2) + 7*log(x**2 - x + 12) + 2*log( - x**2 + x - 12)*log(2)*x**2 + 23*log( - x**2 + x - 12)*log(2) - 2*log( - x**2 + x - 12)*x + log( - x**2 + x - 12))/2