Integrand size = 74, antiderivative size = 22 \[ \int \frac {40-95 x+30 x^2+e^x \left (16-38 x+12 x^2\right )+\left (15-15 x+15 x^2+e^x \left (-10+16 x-16 x^2+6 x^3\right )\right ) \log \left (-1+x-x^2\right )}{1-x+x^2} \, dx=\left (5+2 e^x\right ) (-8+3 x) \log \left (-1+x-x^2\right ) \]
Time = 0.89 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {40-95 x+30 x^2+e^x \left (16-38 x+12 x^2\right )+\left (15-15 x+15 x^2+e^x \left (-10+16 x-16 x^2+6 x^3\right )\right ) \log \left (-1+x-x^2\right )}{1-x+x^2} \, dx=\left (-16 e^x+15 x+6 e^x x\right ) \log \left (-1+x-x^2\right )-40 \log \left (1-x+x^2\right ) \]
Integrate[(40 - 95*x + 30*x^2 + E^x*(16 - 38*x + 12*x^2) + (15 - 15*x + 15 *x^2 + E^x*(-10 + 16*x - 16*x^2 + 6*x^3))*Log[-1 + x - x^2])/(1 - x + x^2) ,x]
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 2.33 (sec) , antiderivative size = 861, normalized size of antiderivative = 39.14, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, 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.
\(\displaystyle \int \frac {30 x^2+e^x \left (12 x^2-38 x+16\right )+\left (15 x^2+e^x \left (6 x^3-16 x^2+16 x-10\right )-15 x+15\right ) \log \left (-x^2+x-1\right )-95 x+40}{x^2-x+1} \, dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {30 x^2}{x^2-x+1}-\frac {95 x}{x^2-x+1}+\frac {40}{x^2-x+1}+\frac {15 x^2 \log \left (-x^2+x-1\right )}{x^2-x+1}-\frac {15 x \log \left (-x^2+x-1\right )}{x^2-x+1}+\frac {15 \log \left (-x^2+x-1\right )}{x^2-x+1}+\frac {2 e^x \left (6 x^2-8 x^2 \log \left (-x^2+x-1\right )+8 x \log \left (-x^2+x-1\right )-5 \log \left (-x^2+x-1\right )+3 x^3 \log \left (-x^2+x-1\right )-19 x+8\right )}{x^2-x+1}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -10 i \sqrt {3} \arctan \left (\frac {1-2 x}{\sqrt {3}}\right )^2-20 \sqrt {3} \log \left (\frac {2 \sqrt {3}}{i (1-2 x)+\sqrt {3}}\right ) \arctan \left (\frac {1-2 x}{\sqrt {3}}\right )-10 \sqrt {3} \log \left (-x^2+x-1\right ) \arctan \left (\frac {1-2 x}{\sqrt {3}}\right )-\frac {5}{4} \left (3+i \sqrt {3}\right ) \log ^2\left (2 x-i \sqrt {3}-1\right )+\frac {5}{4} \left (3-i \sqrt {3}\right ) \log ^2\left (2 x-i \sqrt {3}-1\right )+\frac {5}{4} \left (3+i \sqrt {3}\right ) \log ^2\left (2 x+i \sqrt {3}-1\right )-\frac {5}{4} \left (3-i \sqrt {3}\right ) \log ^2\left (2 x+i \sqrt {3}-1\right )-\frac {5}{2} \left (3+i \sqrt {3}\right ) \log \left (\frac {i \left (-2 x-i \sqrt {3}+1\right )}{2 \sqrt {3}}\right ) \log \left (2 x-i \sqrt {3}-1\right )+\frac {5}{2} \left (3-i \sqrt {3}\right ) \log \left (\frac {i \left (-2 x-i \sqrt {3}+1\right )}{2 \sqrt {3}}\right ) \log \left (2 x-i \sqrt {3}-1\right )+\frac {5}{2} \left (3+i \sqrt {3}\right ) \log \left (-\frac {i \left (-2 x+i \sqrt {3}+1\right )}{2 \sqrt {3}}\right ) \log \left (2 x+i \sqrt {3}-1\right )-\frac {5}{2} \left (3-i \sqrt {3}\right ) \log \left (-\frac {i \left (-2 x+i \sqrt {3}+1\right )}{2 \sqrt {3}}\right ) \log \left (2 x+i \sqrt {3}-1\right )-6 e^x \log \left (-x^2+x-1\right )-2 e^x (5-3 x) \log \left (-x^2+x-1\right )+15 x \log \left (-x^2+x-1\right )+\frac {5}{2} \left (3+i \sqrt {3}\right ) \log \left (2 x-i \sqrt {3}-1\right ) \log \left (-x^2+x-1\right )-\frac {5}{2} \left (3-i \sqrt {3}\right ) \log \left (2 x-i \sqrt {3}-1\right ) \log \left (-x^2+x-1\right )-\frac {5}{2} \left (3+i \sqrt {3}\right ) \log \left (2 x+i \sqrt {3}-1\right ) \log \left (-x^2+x-1\right )+\frac {5}{2} \left (3-i \sqrt {3}\right ) \log \left (2 x+i \sqrt {3}-1\right ) \log \left (-x^2+x-1\right )-40 \log \left (x^2-x+1\right )-10 i \sqrt {3} \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {3}}{i (1-2 x)+\sqrt {3}}\right )-\frac {5}{2} \left (3+i \sqrt {3}\right ) \operatorname {PolyLog}\left (2,-\frac {-2 i x-\sqrt {3}+i}{2 \sqrt {3}}\right )+\frac {5}{2} \left (3-i \sqrt {3}\right ) \operatorname {PolyLog}\left (2,-\frac {-2 i x-\sqrt {3}+i}{2 \sqrt {3}}\right )+\frac {5}{2} \left (3+i \sqrt {3}\right ) \operatorname {PolyLog}\left (2,\frac {-2 i x+\sqrt {3}+i}{2 \sqrt {3}}\right )-\frac {5}{2} \left (3-i \sqrt {3}\right ) \operatorname {PolyLog}\left (2,\frac {-2 i x+\sqrt {3}+i}{2 \sqrt {3}}\right )\) |
Int[(40 - 95*x + 30*x^2 + E^x*(16 - 38*x + 12*x^2) + (15 - 15*x + 15*x^2 + E^x*(-10 + 16*x - 16*x^2 + 6*x^3))*Log[-1 + x - x^2])/(1 - x + x^2),x]
(-10*I)*Sqrt[3]*ArcTan[(1 - 2*x)/Sqrt[3]]^2 - 20*Sqrt[3]*ArcTan[(1 - 2*x)/ Sqrt[3]]*Log[(2*Sqrt[3])/(Sqrt[3] + I*(1 - 2*x))] + (5*(3 - I*Sqrt[3])*Log [((I/2)*(1 - I*Sqrt[3] - 2*x))/Sqrt[3]]*Log[-1 - I*Sqrt[3] + 2*x])/2 - (5* (3 + I*Sqrt[3])*Log[((I/2)*(1 - I*Sqrt[3] - 2*x))/Sqrt[3]]*Log[-1 - I*Sqrt [3] + 2*x])/2 + (5*(3 - I*Sqrt[3])*Log[-1 - I*Sqrt[3] + 2*x]^2)/4 - (5*(3 + I*Sqrt[3])*Log[-1 - I*Sqrt[3] + 2*x]^2)/4 - (5*(3 - I*Sqrt[3])*Log[((-1/ 2*I)*(1 + I*Sqrt[3] - 2*x))/Sqrt[3]]*Log[-1 + I*Sqrt[3] + 2*x])/2 + (5*(3 + I*Sqrt[3])*Log[((-1/2*I)*(1 + I*Sqrt[3] - 2*x))/Sqrt[3]]*Log[-1 + I*Sqrt [3] + 2*x])/2 - (5*(3 - I*Sqrt[3])*Log[-1 + I*Sqrt[3] + 2*x]^2)/4 + (5*(3 + I*Sqrt[3])*Log[-1 + I*Sqrt[3] + 2*x]^2)/4 - 6*E^x*Log[-1 + x - x^2] - 2* E^x*(5 - 3*x)*Log[-1 + x - x^2] + 15*x*Log[-1 + x - x^2] - 10*Sqrt[3]*ArcT an[(1 - 2*x)/Sqrt[3]]*Log[-1 + x - x^2] - (5*(3 - I*Sqrt[3])*Log[-1 - I*Sq rt[3] + 2*x]*Log[-1 + x - x^2])/2 + (5*(3 + I*Sqrt[3])*Log[-1 - I*Sqrt[3] + 2*x]*Log[-1 + x - x^2])/2 + (5*(3 - I*Sqrt[3])*Log[-1 + I*Sqrt[3] + 2*x] *Log[-1 + x - x^2])/2 - (5*(3 + I*Sqrt[3])*Log[-1 + I*Sqrt[3] + 2*x]*Log[- 1 + x - x^2])/2 - 40*Log[1 - x + x^2] - (10*I)*Sqrt[3]*PolyLog[2, 1 - (2*S qrt[3])/(Sqrt[3] + I*(1 - 2*x))] + (5*(3 - I*Sqrt[3])*PolyLog[2, -1/2*(I - Sqrt[3] - (2*I)*x)/Sqrt[3]])/2 - (5*(3 + I*Sqrt[3])*PolyLog[2, -1/2*(I - Sqrt[3] - (2*I)*x)/Sqrt[3]])/2 - (5*(3 - I*Sqrt[3])*PolyLog[2, (I + Sqrt[3 ] - (2*I)*x)/(2*Sqrt[3])])/2 + (5*(3 + I*Sqrt[3])*PolyLog[2, (I + Sqrt[...
3.28.3.3.1 Defintions of rubi rules used
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 = 2.47 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64
method | result | size |
risch | \(\left (6 \,{\mathrm e}^{x} x +15 x -16 \,{\mathrm e}^{x}\right ) \ln \left (-x^{2}+x -1\right )-40 \ln \left (x^{2}-x +1\right )\) | \(36\) |
default | \(-16 \,{\mathrm e}^{x} \ln \left (-x^{2}+x -1\right )+6 \,{\mathrm e}^{x} \ln \left (-x^{2}+x -1\right ) x -40 \ln \left (x^{2}-x +1\right )+15 \ln \left (-x^{2}+x -1\right ) x\) | \(52\) |
norman | \(-40 \ln \left (-x^{2}+x -1\right )-16 \,{\mathrm e}^{x} \ln \left (-x^{2}+x -1\right )+15 \ln \left (-x^{2}+x -1\right ) x +6 \,{\mathrm e}^{x} \ln \left (-x^{2}+x -1\right ) x\) | \(52\) |
parallelrisch | \(-40 \ln \left (-x^{2}+x -1\right )-16 \,{\mathrm e}^{x} \ln \left (-x^{2}+x -1\right )+15 \ln \left (-x^{2}+x -1\right ) x +6 \,{\mathrm e}^{x} \ln \left (-x^{2}+x -1\right ) x\) | \(52\) |
parts | \(-16 \,{\mathrm e}^{x} \ln \left (-x^{2}+x -1\right )+6 \,{\mathrm e}^{x} \ln \left (-x^{2}+x -1\right ) x -40 \ln \left (x^{2}-x +1\right )+15 \ln \left (-x^{2}+x -1\right ) x\) | \(52\) |
int((((6*x^3-16*x^2+16*x-10)*exp(x)+15*x^2-15*x+15)*ln(-x^2+x-1)+(12*x^2-3 8*x+16)*exp(x)+30*x^2-95*x+40)/(x^2-x+1),x,method=_RETURNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {40-95 x+30 x^2+e^x \left (16-38 x+12 x^2\right )+\left (15-15 x+15 x^2+e^x \left (-10+16 x-16 x^2+6 x^3\right )\right ) \log \left (-1+x-x^2\right )}{1-x+x^2} \, dx={\left (2 \, {\left (3 \, x - 8\right )} e^{x} + 15 \, x - 40\right )} \log \left (-x^{2} + x - 1\right ) \]
integrate((((6*x^3-16*x^2+16*x-10)*exp(x)+15*x^2-15*x+15)*log(-x^2+x-1)+(1 2*x^2-38*x+16)*exp(x)+30*x^2-95*x+40)/(x^2-x+1),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (19) = 38\).
Time = 0.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.09 \[ \int \frac {40-95 x+30 x^2+e^x \left (16-38 x+12 x^2\right )+\left (15-15 x+15 x^2+e^x \left (-10+16 x-16 x^2+6 x^3\right )\right ) \log \left (-1+x-x^2\right )}{1-x+x^2} \, dx=15 x \log {\left (- x^{2} + x - 1 \right )} + \left (6 x \log {\left (- x^{2} + x - 1 \right )} - 16 \log {\left (- x^{2} + x - 1 \right )}\right ) e^{x} - 40 \log {\left (x^{2} - x + 1 \right )} \]
integrate((((6*x**3-16*x**2+16*x-10)*exp(x)+15*x**2-15*x+15)*ln(-x**2+x-1) +(12*x**2-38*x+16)*exp(x)+30*x**2-95*x+40)/(x**2-x+1),x)
15*x*log(-x**2 + x - 1) + (6*x*log(-x**2 + x - 1) - 16*log(-x**2 + x - 1)) *exp(x) - 40*log(x**2 - x + 1)
Time = 0.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {40-95 x+30 x^2+e^x \left (16-38 x+12 x^2\right )+\left (15-15 x+15 x^2+e^x \left (-10+16 x-16 x^2+6 x^3\right )\right ) \log \left (-1+x-x^2\right )}{1-x+x^2} \, dx=\frac {1}{2} \, {\left (4 \, {\left (3 \, x - 8\right )} e^{x} + 30 \, x - 15\right )} \log \left (-x^{2} + x - 1\right ) - \frac {65}{2} \, \log \left (x^{2} - x + 1\right ) \]
integrate((((6*x^3-16*x^2+16*x-10)*exp(x)+15*x^2-15*x+15)*log(-x^2+x-1)+(1 2*x^2-38*x+16)*exp(x)+30*x^2-95*x+40)/(x^2-x+1),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (21) = 42\).
Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.32 \[ \int \frac {40-95 x+30 x^2+e^x \left (16-38 x+12 x^2\right )+\left (15-15 x+15 x^2+e^x \left (-10+16 x-16 x^2+6 x^3\right )\right ) \log \left (-1+x-x^2\right )}{1-x+x^2} \, dx=6 \, x e^{x} \log \left (-x^{2} + x - 1\right ) + 15 \, x \log \left (-x^{2} + x - 1\right ) - 16 \, e^{x} \log \left (-x^{2} + x - 1\right ) - 40 \, \log \left (x^{2} - x + 1\right ) \]
integrate((((6*x^3-16*x^2+16*x-10)*exp(x)+15*x^2-15*x+15)*log(-x^2+x-1)+(1 2*x^2-38*x+16)*exp(x)+30*x^2-95*x+40)/(x^2-x+1),x, algorithm=\
6*x*e^x*log(-x^2 + x - 1) + 15*x*log(-x^2 + x - 1) - 16*e^x*log(-x^2 + x - 1) - 40*log(x^2 - x + 1)
Time = 14.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \frac {40-95 x+30 x^2+e^x \left (16-38 x+12 x^2\right )+\left (15-15 x+15 x^2+e^x \left (-10+16 x-16 x^2+6 x^3\right )\right ) \log \left (-1+x-x^2\right )}{1-x+x^2} \, dx=\ln \left (-x^2+x-1\right )\,\left (15\,x+{\mathrm {e}}^x\,\left (6\,x-16\right )\right )-40\,\ln \left (x^2-x+1\right ) \]