Integrand size = 205, antiderivative size = 28 \[ \int \frac {\left (-3 x+6 x^2-x^3+2 x^4+\left (-5 x^3+10 x^4\right ) \log (4)\right ) \log \left (-3-x^2-5 x^2 \log (4)\right )+\log \left (5-x+x^2\right ) \left (-10 x^2+2 x^3-2 x^4+\left (-50 x^2+10 x^3-10 x^4\right ) \log (4)+\left (15-3 x+8 x^2-x^3+x^4+\left (25 x^2-5 x^3+5 x^4\right ) \log (4)\right ) \log \left (-3-x^2-5 x^2 \log (4)\right )\right )}{\left (15-3 x+8 x^2-x^3+x^4+\left (25 x^2-5 x^3+5 x^4\right ) \log (4)\right ) \log ^2\left (-3-x^2-5 x^2 \log (4)\right )} \, dx=\frac {x \log \left (5-x+x^2\right )}{\log \left (-3-x^2-5 x^2 \log (4)\right )} \]
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {\left (-3 x+6 x^2-x^3+2 x^4+\left (-5 x^3+10 x^4\right ) \log (4)\right ) \log \left (-3-x^2-5 x^2 \log (4)\right )+\log \left (5-x+x^2\right ) \left (-10 x^2+2 x^3-2 x^4+\left (-50 x^2+10 x^3-10 x^4\right ) \log (4)+\left (15-3 x+8 x^2-x^3+x^4+\left (25 x^2-5 x^3+5 x^4\right ) \log (4)\right ) \log \left (-3-x^2-5 x^2 \log (4)\right )\right )}{\left (15-3 x+8 x^2-x^3+x^4+\left (25 x^2-5 x^3+5 x^4\right ) \log (4)\right ) \log ^2\left (-3-x^2-5 x^2 \log (4)\right )} \, dx=\frac {x \log \left (5-x+x^2\right )}{\log \left (-3-x^2 (1+5 \log (4))\right )} \]
Integrate[((-3*x + 6*x^2 - x^3 + 2*x^4 + (-5*x^3 + 10*x^4)*Log[4])*Log[-3 - x^2 - 5*x^2*Log[4]] + Log[5 - x + x^2]*(-10*x^2 + 2*x^3 - 2*x^4 + (-50*x ^2 + 10*x^3 - 10*x^4)*Log[4] + (15 - 3*x + 8*x^2 - x^3 + x^4 + (25*x^2 - 5 *x^3 + 5*x^4)*Log[4])*Log[-3 - x^2 - 5*x^2*Log[4]]))/((15 - 3*x + 8*x^2 - x^3 + x^4 + (25*x^2 - 5*x^3 + 5*x^4)*Log[4])*Log[-3 - x^2 - 5*x^2*Log[4]]^ 2),x]
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.
| \(\displaystyle \int \frac {\left (2 x^4-x^3+6 x^2+\left (10 x^4-5 x^3\right ) \log (4)-3 x\right ) \log \left (-x^2-5 x^2 \log (4)-3\right )+\log \left (x^2-x+5\right ) \left (-2 x^4+2 x^3-10 x^2+\left (x^4-x^3+8 x^2+\left (5 x^4-5 x^3+25 x^2\right ) \log (4)-3 x+15\right ) \log \left (-x^2-5 x^2 \log (4)-3\right )+\left (-10 x^4+10 x^3-50 x^2\right ) \log (4)\right )}{\left (x^4-x^3+8 x^2+\left (5 x^4-5 x^3+25 x^2\right ) \log (4)-3 x+15\right ) \log ^2\left (-x^2-5 x^2 \log (4)-3\right )} \, dx\) |
| \(\Big \downarrow \) 2463 |
| \(\displaystyle \int \left (\frac {(-(x (1+5 \log (4)))-1-20 \log (4)) \left (\left (2 x^4-x^3+6 x^2+\left (10 x^4-5 x^3\right ) \log (4)-3 x\right ) \log \left (-x^2-5 x^2 \log (4)-3\right )+\log \left (x^2-x+5\right ) \left (-2 x^4+2 x^3-10 x^2+\left (x^4-x^3+8 x^2+\left (5 x^4-5 x^3+25 x^2\right ) \log (4)-3 x+15\right ) \log \left (-x^2-5 x^2 \log (4)-3\right )+\left (-10 x^4+10 x^3-50 x^2\right ) \log (4)\right )\right )}{\left (x^2-x+5\right ) \left (7+625 \log ^2(4)+115 \log (4)\right ) \log ^2\left (-x^2-5 x^2 \log (4)-3\right )}+\frac {(1+5 \log (4)) (x (1+5 \log (4))+2+25 \log (4)) \left (\left (2 x^4-x^3+6 x^2+\left (10 x^4-5 x^3\right ) \log (4)-3 x\right ) \log \left (-x^2-5 x^2 \log (4)-3\right )+\log \left (x^2-x+5\right ) \left (-2 x^4+2 x^3-10 x^2+\left (x^4-x^3+8 x^2+\left (5 x^4-5 x^3+25 x^2\right ) \log (4)-3 x+15\right ) \log \left (-x^2-5 x^2 \log (4)-3\right )+\left (-10 x^4+10 x^3-50 x^2\right ) \log (4)\right )\right )}{\left (7+625 \log ^2(4)+115 \log (4)\right ) \left (x^2 (1+5 \log (4))+3\right ) \log ^2\left (-x^2-5 x^2 \log (4)-3\right )}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {-\frac {2 x^2 (1+5 \log (4)) \log \left (x^2-x+5\right )}{x^2 (1+5 \log (4))+3}+\frac {(2 x-1) x \log \left (-\left (x^2 (1+5 \log (4))\right )-3\right )}{x^2-x+5}+\log \left (x^2-x+5\right ) \log \left (-\left (x^2 (1+5 \log (4))\right )-3\right )}{\log ^2\left (-\left (x^2 (1+5 \log (4))\right )-3\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2 x^2 (-1-5 \log (4)) \log \left (x^2-x+5\right )}{\left (x^2 (1+5 \log (4))+3\right ) \log ^2\left (-\left (x^2 (1+5 \log (4))\right )-3\right )}+\frac {2 x^2+x^2 \log \left (x^2-x+5\right )-x \log \left (x^2-x+5\right )+5 \log \left (x^2-x+5\right )-x}{\left (x^2-x+5\right ) \log \left (-\left (x^2 (1+5 \log (4))\right )-3\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -2 \int \frac {\log \left (x^2-x+5\right )}{\log ^2\left (-\left ((1+5 \log (4)) x^2\right )-3\right )}dx+i \sqrt {\frac {3}{1+5 \log (4)}} \int \frac {\log \left (x^2-x+5\right )}{\left (i \sqrt {\frac {3}{1+5 \log (4)}}-x\right ) \log ^2\left (-\left ((1+5 \log (4)) x^2\right )-3\right )}dx+i \sqrt {\frac {3}{1+5 \log (4)}} \int \frac {\log \left (x^2-x+5\right )}{\left (x+i \sqrt {\frac {3}{1+5 \log (4)}}\right ) \log ^2\left (-\left ((1+5 \log (4)) x^2\right )-3\right )}dx+2 \int \frac {1}{\log \left (x^2 (-1-5 \log (4))-3\right )}dx-\frac {20 i \int \frac {1}{\left (-2 x+i \sqrt {19}+1\right ) \log \left (x^2 (-1-5 \log (4))-3\right )}dx}{\sqrt {19}}+\frac {1}{19} \left (19-i \sqrt {19}\right ) \int \frac {1}{\left (2 x-i \sqrt {19}-1\right ) \log \left (x^2 (-1-5 \log (4))-3\right )}dx+\frac {1}{19} \left (19+i \sqrt {19}\right ) \int \frac {1}{\left (2 x+i \sqrt {19}-1\right ) \log \left (x^2 (-1-5 \log (4))-3\right )}dx-\frac {20 i \int \frac {1}{\left (2 x+i \sqrt {19}-1\right ) \log \left (x^2 (-1-5 \log (4))-3\right )}dx}{\sqrt {19}}+\int \frac {\log \left (x^2-x+5\right )}{\log \left (-\left ((1+5 \log (4)) x^2\right )-3\right )}dx\) |
Int[((-3*x + 6*x^2 - x^3 + 2*x^4 + (-5*x^3 + 10*x^4)*Log[4])*Log[-3 - x^2 - 5*x^2*Log[4]] + Log[5 - x + x^2]*(-10*x^2 + 2*x^3 - 2*x^4 + (-50*x^2 + 1 0*x^3 - 10*x^4)*Log[4] + (15 - 3*x + 8*x^2 - x^3 + x^4 + (25*x^2 - 5*x^3 + 5*x^4)*Log[4])*Log[-3 - x^2 - 5*x^2*Log[4]]))/((15 - 3*x + 8*x^2 - x^3 + x^4 + (25*x^2 - 5*x^3 + 5*x^4)*Log[4])*Log[-3 - x^2 - 5*x^2*Log[4]]^2),x]
3.9.80.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 14.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04
| method | result | size |
| risch | \(\frac {x \ln \left (x^{2}-x +5\right )}{\ln \left (-10 x^{2} \ln \left (2\right )-x^{2}-3\right )}\) | \(29\) |
| parallelrisch | \(\frac {x \ln \left (x^{2}-x +5\right )}{\ln \left (-10 x^{2} \ln \left (2\right )-x^{2}-3\right )}\) | \(29\) |
int((((2*(5*x^4-5*x^3+25*x^2)*ln(2)+x^4-x^3+8*x^2-3*x+15)*ln(-10*x^2*ln(2) -x^2-3)+2*(-10*x^4+10*x^3-50*x^2)*ln(2)-2*x^4+2*x^3-10*x^2)*ln(x^2-x+5)+(2 *(10*x^4-5*x^3)*ln(2)+2*x^4-x^3+6*x^2-3*x)*ln(-10*x^2*ln(2)-x^2-3))/(2*(5* x^4-5*x^3+25*x^2)*ln(2)+x^4-x^3+8*x^2-3*x+15)/ln(-10*x^2*ln(2)-x^2-3)^2,x, method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-3 x+6 x^2-x^3+2 x^4+\left (-5 x^3+10 x^4\right ) \log (4)\right ) \log \left (-3-x^2-5 x^2 \log (4)\right )+\log \left (5-x+x^2\right ) \left (-10 x^2+2 x^3-2 x^4+\left (-50 x^2+10 x^3-10 x^4\right ) \log (4)+\left (15-3 x+8 x^2-x^3+x^4+\left (25 x^2-5 x^3+5 x^4\right ) \log (4)\right ) \log \left (-3-x^2-5 x^2 \log (4)\right )\right )}{\left (15-3 x+8 x^2-x^3+x^4+\left (25 x^2-5 x^3+5 x^4\right ) \log (4)\right ) \log ^2\left (-3-x^2-5 x^2 \log (4)\right )} \, dx=\frac {x \log \left (x^{2} - x + 5\right )}{\log \left (-10 \, x^{2} \log \left (2\right ) - x^{2} - 3\right )} \]
integrate((((2*(5*x^4-5*x^3+25*x^2)*log(2)+x^4-x^3+8*x^2-3*x+15)*log(-10*x ^2*log(2)-x^2-3)+2*(-10*x^4+10*x^3-50*x^2)*log(2)-2*x^4+2*x^3-10*x^2)*log( x^2-x+5)+(2*(10*x^4-5*x^3)*log(2)+2*x^4-x^3+6*x^2-3*x)*log(-10*x^2*log(2)- x^2-3))/(2*(5*x^4-5*x^3+25*x^2)*log(2)+x^4-x^3+8*x^2-3*x+15)/log(-10*x^2*l og(2)-x^2-3)^2,x, algorithm=\
Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {\left (-3 x+6 x^2-x^3+2 x^4+\left (-5 x^3+10 x^4\right ) \log (4)\right ) \log \left (-3-x^2-5 x^2 \log (4)\right )+\log \left (5-x+x^2\right ) \left (-10 x^2+2 x^3-2 x^4+\left (-50 x^2+10 x^3-10 x^4\right ) \log (4)+\left (15-3 x+8 x^2-x^3+x^4+\left (25 x^2-5 x^3+5 x^4\right ) \log (4)\right ) \log \left (-3-x^2-5 x^2 \log (4)\right )\right )}{\left (15-3 x+8 x^2-x^3+x^4+\left (25 x^2-5 x^3+5 x^4\right ) \log (4)\right ) \log ^2\left (-3-x^2-5 x^2 \log (4)\right )} \, dx=\frac {x \log {\left (x^{2} - x + 5 \right )}}{\log {\left (- 10 x^{2} \log {\left (2 \right )} - x^{2} - 3 \right )}} \]
integrate((((2*(5*x**4-5*x**3+25*x**2)*ln(2)+x**4-x**3+8*x**2-3*x+15)*ln(- 10*x**2*ln(2)-x**2-3)+2*(-10*x**4+10*x**3-50*x**2)*ln(2)-2*x**4+2*x**3-10* x**2)*ln(x**2-x+5)+(2*(10*x**4-5*x**3)*ln(2)+2*x**4-x**3+6*x**2-3*x)*ln(-1 0*x**2*ln(2)-x**2-3))/(2*(5*x**4-5*x**3+25*x**2)*ln(2)+x**4-x**3+8*x**2-3* x+15)/ln(-10*x**2*ln(2)-x**2-3)**2,x)
Time = 0.33 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {\left (-3 x+6 x^2-x^3+2 x^4+\left (-5 x^3+10 x^4\right ) \log (4)\right ) \log \left (-3-x^2-5 x^2 \log (4)\right )+\log \left (5-x+x^2\right ) \left (-10 x^2+2 x^3-2 x^4+\left (-50 x^2+10 x^3-10 x^4\right ) \log (4)+\left (15-3 x+8 x^2-x^3+x^4+\left (25 x^2-5 x^3+5 x^4\right ) \log (4)\right ) \log \left (-3-x^2-5 x^2 \log (4)\right )\right )}{\left (15-3 x+8 x^2-x^3+x^4+\left (25 x^2-5 x^3+5 x^4\right ) \log (4)\right ) \log ^2\left (-3-x^2-5 x^2 \log (4)\right )} \, dx=\frac {x \log \left (x^{2} - x + 5\right )}{\log \left (-x^{2} {\left (10 \, \log \left (2\right ) + 1\right )} - 3\right )} \]
integrate((((2*(5*x^4-5*x^3+25*x^2)*log(2)+x^4-x^3+8*x^2-3*x+15)*log(-10*x ^2*log(2)-x^2-3)+2*(-10*x^4+10*x^3-50*x^2)*log(2)-2*x^4+2*x^3-10*x^2)*log( x^2-x+5)+(2*(10*x^4-5*x^3)*log(2)+2*x^4-x^3+6*x^2-3*x)*log(-10*x^2*log(2)- x^2-3))/(2*(5*x^4-5*x^3+25*x^2)*log(2)+x^4-x^3+8*x^2-3*x+15)/log(-10*x^2*l og(2)-x^2-3)^2,x, algorithm=\
Exception generated. \[ \int \frac {\left (-3 x+6 x^2-x^3+2 x^4+\left (-5 x^3+10 x^4\right ) \log (4)\right ) \log \left (-3-x^2-5 x^2 \log (4)\right )+\log \left (5-x+x^2\right ) \left (-10 x^2+2 x^3-2 x^4+\left (-50 x^2+10 x^3-10 x^4\right ) \log (4)+\left (15-3 x+8 x^2-x^3+x^4+\left (25 x^2-5 x^3+5 x^4\right ) \log (4)\right ) \log \left (-3-x^2-5 x^2 \log (4)\right )\right )}{\left (15-3 x+8 x^2-x^3+x^4+\left (25 x^2-5 x^3+5 x^4\right ) \log (4)\right ) \log ^2\left (-3-x^2-5 x^2 \log (4)\right )} \, dx=\text {Exception raised: TypeError} \]
integrate((((2*(5*x^4-5*x^3+25*x^2)*log(2)+x^4-x^3+8*x^2-3*x+15)*log(-10*x ^2*log(2)-x^2-3)+2*(-10*x^4+10*x^3-50*x^2)*log(2)-2*x^4+2*x^3-10*x^2)*log( x^2-x+5)+(2*(10*x^4-5*x^3)*log(2)+2*x^4-x^3+6*x^2-3*x)*log(-10*x^2*log(2)- x^2-3))/(2*(5*x^4-5*x^3+25*x^2)*log(2)+x^4-x^3+8*x^2-3*x+15)/log(-10*x^2*l og(2)-x^2-3)^2,x, algorithm=\
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Time = 1.15 (sec) , antiderivative size = 199, normalized size of antiderivative = 7.11 \[ \int \frac {\left (-3 x+6 x^2-x^3+2 x^4+\left (-5 x^3+10 x^4\right ) \log (4)\right ) \log \left (-3-x^2-5 x^2 \log (4)\right )+\log \left (5-x+x^2\right ) \left (-10 x^2+2 x^3-2 x^4+\left (-50 x^2+10 x^3-10 x^4\right ) \log (4)+\left (15-3 x+8 x^2-x^3+x^4+\left (25 x^2-5 x^3+5 x^4\right ) \log (4)\right ) \log \left (-3-x^2-5 x^2 \log (4)\right )\right )}{\left (15-3 x+8 x^2-x^3+x^4+\left (25 x^2-5 x^3+5 x^4\right ) \log (4)\right ) \log ^2\left (-3-x^2-5 x^2 \log (4)\right )} \, dx=x+\frac {x\,\ln \left (x^2-x+5\right )-\frac {\ln \left (-10\,x^2\,\ln \left (2\right )-x^2-3\right )\,\left (10\,x^2\,\ln \left (2\right )+x^2+3\right )\,\left (5\,\ln \left (x^2-x+5\right )-x-x\,\ln \left (x^2-x+5\right )+x^2\,\ln \left (x^2-x+5\right )+2\,x^2\right )}{2\,x\,\left (10\,\ln \left (2\right )+1\right )\,\left (x^2-x+5\right )}}{\ln \left (-10\,x^2\,\ln \left (2\right )-x^2-3\right )}-\frac {50\,\ln \left (2\right )+x\,\left (90\,\ln \left (2\right )+3\right )+8}{\left (20\,\ln \left (2\right )+2\right )\,x^2+\left (-20\,\ln \left (2\right )-2\right )\,x+100\,\ln \left (2\right )+10}+\frac {\ln \left (x^2-x+5\right )\,\left (\frac {x^2}{2}+\frac {3}{2\,\left (10\,\ln \left (2\right )+1\right )}\right )}{x} \]
int(-(log(- 10*x^2*log(2) - x^2 - 3)*(3*x + 2*log(2)*(5*x^3 - 10*x^4) - 6* x^2 + x^3 - 2*x^4) + log(x^2 - x + 5)*(2*log(2)*(50*x^2 - 10*x^3 + 10*x^4) - log(- 10*x^2*log(2) - x^2 - 3)*(2*log(2)*(25*x^2 - 5*x^3 + 5*x^4) - 3*x + 8*x^2 - x^3 + x^4 + 15) + 10*x^2 - 2*x^3 + 2*x^4))/(log(- 10*x^2*log(2) - x^2 - 3)^2*(2*log(2)*(25*x^2 - 5*x^3 + 5*x^4) - 3*x + 8*x^2 - x^3 + x^4 + 15)),x)
x + (x*log(x^2 - x + 5) - (log(- 10*x^2*log(2) - x^2 - 3)*(10*x^2*log(2) + x^2 + 3)*(5*log(x^2 - x + 5) - x - x*log(x^2 - x + 5) + x^2*log(x^2 - x + 5) + 2*x^2))/(2*x*(10*log(2) + 1)*(x^2 - x + 5)))/log(- 10*x^2*log(2) - x ^2 - 3) - (50*log(2) + x*(90*log(2) + 3) + 8)/(100*log(2) - x*(20*log(2) + 2) + x^2*(20*log(2) + 2) + 10) + (log(x^2 - x + 5)*(3/(2*(10*log(2) + 1)) + x^2/2))/x