Integrand size = 129, antiderivative size = 30 \[ \int \frac {16-8 x^3+e \left (-64+16 x+4 x^2+32 x^3-8 x^4\right )+\left (4-2 x^3+e \left (-16+4 x+x^2+8 x^3-2 x^4\right )\right ) \log ^2(5)+e^x \left (e \left (-16-4 x^2+8 x^3\right )+e \left (-4-x^2+2 x^3\right ) \log ^2(5)\right )}{-x^2+e^{1+x} x^2+e \left (4 x^2-x^3\right )} \, dx=\left (4+\log ^2(5)\right ) \left (\frac {4}{x}+x^2-\log \left (-4+\frac {1}{e}-e^x+x\right )\right ) \] Output:
(4+ln(5)^2)*(x^2+4/x-ln(1/exp(1)-exp(x)+x-4))
Time = 2.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {16-8 x^3+e \left (-64+16 x+4 x^2+32 x^3-8 x^4\right )+\left (4-2 x^3+e \left (-16+4 x+x^2+8 x^3-2 x^4\right )\right ) \log ^2(5)+e^x \left (e \left (-16-4 x^2+8 x^3\right )+e \left (-4-x^2+2 x^3\right ) \log ^2(5)\right )}{-x^2+e^{1+x} x^2+e \left (4 x^2-x^3\right )} \, dx=\left (4+\log ^2(5)\right ) \left (\frac {4}{x}+x^2-\log \left (1-4 e-e^{1+x}+e x\right )\right ) \] Input:
Integrate[(16 - 8*x^3 + E*(-64 + 16*x + 4*x^2 + 32*x^3 - 8*x^4) + (4 - 2*x ^3 + E*(-16 + 4*x + x^2 + 8*x^3 - 2*x^4))*Log[5]^2 + E^x*(E*(-16 - 4*x^2 + 8*x^3) + E*(-4 - x^2 + 2*x^3)*Log[5]^2))/(-x^2 + E^(1 + x)*x^2 + E*(4*x^2 - x^3)),x]
Output:
(4 + Log[5]^2)*(4/x + x^2 - Log[1 - 4*E - E^(1 + x) + E*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.
\(\displaystyle \int \frac {-8 x^3+e^x \left (e \left (8 x^3-4 x^2-16\right )+e \left (2 x^3-x^2-4\right ) \log ^2(5)\right )+e \left (-8 x^4+32 x^3+4 x^2+16 x-64\right )+\left (-2 x^3+e \left (-2 x^4+8 x^3+x^2+4 x-16\right )+4\right ) \log ^2(5)+16}{e^{x+1} x^2-x^2+e \left (4 x^2-x^3\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {\left (2 \left (x^3-2\right )+e^{x+1} \left (-2 x^3+x^2+4\right )+e \left (2 x^4-8 x^3-x^2-4 x+16\right )\right ) \left (4+\log ^2(5)\right )}{\left (e (x-4)-e^{x+1}+1\right ) x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \left (4+\log ^2(5)\right ) \int \frac {e^{x+1} \left (-2 x^3+x^2+4\right )-2 \left (2-x^3\right )+e \left (2 x^4-8 x^3-x^2-4 x+16\right )}{\left (-e (4-x)-e^{x+1}+1\right ) x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \left (4+\log ^2(5)\right ) \int \left (\frac {e x-5 e+1}{e x-e^{x+1}-4 e+1}+\frac {2 x^3-x^2-4}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \left (4+\log ^2(5)\right ) \left (-(1-5 e) \int \frac {1}{-e x+e^{x+1}+4 e-1}dx+e \int \frac {x}{e x-e^{x+1}-4 e+1}dx+x^2-x+\frac {4}{x}\right )\) |
Input:
Int[(16 - 8*x^3 + E*(-64 + 16*x + 4*x^2 + 32*x^3 - 8*x^4) + (4 - 2*x^3 + E *(-16 + 4*x + x^2 + 8*x^3 - 2*x^4))*Log[5]^2 + E^x*(E*(-16 - 4*x^2 + 8*x^3 ) + E*(-4 - x^2 + 2*x^3)*Log[5]^2))/(-x^2 + E^(1 + x)*x^2 + E*(4*x^2 - x^3 )),x]
Output:
$Aborted
Time = 0.66 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.67
method | result | size |
norman | \(\frac {\left (4+\ln \left (5\right )^{2}\right ) x^{3}+4 \ln \left (5\right )^{2}+16}{x}+\left (-\ln \left (5\right )^{2}-4\right ) \ln \left (x \,{\mathrm e}-{\mathrm e} \,{\mathrm e}^{x}-4 \,{\mathrm e}+1\right )\) | \(50\) |
risch | \(x^{2} \ln \left (5\right )^{2}+4 x^{2}+\frac {4 \ln \left (5\right )^{2}}{x}+\frac {16}{x}-\ln \left (5\right )^{2} \ln \left (-x +4+{\mathrm e}^{x}-{\mathrm e}^{-1}\right )-4 \ln \left (-x +4+{\mathrm e}^{x}-{\mathrm e}^{-1}\right )\) | \(61\) |
parallelrisch | \(-\frac {-x^{3} \ln \left (5\right )^{2}+\ln \left (5\right )^{2} \ln \left (\left (x \,{\mathrm e}-{\mathrm e} \,{\mathrm e}^{x}-4 \,{\mathrm e}+1\right ) {\mathrm e}^{-1}\right ) x -4 x^{3}-4 \ln \left (5\right )^{2}+4 \ln \left (\left (x \,{\mathrm e}-{\mathrm e} \,{\mathrm e}^{x}-4 \,{\mathrm e}+1\right ) {\mathrm e}^{-1}\right ) x -16}{x}\) | \(81\) |
Input:
int((((2*x^3-x^2-4)*exp(1)*ln(5)^2+(8*x^3-4*x^2-16)*exp(1))*exp(x)+((-2*x^ 4+8*x^3+x^2+4*x-16)*exp(1)-2*x^3+4)*ln(5)^2+(-8*x^4+32*x^3+4*x^2+16*x-64)* exp(1)-8*x^3+16)/(x^2*exp(1)*exp(x)+(-x^3+4*x^2)*exp(1)-x^2),x,method=_RET URNVERBOSE)
Output:
((4+ln(5)^2)*x^3+4*ln(5)^2+16)/x+(-ln(5)^2-4)*ln(x*exp(1)-exp(1)*exp(x)-4* exp(1)+1)
Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.57 \[ \int \frac {16-8 x^3+e \left (-64+16 x+4 x^2+32 x^3-8 x^4\right )+\left (4-2 x^3+e \left (-16+4 x+x^2+8 x^3-2 x^4\right )\right ) \log ^2(5)+e^x \left (e \left (-16-4 x^2+8 x^3\right )+e \left (-4-x^2+2 x^3\right ) \log ^2(5)\right )}{-x^2+e^{1+x} x^2+e \left (4 x^2-x^3\right )} \, dx=\frac {4 \, x^{3} + {\left (x^{3} + 4\right )} \log \left (5\right )^{2} - {\left (x \log \left (5\right )^{2} + 4 \, x\right )} \log \left (-{\left (x - 4\right )} e + e^{\left (x + 1\right )} - 1\right ) + 16}{x} \] Input:
integrate((((2*x^3-x^2-4)*exp(1)*log(5)^2+(8*x^3-4*x^2-16)*exp(1))*exp(x)+ ((-2*x^4+8*x^3+x^2+4*x-16)*exp(1)-2*x^3+4)*log(5)^2+(-8*x^4+32*x^3+4*x^2+1 6*x-64)*exp(1)-8*x^3+16)/(x^2*exp(1)*exp(x)+(-x^3+4*x^2)*exp(1)-x^2),x, al gorithm="fricas")
Output:
(4*x^3 + (x^3 + 4)*log(5)^2 - (x*log(5)^2 + 4*x)*log(-(x - 4)*e + e^(x + 1 ) - 1) + 16)/x
Time = 0.16 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.60 \[ \int \frac {16-8 x^3+e \left (-64+16 x+4 x^2+32 x^3-8 x^4\right )+\left (4-2 x^3+e \left (-16+4 x+x^2+8 x^3-2 x^4\right )\right ) \log ^2(5)+e^x \left (e \left (-16-4 x^2+8 x^3\right )+e \left (-4-x^2+2 x^3\right ) \log ^2(5)\right )}{-x^2+e^{1+x} x^2+e \left (4 x^2-x^3\right )} \, dx=x^{2} \left (\log {\left (5 \right )}^{2} + 4\right ) + \left (-4 - \log {\left (5 \right )}^{2}\right ) \log {\left (\frac {- e x - 1 + 4 e}{e} + e^{x} \right )} + \frac {4 \log {\left (5 \right )}^{2} + 16}{x} \] Input:
integrate((((2*x**3-x**2-4)*exp(1)*ln(5)**2+(8*x**3-4*x**2-16)*exp(1))*exp (x)+((-2*x**4+8*x**3+x**2+4*x-16)*exp(1)-2*x**3+4)*ln(5)**2+(-8*x**4+32*x* *3+4*x**2+16*x-64)*exp(1)-8*x**3+16)/(x**2*exp(1)*exp(x)+(-x**3+4*x**2)*ex p(1)-x**2),x)
Output:
x**2*(log(5)**2 + 4) + (-4 - log(5)**2)*log((-E*x - 1 + 4*E)*exp(-1) + exp (x)) + (4*log(5)**2 + 16)/x
Time = 0.15 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.73 \[ \int \frac {16-8 x^3+e \left (-64+16 x+4 x^2+32 x^3-8 x^4\right )+\left (4-2 x^3+e \left (-16+4 x+x^2+8 x^3-2 x^4\right )\right ) \log ^2(5)+e^x \left (e \left (-16-4 x^2+8 x^3\right )+e \left (-4-x^2+2 x^3\right ) \log ^2(5)\right )}{-x^2+e^{1+x} x^2+e \left (4 x^2-x^3\right )} \, dx=-{\left (\log \left (5\right )^{2} + 4\right )} \log \left (-{\left (x e - 4 \, e - e^{\left (x + 1\right )} + 1\right )} e^{\left (-1\right )}\right ) + \frac {{\left (\log \left (5\right )^{2} + 4\right )} x^{3} + 4 \, \log \left (5\right )^{2} + 16}{x} \] Input:
integrate((((2*x^3-x^2-4)*exp(1)*log(5)^2+(8*x^3-4*x^2-16)*exp(1))*exp(x)+ ((-2*x^4+8*x^3+x^2+4*x-16)*exp(1)-2*x^3+4)*log(5)^2+(-8*x^4+32*x^3+4*x^2+1 6*x-64)*exp(1)-8*x^3+16)/(x^2*exp(1)*exp(x)+(-x^3+4*x^2)*exp(1)-x^2),x, al gorithm="maxima")
Output:
-(log(5)^2 + 4)*log(-(x*e - 4*e - e^(x + 1) + 1)*e^(-1)) + ((log(5)^2 + 4) *x^3 + 4*log(5)^2 + 16)/x
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (28) = 56\).
Time = 0.12 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.23 \[ \int \frac {16-8 x^3+e \left (-64+16 x+4 x^2+32 x^3-8 x^4\right )+\left (4-2 x^3+e \left (-16+4 x+x^2+8 x^3-2 x^4\right )\right ) \log ^2(5)+e^x \left (e \left (-16-4 x^2+8 x^3\right )+e \left (-4-x^2+2 x^3\right ) \log ^2(5)\right )}{-x^2+e^{1+x} x^2+e \left (4 x^2-x^3\right )} \, dx=\frac {x^{3} \log \left (5\right )^{2} - x \log \left (5\right )^{2} \log \left (-x e + 4 \, e + e^{\left (x + 1\right )} - 1\right ) + 4 \, x^{3} + 4 \, \log \left (5\right )^{2} - 4 \, x \log \left (-x e + 4 \, e + e^{\left (x + 1\right )} - 1\right ) + 16}{x} \] Input:
integrate((((2*x^3-x^2-4)*exp(1)*log(5)^2+(8*x^3-4*x^2-16)*exp(1))*exp(x)+ ((-2*x^4+8*x^3+x^2+4*x-16)*exp(1)-2*x^3+4)*log(5)^2+(-8*x^4+32*x^3+4*x^2+1 6*x-64)*exp(1)-8*x^3+16)/(x^2*exp(1)*exp(x)+(-x^3+4*x^2)*exp(1)-x^2),x, al gorithm="giac")
Output:
(x^3*log(5)^2 - x*log(5)^2*log(-x*e + 4*e + e^(x + 1) - 1) + 4*x^3 + 4*log (5)^2 - 4*x*log(-x*e + 4*e + e^(x + 1) - 1) + 16)/x
Time = 0.16 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37 \[ \int \frac {16-8 x^3+e \left (-64+16 x+4 x^2+32 x^3-8 x^4\right )+\left (4-2 x^3+e \left (-16+4 x+x^2+8 x^3-2 x^4\right )\right ) \log ^2(5)+e^x \left (e \left (-16-4 x^2+8 x^3\right )+e \left (-4-x^2+2 x^3\right ) \log ^2(5)\right )}{-x^2+e^{1+x} x^2+e \left (4 x^2-x^3\right )} \, dx=\frac {\left ({\ln \left (5\right )}^2+4\right )\,x^3+4\,{\ln \left (5\right )}^2+16}{x}-\ln \left (x+{\mathrm {e}}^{-1}-{\mathrm {e}}^x-4\right )\,\left ({\ln \left (5\right )}^2+4\right ) \] Input:
int((log(5)^2*(exp(1)*(4*x + x^2 + 8*x^3 - 2*x^4 - 16) - 2*x^3 + 4) - exp( x)*(exp(1)*(4*x^2 - 8*x^3 + 16) + exp(1)*log(5)^2*(x^2 - 2*x^3 + 4)) + exp (1)*(16*x + 4*x^2 + 32*x^3 - 8*x^4 - 64) - 8*x^3 + 16)/(exp(1)*(4*x^2 - x^ 3) - x^2 + x^2*exp(1)*exp(x)),x)
Output:
(x^3*(log(5)^2 + 4) + 4*log(5)^2 + 16)/x - log(x + exp(-1) - exp(x) - 4)*( log(5)^2 + 4)
Time = 0.15 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.17 \[ \int \frac {16-8 x^3+e \left (-64+16 x+4 x^2+32 x^3-8 x^4\right )+\left (4-2 x^3+e \left (-16+4 x+x^2+8 x^3-2 x^4\right )\right ) \log ^2(5)+e^x \left (e \left (-16-4 x^2+8 x^3\right )+e \left (-4-x^2+2 x^3\right ) \log ^2(5)\right )}{-x^2+e^{1+x} x^2+e \left (4 x^2-x^3\right )} \, dx=\frac {-\mathrm {log}\left (e^{x} e -e x +4 e -1\right ) \mathrm {log}\left (5\right )^{2} x -4 \,\mathrm {log}\left (e^{x} e -e x +4 e -1\right ) x +\mathrm {log}\left (5\right )^{2} x^{3}+4 \mathrm {log}\left (5\right )^{2}+4 x^{3}+16}{x} \] Input:
int((((2*x^3-x^2-4)*exp(1)*log(5)^2+(8*x^3-4*x^2-16)*exp(1))*exp(x)+((-2*x ^4+8*x^3+x^2+4*x-16)*exp(1)-2*x^3+4)*log(5)^2+(-8*x^4+32*x^3+4*x^2+16*x-64 )*exp(1)-8*x^3+16)/(x^2*exp(1)*exp(x)+(-x^3+4*x^2)*exp(1)-x^2),x)
Output:
( - log(e**x*e - e*x + 4*e - 1)*log(5)**2*x - 4*log(e**x*e - e*x + 4*e - 1 )*x + log(5)**2*x**3 + 4*log(5)**2 + 4*x**3 + 16)/x