Integrand size = 93, antiderivative size = 33 \[ \int \frac {-32 x^3+e^x \left (3-3 x-80 x^2\right ) \log (2)-50 e^{2 x} x \log ^2(2)+e^{4 x} \left (-64 x^2-160 e^x x \log (2)-100 e^{2 x} \log ^2(2)\right )}{16 x^2+40 e^x x \log (2)+25 e^{2 x} \log ^2(2)} \, dx=4-e^{4 x}-x^2-\frac {3}{-5+\frac {1}{1+\frac {e^x \log (2)}{x}}} \] Output:
4-x^2-3/(1/(1+ln(2)*exp(x)/x)-5)-exp(4*x)
Leaf count is larger than twice the leaf count of optimal. \(157\) vs. \(2(33)=66\).
Time = 0.10 (sec) , antiderivative size = 157, normalized size of antiderivative = 4.76 \[ \int \frac {-32 x^3+e^x \left (3-3 x-80 x^2\right ) \log (2)-50 e^{2 x} x \log ^2(2)+e^{4 x} \left (-64 x^2-160 e^x x \log (2)-100 e^{2 x} \log ^2(2)\right )}{16 x^2+40 e^x x \log (2)+25 e^{2 x} \log ^2(2)} \, dx=-\frac {125 e^{4 x} \log ^2(2) \log ^4(32)+125 x^2 \log ^2(2) \log ^4(32)-\frac {5 x \log (8) \log ^5(32)}{4 x+e^x \log (32)}+80 e^{2 x} \left (1-2 x+2 x^2\right ) \log ^2(32) \left (75 \log ^2(2)-20 \log (2) \log (32)+\log ^2(32)\right )-2560 e^x \left (-6+6 x-3 x^2+x^3\right ) \log (32) \left (50 \log ^2(2)-15 \log (2) \log (32)+\log ^2(32)\right )+1024 x^5 \left (125 \log ^2(2)-40 \log (2) \log (32)+3 \log ^2(32)\right )}{5 \log ^6(32)} \] Input:
Integrate[(-32*x^3 + E^x*(3 - 3*x - 80*x^2)*Log[2] - 50*E^(2*x)*x*Log[2]^2 + E^(4*x)*(-64*x^2 - 160*E^x*x*Log[2] - 100*E^(2*x)*Log[2]^2))/(16*x^2 + 40*E^x*x*Log[2] + 25*E^(2*x)*Log[2]^2),x]
Output:
-1/5*(125*E^(4*x)*Log[2]^2*Log[32]^4 + 125*x^2*Log[2]^2*Log[32]^4 - (5*x*L og[8]*Log[32]^5)/(4*x + E^x*Log[32]) + 80*E^(2*x)*(1 - 2*x + 2*x^2)*Log[32 ]^2*(75*Log[2]^2 - 20*Log[2]*Log[32] + Log[32]^2) - 2560*E^x*(-6 + 6*x - 3 *x^2 + x^3)*Log[32]*(50*Log[2]^2 - 15*Log[2]*Log[32] + Log[32]^2) + 1024*x ^5*(125*Log[2]^2 - 40*Log[2]*Log[32] + 3*Log[32]^2))/Log[32]^6
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.
\(\displaystyle \int \frac {-32 x^3+e^{4 x} \left (-64 x^2-100 e^{2 x} \log ^2(2)-160 e^x x \log (2)\right )+e^x \left (-80 x^2-3 x+3\right ) \log (2)-50 e^{2 x} x \log ^2(2)}{16 x^2+25 e^{2 x} \log ^2(2)+40 e^x x \log (2)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-32 x^3+e^{4 x} \left (-64 x^2-100 e^{2 x} \log ^2(2)-160 e^x x \log (2)\right )+e^x \left (-80 x^2-3 x+3\right ) \log (2)-50 e^{2 x} x \log ^2(2)}{\left (4 x+e^x \log (32)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {614400 x^5 \log ^2(2) \left (1-\frac {\log (32) \log (32768)}{75 \log ^2(2)}\right )-x \log (8) \log ^5(32)+\log (8) \log ^5(32)}{\log ^6(32) \left (4 x+e^x \log (32)\right )}+\frac {512 e^x x^3 \left (50 \log ^2(2)+\log ^2(32)-15 \log (2) \log (32)\right )}{\log ^5(32)}+\frac {2 x \left (-512 x^3 (25 \log (2)-3 \log (32)) (5 \log (2)-\log (32))-25 \log ^2(2) \log ^4(32)\right )}{\log ^6(32)}-\frac {64 e^{2 x} x^2 \left (75 \log ^2(2)+\log ^2(32)-20 \log (2) \log (32)\right )}{\log ^4(32)}-\frac {100 e^{4 x} \log ^2(2)}{\log ^2(32)}+\frac {4 (x-1) x \log (8)}{\log (32) \left (4 x+e^x \log (32)\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 \log (8) \int \frac {x^2}{\left (4 x+e^x \log (32)\right )^2}dx}{\log (32)}-\frac {4 \log (8) \int \frac {x}{\left (4 x+e^x \log (32)\right )^2}dx}{\log (32)}+\frac {\log (8) \int \frac {1}{4 x+e^x \log (32)}dx}{\log (32)}-\frac {\log (8) \int \frac {x}{4 x+e^x \log (32)}dx}{\log (32)}-\frac {25 x^2 \log ^2(2)}{\log ^2(32)}-\frac {25 e^{4 x} \log ^2(2)}{\log ^2(32)}\) |
Input:
Int[(-32*x^3 + E^x*(3 - 3*x - 80*x^2)*Log[2] - 50*E^(2*x)*x*Log[2]^2 + E^( 4*x)*(-64*x^2 - 160*E^x*x*Log[2] - 100*E^(2*x)*Log[2]^2))/(16*x^2 + 40*E^x *x*Log[2] + 25*E^(2*x)*Log[2]^2),x]
Output:
$Aborted
Time = 0.44 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85
method | result | size |
risch | \(-{\mathrm e}^{4 x}-x^{2}+\frac {3 x}{5 \left (5 \,{\mathrm e}^{x} \ln \left (2\right )+4 x \right )}\) | \(28\) |
norman | \(\frac {-\frac {3 \,{\mathrm e}^{x} \ln \left (2\right )}{4}-4 x^{3}-4 x \,{\mathrm e}^{4 x}-5 \ln \left (2\right ) {\mathrm e}^{5 x}-5 x^{2} \ln \left (2\right ) {\mathrm e}^{x}}{5 \,{\mathrm e}^{x} \ln \left (2\right )+4 x}\) | \(50\) |
parallelrisch | \(-\frac {20 x^{2} \ln \left (2\right ) {\mathrm e}^{x}+20 \,{\mathrm e}^{x} \ln \left (2\right ) {\mathrm e}^{4 x}+16 x^{3}+3 \,{\mathrm e}^{x} \ln \left (2\right )+16 x \,{\mathrm e}^{4 x}}{4 \left (5 \,{\mathrm e}^{x} \ln \left (2\right )+4 x \right )}\) | \(53\) |
Input:
int(((-100*ln(2)^2*exp(x)^2-160*x*ln(2)*exp(x)-64*x^2)*exp(4*x)-50*x*ln(2) ^2*exp(x)^2+(-80*x^2-3*x+3)*ln(2)*exp(x)-32*x^3)/(25*ln(2)^2*exp(x)^2+40*x *ln(2)*exp(x)+16*x^2),x,method=_RETURNVERBOSE)
Output:
-exp(4*x)-x^2+3/5*x/(5*exp(x)*ln(2)+4*x)
Time = 0.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.42 \[ \int \frac {-32 x^3+e^x \left (3-3 x-80 x^2\right ) \log (2)-50 e^{2 x} x \log ^2(2)+e^{4 x} \left (-64 x^2-160 e^x x \log (2)-100 e^{2 x} \log ^2(2)\right )}{16 x^2+40 e^x x \log (2)+25 e^{2 x} \log ^2(2)} \, dx=-\frac {25 \, x^{2} e^{x} \log \left (2\right ) + 20 \, x^{3} + 20 \, x e^{\left (4 \, x\right )} + 25 \, e^{\left (5 \, x\right )} \log \left (2\right ) - 3 \, x}{5 \, {\left (5 \, e^{x} \log \left (2\right ) + 4 \, x\right )}} \] Input:
integrate(((-100*log(2)^2*exp(x)^2-160*x*log(2)*exp(x)-64*x^2)*exp(4*x)-50 *x*log(2)^2*exp(x)^2+(-80*x^2-3*x+3)*log(2)*exp(x)-32*x^3)/(25*log(2)^2*ex p(x)^2+40*x*log(2)*exp(x)+16*x^2),x, algorithm="fricas")
Output:
-1/5*(25*x^2*e^x*log(2) + 20*x^3 + 20*x*e^(4*x) + 25*e^(5*x)*log(2) - 3*x) /(5*e^x*log(2) + 4*x)
Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.67 \[ \int \frac {-32 x^3+e^x \left (3-3 x-80 x^2\right ) \log (2)-50 e^{2 x} x \log ^2(2)+e^{4 x} \left (-64 x^2-160 e^x x \log (2)-100 e^{2 x} \log ^2(2)\right )}{16 x^2+40 e^x x \log (2)+25 e^{2 x} \log ^2(2)} \, dx=- x^{2} + \frac {3 x}{20 x + 25 e^{x} \log {\left (2 \right )}} - e^{4 x} \] Input:
integrate(((-100*ln(2)**2*exp(x)**2-160*x*ln(2)*exp(x)-64*x**2)*exp(4*x)-5 0*x*ln(2)**2*exp(x)**2+(-80*x**2-3*x+3)*ln(2)*exp(x)-32*x**3)/(25*ln(2)**2 *exp(x)**2+40*x*ln(2)*exp(x)+16*x**2),x)
Output:
-x**2 + 3*x/(20*x + 25*exp(x)*log(2)) - exp(4*x)
Time = 0.17 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.42 \[ \int \frac {-32 x^3+e^x \left (3-3 x-80 x^2\right ) \log (2)-50 e^{2 x} x \log ^2(2)+e^{4 x} \left (-64 x^2-160 e^x x \log (2)-100 e^{2 x} \log ^2(2)\right )}{16 x^2+40 e^x x \log (2)+25 e^{2 x} \log ^2(2)} \, dx=-\frac {25 \, x^{2} e^{x} \log \left (2\right ) + 20 \, x^{3} + 20 \, x e^{\left (4 \, x\right )} + 25 \, e^{\left (5 \, x\right )} \log \left (2\right ) - 3 \, x}{5 \, {\left (5 \, e^{x} \log \left (2\right ) + 4 \, x\right )}} \] Input:
integrate(((-100*log(2)^2*exp(x)^2-160*x*log(2)*exp(x)-64*x^2)*exp(4*x)-50 *x*log(2)^2*exp(x)^2+(-80*x^2-3*x+3)*log(2)*exp(x)-32*x^3)/(25*log(2)^2*ex p(x)^2+40*x*log(2)*exp(x)+16*x^2),x, algorithm="maxima")
Output:
-1/5*(25*x^2*e^x*log(2) + 20*x^3 + 20*x*e^(4*x) + 25*e^(5*x)*log(2) - 3*x) /(5*e^x*log(2) + 4*x)
Time = 0.12 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.42 \[ \int \frac {-32 x^3+e^x \left (3-3 x-80 x^2\right ) \log (2)-50 e^{2 x} x \log ^2(2)+e^{4 x} \left (-64 x^2-160 e^x x \log (2)-100 e^{2 x} \log ^2(2)\right )}{16 x^2+40 e^x x \log (2)+25 e^{2 x} \log ^2(2)} \, dx=-\frac {25 \, x^{2} e^{x} \log \left (2\right ) + 20 \, x^{3} + 20 \, x e^{\left (4 \, x\right )} + 25 \, e^{\left (5 \, x\right )} \log \left (2\right ) - 3 \, x}{5 \, {\left (5 \, e^{x} \log \left (2\right ) + 4 \, x\right )}} \] Input:
integrate(((-100*log(2)^2*exp(x)^2-160*x*log(2)*exp(x)-64*x^2)*exp(4*x)-50 *x*log(2)^2*exp(x)^2+(-80*x^2-3*x+3)*log(2)*exp(x)-32*x^3)/(25*log(2)^2*ex p(x)^2+40*x*log(2)*exp(x)+16*x^2),x, algorithm="giac")
Output:
-1/5*(25*x^2*e^x*log(2) + 20*x^3 + 20*x*e^(4*x) + 25*e^(5*x)*log(2) - 3*x) /(5*e^x*log(2) + 4*x)
Time = 1.92 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.42 \[ \int \frac {-32 x^3+e^x \left (3-3 x-80 x^2\right ) \log (2)-50 e^{2 x} x \log ^2(2)+e^{4 x} \left (-64 x^2-160 e^x x \log (2)-100 e^{2 x} \log ^2(2)\right )}{16 x^2+40 e^x x \log (2)+25 e^{2 x} \log ^2(2)} \, dx=-\frac {5\,{\mathrm {e}}^{5\,x}\,\ln \left (32\right )-3\,x+20\,x\,{\mathrm {e}}^{4\,x}+20\,x^3+5\,x^2\,{\mathrm {e}}^x\,\ln \left (32\right )}{5\,\left (4\,x+5\,{\mathrm {e}}^x\,\ln \left (2\right )\right )} \] Input:
int(-(exp(4*x)*(100*exp(2*x)*log(2)^2 + 64*x^2 + 160*x*exp(x)*log(2)) + 32 *x^3 + exp(x)*log(2)*(3*x + 80*x^2 - 3) + 50*x*exp(2*x)*log(2)^2)/(25*exp( 2*x)*log(2)^2 + 16*x^2 + 40*x*exp(x)*log(2)),x)
Output:
-(5*exp(5*x)*log(32) - 3*x + 20*x*exp(4*x) + 20*x^3 + 5*x^2*exp(x)*log(32) )/(5*(4*x + 5*exp(x)*log(2)))
Time = 0.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.64 \[ \int \frac {-32 x^3+e^x \left (3-3 x-80 x^2\right ) \log (2)-50 e^{2 x} x \log ^2(2)+e^{4 x} \left (-64 x^2-160 e^x x \log (2)-100 e^{2 x} \log ^2(2)\right )}{16 x^2+40 e^x x \log (2)+25 e^{2 x} \log ^2(2)} \, dx=\frac {-20 e^{5 x} \mathrm {log}\left (2\right )-16 e^{4 x} x -20 e^{x} \mathrm {log}\left (2\right ) x^{2}-3 e^{x} \mathrm {log}\left (2\right )-16 x^{3}}{20 e^{x} \mathrm {log}\left (2\right )+16 x} \] Input:
int(((-100*log(2)^2*exp(x)^2-160*x*log(2)*exp(x)-64*x^2)*exp(4*x)-50*x*log (2)^2*exp(x)^2+(-80*x^2-3*x+3)*log(2)*exp(x)-32*x^3)/(25*log(2)^2*exp(x)^2 +40*x*log(2)*exp(x)+16*x^2),x)
Output:
( - 20*e**(5*x)*log(2) - 16*e**(4*x)*x - 20*e**x*log(2)*x**2 - 3*e**x*log( 2) - 16*x**3)/(4*(5*e**x*log(2) + 4*x))