Integrand size = 190, antiderivative size = 24 \[ \int \frac {-32-32 e^5-32 x+e^x \left (e^{10} (-160+40 e)-160 x-160 x^2+e \left (40 x+40 x^2\right )+e^5 (-160-320 x+e (40+80 x))\right )}{16 e^5+16 x+e^x \left ((160-40 e) e^{10}+160 x^2-40 e x^2+e^5 (320 x-80 e x)\right )+e^{2 x} \left (e^{15} \left (450-200 e+25 e^2\right )+450 x^3-200 e x^3+25 e^2 x^3+e^{10} \left (1350 x-600 e x+75 e^2 x\right )+e^5 \left (1350 x^2-600 e x^2+75 e^2 x^2\right )\right )} \, dx=\log \left (2+\left (-4+e-\frac {4 e^{-x}}{5 \left (e^5+x\right )}\right )^2\right ) \] Output:
ln((exp(1)-4/5/(exp(5)+x)/exp(x)-4)^2+2)
Leaf count is larger than twice the leaf count of optimal. \(144\) vs. \(2(24)=48\).
Time = 1.30 (sec) , antiderivative size = 144, normalized size of antiderivative = 6.00 \[ \int \frac {-32-32 e^5-32 x+e^x \left (e^{10} (-160+40 e)-160 x-160 x^2+e \left (40 x+40 x^2\right )+e^5 (-160-320 x+e (40+80 x))\right )}{16 e^5+16 x+e^x \left ((160-40 e) e^{10}+160 x^2-40 e x^2+e^5 (320 x-80 e x)\right )+e^{2 x} \left (e^{15} \left (450-200 e+25 e^2\right )+450 x^3-200 e x^3+25 e^2 x^3+e^{10} \left (1350 x-600 e x+75 e^2 x\right )+e^5 \left (1350 x^2-600 e x^2+75 e^2 x^2\right )\right )} \, dx=8 \left (-\frac {x}{4}-\frac {1}{4} \log \left (e^5+x\right )+\frac {1}{8} \log \left (16+160 e^{5+x}-40 e^{6+x}+450 e^{10+2 x}-200 e^{11+2 x}+25 e^{12+2 x}+160 e^x x-40 e^{1+x} x+900 e^{5+2 x} x-400 e^{6+2 x} x+50 e^{7+2 x} x+450 e^{2 x} x^2-200 e^{1+2 x} x^2+25 e^{2+2 x} x^2\right )\right ) \] Input:
Integrate[(-32 - 32*E^5 - 32*x + E^x*(E^10*(-160 + 40*E) - 160*x - 160*x^2 + E*(40*x + 40*x^2) + E^5*(-160 - 320*x + E*(40 + 80*x))))/(16*E^5 + 16*x + E^x*((160 - 40*E)*E^10 + 160*x^2 - 40*E*x^2 + E^5*(320*x - 80*E*x)) + E ^(2*x)*(E^15*(450 - 200*E + 25*E^2) + 450*x^3 - 200*E*x^3 + 25*E^2*x^3 + E ^10*(1350*x - 600*E*x + 75*E^2*x) + E^5*(1350*x^2 - 600*E*x^2 + 75*E^2*x^2 ))),x]
Output:
8*(-1/4*x - Log[E^5 + x]/4 + Log[16 + 160*E^(5 + x) - 40*E^(6 + x) + 450*E ^(10 + 2*x) - 200*E^(11 + 2*x) + 25*E^(12 + 2*x) + 160*E^x*x - 40*E^(1 + x )*x + 900*E^(5 + 2*x)*x - 400*E^(6 + 2*x)*x + 50*E^(7 + 2*x)*x + 450*E^(2* x)*x^2 - 200*E^(1 + 2*x)*x^2 + 25*E^(2 + 2*x)*x^2]/8)
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.
| \(\displaystyle \int \frac {e^x \left (-160 x^2+e \left (40 x^2+40 x\right )-160 x+e^5 (-320 x+e (80 x+40)-160)+e^{10} (40 e-160)\right )-32 x-32 e^5-32}{e^x \left (-40 e x^2+160 x^2+e^5 (320 x-80 e x)+(160-40 e) e^{10}\right )+e^{2 x} \left (25 e^2 x^3-200 e x^3+450 x^3+e^5 \left (75 e^2 x^2-600 e x^2+1350 x^2\right )+e^{10} \left (75 e^2 x-600 e x+1350 x\right )+e^{15} \left (450-200 e+25 e^2\right )\right )+16 x+16 e^5} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {8 \left (x+e^5+1\right ) \left (-20 \left (1-\frac {e}{4}\right ) e^x x-20 \left (1-\frac {e}{4}\right ) e^{x+5}-4\right )}{\left (x+e^5\right ) \left (450 \left (1+\frac {1}{18} (e-8) e\right ) e^{2 x} x^2+160 \left (1-\frac {e}{4}\right ) e^x x+900 \left (1+\frac {1}{18} (e-8) e\right ) e^{2 x+5} x+160 \left (1-\frac {e}{4}\right ) e^{x+5}+450 \left (1+\frac {1}{18} (e-8) e\right ) e^{2 x+10}+16\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 8 \int -\frac {\left (x+e^5+1\right ) \left (5 (4-e) e^x x+5 (4-e) e^{x+5}+4\right )}{\left (x+e^5\right ) \left (25 e^{2 x} (18-(8-e) e) x^2+40 (4-e) e^x x+50 e^{2 x+5} (18-(8-e) e) x+40 (4-e) e^{x+5}+25 e^{2 x+10} (18-(8-e) e)+16\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -8 \int \frac {\left (x+e^5+1\right ) \left (5 (4-e) e^x x+5 (4-e) e^{x+5}+4\right )}{\left (x+e^5\right ) \left (25 e^{2 x} \left (18-8 e+e^2\right ) x^2+40 (4-e) e^x x+50 e^{2 x+5} \left (18-8 e+e^2\right ) x+40 (4-e) e^{x+5}+25 e^{2 x+10} \left (18-8 e+e^2\right )+16\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -8 \int \left (\frac {20 \left (1-\frac {e}{4}\right ) e^x x+20 \left (1-\frac {e}{4}\right ) e^{x+5}+4}{\left (x+e^5\right ) \left (450 e^{2 x} \left (1+\frac {1}{18} (-8+e) e\right ) x^2+160 \left (1-\frac {e}{4}\right ) e^x x+900 e^{2 x+5} \left (1+\frac {1}{18} (-8+e) e\right ) x+160 \left (1-\frac {e}{4}\right ) e^{x+5}+450 e^{2 x+10} \left (1+\frac {1}{18} (-8+e) e\right )+16\right )}+\frac {20 \left (1-\frac {e}{4}\right ) e^x x+20 \left (1-\frac {e}{4}\right ) e^{x+5}+4}{450 e^{2 x} \left (1+\frac {1}{18} (-8+e) e\right ) x^2+160 \left (1-\frac {e}{4}\right ) e^x x+900 e^{2 x+5} \left (1+\frac {1}{18} (-8+e) e\right ) x+160 \left (1-\frac {e}{4}\right ) e^{x+5}+450 e^{2 x+10} \left (1+\frac {1}{18} (-8+e) e\right )+16}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -8 \left (4 \int \frac {1}{450 e^{2 x} \left (1+\frac {1}{18} (-8+e) e\right ) x^2+160 \left (1-\frac {e}{4}\right ) e^x x+900 e^{2 x+5} \left (1+\frac {1}{18} (-8+e) e\right ) x+160 \left (1-\frac {e}{4}\right ) e^{x+5}+450 e^{2 x+10} \left (1+\frac {1}{18} (-8+e) e\right )+16}dx+5 (4-e) \int \frac {e^x}{450 e^{2 x} \left (1+\frac {1}{18} (-8+e) e\right ) x^2+160 \left (1-\frac {e}{4}\right ) e^x x+900 e^{2 x+5} \left (1+\frac {1}{18} (-8+e) e\right ) x+160 \left (1-\frac {e}{4}\right ) e^{x+5}+450 e^{2 x+10} \left (1+\frac {1}{18} (-8+e) e\right )+16}dx+5 (4-e) \int \frac {e^{x+5}}{450 e^{2 x} \left (1+\frac {1}{18} (-8+e) e\right ) x^2+160 \left (1-\frac {e}{4}\right ) e^x x+900 e^{2 x+5} \left (1+\frac {1}{18} (-8+e) e\right ) x+160 \left (1-\frac {e}{4}\right ) e^{x+5}+450 e^{2 x+10} \left (1+\frac {1}{18} (-8+e) e\right )+16}dx+5 (4-e) \int \frac {e^{x+5}}{\left (-x-e^5\right ) \left (450 e^{2 x} \left (1+\frac {1}{18} (-8+e) e\right ) x^2+160 \left (1-\frac {e}{4}\right ) e^x x+900 e^{2 x+5} \left (1+\frac {1}{18} (-8+e) e\right ) x+160 \left (1-\frac {e}{4}\right ) e^{x+5}+450 e^{2 x+10} \left (1+\frac {1}{18} (-8+e) e\right )+16\right )}dx+5 (4-e) \int \frac {e^x x}{450 e^{2 x} \left (1+\frac {1}{18} (-8+e) e\right ) x^2+160 \left (1-\frac {e}{4}\right ) e^x x+900 e^{2 x+5} \left (1+\frac {1}{18} (-8+e) e\right ) x+160 \left (1-\frac {e}{4}\right ) e^{x+5}+450 e^{2 x+10} \left (1+\frac {1}{18} (-8+e) e\right )+16}dx+4 \int \frac {1}{\left (x+e^5\right ) \left (450 e^{2 x} \left (1+\frac {1}{18} (-8+e) e\right ) x^2+160 \left (1-\frac {e}{4}\right ) e^x x+900 e^{2 x+5} \left (1+\frac {1}{18} (-8+e) e\right ) x+160 \left (1-\frac {e}{4}\right ) e^{x+5}+450 e^{2 x+10} \left (1+\frac {1}{18} (-8+e) e\right )+16\right )}dx+5 (4-e) \int \frac {e^{x+5}}{\left (x+e^5\right ) \left (450 e^{2 x} \left (1+\frac {1}{18} (-8+e) e\right ) x^2+160 \left (1-\frac {e}{4}\right ) e^x x+900 e^{2 x+5} \left (1+\frac {1}{18} (-8+e) e\right ) x+160 \left (1-\frac {e}{4}\right ) e^{x+5}+450 e^{2 x+10} \left (1+\frac {1}{18} (-8+e) e\right )+16\right )}dx\right )\) |
Input:
Int[(-32 - 32*E^5 - 32*x + E^x*(E^10*(-160 + 40*E) - 160*x - 160*x^2 + E*( 40*x + 40*x^2) + E^5*(-160 - 320*x + E*(40 + 80*x))))/(16*E^5 + 16*x + E^x *((160 - 40*E)*E^10 + 160*x^2 - 40*E*x^2 + E^5*(320*x - 80*E*x)) + E^(2*x) *(E^15*(450 - 200*E + 25*E^2) + 450*x^3 - 200*E*x^3 + 25*E^2*x^3 + E^10*(1 350*x - 600*E*x + 75*E^2*x) + E^5*(1350*x^2 - 600*E*x^2 + 75*E^2*x^2))),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(91\) vs. \(2(21)=42\).
Time = 0.86 (sec) , antiderivative size = 92, normalized size of antiderivative = 3.83
| method | result | size |
| risch | \(-2 x +\ln \left ({\mathrm e}^{2 x}-\frac {8 \left ({\mathrm e}-4\right ) {\mathrm e}^{x}}{5 \left ({\mathrm e}^{7}-8 \,{\mathrm e}^{6}+18 \,{\mathrm e}^{5}+{\mathrm e}^{2} x -8 x \,{\mathrm e}+18 x \right )}+\frac {16}{25 \left ({\mathrm e}^{12}-8 \,{\mathrm e}^{11}+18 \,{\mathrm e}^{10}+2 x \,{\mathrm e}^{7}-16 x \,{\mathrm e}^{6}+36 x \,{\mathrm e}^{5}+x^{2} {\mathrm e}^{2}-8 x^{2} {\mathrm e}+18 x^{2}\right )}\right )\) | \(92\) |
| norman | \(-2 x -2 \ln \left ({\mathrm e}^{5}+x \right )+\ln \left (25 \,{\mathrm e}^{2} {\mathrm e}^{10} {\mathrm e}^{2 x}+50 \,{\mathrm e}^{5} {\mathrm e}^{2} {\mathrm e}^{2 x} x +25 \,{\mathrm e}^{2} {\mathrm e}^{2 x} x^{2}-200 \,{\mathrm e}^{10} {\mathrm e} \,{\mathrm e}^{2 x}-400 \,{\mathrm e}^{5} {\mathrm e} \,{\mathrm e}^{2 x} x -200 \,{\mathrm e} \,{\mathrm e}^{2 x} x^{2}+450 \,{\mathrm e}^{10} {\mathrm e}^{2 x}+900 \,{\mathrm e}^{5} {\mathrm e}^{2 x} x +450 \,{\mathrm e}^{2 x} x^{2}-40 \,{\mathrm e}^{5} {\mathrm e} \,{\mathrm e}^{x}-40 x \,{\mathrm e} \,{\mathrm e}^{x}+160 \,{\mathrm e}^{5} {\mathrm e}^{x}+160 \,{\mathrm e}^{x} x +16\right )\) | \(143\) |
| parallelrisch | \(-2 x +\ln \left (\frac {25 \,{\mathrm e}^{2} {\mathrm e}^{10} {\mathrm e}^{2 x}+50 \,{\mathrm e}^{5} {\mathrm e}^{2} {\mathrm e}^{2 x} x +25 \,{\mathrm e}^{2} {\mathrm e}^{2 x} x^{2}-200 \,{\mathrm e}^{10} {\mathrm e} \,{\mathrm e}^{2 x}-400 \,{\mathrm e}^{5} {\mathrm e} \,{\mathrm e}^{2 x} x -200 \,{\mathrm e} \,{\mathrm e}^{2 x} x^{2}+450 \,{\mathrm e}^{10} {\mathrm e}^{2 x}+900 \,{\mathrm e}^{5} {\mathrm e}^{2 x} x +450 \,{\mathrm e}^{2 x} x^{2}-40 \,{\mathrm e}^{5} {\mathrm e} \,{\mathrm e}^{x}-40 x \,{\mathrm e} \,{\mathrm e}^{x}+160 \,{\mathrm e}^{5} {\mathrm e}^{x}+160 \,{\mathrm e}^{x} x +16}{25 \,{\mathrm e}^{2}-200 \,{\mathrm e}+450}\right )-2 \ln \left ({\mathrm e}^{5}+x \right )\) | \(157\) |
Input:
int((((40*exp(1)-160)*exp(5)^2+((80*x+40)*exp(1)-320*x-160)*exp(5)+(40*x^2 +40*x)*exp(1)-160*x^2-160*x)*exp(x)-32*exp(5)-32*x-32)/(((25*exp(1)^2-200* exp(1)+450)*exp(5)^3+(75*x*exp(1)^2-600*x*exp(1)+1350*x)*exp(5)^2+(75*x^2* exp(1)^2-600*x^2*exp(1)+1350*x^2)*exp(5)+25*x^3*exp(1)^2-200*x^3*exp(1)+45 0*x^3)*exp(x)^2+((-40*exp(1)+160)*exp(5)^2+(-80*x*exp(1)+320*x)*exp(5)-40* x^2*exp(1)+160*x^2)*exp(x)+16*exp(5)+16*x),x,method=_RETURNVERBOSE)
Output:
-2*x+ln(exp(2*x)-8/5*(exp(1)-4)/(exp(7)-8*exp(6)+18*exp(5)+exp(2)*x-8*x*ex p(1)+18*x)*exp(x)+16/25/(exp(12)-8*exp(11)+18*exp(10)+2*x*exp(7)-16*x*exp( 6)+36*x*exp(5)+x^2*exp(2)-8*x^2*exp(1)+18*x^2))
Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (25) = 50\).
Time = 0.12 (sec) , antiderivative size = 89, normalized size of antiderivative = 3.71 \[ \int \frac {-32-32 e^5-32 x+e^x \left (e^{10} (-160+40 e)-160 x-160 x^2+e \left (40 x+40 x^2\right )+e^5 (-160-320 x+e (40+80 x))\right )}{16 e^5+16 x+e^x \left ((160-40 e) e^{10}+160 x^2-40 e x^2+e^5 (320 x-80 e x)\right )+e^{2 x} \left (e^{15} \left (450-200 e+25 e^2\right )+450 x^3-200 e x^3+25 e^2 x^3+e^{10} \left (1350 x-600 e x+75 e^2 x\right )+e^5 \left (1350 x^2-600 e x^2+75 e^2 x^2\right )\right )} \, dx=-2 \, x + \log \left (\frac {25 \, {\left (x^{2} e^{2} - 8 \, x^{2} e + 18 \, x^{2} + 2 \, x e^{7} - 16 \, x e^{6} + 36 \, x e^{5} + e^{12} - 8 \, e^{11} + 18 \, e^{10}\right )} e^{\left (2 \, x\right )} - 40 \, {\left (x e - 4 \, x + e^{6} - 4 \, e^{5}\right )} e^{x} + 16}{x^{2} + 2 \, x e^{5} + e^{10}}\right ) \] Input:
integrate((((40*exp(1)-160)*exp(5)^2+((80*x+40)*exp(1)-320*x-160)*exp(5)+( 40*x^2+40*x)*exp(1)-160*x^2-160*x)*exp(x)-32*exp(5)-32*x-32)/(((25*exp(1)^ 2-200*exp(1)+450)*exp(5)^3+(75*x*exp(1)^2-600*exp(1)*x+1350*x)*exp(5)^2+(7 5*x^2*exp(1)^2-600*x^2*exp(1)+1350*x^2)*exp(5)+25*x^3*exp(1)^2-200*x^3*exp (1)+450*x^3)*exp(x)^2+((-40*exp(1)+160)*exp(5)^2+(-80*exp(1)*x+320*x)*exp( 5)-40*x^2*exp(1)+160*x^2)*exp(x)+16*exp(5)+16*x),x, algorithm="fricas")
Output:
-2*x + log((25*(x^2*e^2 - 8*x^2*e + 18*x^2 + 2*x*e^7 - 16*x*e^6 + 36*x*e^5 + e^12 - 8*e^11 + 18*e^10)*e^(2*x) - 40*(x*e - 4*x + e^6 - 4*e^5)*e^x + 1 6)/(x^2 + 2*x*e^5 + e^10))
Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (20) = 40\).
Time = 3.82 (sec) , antiderivative size = 110, normalized size of antiderivative = 4.58 \[ \int \frac {-32-32 e^5-32 x+e^x \left (e^{10} (-160+40 e)-160 x-160 x^2+e \left (40 x+40 x^2\right )+e^5 (-160-320 x+e (40+80 x))\right )}{16 e^5+16 x+e^x \left ((160-40 e) e^{10}+160 x^2-40 e x^2+e^5 (320 x-80 e x)\right )+e^{2 x} \left (e^{15} \left (450-200 e+25 e^2\right )+450 x^3-200 e x^3+25 e^2 x^3+e^{10} \left (1350 x-600 e x+75 e^2 x\right )+e^5 \left (1350 x^2-600 e x^2+75 e^2 x^2\right )\right )} \, dx=- 2 x + \log {\left (e^{2 x} + \frac {16}{- 200 e x^{2} + 25 x^{2} e^{2} + 450 x^{2} - 400 x e^{6} + 50 x e^{7} + 900 x e^{5} - 200 e^{11} + 25 e^{12} + 450 e^{10}} + \frac {\left (32 - 8 e\right ) e^{x}}{- 40 e x + 5 x e^{2} + 90 x - 40 e^{6} + 5 e^{7} + 90 e^{5}} \right )} \] Input:
integrate((((40*exp(1)-160)*exp(5)**2+((80*x+40)*exp(1)-320*x-160)*exp(5)+ (40*x**2+40*x)*exp(1)-160*x**2-160*x)*exp(x)-32*exp(5)-32*x-32)/(((25*exp( 1)**2-200*exp(1)+450)*exp(5)**3+(75*x*exp(1)**2-600*exp(1)*x+1350*x)*exp(5 )**2+(75*x**2*exp(1)**2-600*x**2*exp(1)+1350*x**2)*exp(5)+25*x**3*exp(1)** 2-200*x**3*exp(1)+450*x**3)*exp(x)**2+((-40*exp(1)+160)*exp(5)**2+(-80*exp (1)*x+320*x)*exp(5)-40*x**2*exp(1)+160*x**2)*exp(x)+16*exp(5)+16*x),x)
Output:
-2*x + log(exp(2*x) + 16/(-200*E*x**2 + 25*x**2*exp(2) + 450*x**2 - 400*x* exp(6) + 50*x*exp(7) + 900*x*exp(5) - 200*exp(11) + 25*exp(12) + 450*exp(1 0)) + (32 - 8*E)*exp(x)/(-40*E*x + 5*x*exp(2) + 90*x - 40*exp(6) + 5*exp(7 ) + 90*exp(5)))
Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (25) = 50\).
Time = 0.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 4.50 \[ \int \frac {-32-32 e^5-32 x+e^x \left (e^{10} (-160+40 e)-160 x-160 x^2+e \left (40 x+40 x^2\right )+e^5 (-160-320 x+e (40+80 x))\right )}{16 e^5+16 x+e^x \left ((160-40 e) e^{10}+160 x^2-40 e x^2+e^5 (320 x-80 e x)\right )+e^{2 x} \left (e^{15} \left (450-200 e+25 e^2\right )+450 x^3-200 e x^3+25 e^2 x^3+e^{10} \left (1350 x-600 e x+75 e^2 x\right )+e^5 \left (1350 x^2-600 e x^2+75 e^2 x^2\right )\right )} \, dx=-2 \, x + \log \left (\frac {25 \, {\left (x^{2} {\left (e^{2} - 8 \, e + 18\right )} + 2 \, x {\left (e^{7} - 8 \, e^{6} + 18 \, e^{5}\right )} + e^{12} - 8 \, e^{11} + 18 \, e^{10}\right )} e^{\left (2 \, x\right )} - 40 \, {\left (x {\left (e - 4\right )} + e^{6} - 4 \, e^{5}\right )} e^{x} + 16}{25 \, {\left (x^{2} {\left (e^{2} - 8 \, e + 18\right )} + 2 \, x {\left (e^{7} - 8 \, e^{6} + 18 \, e^{5}\right )} + e^{12} - 8 \, e^{11} + 18 \, e^{10}\right )}}\right ) \] Input:
integrate((((40*exp(1)-160)*exp(5)^2+((80*x+40)*exp(1)-320*x-160)*exp(5)+( 40*x^2+40*x)*exp(1)-160*x^2-160*x)*exp(x)-32*exp(5)-32*x-32)/(((25*exp(1)^ 2-200*exp(1)+450)*exp(5)^3+(75*x*exp(1)^2-600*exp(1)*x+1350*x)*exp(5)^2+(7 5*x^2*exp(1)^2-600*x^2*exp(1)+1350*x^2)*exp(5)+25*x^3*exp(1)^2-200*x^3*exp (1)+450*x^3)*exp(x)^2+((-40*exp(1)+160)*exp(5)^2+(-80*exp(1)*x+320*x)*exp( 5)-40*x^2*exp(1)+160*x^2)*exp(x)+16*exp(5)+16*x),x, algorithm="maxima")
Output:
-2*x + log(1/25*(25*(x^2*(e^2 - 8*e + 18) + 2*x*(e^7 - 8*e^6 + 18*e^5) + e ^12 - 8*e^11 + 18*e^10)*e^(2*x) - 40*(x*(e - 4) + e^6 - 4*e^5)*e^x + 16)/( x^2*(e^2 - 8*e + 18) + 2*x*(e^7 - 8*e^6 + 18*e^5) + e^12 - 8*e^11 + 18*e^1 0))
Timed out. \[ \int \frac {-32-32 e^5-32 x+e^x \left (e^{10} (-160+40 e)-160 x-160 x^2+e \left (40 x+40 x^2\right )+e^5 (-160-320 x+e (40+80 x))\right )}{16 e^5+16 x+e^x \left ((160-40 e) e^{10}+160 x^2-40 e x^2+e^5 (320 x-80 e x)\right )+e^{2 x} \left (e^{15} \left (450-200 e+25 e^2\right )+450 x^3-200 e x^3+25 e^2 x^3+e^{10} \left (1350 x-600 e x+75 e^2 x\right )+e^5 \left (1350 x^2-600 e x^2+75 e^2 x^2\right )\right )} \, dx=\text {Timed out} \] Input:
integrate((((40*exp(1)-160)*exp(5)^2+((80*x+40)*exp(1)-320*x-160)*exp(5)+( 40*x^2+40*x)*exp(1)-160*x^2-160*x)*exp(x)-32*exp(5)-32*x-32)/(((25*exp(1)^ 2-200*exp(1)+450)*exp(5)^3+(75*x*exp(1)^2-600*exp(1)*x+1350*x)*exp(5)^2+(7 5*x^2*exp(1)^2-600*x^2*exp(1)+1350*x^2)*exp(5)+25*x^3*exp(1)^2-200*x^3*exp (1)+450*x^3)*exp(x)^2+((-40*exp(1)+160)*exp(5)^2+(-80*exp(1)*x+320*x)*exp( 5)-40*x^2*exp(1)+160*x^2)*exp(x)+16*exp(5)+16*x),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {-32-32 e^5-32 x+e^x \left (e^{10} (-160+40 e)-160 x-160 x^2+e \left (40 x+40 x^2\right )+e^5 (-160-320 x+e (40+80 x))\right )}{16 e^5+16 x+e^x \left ((160-40 e) e^{10}+160 x^2-40 e x^2+e^5 (320 x-80 e x)\right )+e^{2 x} \left (e^{15} \left (450-200 e+25 e^2\right )+450 x^3-200 e x^3+25 e^2 x^3+e^{10} \left (1350 x-600 e x+75 e^2 x\right )+e^5 \left (1350 x^2-600 e x^2+75 e^2 x^2\right )\right )} \, dx=\int -\frac {32\,x+32\,{\mathrm {e}}^5+{\mathrm {e}}^x\,\left (160\,x-\mathrm {e}\,\left (40\,x^2+40\,x\right )+160\,x^2-{\mathrm {e}}^{10}\,\left (40\,\mathrm {e}-160\right )+{\mathrm {e}}^5\,\left (320\,x-\mathrm {e}\,\left (80\,x+40\right )+160\right )\right )+32}{16\,x+16\,{\mathrm {e}}^5+{\mathrm {e}}^{2\,x}\,\left ({\mathrm {e}}^{10}\,\left (1350\,x-600\,x\,\mathrm {e}+75\,x\,{\mathrm {e}}^2\right )+{\mathrm {e}}^5\,\left (75\,x^2\,{\mathrm {e}}^2-600\,x^2\,\mathrm {e}+1350\,x^2\right )-200\,x^3\,\mathrm {e}+25\,x^3\,{\mathrm {e}}^2+{\mathrm {e}}^{15}\,\left (25\,{\mathrm {e}}^2-200\,\mathrm {e}+450\right )+450\,x^3\right )+{\mathrm {e}}^x\,\left ({\mathrm {e}}^5\,\left (320\,x-80\,x\,\mathrm {e}\right )-40\,x^2\,\mathrm {e}+160\,x^2-{\mathrm {e}}^{10}\,\left (40\,\mathrm {e}-160\right )\right )} \,d x \] Input:
int(-(32*x + 32*exp(5) + exp(x)*(160*x - exp(1)*(40*x + 40*x^2) + 160*x^2 - exp(10)*(40*exp(1) - 160) + exp(5)*(320*x - exp(1)*(80*x + 40) + 160)) + 32)/(16*x + 16*exp(5) + exp(2*x)*(exp(10)*(1350*x - 600*x*exp(1) + 75*x*e xp(2)) + exp(5)*(75*x^2*exp(2) - 600*x^2*exp(1) + 1350*x^2) - 200*x^3*exp( 1) + 25*x^3*exp(2) + exp(15)*(25*exp(2) - 200*exp(1) + 450) + 450*x^3) + e xp(x)*(exp(5)*(320*x - 80*x*exp(1)) - 40*x^2*exp(1) + 160*x^2 - exp(10)*(4 0*exp(1) - 160))),x)
Output:
int(-(32*x + 32*exp(5) + exp(x)*(160*x - exp(1)*(40*x + 40*x^2) + 160*x^2 - exp(10)*(40*exp(1) - 160) + exp(5)*(320*x - exp(1)*(80*x + 40) + 160)) + 32)/(16*x + 16*exp(5) + exp(2*x)*(exp(10)*(1350*x - 600*x*exp(1) + 75*x*e xp(2)) + exp(5)*(75*x^2*exp(2) - 600*x^2*exp(1) + 1350*x^2) - 200*x^3*exp( 1) + 25*x^3*exp(2) + exp(15)*(25*exp(2) - 200*exp(1) + 450) + 450*x^3) + e xp(x)*(exp(5)*(320*x - 80*x*exp(1)) - 40*x^2*exp(1) + 160*x^2 - exp(10)*(4 0*exp(1) - 160))), x)
Time = 0.18 (sec) , antiderivative size = 141, normalized size of antiderivative = 5.88 \[ \int \frac {-32-32 e^5-32 x+e^x \left (e^{10} (-160+40 e)-160 x-160 x^2+e \left (40 x+40 x^2\right )+e^5 (-160-320 x+e (40+80 x))\right )}{16 e^5+16 x+e^x \left ((160-40 e) e^{10}+160 x^2-40 e x^2+e^5 (320 x-80 e x)\right )+e^{2 x} \left (e^{15} \left (450-200 e+25 e^2\right )+450 x^3-200 e x^3+25 e^2 x^3+e^{10} \left (1350 x-600 e x+75 e^2 x\right )+e^5 \left (1350 x^2-600 e x^2+75 e^2 x^2\right )\right )} \, dx=-2 \,\mathrm {log}\left (e^{5}+x \right )+\mathrm {log}\left (25 e^{2 x} e^{12}-200 e^{2 x} e^{11}+450 e^{2 x} e^{10}+50 e^{2 x} e^{7} x -400 e^{2 x} e^{6} x +900 e^{2 x} e^{5} x +25 e^{2 x} e^{2} x^{2}-200 e^{2 x} e \,x^{2}+450 e^{2 x} x^{2}-40 e^{x} e^{6}+160 e^{x} e^{5}-40 e^{x} e x +160 e^{x} x +16\right )-2 x \] Input:
int((((40*exp(1)-160)*exp(5)^2+((80*x+40)*exp(1)-320*x-160)*exp(5)+(40*x^2 +40*x)*exp(1)-160*x^2-160*x)*exp(x)-32*exp(5)-32*x-32)/(((25*exp(1)^2-200* exp(1)+450)*exp(5)^3+(75*x*exp(1)^2-600*exp(1)*x+1350*x)*exp(5)^2+(75*x^2* exp(1)^2-600*x^2*exp(1)+1350*x^2)*exp(5)+25*x^3*exp(1)^2-200*x^3*exp(1)+45 0*x^3)*exp(x)^2+((-40*exp(1)+160)*exp(5)^2+(-80*exp(1)*x+320*x)*exp(5)-40* x^2*exp(1)+160*x^2)*exp(x)+16*exp(5)+16*x),x)
Output:
- 2*log(e**5 + x) + log(25*e**(2*x)*e**12 - 200*e**(2*x)*e**11 + 450*e**( 2*x)*e**10 + 50*e**(2*x)*e**7*x - 400*e**(2*x)*e**6*x + 900*e**(2*x)*e**5* x + 25*e**(2*x)*e**2*x**2 - 200*e**(2*x)*e*x**2 + 450*e**(2*x)*x**2 - 40*e **x*e**6 + 160*e**x*e**5 - 40*e**x*e*x + 160*e**x*x + 16) - 2*x