Integrand size = 225, antiderivative size = 26 \[ \int \frac {48 x^2+e^3 \left (80 x-24 x^2\right )+e^6 \left (25-20 x+3 x^2\right )+\left (e^3 (40-16 x)+64 x\right ) \log (4)+16 \log ^2(4)+e^{9+e^{x^2}} \left (-24 x^2+e^3 \left (-20 x+6 x^2\right )+\left (-32 x+e^3 (-10+4 x)\right ) \log (4)-8 \log ^2(4)+e^{x^2} \left (-16 x^4+e^3 \left (-20 x^3+4 x^4\right )+\left (-32 x^3+e^3 \left (-20 x^2+4 x^3\right )\right ) \log (4)-16 x^2 \log ^2(4)\right )\right )+e^{18+2 e^{x^2}} \left (3 x^2+4 x \log (4)+\log ^2(4)+e^{x^2} \left (4 x^4+8 x^3 \log (4)+4 x^2 \log ^2(4)\right )\right )}{e^6} \, dx=x \left (-5+x+\frac {\left (-4+e^{9+e^{x^2}}\right ) (x+\log (4))}{e^3}\right )^2 \] Output:
x*(x+(x+2*ln(2))/exp(3)*(exp(exp(x^2)+9)-4)-5)^2
Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {48 x^2+e^3 \left (80 x-24 x^2\right )+e^6 \left (25-20 x+3 x^2\right )+\left (e^3 (40-16 x)+64 x\right ) \log (4)+16 \log ^2(4)+e^{9+e^{x^2}} \left (-24 x^2+e^3 \left (-20 x+6 x^2\right )+\left (-32 x+e^3 (-10+4 x)\right ) \log (4)-8 \log ^2(4)+e^{x^2} \left (-16 x^4+e^3 \left (-20 x^3+4 x^4\right )+\left (-32 x^3+e^3 \left (-20 x^2+4 x^3\right )\right ) \log (4)-16 x^2 \log ^2(4)\right )\right )+e^{18+2 e^{x^2}} \left (3 x^2+4 x \log (4)+\log ^2(4)+e^{x^2} \left (4 x^4+8 x^3 \log (4)+4 x^2 \log ^2(4)\right )\right )}{e^6} \, dx=\frac {x \left (e^3 (-5+x)-4 (x+\log (4))+e^{9+e^{x^2}} (x+\log (4))\right )^2}{e^6} \] Input:
Integrate[(48*x^2 + E^3*(80*x - 24*x^2) + E^6*(25 - 20*x + 3*x^2) + (E^3*( 40 - 16*x) + 64*x)*Log[4] + 16*Log[4]^2 + E^(9 + E^x^2)*(-24*x^2 + E^3*(-2 0*x + 6*x^2) + (-32*x + E^3*(-10 + 4*x))*Log[4] - 8*Log[4]^2 + E^x^2*(-16* x^4 + E^3*(-20*x^3 + 4*x^4) + (-32*x^3 + E^3*(-20*x^2 + 4*x^3))*Log[4] - 1 6*x^2*Log[4]^2)) + E^(18 + 2*E^x^2)*(3*x^2 + 4*x*Log[4] + Log[4]^2 + E^x^2 *(4*x^4 + 8*x^3*Log[4] + 4*x^2*Log[4]^2)))/E^6,x]
Output:
(x*(E^3*(-5 + x) - 4*(x + Log[4]) + E^(9 + E^x^2)*(x + Log[4]))^2)/E^6
Leaf count is larger than twice the leaf count of optimal. \(184\) vs. \(2(26)=52\).
Time = 0.50 (sec) , antiderivative size = 184, normalized size of antiderivative = 7.08, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {27, 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.
| \(\displaystyle \int \frac {48 x^2+e^3 \left (80 x-24 x^2\right )+e^6 \left (3 x^2-20 x+25\right )+e^{e^{x^2}+9} \left (-24 x^2+e^3 \left (6 x^2-20 x\right )+e^{x^2} \left (-16 x^4-16 x^2 \log ^2(4)+e^3 \left (4 x^4-20 x^3\right )+\left (e^3 \left (4 x^3-20 x^2\right )-32 x^3\right ) \log (4)\right )+\left (e^3 (4 x-10)-32 x\right ) \log (4)-8 \log ^2(4)\right )+e^{2 e^{x^2}+18} \left (3 x^2+e^{x^2} \left (4 x^4+8 x^3 \log (4)+4 x^2 \log ^2(4)\right )+4 x \log (4)+\log ^2(4)\right )+\left (e^3 (40-16 x)+64 x\right ) \log (4)+16 \log ^2(4)}{e^6} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \left (48 x^2+8 e^3 \left (10 x-3 x^2\right )+e^6 \left (3 x^2-20 x+25\right )+e^{18+2 e^{x^2}} \left (3 x^2+4 \log (4) x+4 e^{x^2} \left (x^4+2 \log (4) x^3+\log ^2(4) x^2\right )+\log ^2(4)\right )-2 e^{9+e^{x^2}} \left (12 x^2+e^3 \left (10 x-3 x^2\right )+2 e^{x^2} \left (4 x^4+4 \log ^2(4) x^2+e^3 \left (5 x^3-x^4\right )+\left (8 x^3+e^3 \left (5 x^2-x^3\right )\right ) \log (4)\right )+4 \log ^2(4)+\left (e^3 (5-2 x)+16 x\right ) \log (4)\right )+16 \log ^2(4)+8 \left (e^3 (5-2 x)+8 x\right ) \log (4)\right )dx}{e^6}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^6 x^3-8 e^3 x^3+16 x^3-10 e^6 x^2+40 e^3 x^2+\frac {e^{2 e^{x^2}+18} \left (x^4+2 x^3 \log (4)+x^2 \log ^2(4)\right )}{x}-\frac {2 e^{e^{x^2}+9} \left (4 x^4+4 x^2 \log ^2(4)+e^3 \left (5 x^3-x^4\right )+\left (8 x^3+e^3 \left (5 x^2-x^3\right )\right ) \log (4)\right )}{x}+25 e^6 x+16 x \log ^2(4)+\frac {2 \left (e^3 (5-2 x)+8 x\right )^2 \log (4)}{4-e^3}}{e^6}\) |
Input:
Int[(48*x^2 + E^3*(80*x - 24*x^2) + E^6*(25 - 20*x + 3*x^2) + (E^3*(40 - 1 6*x) + 64*x)*Log[4] + 16*Log[4]^2 + E^(9 + E^x^2)*(-24*x^2 + E^3*(-20*x + 6*x^2) + (-32*x + E^3*(-10 + 4*x))*Log[4] - 8*Log[4]^2 + E^x^2*(-16*x^4 + E^3*(-20*x^3 + 4*x^4) + (-32*x^3 + E^3*(-20*x^2 + 4*x^3))*Log[4] - 16*x^2* Log[4]^2)) + E^(18 + 2*E^x^2)*(3*x^2 + 4*x*Log[4] + Log[4]^2 + E^x^2*(4*x^ 4 + 8*x^3*Log[4] + 4*x^2*Log[4]^2)))/E^6,x]
Output:
(25*E^6*x + 40*E^3*x^2 - 10*E^6*x^2 + 16*x^3 - 8*E^3*x^3 + E^6*x^3 + (2*(E ^3*(5 - 2*x) + 8*x)^2*Log[4])/(4 - E^3) + 16*x*Log[4]^2 + (E^(18 + 2*E^x^2 )*(x^4 + 2*x^3*Log[4] + x^2*Log[4]^2))/x - (2*E^(9 + E^x^2)*(4*x^4 + E^3*( 5*x^3 - x^4) + (8*x^3 + E^3*(5*x^2 - x^3))*Log[4] + 4*x^2*Log[4]^2))/x)/E^ 6
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Leaf count of result is larger than twice the leaf count of optimal. \(170\) vs. \(2(27)=54\).
Time = 1.43 (sec) , antiderivative size = 171, normalized size of antiderivative = 6.58
| method | result | size |
| risch | \({\mathrm e}^{6} {\mathrm e}^{-6} x^{3}-8 \,{\mathrm e}^{3} {\mathrm e}^{-6} x^{3}+16 \,{\mathrm e}^{-6} x^{3}-16 \ln \left (2\right ) {\mathrm e}^{3} {\mathrm e}^{-6} x^{2}+64 \ln \left (2\right ) {\mathrm e}^{-6} x^{2}+40 \,{\mathrm e}^{3} {\mathrm e}^{-6} x^{2}-10 \,{\mathrm e}^{6} {\mathrm e}^{-6} x^{2}+64 \ln \left (2\right )^{2} {\mathrm e}^{-6} x +80 \ln \left (2\right ) {\mathrm e}^{3} {\mathrm e}^{-6} x +25 \,{\mathrm e}^{6} {\mathrm e}^{-6} x +\left (4 \ln \left (2\right )^{2}+4 x \ln \left (2\right )+x^{2}\right ) x \,{\mathrm e}^{12+2 \,{\mathrm e}^{x^{2}}}-2 \left (-2 x \,{\mathrm e}^{3} \ln \left (2\right )-x^{2} {\mathrm e}^{3}+16 \ln \left (2\right )^{2}+10 \,{\mathrm e}^{3} \ln \left (2\right )+16 x \ln \left (2\right )+5 x \,{\mathrm e}^{3}+4 x^{2}\right ) x \,{\mathrm e}^{{\mathrm e}^{x^{2}}+3}\) | \(171\) |
| default | \({\mathrm e}^{-6} \left (x^{3} {\mathrm e}^{2 \,{\mathrm e}^{x^{2}}+18}+4 \ln \left (2\right ) x^{2} {\mathrm e}^{2 \,{\mathrm e}^{x^{2}}+18}+4 \ln \left (2\right )^{2} x \,{\mathrm e}^{2 \,{\mathrm e}^{x^{2}}+18}+\left (-32 \ln \left (2\right )^{2}-20 \,{\mathrm e}^{3} \ln \left (2\right )\right ) x \,{\mathrm e}^{{\mathrm e}^{x^{2}}+9}+\left (2 \,{\mathrm e}^{3}-8\right ) x^{3} {\mathrm e}^{{\mathrm e}^{x^{2}}+9}+\left (4 \,{\mathrm e}^{3} \ln \left (2\right )-32 \ln \left (2\right )-10 \,{\mathrm e}^{3}\right ) x^{2} {\mathrm e}^{{\mathrm e}^{x^{2}}+9}+8 \,{\mathrm e}^{3} \left (-x^{3}+5 x^{2}\right )+{\mathrm e}^{6} \left (x^{3}-10 x^{2}+25 x \right )+16 x^{3}+64 x \ln \left (2\right )^{2}+2 \ln \left (2\right ) \left (-8 x^{2} {\mathrm e}^{3}+40 x \,{\mathrm e}^{3}+32 x^{2}\right )\right )\) | \(183\) |
| norman | \(\left (\left ({\mathrm e}^{6}-8 \,{\mathrm e}^{3}+16\right ) {\mathrm e}^{-3} x^{3}+\left (64 \ln \left (2\right )^{2}+25 \,{\mathrm e}^{6}+80 \,{\mathrm e}^{3} \ln \left (2\right )\right ) {\mathrm e}^{-3} x +{\mathrm e}^{-3} x^{3} {\mathrm e}^{2 \,{\mathrm e}^{x^{2}}+18}-2 \left (8 \,{\mathrm e}^{3} \ln \left (2\right )+5 \,{\mathrm e}^{6}-32 \ln \left (2\right )-20 \,{\mathrm e}^{3}\right ) {\mathrm e}^{-3} x^{2}+2 \left ({\mathrm e}^{3}-4\right ) {\mathrm e}^{-3} x^{3} {\mathrm e}^{{\mathrm e}^{x^{2}}+9}+2 \left (2 \,{\mathrm e}^{3} \ln \left (2\right )-16 \ln \left (2\right )-5 \,{\mathrm e}^{3}\right ) {\mathrm e}^{-3} x^{2} {\mathrm e}^{{\mathrm e}^{x^{2}}+9}+4 \,{\mathrm e}^{-3} \ln \left (2\right ) x^{2} {\mathrm e}^{2 \,{\mathrm e}^{x^{2}}+18}+4 \,{\mathrm e}^{-3} \ln \left (2\right )^{2} x \,{\mathrm e}^{2 \,{\mathrm e}^{x^{2}}+18}-4 \ln \left (2\right ) \left (8 \ln \left (2\right )+5 \,{\mathrm e}^{3}\right ) {\mathrm e}^{-3} x \,{\mathrm e}^{{\mathrm e}^{x^{2}}+9}\right ) {\mathrm e}^{-3}\) | \(213\) |
| parallelrisch | \({\mathrm e}^{-6} \left (4 \ln \left (2\right )^{2} x \,{\mathrm e}^{2 \,{\mathrm e}^{x^{2}}+18}+4 \ln \left (2\right ) {\mathrm e}^{3} x^{2} {\mathrm e}^{{\mathrm e}^{x^{2}}+9}+4 \ln \left (2\right ) x^{2} {\mathrm e}^{2 \,{\mathrm e}^{x^{2}}+18}+x^{3} {\mathrm e}^{6}+2 x^{3} {\mathrm e}^{{\mathrm e}^{x^{2}}+9} {\mathrm e}^{3}+x^{3} {\mathrm e}^{2 \,{\mathrm e}^{x^{2}}+18}-32 \ln \left (2\right )^{2} x \,{\mathrm e}^{{\mathrm e}^{x^{2}}+9}-16 x^{2} {\mathrm e}^{3} \ln \left (2\right )-20 \ln \left (2\right ) {\mathrm e}^{{\mathrm e}^{x^{2}}+9} {\mathrm e}^{3} x -32 \ln \left (2\right ) x^{2} {\mathrm e}^{{\mathrm e}^{x^{2}}+9}-10 x^{2} {\mathrm e}^{6}-8 x^{3} {\mathrm e}^{3}-10 \,{\mathrm e}^{{\mathrm e}^{x^{2}}+9} {\mathrm e}^{3} x^{2}-8 x^{3} {\mathrm e}^{{\mathrm e}^{x^{2}}+9}+80 x \,{\mathrm e}^{3} \ln \left (2\right )+64 x^{2} \ln \left (2\right )+25 x \,{\mathrm e}^{6}+40 x^{2} {\mathrm e}^{3}+16 x^{3}+64 x \ln \left (2\right )^{2}\right )\) | \(223\) |
Input:
int((((16*x^2*ln(2)^2+16*x^3*ln(2)+4*x^4)*exp(x^2)+4*ln(2)^2+8*x*ln(2)+3*x ^2)*exp(exp(x^2)+9)^2+((-64*x^2*ln(2)^2+2*((4*x^3-20*x^2)*exp(3)-32*x^3)*l n(2)+(4*x^4-20*x^3)*exp(3)-16*x^4)*exp(x^2)-32*ln(2)^2+2*((4*x-10)*exp(3)- 32*x)*ln(2)+(6*x^2-20*x)*exp(3)-24*x^2)*exp(exp(x^2)+9)+64*ln(2)^2+2*((-16 *x+40)*exp(3)+64*x)*ln(2)+(3*x^2-20*x+25)*exp(3)^2+(-24*x^2+80*x)*exp(3)+4 8*x^2)/exp(3)^2,x,method=_RETURNVERBOSE)
Output:
exp(3)^2*exp(-6)*x^3-8*exp(3)*exp(-6)*x^3+16*exp(-6)*x^3-16*ln(2)*exp(3)*e xp(-6)*x^2+64*ln(2)*exp(-6)*x^2+40*exp(3)*exp(-6)*x^2-10*exp(6)*exp(-6)*x^ 2+64*ln(2)^2*exp(-6)*x+80*ln(2)*exp(3)*exp(-6)*x+25*exp(3)^2*exp(-6)*x+(4* ln(2)^2+4*x*ln(2)+x^2)*x*exp(12+2*exp(x^2))-2*(-2*x*exp(3)*ln(2)-x^2*exp(3 )+16*ln(2)^2+10*exp(3)*ln(2)+16*x*ln(2)+5*x*exp(3)+4*x^2)*x*exp(exp(x^2)+3 )
Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (25) = 50\).
Time = 0.10 (sec) , antiderivative size = 149, normalized size of antiderivative = 5.73 \[ \int \frac {48 x^2+e^3 \left (80 x-24 x^2\right )+e^6 \left (25-20 x+3 x^2\right )+\left (e^3 (40-16 x)+64 x\right ) \log (4)+16 \log ^2(4)+e^{9+e^{x^2}} \left (-24 x^2+e^3 \left (-20 x+6 x^2\right )+\left (-32 x+e^3 (-10+4 x)\right ) \log (4)-8 \log ^2(4)+e^{x^2} \left (-16 x^4+e^3 \left (-20 x^3+4 x^4\right )+\left (-32 x^3+e^3 \left (-20 x^2+4 x^3\right )\right ) \log (4)-16 x^2 \log ^2(4)\right )\right )+e^{18+2 e^{x^2}} \left (3 x^2+4 x \log (4)+\log ^2(4)+e^{x^2} \left (4 x^4+8 x^3 \log (4)+4 x^2 \log ^2(4)\right )\right )}{e^6} \, dx={\left (16 \, x^{3} + 64 \, x \log \left (2\right )^{2} + {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} e^{6} - 8 \, {\left (x^{3} - 5 \, x^{2}\right )} e^{3} + {\left (x^{3} + 4 \, x^{2} \log \left (2\right ) + 4 \, x \log \left (2\right )^{2}\right )} e^{\left (2 \, e^{\left (x^{2}\right )} + 18\right )} - 2 \, {\left (4 \, x^{3} + 16 \, x \log \left (2\right )^{2} - {\left (x^{3} - 5 \, x^{2}\right )} e^{3} + 2 \, {\left (8 \, x^{2} - {\left (x^{2} - 5 \, x\right )} e^{3}\right )} \log \left (2\right )\right )} e^{\left (e^{\left (x^{2}\right )} + 9\right )} + 16 \, {\left (4 \, x^{2} - {\left (x^{2} - 5 \, x\right )} e^{3}\right )} \log \left (2\right )\right )} e^{\left (-6\right )} \] Input:
integrate((((16*x^2*log(2)^2+16*x^3*log(2)+4*x^4)*exp(x^2)+4*log(2)^2+8*x* log(2)+3*x^2)*exp(exp(x^2)+9)^2+((-64*x^2*log(2)^2+2*((4*x^3-20*x^2)*exp(3 )-32*x^3)*log(2)+(4*x^4-20*x^3)*exp(3)-16*x^4)*exp(x^2)-32*log(2)^2+2*((4* x-10)*exp(3)-32*x)*log(2)+(6*x^2-20*x)*exp(3)-24*x^2)*exp(exp(x^2)+9)+64*l og(2)^2+2*((-16*x+40)*exp(3)+64*x)*log(2)+(3*x^2-20*x+25)*exp(3)^2+(-24*x^ 2+80*x)*exp(3)+48*x^2)/exp(3)^2,x, algorithm="fricas")
Output:
(16*x^3 + 64*x*log(2)^2 + (x^3 - 10*x^2 + 25*x)*e^6 - 8*(x^3 - 5*x^2)*e^3 + (x^3 + 4*x^2*log(2) + 4*x*log(2)^2)*e^(2*e^(x^2) + 18) - 2*(4*x^3 + 16*x *log(2)^2 - (x^3 - 5*x^2)*e^3 + 2*(8*x^2 - (x^2 - 5*x)*e^3)*log(2))*e^(e^( x^2) + 9) + 16*(4*x^2 - (x^2 - 5*x)*e^3)*log(2))*e^(-6)
Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (26) = 52\).
Time = 0.97 (sec) , antiderivative size = 196, normalized size of antiderivative = 7.54 \[ \int \frac {48 x^2+e^3 \left (80 x-24 x^2\right )+e^6 \left (25-20 x+3 x^2\right )+\left (e^3 (40-16 x)+64 x\right ) \log (4)+16 \log ^2(4)+e^{9+e^{x^2}} \left (-24 x^2+e^3 \left (-20 x+6 x^2\right )+\left (-32 x+e^3 (-10+4 x)\right ) \log (4)-8 \log ^2(4)+e^{x^2} \left (-16 x^4+e^3 \left (-20 x^3+4 x^4\right )+\left (-32 x^3+e^3 \left (-20 x^2+4 x^3\right )\right ) \log (4)-16 x^2 \log ^2(4)\right )\right )+e^{18+2 e^{x^2}} \left (3 x^2+4 x \log (4)+\log ^2(4)+e^{x^2} \left (4 x^4+8 x^3 \log (4)+4 x^2 \log ^2(4)\right )\right )}{e^6} \, dx=\frac {x^{3} \left (- 8 e^{3} + 16 + e^{6}\right )}{e^{6}} + \frac {x^{2} \left (- 10 e^{6} - 16 e^{3} \log {\left (2 \right )} + 64 \log {\left (2 \right )} + 40 e^{3}\right )}{e^{6}} + \frac {x \left (64 \log {\left (2 \right )}^{2} + 80 e^{3} \log {\left (2 \right )} + 25 e^{6}\right )}{e^{6}} + \frac {\left (x^{3} e^{6} + 4 x^{2} e^{6} \log {\left (2 \right )} + 4 x e^{6} \log {\left (2 \right )}^{2}\right ) e^{2 e^{x^{2}} + 18} + \left (- 8 x^{3} e^{6} + 2 x^{3} e^{9} - 10 x^{2} e^{9} - 32 x^{2} e^{6} \log {\left (2 \right )} + 4 x^{2} e^{9} \log {\left (2 \right )} - 20 x e^{9} \log {\left (2 \right )} - 32 x e^{6} \log {\left (2 \right )}^{2}\right ) e^{e^{x^{2}} + 9}}{e^{12}} \] Input:
integrate((((16*x**2*ln(2)**2+16*x**3*ln(2)+4*x**4)*exp(x**2)+4*ln(2)**2+8 *x*ln(2)+3*x**2)*exp(exp(x**2)+9)**2+((-64*x**2*ln(2)**2+2*((4*x**3-20*x** 2)*exp(3)-32*x**3)*ln(2)+(4*x**4-20*x**3)*exp(3)-16*x**4)*exp(x**2)-32*ln( 2)**2+2*((4*x-10)*exp(3)-32*x)*ln(2)+(6*x**2-20*x)*exp(3)-24*x**2)*exp(exp (x**2)+9)+64*ln(2)**2+2*((-16*x+40)*exp(3)+64*x)*ln(2)+(3*x**2-20*x+25)*ex p(3)**2+(-24*x**2+80*x)*exp(3)+48*x**2)/exp(3)**2,x)
Output:
x**3*(-8*exp(3) + 16 + exp(6))*exp(-6) + x**2*(-10*exp(6) - 16*exp(3)*log( 2) + 64*log(2) + 40*exp(3))*exp(-6) + x*(64*log(2)**2 + 80*exp(3)*log(2) + 25*exp(6))*exp(-6) + ((x**3*exp(6) + 4*x**2*exp(6)*log(2) + 4*x*exp(6)*lo g(2)**2)*exp(2*exp(x**2) + 18) + (-8*x**3*exp(6) + 2*x**3*exp(9) - 10*x**2 *exp(9) - 32*x**2*exp(6)*log(2) + 4*x**2*exp(9)*log(2) - 20*x*exp(9)*log(2 ) - 32*x*exp(6)*log(2)**2)*exp(exp(x**2) + 9))*exp(-12)
Leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (25) = 50\).
Time = 0.12 (sec) , antiderivative size = 155, normalized size of antiderivative = 5.96 \[ \int \frac {48 x^2+e^3 \left (80 x-24 x^2\right )+e^6 \left (25-20 x+3 x^2\right )+\left (e^3 (40-16 x)+64 x\right ) \log (4)+16 \log ^2(4)+e^{9+e^{x^2}} \left (-24 x^2+e^3 \left (-20 x+6 x^2\right )+\left (-32 x+e^3 (-10+4 x)\right ) \log (4)-8 \log ^2(4)+e^{x^2} \left (-16 x^4+e^3 \left (-20 x^3+4 x^4\right )+\left (-32 x^3+e^3 \left (-20 x^2+4 x^3\right )\right ) \log (4)-16 x^2 \log ^2(4)\right )\right )+e^{18+2 e^{x^2}} \left (3 x^2+4 x \log (4)+\log ^2(4)+e^{x^2} \left (4 x^4+8 x^3 \log (4)+4 x^2 \log ^2(4)\right )\right )}{e^6} \, dx={\left (16 \, x^{3} + 64 \, x \log \left (2\right )^{2} + {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} e^{6} - 8 \, {\left (x^{3} - 5 \, x^{2}\right )} e^{3} + {\left (x^{3} e^{18} + 4 \, x^{2} e^{18} \log \left (2\right ) + 4 \, x e^{18} \log \left (2\right )^{2}\right )} e^{\left (2 \, e^{\left (x^{2}\right )}\right )} + 2 \, {\left (x^{3} {\left (e^{12} - 4 \, e^{9}\right )} + {\left ({\left (2 \, \log \left (2\right ) - 5\right )} e^{12} - 16 \, e^{9} \log \left (2\right )\right )} x^{2} - 2 \, {\left (8 \, e^{9} \log \left (2\right )^{2} + 5 \, e^{12} \log \left (2\right )\right )} x\right )} e^{\left (e^{\left (x^{2}\right )}\right )} + 16 \, {\left (4 \, x^{2} - {\left (x^{2} - 5 \, x\right )} e^{3}\right )} \log \left (2\right )\right )} e^{\left (-6\right )} \] Input:
integrate((((16*x^2*log(2)^2+16*x^3*log(2)+4*x^4)*exp(x^2)+4*log(2)^2+8*x* log(2)+3*x^2)*exp(exp(x^2)+9)^2+((-64*x^2*log(2)^2+2*((4*x^3-20*x^2)*exp(3 )-32*x^3)*log(2)+(4*x^4-20*x^3)*exp(3)-16*x^4)*exp(x^2)-32*log(2)^2+2*((4* x-10)*exp(3)-32*x)*log(2)+(6*x^2-20*x)*exp(3)-24*x^2)*exp(exp(x^2)+9)+64*l og(2)^2+2*((-16*x+40)*exp(3)+64*x)*log(2)+(3*x^2-20*x+25)*exp(3)^2+(-24*x^ 2+80*x)*exp(3)+48*x^2)/exp(3)^2,x, algorithm="maxima")
Output:
(16*x^3 + 64*x*log(2)^2 + (x^3 - 10*x^2 + 25*x)*e^6 - 8*(x^3 - 5*x^2)*e^3 + (x^3*e^18 + 4*x^2*e^18*log(2) + 4*x*e^18*log(2)^2)*e^(2*e^(x^2)) + 2*(x^ 3*(e^12 - 4*e^9) + ((2*log(2) - 5)*e^12 - 16*e^9*log(2))*x^2 - 2*(8*e^9*lo g(2)^2 + 5*e^12*log(2))*x)*e^(e^(x^2)) + 16*(4*x^2 - (x^2 - 5*x)*e^3)*log( 2))*e^(-6)
Leaf count of result is larger than twice the leaf count of optimal. 229 vs. \(2 (25) = 50\).
Time = 0.14 (sec) , antiderivative size = 229, normalized size of antiderivative = 8.81 \[ \int \frac {48 x^2+e^3 \left (80 x-24 x^2\right )+e^6 \left (25-20 x+3 x^2\right )+\left (e^3 (40-16 x)+64 x\right ) \log (4)+16 \log ^2(4)+e^{9+e^{x^2}} \left (-24 x^2+e^3 \left (-20 x+6 x^2\right )+\left (-32 x+e^3 (-10+4 x)\right ) \log (4)-8 \log ^2(4)+e^{x^2} \left (-16 x^4+e^3 \left (-20 x^3+4 x^4\right )+\left (-32 x^3+e^3 \left (-20 x^2+4 x^3\right )\right ) \log (4)-16 x^2 \log ^2(4)\right )\right )+e^{18+2 e^{x^2}} \left (3 x^2+4 x \log (4)+\log ^2(4)+e^{x^2} \left (4 x^4+8 x^3 \log (4)+4 x^2 \log ^2(4)\right )\right )}{e^6} \, dx={\left (x^{3} e^{\left (2 \, e^{\left (x^{2}\right )} + 18\right )} + 4 \, x^{2} e^{\left (2 \, e^{\left (x^{2}\right )} + 18\right )} \log \left (2\right ) + 4 \, x e^{\left (2 \, e^{\left (x^{2}\right )} + 18\right )} \log \left (2\right )^{2} + 16 \, x^{3} + 64 \, x \log \left (2\right )^{2} + {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} e^{6} - 8 \, {\left (x^{3} - 5 \, x^{2}\right )} e^{3} + 2 \, {\left (x^{3} e^{\left (x^{2} + e^{\left (x^{2}\right )} + 12\right )} - 4 \, x^{3} e^{\left (x^{2} + e^{\left (x^{2}\right )} + 9\right )} + 2 \, x^{2} e^{\left (x^{2} + e^{\left (x^{2}\right )} + 12\right )} \log \left (2\right ) - 16 \, x^{2} e^{\left (x^{2} + e^{\left (x^{2}\right )} + 9\right )} \log \left (2\right ) - 16 \, x e^{\left (x^{2} + e^{\left (x^{2}\right )} + 9\right )} \log \left (2\right )^{2} - 5 \, x^{2} e^{\left (x^{2} + e^{\left (x^{2}\right )} + 12\right )} - 10 \, x e^{\left (x^{2} + e^{\left (x^{2}\right )} + 12\right )} \log \left (2\right )\right )} e^{\left (-x^{2}\right )} + 16 \, {\left (4 \, x^{2} - {\left (x^{2} - 5 \, x\right )} e^{3}\right )} \log \left (2\right )\right )} e^{\left (-6\right )} \] Input:
integrate((((16*x^2*log(2)^2+16*x^3*log(2)+4*x^4)*exp(x^2)+4*log(2)^2+8*x* log(2)+3*x^2)*exp(exp(x^2)+9)^2+((-64*x^2*log(2)^2+2*((4*x^3-20*x^2)*exp(3 )-32*x^3)*log(2)+(4*x^4-20*x^3)*exp(3)-16*x^4)*exp(x^2)-32*log(2)^2+2*((4* x-10)*exp(3)-32*x)*log(2)+(6*x^2-20*x)*exp(3)-24*x^2)*exp(exp(x^2)+9)+64*l og(2)^2+2*((-16*x+40)*exp(3)+64*x)*log(2)+(3*x^2-20*x+25)*exp(3)^2+(-24*x^ 2+80*x)*exp(3)+48*x^2)/exp(3)^2,x, algorithm="giac")
Output:
(x^3*e^(2*e^(x^2) + 18) + 4*x^2*e^(2*e^(x^2) + 18)*log(2) + 4*x*e^(2*e^(x^ 2) + 18)*log(2)^2 + 16*x^3 + 64*x*log(2)^2 + (x^3 - 10*x^2 + 25*x)*e^6 - 8 *(x^3 - 5*x^2)*e^3 + 2*(x^3*e^(x^2 + e^(x^2) + 12) - 4*x^3*e^(x^2 + e^(x^2 ) + 9) + 2*x^2*e^(x^2 + e^(x^2) + 12)*log(2) - 16*x^2*e^(x^2 + e^(x^2) + 9 )*log(2) - 16*x*e^(x^2 + e^(x^2) + 9)*log(2)^2 - 5*x^2*e^(x^2 + e^(x^2) + 12) - 10*x*e^(x^2 + e^(x^2) + 12)*log(2))*e^(-x^2) + 16*(4*x^2 - (x^2 - 5* x)*e^3)*log(2))*e^(-6)
Time = 3.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {48 x^2+e^3 \left (80 x-24 x^2\right )+e^6 \left (25-20 x+3 x^2\right )+\left (e^3 (40-16 x)+64 x\right ) \log (4)+16 \log ^2(4)+e^{9+e^{x^2}} \left (-24 x^2+e^3 \left (-20 x+6 x^2\right )+\left (-32 x+e^3 (-10+4 x)\right ) \log (4)-8 \log ^2(4)+e^{x^2} \left (-16 x^4+e^3 \left (-20 x^3+4 x^4\right )+\left (-32 x^3+e^3 \left (-20 x^2+4 x^3\right )\right ) \log (4)-16 x^2 \log ^2(4)\right )\right )+e^{18+2 e^{x^2}} \left (3 x^2+4 x \log (4)+\log ^2(4)+e^{x^2} \left (4 x^4+8 x^3 \log (4)+4 x^2 \log ^2(4)\right )\right )}{e^6} \, dx=x\,{\mathrm {e}}^{-6}\,{\left (4\,x+5\,{\mathrm {e}}^3+8\,\ln \left (2\right )-x\,{\mathrm {e}}^3-2\,{\mathrm {e}}^9\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}}\,\ln \left (2\right )-x\,{\mathrm {e}}^9\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}}\right )}^2 \] Input:
int(exp(-6)*(exp(6)*(3*x^2 - 20*x + 25) - exp(exp(x^2) + 9)*(exp(3)*(20*x - 6*x^2) + exp(x^2)*(64*x^2*log(2)^2 + exp(3)*(20*x^3 - 4*x^4) + 16*x^4 + 2*log(2)*(exp(3)*(20*x^2 - 4*x^3) + 32*x^3)) + 32*log(2)^2 + 24*x^2 + 2*lo g(2)*(32*x - exp(3)*(4*x - 10))) + exp(3)*(80*x - 24*x^2) + exp(2*exp(x^2) + 18)*(8*x*log(2) + 4*log(2)^2 + 3*x^2 + exp(x^2)*(16*x^2*log(2)^2 + 16*x ^3*log(2) + 4*x^4)) + 64*log(2)^2 + 48*x^2 + 2*log(2)*(64*x - exp(3)*(16*x - 40))),x)
Output:
x*exp(-6)*(4*x + 5*exp(3) + 8*log(2) - x*exp(3) - 2*exp(9)*exp(exp(x^2))*l og(2) - x*exp(9)*exp(exp(x^2)))^2
Time = 0.21 (sec) , antiderivative size = 223, normalized size of antiderivative = 8.58 \[ \int \frac {48 x^2+e^3 \left (80 x-24 x^2\right )+e^6 \left (25-20 x+3 x^2\right )+\left (e^3 (40-16 x)+64 x\right ) \log (4)+16 \log ^2(4)+e^{9+e^{x^2}} \left (-24 x^2+e^3 \left (-20 x+6 x^2\right )+\left (-32 x+e^3 (-10+4 x)\right ) \log (4)-8 \log ^2(4)+e^{x^2} \left (-16 x^4+e^3 \left (-20 x^3+4 x^4\right )+\left (-32 x^3+e^3 \left (-20 x^2+4 x^3\right )\right ) \log (4)-16 x^2 \log ^2(4)\right )\right )+e^{18+2 e^{x^2}} \left (3 x^2+4 x \log (4)+\log ^2(4)+e^{x^2} \left (4 x^4+8 x^3 \log (4)+4 x^2 \log ^2(4)\right )\right )}{e^6} \, dx=\frac {x \left (4 e^{2 e^{x^{2}}} \mathrm {log}\left (2\right )^{2} e^{18}+4 e^{2 e^{x^{2}}} \mathrm {log}\left (2\right ) e^{18} x +e^{2 e^{x^{2}}} e^{18} x^{2}-32 e^{e^{x^{2}}} \mathrm {log}\left (2\right )^{2} e^{9}+4 e^{e^{x^{2}}} \mathrm {log}\left (2\right ) e^{12} x -20 e^{e^{x^{2}}} \mathrm {log}\left (2\right ) e^{12}-32 e^{e^{x^{2}}} \mathrm {log}\left (2\right ) e^{9} x +2 e^{e^{x^{2}}} e^{12} x^{2}-10 e^{e^{x^{2}}} e^{12} x -8 e^{e^{x^{2}}} e^{9} x^{2}+64 \mathrm {log}\left (2\right )^{2}-16 \,\mathrm {log}\left (2\right ) e^{3} x +80 \,\mathrm {log}\left (2\right ) e^{3}+64 \,\mathrm {log}\left (2\right ) x +e^{6} x^{2}-10 e^{6} x +25 e^{6}-8 e^{3} x^{2}+40 e^{3} x +16 x^{2}\right )}{e^{6}} \] Input:
int((((16*x^2*log(2)^2+16*x^3*log(2)+4*x^4)*exp(x^2)+4*log(2)^2+8*x*log(2) +3*x^2)*exp(exp(x^2)+9)^2+((-64*x^2*log(2)^2+2*((4*x^3-20*x^2)*exp(3)-32*x ^3)*log(2)+(4*x^4-20*x^3)*exp(3)-16*x^4)*exp(x^2)-32*log(2)^2+2*((4*x-10)* exp(3)-32*x)*log(2)+(6*x^2-20*x)*exp(3)-24*x^2)*exp(exp(x^2)+9)+64*log(2)^ 2+2*((-16*x+40)*exp(3)+64*x)*log(2)+(3*x^2-20*x+25)*exp(3)^2+(-24*x^2+80*x )*exp(3)+48*x^2)/exp(3)^2,x)
Output:
(x*(4*e**(2*e**(x**2))*log(2)**2*e**18 + 4*e**(2*e**(x**2))*log(2)*e**18*x + e**(2*e**(x**2))*e**18*x**2 - 32*e**(e**(x**2))*log(2)**2*e**9 + 4*e**( e**(x**2))*log(2)*e**12*x - 20*e**(e**(x**2))*log(2)*e**12 - 32*e**(e**(x* *2))*log(2)*e**9*x + 2*e**(e**(x**2))*e**12*x**2 - 10*e**(e**(x**2))*e**12 *x - 8*e**(e**(x**2))*e**9*x**2 + 64*log(2)**2 - 16*log(2)*e**3*x + 80*log (2)*e**3 + 64*log(2)*x + e**6*x**2 - 10*e**6*x + 25*e**6 - 8*e**3*x**2 + 4 0*e**3*x + 16*x**2))/e**6