\(\int \frac {e^{\frac {(100 x-40 x^2+4 x^3+e^4 (25 x-10 x^2+x^3)) \log ^2(\frac {e}{3})+(-40 x+8 x^2+e^4 (-10 x+2 x^2)) \log (\frac {e}{3}) \log (x)+(4 x+e^4 x) \log ^2(x)}{\log ^2(\frac {e}{3})}} ((100-80 x+12 x^2+e^4 (25-20 x+3 x^2)) \log ^2(\frac {e}{3})+(8+2 e^4) \log (x)+(4+e^4) \log ^2(x)+\log (\frac {e}{3}) (-40+8 x+e^4 (-10+2 x)+(-40+16 x+e^4 (-10+4 x)) \log (x)))}{\log ^2(\frac {e}{3})} \, dx\) [2900]

Optimal result

Integrand size = 197, antiderivative size = 25 \[ \int \frac {e^{\frac {\left (100 x-40 x^2+4 x^3+e^4 \left (25 x-10 x^2+x^3\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (-40 x+8 x^2+e^4 \left (-10 x+2 x^2\right )\right ) \log \left (\frac {e}{3}\right ) \log (x)+\left (4 x+e^4 x\right ) \log ^2(x)}{\log ^2\left (\frac {e}{3}\right )}} \left (\left (100-80 x+12 x^2+e^4 \left (25-20 x+3 x^2\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (8+2 e^4\right ) \log (x)+\left (4+e^4\right ) \log ^2(x)+\log \left (\frac {e}{3}\right ) \left (-40+8 x+e^4 (-10+2 x)+\left (-40+16 x+e^4 (-10+4 x)\right ) \log (x)\right )\right )}{\log ^2\left (\frac {e}{3}\right )} \, dx=e^{\left (4+e^4\right ) x \left (-5+x+\frac {\log (x)}{\log \left (\frac {e}{3}\right )}\right )^2} \] Output:

exp((4+exp(4))*x*(ln(x)/ln(1/3*exp(1))+x-5)^2)
 

Mathematica [F]

\[ \int \frac {e^{\frac {\left (100 x-40 x^2+4 x^3+e^4 \left (25 x-10 x^2+x^3\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (-40 x+8 x^2+e^4 \left (-10 x+2 x^2\right )\right ) \log \left (\frac {e}{3}\right ) \log (x)+\left (4 x+e^4 x\right ) \log ^2(x)}{\log ^2\left (\frac {e}{3}\right )}} \left (\left (100-80 x+12 x^2+e^4 \left (25-20 x+3 x^2\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (8+2 e^4\right ) \log (x)+\left (4+e^4\right ) \log ^2(x)+\log \left (\frac {e}{3}\right ) \left (-40+8 x+e^4 (-10+2 x)+\left (-40+16 x+e^4 (-10+4 x)\right ) \log (x)\right )\right )}{\log ^2\left (\frac {e}{3}\right )} \, dx=\int \frac {e^{\frac {\left (100 x-40 x^2+4 x^3+e^4 \left (25 x-10 x^2+x^3\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (-40 x+8 x^2+e^4 \left (-10 x+2 x^2\right )\right ) \log \left (\frac {e}{3}\right ) \log (x)+\left (4 x+e^4 x\right ) \log ^2(x)}{\log ^2\left (\frac {e}{3}\right )}} \left (\left (100-80 x+12 x^2+e^4 \left (25-20 x+3 x^2\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (8+2 e^4\right ) \log (x)+\left (4+e^4\right ) \log ^2(x)+\log \left (\frac {e}{3}\right ) \left (-40+8 x+e^4 (-10+2 x)+\left (-40+16 x+e^4 (-10+4 x)\right ) \log (x)\right )\right )}{\log ^2\left (\frac {e}{3}\right )} \, dx \] Input:

Integrate[(E^(((100*x - 40*x^2 + 4*x^3 + E^4*(25*x - 10*x^2 + x^3))*Log[E/ 
3]^2 + (-40*x + 8*x^2 + E^4*(-10*x + 2*x^2))*Log[E/3]*Log[x] + (4*x + E^4* 
x)*Log[x]^2)/Log[E/3]^2)*((100 - 80*x + 12*x^2 + E^4*(25 - 20*x + 3*x^2))* 
Log[E/3]^2 + (8 + 2*E^4)*Log[x] + (4 + E^4)*Log[x]^2 + Log[E/3]*(-40 + 8*x 
 + E^4*(-10 + 2*x) + (-40 + 16*x + E^4*(-10 + 4*x))*Log[x])))/Log[E/3]^2,x 
]
 

Output:

Integrate[E^(((100*x - 40*x^2 + 4*x^3 + E^4*(25*x - 10*x^2 + x^3))*Log[E/3 
]^2 + (-40*x + 8*x^2 + E^4*(-10*x + 2*x^2))*Log[E/3]*Log[x] + (4*x + E^4*x 
)*Log[x]^2)/Log[E/3]^2)*((100 - 80*x + 12*x^2 + E^4*(25 - 20*x + 3*x^2))*L 
og[E/3]^2 + (8 + 2*E^4)*Log[x] + (4 + E^4)*Log[x]^2 + Log[E/3]*(-40 + 8*x 
+ E^4*(-10 + 2*x) + (-40 + 16*x + E^4*(-10 + 4*x))*Log[x])), x]/Log[E/3]^2
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(97\) vs. \(2(25)=50\).

Time = 2.29 (sec) , antiderivative size = 97, normalized size of antiderivative = 3.88, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {27, 2726}

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.

\(\displaystyle \int \frac {\left (\left (12 x^2+e^4 \left (3 x^2-20 x+25\right )-80 x+100\right ) \log ^2\left (\frac {e}{3}\right )+\left (4+e^4\right ) \log ^2(x)+\left (8+2 e^4\right ) \log (x)+\log \left (\frac {e}{3}\right ) \left (8 x+e^4 (2 x-10)+\left (16 x+e^4 (4 x-10)-40\right ) \log (x)-40\right )\right ) \exp \left (\frac {\left (8 x^2+e^4 \left (2 x^2-10 x\right )-40 x\right ) \log \left (\frac {e}{3}\right ) \log (x)+\left (4 x^3-40 x^2+e^4 \left (x^3-10 x^2+25 x\right )+100 x\right ) \log ^2\left (\frac {e}{3}\right )+\left (e^4 x+4 x\right ) \log ^2(x)}{\log ^2\left (\frac {e}{3}\right )}\right )}{\log ^2\left (\frac {e}{3}\right )} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \exp \left (\frac {\left (4+e^4\right ) x \log ^2(x)+\left (4 x^3-40 x^2+100 x+e^4 \left (x^3-10 x^2+25 x\right )\right ) (1-\log (3))^2}{(1-\log (3))^2}\right ) x^{-\frac {-8 x^2+40 x+e^4 \left (10 x-2 x^2\right )}{\log \left (\frac {e}{3}\right )}} \left (\left (4+e^4\right ) \log ^2(x)+2 \left (4+e^4\right ) \log (x)-2 \log \left (\frac {e}{3}\right ) \left (e^4 (5-x)-4 x+\left (e^4 (5-2 x)-8 x+20\right ) \log (x)+20\right )+\left (12 x^2-80 x+e^4 \left (3 x^2-20 x+25\right )+100\right ) (1-\log (3))^2\right )dx}{(1-\log (3))^2}\)

\(\Big \downarrow \) 2726

\(\displaystyle x^{-\frac {2 \left (-4 x^2+e^4 \left (5 x-x^2\right )+20 x\right )}{\log \left (\frac {e}{3}\right )}} \exp \left (\frac {\left (4 x^3-40 x^2+e^4 \left (x^3-10 x^2+25 x\right )+100 x\right ) (1-\log (3))^2+\left (4+e^4\right ) x \log ^2(x)}{(1-\log (3))^2}\right )\)

Input:

Int[(E^(((100*x - 40*x^2 + 4*x^3 + E^4*(25*x - 10*x^2 + x^3))*Log[E/3]^2 + 
 (-40*x + 8*x^2 + E^4*(-10*x + 2*x^2))*Log[E/3]*Log[x] + (4*x + E^4*x)*Log 
[x]^2)/Log[E/3]^2)*((100 - 80*x + 12*x^2 + E^4*(25 - 20*x + 3*x^2))*Log[E/ 
3]^2 + (8 + 2*E^4)*Log[x] + (4 + E^4)*Log[x]^2 + Log[E/3]*(-40 + 8*x + E^4 
*(-10 + 2*x) + (-40 + 16*x + E^4*(-10 + 4*x))*Log[x])))/Log[E/3]^2,x]
 

Output:

E^(((100*x - 40*x^2 + 4*x^3 + E^4*(25*x - 10*x^2 + x^3))*(1 - Log[3])^2 + 
(4 + E^4)*x*Log[x]^2)/(1 - Log[3])^2)/x^((2*(20*x - 4*x^2 + E^4*(5*x - x^2 
)))/Log[E/3])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2726

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, 
 x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(89\) vs. \(2(22)=44\).

Time = 1.12 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.60

method	result	size
default	\({\mathrm e}^{\frac {\left (\left (x^{3}-10 x^{2}+25 x \right ) {\mathrm e}^{4}+4 x^{3}-40 x^{2}+100 x \right ) \ln \left (\frac {{\mathrm e}}{3}\right )^{2}+\left (\left (2 x^{2}-10 x \right ) {\mathrm e}^{4}+8 x^{2}-40 x \right ) \ln \left (x \right ) \ln \left (\frac {{\mathrm e}}{3}\right )+\left (x \,{\mathrm e}^{4}+4 x \right ) \ln \left (x \right )^{2}}{\ln \left (\frac {{\mathrm e}}{3}\right )^{2}}}\)	\(90\)
norman	\({\mathrm e}^{\frac {\left (\left (x^{3}-10 x^{2}+25 x \right ) {\mathrm e}^{4}+4 x^{3}-40 x^{2}+100 x \right ) \ln \left (\frac {{\mathrm e}}{3}\right )^{2}+\left (\left (2 x^{2}-10 x \right ) {\mathrm e}^{4}+8 x^{2}-40 x \right ) \ln \left (x \right ) \ln \left (\frac {{\mathrm e}}{3}\right )+\left (x \,{\mathrm e}^{4}+4 x \right ) \ln \left (x \right )^{2}}{\ln \left (\frac {{\mathrm e}}{3}\right )^{2}}}\)	\(90\)
parallelrisch	\({\mathrm e}^{\frac {\left (\left (x^{3}-10 x^{2}+25 x \right ) {\mathrm e}^{4}+4 x^{3}-40 x^{2}+100 x \right ) \ln \left (\frac {{\mathrm e}}{3}\right )^{2}+\left (\left (2 x^{2}-10 x \right ) {\mathrm e}^{4}+8 x^{2}-40 x \right ) \ln \left (x \right ) \ln \left (\frac {{\mathrm e}}{3}\right )+\left (x \,{\mathrm e}^{4}+4 x \right ) \ln \left (x \right )^{2}}{\ln \left (\frac {{\mathrm e}}{3}\right )^{2}}}\)	\(90\)
risch	\(\frac {{\mathrm e}^{\frac {x \left (x \ln \left (3\right )-5 \ln \left (3\right )-\ln \left (x \right )-x +5\right )^{2} \left (4+{\mathrm e}^{4}\right )}{\left (\ln \left (3\right )-1\right )^{2}}} \ln \left (3\right )^{2}}{\left (1-\ln \left (3\right )\right )^{2}}-\frac {2 \,{\mathrm e}^{\frac {x \left (x \ln \left (3\right )-5 \ln \left (3\right )-\ln \left (x \right )-x +5\right )^{2} \left (4+{\mathrm e}^{4}\right )}{\left (\ln \left (3\right )-1\right )^{2}}} \ln \left (3\right )}{\left (1-\ln \left (3\right )\right )^{2}}+\frac {{\mathrm e}^{\frac {x \left (x \ln \left (3\right )-5 \ln \left (3\right )-\ln \left (x \right )-x +5\right )^{2} \left (4+{\mathrm e}^{4}\right )}{\left (\ln \left (3\right )-1\right )^{2}}}}{\left (1-\ln \left (3\right )\right )^{2}}\)	\(132\)

Input:

int((((3*x^2-20*x+25)*exp(4)+12*x^2-80*x+100)*ln(1/3*exp(1))^2+(((4*x-10)* 
exp(4)+16*x-40)*ln(x)+(2*x-10)*exp(4)+8*x-40)*ln(1/3*exp(1))+(4+exp(4))*ln 
(x)^2+(2*exp(4)+8)*ln(x))*exp((((x^3-10*x^2+25*x)*exp(4)+4*x^3-40*x^2+100* 
x)*ln(1/3*exp(1))^2+((2*x^2-10*x)*exp(4)+8*x^2-40*x)*ln(x)*ln(1/3*exp(1))+ 
(x*exp(4)+4*x)*ln(x)^2)/ln(1/3*exp(1))^2)/ln(1/3*exp(1))^2,x,method=_RETUR 
NVERBOSE)
 

Output:

exp((((x^3-10*x^2+25*x)*exp(4)+4*x^3-40*x^2+100*x)*ln(1/3*exp(1))^2+((2*x^ 
2-10*x)*exp(4)+8*x^2-40*x)*ln(x)*ln(1/3*exp(1))+(x*exp(4)+4*x)*ln(x)^2)/ln 
(1/3*exp(1))^2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (22) = 44\).

Time = 0.09 (sec) , antiderivative size = 169, normalized size of antiderivative = 6.76 \[ \int \frac {e^{\frac {\left (100 x-40 x^2+4 x^3+e^4 \left (25 x-10 x^2+x^3\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (-40 x+8 x^2+e^4 \left (-10 x+2 x^2\right )\right ) \log \left (\frac {e}{3}\right ) \log (x)+\left (4 x+e^4 x\right ) \log ^2(x)}{\log ^2\left (\frac {e}{3}\right )}} \left (\left (100-80 x+12 x^2+e^4 \left (25-20 x+3 x^2\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (8+2 e^4\right ) \log (x)+\left (4+e^4\right ) \log ^2(x)+\log \left (\frac {e}{3}\right ) \left (-40+8 x+e^4 (-10+2 x)+\left (-40+16 x+e^4 (-10+4 x)\right ) \log (x)\right )\right )}{\log ^2\left (\frac {e}{3}\right )} \, dx=e^{\left (\frac {4 \, x^{3} + {\left (4 \, x^{3} - 40 \, x^{2} + {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} e^{4} + 100 \, x\right )} \log \left (3\right )^{2} + {\left (x e^{4} + 4 \, x\right )} \log \left (x\right )^{2} - 40 \, x^{2} + {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} e^{4} - 2 \, {\left (4 \, x^{3} - 40 \, x^{2} + {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} e^{4} + 100 \, x\right )} \log \left (3\right ) + 2 \, {\left (4 \, x^{2} + {\left (x^{2} - 5 \, x\right )} e^{4} - {\left (4 \, x^{2} + {\left (x^{2} - 5 \, x\right )} e^{4} - 20 \, x\right )} \log \left (3\right ) - 20 \, x\right )} \log \left (x\right ) + 100 \, x}{\log \left (3\right )^{2} - 2 \, \log \left (3\right ) + 1}\right )} \] Input:

integrate((((3*x^2-20*x+25)*exp(4)+12*x^2-80*x+100)*log(1/3*exp(1))^2+(((4 
*x-10)*exp(4)+16*x-40)*log(x)+(2*x-10)*exp(4)+8*x-40)*log(1/3*exp(1))+(4+e 
xp(4))*log(x)^2+(2*exp(4)+8)*log(x))*exp((((x^3-10*x^2+25*x)*exp(4)+4*x^3- 
40*x^2+100*x)*log(1/3*exp(1))^2+((2*x^2-10*x)*exp(4)+8*x^2-40*x)*log(x)*lo 
g(1/3*exp(1))+(x*exp(4)+4*x)*log(x)^2)/log(1/3*exp(1))^2)/log(1/3*exp(1))^ 
2,x, algorithm="fricas")
 

Output:

e^((4*x^3 + (4*x^3 - 40*x^2 + (x^3 - 10*x^2 + 25*x)*e^4 + 100*x)*log(3)^2 
+ (x*e^4 + 4*x)*log(x)^2 - 40*x^2 + (x^3 - 10*x^2 + 25*x)*e^4 - 2*(4*x^3 - 
 40*x^2 + (x^3 - 10*x^2 + 25*x)*e^4 + 100*x)*log(3) + 2*(4*x^2 + (x^2 - 5* 
x)*e^4 - (4*x^2 + (x^2 - 5*x)*e^4 - 20*x)*log(3) - 20*x)*log(x) + 100*x)/( 
log(3)^2 - 2*log(3) + 1))
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (22) = 44\).

Time = 0.89 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.60 \[ \int \frac {e^{\frac {\left (100 x-40 x^2+4 x^3+e^4 \left (25 x-10 x^2+x^3\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (-40 x+8 x^2+e^4 \left (-10 x+2 x^2\right )\right ) \log \left (\frac {e}{3}\right ) \log (x)+\left (4 x+e^4 x\right ) \log ^2(x)}{\log ^2\left (\frac {e}{3}\right )}} \left (\left (100-80 x+12 x^2+e^4 \left (25-20 x+3 x^2\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (8+2 e^4\right ) \log (x)+\left (4+e^4\right ) \log ^2(x)+\log \left (\frac {e}{3}\right ) \left (-40+8 x+e^4 (-10+2 x)+\left (-40+16 x+e^4 (-10+4 x)\right ) \log (x)\right )\right )}{\log ^2\left (\frac {e}{3}\right )} \, dx=e^{\frac {\left (4 x + x e^{4}\right ) \log {\left (x \right )}^{2} + \left (8 x^{2} - 40 x + \left (2 x^{2} - 10 x\right ) e^{4}\right ) \log {\left (\frac {e}{3} \right )} \log {\left (x \right )} + \left (4 x^{3} - 40 x^{2} + 100 x + \left (x^{3} - 10 x^{2} + 25 x\right ) e^{4}\right ) \log {\left (\frac {e}{3} \right )}^{2}}{\log {\left (\frac {e}{3} \right )}^{2}}} \] Input:

integrate((((3*x**2-20*x+25)*exp(4)+12*x**2-80*x+100)*ln(1/3*exp(1))**2+(( 
(4*x-10)*exp(4)+16*x-40)*ln(x)+(2*x-10)*exp(4)+8*x-40)*ln(1/3*exp(1))+(4+e 
xp(4))*ln(x)**2+(2*exp(4)+8)*ln(x))*exp((((x**3-10*x**2+25*x)*exp(4)+4*x** 
3-40*x**2+100*x)*ln(1/3*exp(1))**2+((2*x**2-10*x)*exp(4)+8*x**2-40*x)*ln(x 
)*ln(1/3*exp(1))+(x*exp(4)+4*x)*ln(x)**2)/ln(1/3*exp(1))**2)/ln(1/3*exp(1) 
)**2,x)
 

Output:

exp(((4*x + x*exp(4))*log(x)**2 + (8*x**2 - 40*x + (2*x**2 - 10*x)*exp(4)) 
*log(E/3)*log(x) + (4*x**3 - 40*x**2 + 100*x + (x**3 - 10*x**2 + 25*x)*exp 
(4))*log(E/3)**2)/log(E/3)**2)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (22) = 44\).

Time = 0.40 (sec) , antiderivative size = 142, normalized size of antiderivative = 5.68 \[ \int \frac {e^{\frac {\left (100 x-40 x^2+4 x^3+e^4 \left (25 x-10 x^2+x^3\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (-40 x+8 x^2+e^4 \left (-10 x+2 x^2\right )\right ) \log \left (\frac {e}{3}\right ) \log (x)+\left (4 x+e^4 x\right ) \log ^2(x)}{\log ^2\left (\frac {e}{3}\right )}} \left (\left (100-80 x+12 x^2+e^4 \left (25-20 x+3 x^2\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (8+2 e^4\right ) \log (x)+\left (4+e^4\right ) \log ^2(x)+\log \left (\frac {e}{3}\right ) \left (-40+8 x+e^4 (-10+2 x)+\left (-40+16 x+e^4 (-10+4 x)\right ) \log (x)\right )\right )}{\log ^2\left (\frac {e}{3}\right )} \, dx=\frac {{\left (\log \left (3\right )^{2} - 2 \, \log \left (3\right ) + 1\right )} e^{\left (x^{3} e^{4} + 4 \, x^{3} - 10 \, x^{2} e^{4} - \frac {2 \, x^{2} e^{4} \log \left (x\right )}{\log \left (3\right ) - 1} + \frac {x e^{4} \log \left (x\right )^{2}}{\log \left (3\right )^{2} - 2 \, \log \left (3\right ) + 1} - 40 \, x^{2} + 25 \, x e^{4} - \frac {8 \, x^{2} \log \left (x\right )}{\log \left (3\right ) - 1} + \frac {10 \, x e^{4} \log \left (x\right )}{\log \left (3\right ) - 1} + \frac {4 \, x \log \left (x\right )^{2}}{\log \left (3\right )^{2} - 2 \, \log \left (3\right ) + 1} + 100 \, x + \frac {40 \, x \log \left (x\right )}{\log \left (3\right ) - 1}\right )}}{\log \left (\frac {1}{3} \, e\right )^{2}} \] Input:

integrate((((3*x^2-20*x+25)*exp(4)+12*x^2-80*x+100)*log(1/3*exp(1))^2+(((4 
*x-10)*exp(4)+16*x-40)*log(x)+(2*x-10)*exp(4)+8*x-40)*log(1/3*exp(1))+(4+e 
xp(4))*log(x)^2+(2*exp(4)+8)*log(x))*exp((((x^3-10*x^2+25*x)*exp(4)+4*x^3- 
40*x^2+100*x)*log(1/3*exp(1))^2+((2*x^2-10*x)*exp(4)+8*x^2-40*x)*log(x)*lo 
g(1/3*exp(1))+(x*exp(4)+4*x)*log(x)^2)/log(1/3*exp(1))^2)/log(1/3*exp(1))^ 
2,x, algorithm="maxima")
 

Output:

(log(3)^2 - 2*log(3) + 1)*e^(x^3*e^4 + 4*x^3 - 10*x^2*e^4 - 2*x^2*e^4*log( 
x)/(log(3) - 1) + x*e^4*log(x)^2/(log(3)^2 - 2*log(3) + 1) - 40*x^2 + 25*x 
*e^4 - 8*x^2*log(x)/(log(3) - 1) + 10*x*e^4*log(x)/(log(3) - 1) + 4*x*log( 
x)^2/(log(3)^2 - 2*log(3) + 1) + 100*x + 40*x*log(x)/(log(3) - 1))/log(1/3 
*e)^2
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 690 vs. \(2 (22) = 44\).

Time = 0.84 (sec) , antiderivative size = 690, normalized size of antiderivative = 27.60 \[ \int \frac {e^{\frac {\left (100 x-40 x^2+4 x^3+e^4 \left (25 x-10 x^2+x^3\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (-40 x+8 x^2+e^4 \left (-10 x+2 x^2\right )\right ) \log \left (\frac {e}{3}\right ) \log (x)+\left (4 x+e^4 x\right ) \log ^2(x)}{\log ^2\left (\frac {e}{3}\right )}} \left (\left (100-80 x+12 x^2+e^4 \left (25-20 x+3 x^2\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (8+2 e^4\right ) \log (x)+\left (4+e^4\right ) \log ^2(x)+\log \left (\frac {e}{3}\right ) \left (-40+8 x+e^4 (-10+2 x)+\left (-40+16 x+e^4 (-10+4 x)\right ) \log (x)\right )\right )}{\log ^2\left (\frac {e}{3}\right )} \, dx=\text {Too large to display} \] Input:

integrate((((3*x^2-20*x+25)*exp(4)+12*x^2-80*x+100)*log(1/3*exp(1))^2+(((4 
*x-10)*exp(4)+16*x-40)*log(x)+(2*x-10)*exp(4)+8*x-40)*log(1/3*exp(1))+(4+e 
xp(4))*log(x)^2+(2*exp(4)+8)*log(x))*exp((((x^3-10*x^2+25*x)*exp(4)+4*x^3- 
40*x^2+100*x)*log(1/3*exp(1))^2+((2*x^2-10*x)*exp(4)+8*x^2-40*x)*log(x)*lo 
g(1/3*exp(1))+(x*exp(4)+4*x)*log(x)^2)/log(1/3*exp(1))^2)/log(1/3*exp(1))^ 
2,x, algorithm="giac")
 

Output:

(e^((x^3*e^4*log(3)^2 - 2*x^3*e^4*log(3) + 4*x^3*log(3)^2 - 10*x^2*e^4*log 
(3)^2 - 2*x^2*e^4*log(3)*log(x) + x^3*e^4 - 8*x^3*log(3) + 20*x^2*e^4*log( 
3) - 40*x^2*log(3)^2 + 25*x*e^4*log(3)^2 + 2*x^2*e^4*log(x) - 8*x^2*log(3) 
*log(x) + 10*x*e^4*log(3)*log(x) + x*e^4*log(x)^2 + 4*x^3 - 10*x^2*e^4 + 8 
0*x^2*log(3) - 50*x*e^4*log(3) + 100*x*log(3)^2 + 8*x^2*log(x) - 10*x*e^4* 
log(x) + 40*x*log(3)*log(x) + 4*x*log(x)^2 - 40*x^2 + 25*x*e^4 - 200*x*log 
(3) - 40*x*log(x) + 100*x)/(log(3)^2 - 2*log(3) + 1))*log(3)^2 - 2*e^((x^3 
*e^4*log(3)^2 - 2*x^3*e^4*log(3) + 4*x^3*log(3)^2 - 10*x^2*e^4*log(3)^2 - 
2*x^2*e^4*log(3)*log(x) + x^3*e^4 - 8*x^3*log(3) + 20*x^2*e^4*log(3) - 40* 
x^2*log(3)^2 + 25*x*e^4*log(3)^2 + 2*x^2*e^4*log(x) - 8*x^2*log(3)*log(x) 
+ 10*x*e^4*log(3)*log(x) + x*e^4*log(x)^2 + 4*x^3 - 10*x^2*e^4 + 80*x^2*lo 
g(3) - 50*x*e^4*log(3) + 100*x*log(3)^2 + 8*x^2*log(x) - 10*x*e^4*log(x) + 
 40*x*log(3)*log(x) + 4*x*log(x)^2 - 40*x^2 + 25*x*e^4 - 200*x*log(3) - 40 
*x*log(x) + 100*x)/(log(3)^2 - 2*log(3) + 1))*log(3) + e^((x^3*e^4*log(3)^ 
2 - 2*x^3*e^4*log(3) + 4*x^3*log(3)^2 - 10*x^2*e^4*log(3)^2 - 2*x^2*e^4*lo 
g(3)*log(x) + x^3*e^4 - 8*x^3*log(3) + 20*x^2*e^4*log(3) - 40*x^2*log(3)^2 
 + 25*x*e^4*log(3)^2 + 2*x^2*e^4*log(x) - 8*x^2*log(3)*log(x) + 10*x*e^4*l 
og(3)*log(x) + x*e^4*log(x)^2 + 4*x^3 - 10*x^2*e^4 + 80*x^2*log(3) - 50*x* 
e^4*log(3) + 100*x*log(3)^2 + 8*x^2*log(x) - 10*x*e^4*log(x) + 40*x*log(3) 
*log(x) + 4*x*log(x)^2 - 40*x^2 + 25*x*e^4 - 200*x*log(3) - 40*x*log(x)...
 

Mupad [B] (verification not implemented)

Time = 4.44 (sec) , antiderivative size = 362, normalized size of antiderivative = 14.48 \[ \int \frac {e^{\frac {\left (100 x-40 x^2+4 x^3+e^4 \left (25 x-10 x^2+x^3\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (-40 x+8 x^2+e^4 \left (-10 x+2 x^2\right )\right ) \log \left (\frac {e}{3}\right ) \log (x)+\left (4 x+e^4 x\right ) \log ^2(x)}{\log ^2\left (\frac {e}{3}\right )}} \left (\left (100-80 x+12 x^2+e^4 \left (25-20 x+3 x^2\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (8+2 e^4\right ) \log (x)+\left (4+e^4\right ) \log ^2(x)+\log \left (\frac {e}{3}\right ) \left (-40+8 x+e^4 (-10+2 x)+\left (-40+16 x+e^4 (-10+4 x)\right ) \log (x)\right )\right )}{\log ^2\left (\frac {e}{3}\right )} \, dx={\left (\frac {1}{9}\right )}^{\frac {100\,x+25\,x\,{\mathrm {e}}^4-10\,x^2\,{\mathrm {e}}^4+x^3\,{\mathrm {e}}^4-40\,x^2+4\,x^3}{{\ln \left (3\right )}^2-2\,\ln \left (3\right )+1}}\,x^{\frac {2\,\left (20\,x+5\,x\,{\mathrm {e}}^4-x^2\,{\mathrm {e}}^4-4\,x^2\right )}{\ln \left (3\right )-1}}\,{\mathrm {e}}^{\frac {x^3\,{\mathrm {e}}^4\,{\ln \left (3\right )}^2}{{\ln \left (3\right )}^2-2\,\ln \left (3\right )+1}}\,{\mathrm {e}}^{-\frac {10\,x^2\,{\mathrm {e}}^4\,{\ln \left (3\right )}^2}{{\ln \left (3\right )}^2-2\,\ln \left (3\right )+1}}\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^4\,{\ln \left (x\right )}^2}{{\ln \left (3\right )}^2-2\,\ln \left (3\right )+1}}\,{\mathrm {e}}^{\frac {x^3\,{\mathrm {e}}^4}{{\ln \left (3\right )}^2-2\,\ln \left (3\right )+1}}\,{\mathrm {e}}^{-\frac {10\,x^2\,{\mathrm {e}}^4}{{\ln \left (3\right )}^2-2\,\ln \left (3\right )+1}}\,{\mathrm {e}}^{\frac {100\,x\,{\ln \left (3\right )}^2}{{\ln \left (3\right )}^2-2\,\ln \left (3\right )+1}}\,{\mathrm {e}}^{\frac {100\,x}{{\ln \left (3\right )}^2-2\,\ln \left (3\right )+1}}\,{\mathrm {e}}^{\frac {4\,x^3\,{\ln \left (3\right )}^2}{{\ln \left (3\right )}^2-2\,\ln \left (3\right )+1}}\,{\mathrm {e}}^{-\frac {40\,x^2\,{\ln \left (3\right )}^2}{{\ln \left (3\right )}^2-2\,\ln \left (3\right )+1}}\,{\mathrm {e}}^{\frac {25\,x\,{\mathrm {e}}^4\,{\ln \left (3\right )}^2}{{\ln \left (3\right )}^2-2\,\ln \left (3\right )+1}}\,{\mathrm {e}}^{\frac {4\,x\,{\ln \left (x\right )}^2}{{\ln \left (3\right )}^2-2\,\ln \left (3\right )+1}}\,{\mathrm {e}}^{\frac {4\,x^3}{{\ln \left (3\right )}^2-2\,\ln \left (3\right )+1}}\,{\mathrm {e}}^{-\frac {40\,x^2}{{\ln \left (3\right )}^2-2\,\ln \left (3\right )+1}}\,{\mathrm {e}}^{\frac {25\,x\,{\mathrm {e}}^4}{{\ln \left (3\right )}^2-2\,\ln \left (3\right )+1}} \] Input:

int((exp((log(x)^2*(4*x + x*exp(4)) + log(exp(1)/3)^2*(100*x + exp(4)*(25* 
x - 10*x^2 + x^3) - 40*x^2 + 4*x^3) - log(exp(1)/3)*log(x)*(40*x + exp(4)* 
(10*x - 2*x^2) - 8*x^2))/log(exp(1)/3)^2)*(log(exp(1)/3)*(8*x + log(x)*(16 
*x + exp(4)*(4*x - 10) - 40) + exp(4)*(2*x - 10) - 40) + log(x)*(2*exp(4) 
+ 8) + log(x)^2*(exp(4) + 4) + log(exp(1)/3)^2*(exp(4)*(3*x^2 - 20*x + 25) 
 - 80*x + 12*x^2 + 100)))/log(exp(1)/3)^2,x)
 

Output:

(1/9)^((100*x + 25*x*exp(4) - 10*x^2*exp(4) + x^3*exp(4) - 40*x^2 + 4*x^3) 
/(log(3)^2 - 2*log(3) + 1))*x^((2*(20*x + 5*x*exp(4) - x^2*exp(4) - 4*x^2) 
)/(log(3) - 1))*exp((x^3*exp(4)*log(3)^2)/(log(3)^2 - 2*log(3) + 1))*exp(- 
(10*x^2*exp(4)*log(3)^2)/(log(3)^2 - 2*log(3) + 1))*exp((x*exp(4)*log(x)^2 
)/(log(3)^2 - 2*log(3) + 1))*exp((x^3*exp(4))/(log(3)^2 - 2*log(3) + 1))*e 
xp(-(10*x^2*exp(4))/(log(3)^2 - 2*log(3) + 1))*exp((100*x*log(3)^2)/(log(3 
)^2 - 2*log(3) + 1))*exp((100*x)/(log(3)^2 - 2*log(3) + 1))*exp((4*x^3*log 
(3)^2)/(log(3)^2 - 2*log(3) + 1))*exp(-(40*x^2*log(3)^2)/(log(3)^2 - 2*log 
(3) + 1))*exp((25*x*exp(4)*log(3)^2)/(log(3)^2 - 2*log(3) + 1))*exp((4*x*l 
og(x)^2)/(log(3)^2 - 2*log(3) + 1))*exp((4*x^3)/(log(3)^2 - 2*log(3) + 1)) 
*exp(-(40*x^2)/(log(3)^2 - 2*log(3) + 1))*exp((25*x*exp(4))/(log(3)^2 - 2* 
log(3) + 1))
 

Reduce [F]

\[ \int \frac {e^{\frac {\left (100 x-40 x^2+4 x^3+e^4 \left (25 x-10 x^2+x^3\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (-40 x+8 x^2+e^4 \left (-10 x+2 x^2\right )\right ) \log \left (\frac {e}{3}\right ) \log (x)+\left (4 x+e^4 x\right ) \log ^2(x)}{\log ^2\left (\frac {e}{3}\right )}} \left (\left (100-80 x+12 x^2+e^4 \left (25-20 x+3 x^2\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (8+2 e^4\right ) \log (x)+\left (4+e^4\right ) \log ^2(x)+\log \left (\frac {e}{3}\right ) \left (-40+8 x+e^4 (-10+2 x)+\left (-40+16 x+e^4 (-10+4 x)\right ) \log (x)\right )\right )}{\log ^2\left (\frac {e}{3}\right )} \, dx=\int \frac {\left (\left (\left (3 x^{2}-20 x +25\right ) {\mathrm e}^{4}+12 x^{2}-80 x +100\right ) \mathrm {log}\left (\frac {{\mathrm e}}{3}\right )^{2}+\left (\left (\left (4 x -10\right ) {\mathrm e}^{4}+16 x -40\right ) \mathrm {log}\left (x \right )+\left (2 x -10\right ) {\mathrm e}^{4}+8 x -40\right ) \mathrm {log}\left (\frac {{\mathrm e}}{3}\right )+\left (4+{\mathrm e}^{4}\right ) \mathrm {log}\left (x \right )^{2}+\left (2 \,{\mathrm e}^{4}+8\right ) \mathrm {log}\left (x \right )\right ) {\mathrm e}^{\frac {\left (\left (x^{3}-10 x^{2}+25 x \right ) {\mathrm e}^{4}+4 x^{3}-40 x^{2}+100 x \right ) \mathrm {log}\left (\frac {{\mathrm e}}{3}\right )^{2}+\left (\left (2 x^{2}-10 x \right ) {\mathrm e}^{4}+8 x^{2}-40 x \right ) \mathrm {log}\left (x \right ) \mathrm {log}\left (\frac {{\mathrm e}}{3}\right )+\left (x \,{\mathrm e}^{4}+4 x \right ) \mathrm {log}\left (x \right )^{2}}{\mathrm {log}\left (\frac {{\mathrm e}}{3}\right )^{2}}}}{\mathrm {log}\left (\frac {{\mathrm e}}{3}\right )^{2}}d x \] Input:

int((((3*x^2-20*x+25)*exp(4)+12*x^2-80*x+100)*log(1/3*exp(1))^2+(((4*x-10) 
*exp(4)+16*x-40)*log(x)+(2*x-10)*exp(4)+8*x-40)*log(1/3*exp(1))+(4+exp(4)) 
*log(x)^2+(2*exp(4)+8)*log(x))*exp((((x^3-10*x^2+25*x)*exp(4)+4*x^3-40*x^2 
+100*x)*log(1/3*exp(1))^2+((2*x^2-10*x)*exp(4)+8*x^2-40*x)*log(x)*log(1/3* 
exp(1))+(x*exp(4)+4*x)*log(x)^2)/log(1/3*exp(1))^2)/log(1/3*exp(1))^2,x)
 

Output:

int((((3*x^2-20*x+25)*exp(4)+12*x^2-80*x+100)*log(1/3*exp(1))^2+(((4*x-10) 
*exp(4)+16*x-40)*log(x)+(2*x-10)*exp(4)+8*x-40)*log(1/3*exp(1))+(4+exp(4)) 
*log(x)^2+(2*exp(4)+8)*log(x))*exp((((x^3-10*x^2+25*x)*exp(4)+4*x^3-40*x^2 
+100*x)*log(1/3*exp(1))^2+((2*x^2-10*x)*exp(4)+8*x^2-40*x)*log(x)*log(1/3* 
exp(1))+(x*exp(4)+4*x)*log(x)^2)/log(1/3*exp(1))^2)/log(1/3*exp(1))^2,x)