3.11.35 \(\int \frac {e^{2 e^4} x^2-10 e^{3 e^4} x^2+25 e^{4 e^4} x^2-2 x \log (36)+10 e^{e^4} x \log (36)}{e^{4 e^4} x^2-10 e^{5 e^4} x^2+25 e^{6 e^4} x^2-2 e^{2 e^4} x \log (36)+10 e^{3 e^4} x \log (36)+\log ^2(36)} \, dx\) [1035]

3.11.35.1 Optimal result
3.11.35.2 Mathematica [B] (verified)
3.11.35.3 Rubi [B] (verified)
3.11.35.4 Maple [A] (verified)
3.11.35.5 Fricas [B] (verification not implemented)
3.11.35.6 Sympy [B] (verification not implemented)
3.11.35.7 Maxima [B] (verification not implemented)
3.11.35.8 Giac [F(-2)]
3.11.35.9 Mupad [B] (verification not implemented)

3.11.35.1 Optimal result

Integrand size = 118, antiderivative size = 28 \[ \int \frac {e^{2 e^4} x^2-10 e^{3 e^4} x^2+25 e^{4 e^4} x^2-2 x \log (36)+10 e^{e^4} x \log (36)}{e^{4 e^4} x^2-10 e^{5 e^4} x^2+25 e^{6 e^4} x^2-2 e^{2 e^4} x \log (36)+10 e^{3 e^4} x \log (36)+\log ^2(36)} \, dx=\frac {x}{e^{2 e^4}-\frac {\log (36)}{x-5 e^{e^4} x}} \]

output
x/(exp(exp(4))^2-2*ln(6)/(x-5*x*exp(exp(4))))
 
3.11.35.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(118\) vs. \(2(28)=56\).

Time = 0.07 (sec) , antiderivative size = 118, normalized size of antiderivative = 4.21 \[ \int \frac {e^{2 e^4} x^2-10 e^{3 e^4} x^2+25 e^{4 e^4} x^2-2 x \log (36)+10 e^{e^4} x \log (36)}{e^{4 e^4} x^2-10 e^{5 e^4} x^2+25 e^{6 e^4} x^2-2 e^{2 e^4} x \log (36)+10 e^{3 e^4} x \log (36)+\log ^2(36)} \, dx=\frac {e^{-4 e^4} \left (-1+5 e^{e^4}\right ) \left (e^{4 e^4} x^2-10 e^{5 e^4} x^2+25 e^{6 e^4} x^2-2 e^{2 e^4} x \log (36)+10 e^{3 e^4} x \log (36)+\log (36) \log (1296)\right )}{\left (1-5 e^{e^4}\right )^2 \left (-e^{2 e^4} x+5 e^{3 e^4} x+\log (36)\right )} \]

input
Integrate[(E^(2*E^4)*x^2 - 10*E^(3*E^4)*x^2 + 25*E^(4*E^4)*x^2 - 2*x*Log[3 
6] + 10*E^E^4*x*Log[36])/(E^(4*E^4)*x^2 - 10*E^(5*E^4)*x^2 + 25*E^(6*E^4)* 
x^2 - 2*E^(2*E^4)*x*Log[36] + 10*E^(3*E^4)*x*Log[36] + Log[36]^2),x]
 
output
((-1 + 5*E^E^4)*(E^(4*E^4)*x^2 - 10*E^(5*E^4)*x^2 + 25*E^(6*E^4)*x^2 - 2*E 
^(2*E^4)*x*Log[36] + 10*E^(3*E^4)*x*Log[36] + Log[36]*Log[1296]))/(E^(4*E^ 
4)*(1 - 5*E^E^4)^2*(-(E^(2*E^4)*x) + 5*E^(3*E^4)*x + Log[36]))
 
3.11.35.3 Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(88\) vs. \(2(28)=56\).

Time = 0.34 (sec) , antiderivative size = 88, normalized size of antiderivative = 3.14, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.076, Rules used = {6, 6, 6, 6, 6, 6, 1294, 1107, 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.

\(\displaystyle \int \frac {25 e^{4 e^4} x^2-10 e^{3 e^4} x^2+e^{2 e^4} x^2+10 e^{e^4} x \log (36)-2 x \log (36)}{25 e^{6 e^4} x^2-10 e^{5 e^4} x^2+e^{4 e^4} x^2+10 e^{3 e^4} x \log (36)-2 e^{2 e^4} x \log (36)+\log ^2(36)} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {\left (e^{2 e^4}-10 e^{3 e^4}\right ) x^2+25 e^{4 e^4} x^2+10 e^{e^4} x \log (36)-2 x \log (36)}{25 e^{6 e^4} x^2-10 e^{5 e^4} x^2+e^{4 e^4} x^2+10 e^{3 e^4} x \log (36)-2 e^{2 e^4} x \log (36)+\log ^2(36)}dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {\left (e^{2 e^4}-10 e^{3 e^4}+25 e^{4 e^4}\right ) x^2+10 e^{e^4} x \log (36)-2 x \log (36)}{25 e^{6 e^4} x^2-10 e^{5 e^4} x^2+e^{4 e^4} x^2+10 e^{3 e^4} x \log (36)-2 e^{2 e^4} x \log (36)+\log ^2(36)}dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {\left (e^{2 e^4}-10 e^{3 e^4}+25 e^{4 e^4}\right ) x^2+\left (10 e^{e^4}-2\right ) x \log (36)}{25 e^{6 e^4} x^2-10 e^{5 e^4} x^2+e^{4 e^4} x^2+10 e^{3 e^4} x \log (36)-2 e^{2 e^4} x \log (36)+\log ^2(36)}dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {\left (e^{2 e^4}-10 e^{3 e^4}+25 e^{4 e^4}\right ) x^2+\left (10 e^{e^4}-2\right ) x \log (36)}{\left (e^{4 e^4}-10 e^{5 e^4}\right ) x^2+25 e^{6 e^4} x^2+10 e^{3 e^4} x \log (36)-2 e^{2 e^4} x \log (36)+\log ^2(36)}dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {\left (e^{2 e^4}-10 e^{3 e^4}+25 e^{4 e^4}\right ) x^2+\left (10 e^{e^4}-2\right ) x \log (36)}{\left (e^{4 e^4}-10 e^{5 e^4}+25 e^{6 e^4}\right ) x^2+10 e^{3 e^4} x \log (36)-2 e^{2 e^4} x \log (36)+\log ^2(36)}dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {\left (e^{2 e^4}-10 e^{3 e^4}+25 e^{4 e^4}\right ) x^2+\left (10 e^{e^4}-2\right ) x \log (36)}{\left (e^{4 e^4}-10 e^{5 e^4}+25 e^{6 e^4}\right ) x^2+\left (10 e^{3 e^4}-2 e^{2 e^4}\right ) x \log (36)+\log ^2(36)}dx\)

\(\Big \downarrow \) 1294

\(\displaystyle e^{4 e^4} \left (1-5 e^{e^4}\right )^2 \int \frac {e^{2 e^4} \left (1-5 e^{e^4}\right )^2 x^2-2 \left (1-5 e^{e^4}\right ) x \log (36)}{\left (e^{4 e^4} \left (1-5 e^{e^4}\right )^2 x-e^{2 e^4} \left (1-5 e^{e^4}\right ) \log (36)\right )^2}dx\)

\(\Big \downarrow \) 1107

\(\displaystyle e^{4 e^4} \left (1-5 e^{e^4}\right )^2 \int \left (\frac {e^{-6 e^4}}{\left (-1+5 e^{e^4}\right )^2}-\frac {e^{-6 e^4} \log ^2(36)}{\left (1-5 e^{e^4}\right )^2 \left (e^{2 e^4} \left (1-5 e^{e^4}\right ) x-\log (36)\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle e^{4 e^4} \left (1-5 e^{e^4}\right )^2 \left (\frac {e^{-6 e^4} x}{\left (1-5 e^{e^4}\right )^2}+\frac {e^{-8 e^4} \log ^2(36)}{\left (1-5 e^{e^4}\right )^3 \left (e^{2 e^4} \left (1-5 e^{e^4}\right ) x-\log (36)\right )}\right )\)

input
Int[(E^(2*E^4)*x^2 - 10*E^(3*E^4)*x^2 + 25*E^(4*E^4)*x^2 - 2*x*Log[36] + 1 
0*E^E^4*x*Log[36])/(E^(4*E^4)*x^2 - 10*E^(5*E^4)*x^2 + 25*E^(6*E^4)*x^2 - 
2*E^(2*E^4)*x*Log[36] + 10*E^(3*E^4)*x*Log[36] + Log[36]^2),x]
 
output
E^(4*E^4)*(1 - 5*E^E^4)^2*(x/(E^(6*E^4)*(1 - 5*E^E^4)^2) + Log[36]^2/(E^(8 
*E^4)*(1 - 5*E^E^4)^3*(E^(2*E^4)*(1 - 5*E^E^4)*x - Log[36])))
 

3.11.35.3.1 Defintions of rubi rules used

rule 6
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v 
+ (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] &&  !FreeQ[Fx, x]
 

rule 1107
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; F 
reeQ[{a, b, c, d, e, m}, x] && EqQ[2*c*d - b*e, 0] && IGtQ[p, 0] &&  !(EqQ[ 
m, 3] && NeQ[p, 1])
 

rule 1294
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.)*((d_.) + (e_.)*(x_) + (f_.)*(x 
_)^2)^(q_.), x_Symbol] :> Simp[1/c^p   Int[(b/2 + c*x)^(2*p)*(d + e*x + f*x 
^2)^q, x], x] /; FreeQ[{a, b, c, d, e, f, q}, x] && EqQ[b^2 - 4*a*c, 0] && 
IntegerQ[p]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 
3.11.35.4 Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25

method result size
gosper \(\frac {x^{2} \left (5 \,{\mathrm e}^{{\mathrm e}^{4}}-1\right )}{5 x \,{\mathrm e}^{3 \,{\mathrm e}^{4}}-x \,{\mathrm e}^{2 \,{\mathrm e}^{4}}+2 \ln \left (6\right )}\) \(35\)
norman \(\frac {x^{2} \left (5 \,{\mathrm e}^{{\mathrm e}^{4}}-1\right )}{5 x \,{\mathrm e}^{3 \,{\mathrm e}^{4}}-x \,{\mathrm e}^{2 \,{\mathrm e}^{4}}+2 \ln \left (6\right )}\) \(35\)
parallelrisch \(\frac {10 x^{2} \ln \left (6\right ) {\mathrm e}^{{\mathrm e}^{4}}-2 x^{2} \ln \left (6\right )}{2 \ln \left (6\right ) \left (5 x \,{\mathrm e}^{3 \,{\mathrm e}^{4}}-x \,{\mathrm e}^{2 \,{\mathrm e}^{4}}+2 \ln \left (6\right )\right )}\) \(48\)
risch \(x \,{\mathrm e}^{-2 \,{\mathrm e}^{4}}+\frac {2 \,{\mathrm e}^{-4 \,{\mathrm e}^{4}} \ln \left (3\right )^{2}}{\left (5 \,{\mathrm e}^{{\mathrm e}^{4}}-1\right ) \left (\frac {5 x \,{\mathrm e}^{3 \,{\mathrm e}^{4}}}{2}-\frac {x \,{\mathrm e}^{2 \,{\mathrm e}^{4}}}{2}+\ln \left (3\right )+\ln \left (2\right )\right )}+\frac {4 \,{\mathrm e}^{-4 \,{\mathrm e}^{4}} \ln \left (2\right ) \ln \left (3\right )}{\left (5 \,{\mathrm e}^{{\mathrm e}^{4}}-1\right ) \left (\frac {5 x \,{\mathrm e}^{3 \,{\mathrm e}^{4}}}{2}-\frac {x \,{\mathrm e}^{2 \,{\mathrm e}^{4}}}{2}+\ln \left (3\right )+\ln \left (2\right )\right )}+\frac {2 \,{\mathrm e}^{-4 \,{\mathrm e}^{4}} \ln \left (2\right )^{2}}{\left (5 \,{\mathrm e}^{{\mathrm e}^{4}}-1\right ) \left (\frac {5 x \,{\mathrm e}^{3 \,{\mathrm e}^{4}}}{2}-\frac {x \,{\mathrm e}^{2 \,{\mathrm e}^{4}}}{2}+\ln \left (3\right )+\ln \left (2\right )\right )}\) \(138\)
meijerg \(\frac {8 \,{\mathrm e}^{-2 \,{\mathrm e}^{4}} \left (\frac {25 \,{\mathrm e}^{4 \,{\mathrm e}^{4}}}{4}-\frac {5 \,{\mathrm e}^{3 \,{\mathrm e}^{4}}}{2}+\frac {{\mathrm e}^{2 \,{\mathrm e}^{4}}}{4}\right ) \ln \left (6\right ) \left (\frac {x \,{\mathrm e}^{2 \,{\mathrm e}^{4}} \left (5 \,{\mathrm e}^{{\mathrm e}^{4}}-1\right ) \left (\frac {3 x \,{\mathrm e}^{2 \,{\mathrm e}^{4}} \left (5 \,{\mathrm e}^{{\mathrm e}^{4}}-1\right )}{2 \ln \left (6\right )}+6\right )}{6 \ln \left (6\right ) \left (1+\frac {x \,{\mathrm e}^{2 \,{\mathrm e}^{4}} \left (5 \,{\mathrm e}^{{\mathrm e}^{4}}-1\right )}{2 \ln \left (6\right )}\right )}-2 \ln \left (1+\frac {x \,{\mathrm e}^{2 \,{\mathrm e}^{4}} \left (5 \,{\mathrm e}^{{\mathrm e}^{4}}-1\right )}{2 \ln \left (6\right )}\right )\right )}{\left (25 \,{\mathrm e}^{6 \,{\mathrm e}^{4}}-10 \,{\mathrm e}^{5 \,{\mathrm e}^{4}}+{\mathrm e}^{4 \,{\mathrm e}^{4}}\right ) \left (5 \,{\mathrm e}^{{\mathrm e}^{4}}-1\right )}+\frac {4 \left (5 \ln \left (6\right ) {\mathrm e}^{{\mathrm e}^{4}}-\ln \left (6\right )\right ) \left (-\frac {x \,{\mathrm e}^{2 \,{\mathrm e}^{4}} \left (5 \,{\mathrm e}^{{\mathrm e}^{4}}-1\right )}{2 \ln \left (6\right ) \left (1+\frac {x \,{\mathrm e}^{2 \,{\mathrm e}^{4}} \left (5 \,{\mathrm e}^{{\mathrm e}^{4}}-1\right )}{2 \ln \left (6\right )}\right )}+\ln \left (1+\frac {x \,{\mathrm e}^{2 \,{\mathrm e}^{4}} \left (5 \,{\mathrm e}^{{\mathrm e}^{4}}-1\right )}{2 \ln \left (6\right )}\right )\right )}{25 \,{\mathrm e}^{6 \,{\mathrm e}^{4}}-10 \,{\mathrm e}^{5 \,{\mathrm e}^{4}}+{\mathrm e}^{4 \,{\mathrm e}^{4}}}\) \(253\)

input
int((25*x^2*exp(exp(4))^4-10*x^2*exp(exp(4))^3+x^2*exp(exp(4))^2+20*x*ln(6 
)*exp(exp(4))-4*x*ln(6))/(25*x^2*exp(exp(4))^6-10*x^2*exp(exp(4))^5+x^2*ex 
p(exp(4))^4+20*x*ln(6)*exp(exp(4))^3-4*x*ln(6)*exp(exp(4))^2+4*ln(6)^2),x, 
method=_RETURNVERBOSE)
 
output
x^2*(5*exp(exp(4))-1)/(5*x*exp(exp(4))^3-x*exp(exp(4))^2+2*ln(6))
 
3.11.35.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 101 vs. \(2 (26) = 52\).

Time = 0.24 (sec) , antiderivative size = 101, normalized size of antiderivative = 3.61 \[ \int \frac {e^{2 e^4} x^2-10 e^{3 e^4} x^2+25 e^{4 e^4} x^2-2 x \log (36)+10 e^{e^4} x \log (36)}{e^{4 e^4} x^2-10 e^{5 e^4} x^2+25 e^{6 e^4} x^2-2 e^{2 e^4} x \log (36)+10 e^{3 e^4} x \log (36)+\log ^2(36)} \, dx=\frac {25 \, x^{2} e^{\left (6 \, e^{4}\right )} - 10 \, x^{2} e^{\left (5 \, e^{4}\right )} + x^{2} e^{\left (4 \, e^{4}\right )} + 10 \, x e^{\left (3 \, e^{4}\right )} \log \left (6\right ) - 2 \, x e^{\left (2 \, e^{4}\right )} \log \left (6\right ) + 4 \, \log \left (6\right )^{2}}{25 \, x e^{\left (8 \, e^{4}\right )} - 10 \, x e^{\left (7 \, e^{4}\right )} + x e^{\left (6 \, e^{4}\right )} + 10 \, e^{\left (5 \, e^{4}\right )} \log \left (6\right ) - 2 \, e^{\left (4 \, e^{4}\right )} \log \left (6\right )} \]

input
integrate((25*x^2*exp(exp(4))^4-10*x^2*exp(exp(4))^3+x^2*exp(exp(4))^2+20* 
x*log(6)*exp(exp(4))-4*x*log(6))/(25*x^2*exp(exp(4))^6-10*x^2*exp(exp(4))^ 
5+x^2*exp(exp(4))^4+20*x*log(6)*exp(exp(4))^3-4*x*log(6)*exp(exp(4))^2+4*l 
og(6)^2),x, algorithm=\
 
output
(25*x^2*e^(6*e^4) - 10*x^2*e^(5*e^4) + x^2*e^(4*e^4) + 10*x*e^(3*e^4)*log( 
6) - 2*x*e^(2*e^4)*log(6) + 4*log(6)^2)/(25*x*e^(8*e^4) - 10*x*e^(7*e^4) + 
 x*e^(6*e^4) + 10*e^(5*e^4)*log(6) - 2*e^(4*e^4)*log(6))
 
3.11.35.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (22) = 44\).

Time = 0.33 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.25 \[ \int \frac {e^{2 e^4} x^2-10 e^{3 e^4} x^2+25 e^{4 e^4} x^2-2 x \log (36)+10 e^{e^4} x \log (36)}{e^{4 e^4} x^2-10 e^{5 e^4} x^2+25 e^{6 e^4} x^2-2 e^{2 e^4} x \log (36)+10 e^{3 e^4} x \log (36)+\log ^2(36)} \, dx=\frac {x}{e^{2 e^{4}}} + \frac {4 \log {\left (6 \right )}^{2}}{x \left (- 10 e^{7 e^{4}} + e^{6 e^{4}} + 25 e^{8 e^{4}}\right ) - 2 e^{4 e^{4}} \log {\left (6 \right )} + 10 e^{5 e^{4}} \log {\left (6 \right )}} \]

input
integrate((25*x**2*exp(exp(4))**4-10*x**2*exp(exp(4))**3+x**2*exp(exp(4))* 
*2+20*x*ln(6)*exp(exp(4))-4*x*ln(6))/(25*x**2*exp(exp(4))**6-10*x**2*exp(e 
xp(4))**5+x**2*exp(exp(4))**4+20*x*ln(6)*exp(exp(4))**3-4*x*ln(6)*exp(exp( 
4))**2+4*ln(6)**2),x)
 
output
x*exp(-2*exp(4)) + 4*log(6)**2/(x*(-10*exp(7*exp(4)) + exp(6*exp(4)) + 25* 
exp(8*exp(4))) - 2*exp(4*exp(4))*log(6) + 10*exp(5*exp(4))*log(6))
 
3.11.35.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (26) = 52\).

Time = 0.19 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.07 \[ \int \frac {e^{2 e^4} x^2-10 e^{3 e^4} x^2+25 e^{4 e^4} x^2-2 x \log (36)+10 e^{e^4} x \log (36)}{e^{4 e^4} x^2-10 e^{5 e^4} x^2+25 e^{6 e^4} x^2-2 e^{2 e^4} x \log (36)+10 e^{3 e^4} x \log (36)+\log ^2(36)} \, dx=x e^{\left (-2 \, e^{4}\right )} + \frac {4 \, \log \left (6\right )^{2}}{x {\left (25 \, e^{\left (8 \, e^{4}\right )} - 10 \, e^{\left (7 \, e^{4}\right )} + e^{\left (6 \, e^{4}\right )}\right )} + 2 \, {\left (5 \, e^{\left (5 \, e^{4}\right )} - e^{\left (4 \, e^{4}\right )}\right )} \log \left (6\right )} \]

input
integrate((25*x^2*exp(exp(4))^4-10*x^2*exp(exp(4))^3+x^2*exp(exp(4))^2+20* 
x*log(6)*exp(exp(4))-4*x*log(6))/(25*x^2*exp(exp(4))^6-10*x^2*exp(exp(4))^ 
5+x^2*exp(exp(4))^4+20*x*log(6)*exp(exp(4))^3-4*x*log(6)*exp(exp(4))^2+4*l 
og(6)^2),x, algorithm=\
 
output
x*e^(-2*e^4) + 4*log(6)^2/(x*(25*e^(8*e^4) - 10*e^(7*e^4) + e^(6*e^4)) + 2 
*(5*e^(5*e^4) - e^(4*e^4))*log(6))
 
3.11.35.8 Giac [F(-2)]

Exception generated. \[ \int \frac {e^{2 e^4} x^2-10 e^{3 e^4} x^2+25 e^{4 e^4} x^2-2 x \log (36)+10 e^{e^4} x \log (36)}{e^{4 e^4} x^2-10 e^{5 e^4} x^2+25 e^{6 e^4} x^2-2 e^{2 e^4} x \log (36)+10 e^{3 e^4} x \log (36)+\log ^2(36)} \, dx=\text {Exception raised: NotImplementedError} \]

input
integrate((25*x^2*exp(exp(4))^4-10*x^2*exp(exp(4))^3+x^2*exp(exp(4))^2+20* 
x*log(6)*exp(exp(4))-4*x*log(6))/(25*x^2*exp(exp(4))^6-10*x^2*exp(exp(4))^ 
5+x^2*exp(exp(4))^4+20*x*log(6)*exp(exp(4))^3-4*x*log(6)*exp(exp(4))^2+4*l 
og(6)^2),x, algorithm=\
 
output
Exception raised: NotImplementedError >> unable to parse Giac output: (250 
*exp(6*exp(4))*ln(6)*exp(exp(4))-50*exp(6*exp(4))*ln(6)-100*exp(5*exp(4))* 
ln(6)*exp(exp(4))+20*exp(5*exp(4))*ln(6)-250*exp(4*exp(4))*exp(3*exp(4))*l 
n(6)+50*exp(4*exp(4
 
3.11.35.9 Mupad [B] (verification not implemented)

Time = 11.80 (sec) , antiderivative size = 95, normalized size of antiderivative = 3.39 \[ \int \frac {e^{2 e^4} x^2-10 e^{3 e^4} x^2+25 e^{4 e^4} x^2-2 x \log (36)+10 e^{e^4} x \log (36)}{e^{4 e^4} x^2-10 e^{5 e^4} x^2+25 e^{6 e^4} x^2-2 e^{2 e^4} x \log (36)+10 e^{3 e^4} x \log (36)+\log ^2(36)} \, dx=x\,{\mathrm {e}}^{-2\,{\mathrm {e}}^4}-\frac {\mathrm {atan}\left (\frac {\ln \left (6\right )\,2{}\mathrm {i}-x\,{\mathrm {e}}^{2\,{\mathrm {e}}^4}\,1{}\mathrm {i}+x\,{\mathrm {e}}^{3\,{\mathrm {e}}^4}\,5{}\mathrm {i}}{\sqrt {2\,\ln \left (6\right )+\ln \left (36\right )}\,\sqrt {2\,\ln \left (6\right )-\ln \left (36\right )}}\right )\,{\mathrm {e}}^{-4\,{\mathrm {e}}^4}\,{\ln \left (6\right )}^2\,4{}\mathrm {i}}{\left (5\,{\mathrm {e}}^{{\mathrm {e}}^4}-1\right )\,\sqrt {2\,\ln \left (6\right )+\ln \left (36\right )}\,\sqrt {2\,\ln \left (6\right )-\ln \left (36\right )}} \]

input
int((x^2*exp(2*exp(4)) - 4*x*log(6) - 10*x^2*exp(3*exp(4)) + 25*x^2*exp(4* 
exp(4)) + 20*x*exp(exp(4))*log(6))/(x^2*exp(4*exp(4)) - 10*x^2*exp(5*exp(4 
)) + 25*x^2*exp(6*exp(4)) + 4*log(6)^2 - 4*x*exp(2*exp(4))*log(6) + 20*x*e 
xp(3*exp(4))*log(6)),x)
 
output
x*exp(-2*exp(4)) - (atan((log(6)*2i - x*exp(2*exp(4))*1i + x*exp(3*exp(4)) 
*5i)/((2*log(6) + log(36))^(1/2)*(2*log(6) - log(36))^(1/2)))*exp(-4*exp(4 
))*log(6)^2*4i)/((5*exp(exp(4)) - 1)*(2*log(6) + log(36))^(1/2)*(2*log(6) 
- log(36))^(1/2))