3.77.62 \(\int \frac {20 e^5+20 e^{5+x}+(20+20 e^x) \log (-e^{e^3}+e^x+x)}{e^x (-16+40 e^{10}-25 e^{20})+e^{e^3} (16-40 e^{10}+25 e^{20})-16 x+40 e^{10} x-25 e^{20} x+(e^x (80 e^5-100 e^{15})+e^{e^3} (-80 e^5+100 e^{15})+80 e^5 x-100 e^{15} x) \log (-e^{e^3}+e^x+x)+(e^x (40-150 e^{10})+e^{e^3} (-40+150 e^{10})+40 x-150 e^{10} x) \log ^2(-e^{e^3}+e^x+x)+(100 e^{5+e^3}-100 e^{5+x}-100 e^5 x) \log ^3(-e^{e^3}+e^x+x)+(25 e^{e^3}-25 e^x-25 x) \log ^4(-e^{e^3}+e^x+x)} \, dx\)

Optimal. Leaf size=29 \[ -4+\frac {2}{-4+5 \left (e^5+\log \left (-e^{e^3}+e^x+x\right )\right )^2} \]

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Rubi [A]  time = 0.50, antiderivative size = 46, normalized size of antiderivative = 1.59, number of steps used = 5, number of rules used = 4, integrand size = 267, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {6, 6688, 12, 6686} \begin {gather*} -\frac {2}{-5 \log ^2\left (x+e^x-e^{e^3}\right )-10 e^5 \log \left (x+e^x-e^{e^3}\right )-5 e^{10}+4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20*E^5 + 20*E^(5 + x) + (20 + 20*E^x)*Log[-E^E^3 + E^x + x])/(E^x*(-16 + 40*E^10 - 25*E^20) + E^E^3*(16 -
40*E^10 + 25*E^20) - 16*x + 40*E^10*x - 25*E^20*x + (E^x*(80*E^5 - 100*E^15) + E^E^3*(-80*E^5 + 100*E^15) + 8
0*E^5*x - 100*E^15*x)*Log[-E^E^3 + E^x + x] + (E^x*(40 - 150*E^10) + E^E^3*(-40 + 150*E^10) + 40*x - 150*E^10*
x)*Log[-E^E^3 + E^x + x]^2 + (100*E^(5 + E^3) - 100*E^(5 + x) - 100*E^5*x)*Log[-E^E^3 + E^x + x]^3 + (25*E^E^3
- 25*E^x - 25*x)*Log[-E^E^3 + E^x + x]^4),x]

[Out]

-2/(4 - 5*E^10 - 10*E^5*Log[-E^E^3 + E^x + x] - 5*Log[-E^E^3 + E^x + x]^2)

Rule 6

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {20 e^5+20 e^{5+x}+\left (20+20 e^x\right ) \log \left (-e^{e^3}+e^x+x\right )}{e^x \left (-16+40 e^{10}-25 e^{20}\right )+e^{e^3} \left (16-40 e^{10}+25 e^{20}\right )-25 e^{20} x+\left (-16+40 e^{10}\right ) x+\left (e^x \left (80 e^5-100 e^{15}\right )+e^{e^3} \left (-80 e^5+100 e^{15}\right )+80 e^5 x-100 e^{15} x\right ) \log \left (-e^{e^3}+e^x+x\right )+\left (e^x \left (40-150 e^{10}\right )+e^{e^3} \left (-40+150 e^{10}\right )+40 x-150 e^{10} x\right ) \log ^2\left (-e^{e^3}+e^x+x\right )+\left (100 e^{5+e^3}-100 e^{5+x}-100 e^5 x\right ) \log ^3\left (-e^{e^3}+e^x+x\right )+\left (25 e^{e^3}-25 e^x-25 x\right ) \log ^4\left (-e^{e^3}+e^x+x\right )} \, dx\\ &=\int \frac {20 e^5+20 e^{5+x}+\left (20+20 e^x\right ) \log \left (-e^{e^3}+e^x+x\right )}{e^x \left (-16+40 e^{10}-25 e^{20}\right )+e^{e^3} \left (16-40 e^{10}+25 e^{20}\right )+\left (-16+40 e^{10}-25 e^{20}\right ) x+\left (e^x \left (80 e^5-100 e^{15}\right )+e^{e^3} \left (-80 e^5+100 e^{15}\right )+80 e^5 x-100 e^{15} x\right ) \log \left (-e^{e^3}+e^x+x\right )+\left (e^x \left (40-150 e^{10}\right )+e^{e^3} \left (-40+150 e^{10}\right )+40 x-150 e^{10} x\right ) \log ^2\left (-e^{e^3}+e^x+x\right )+\left (100 e^{5+e^3}-100 e^{5+x}-100 e^5 x\right ) \log ^3\left (-e^{e^3}+e^x+x\right )+\left (25 e^{e^3}-25 e^x-25 x\right ) \log ^4\left (-e^{e^3}+e^x+x\right )} \, dx\\ &=\int \frac {20 \left (1+e^x\right ) \left (e^5+\log \left (-e^{e^3}+e^x+x\right )\right )}{\left (e^{e^3}-e^x-x\right ) \left (4 \left (1-\frac {5 e^{10}}{4}\right )-10 e^5 \log \left (-e^{e^3}+e^x+x\right )-5 \log ^2\left (-e^{e^3}+e^x+x\right )\right )^2} \, dx\\ &=20 \int \frac {\left (1+e^x\right ) \left (e^5+\log \left (-e^{e^3}+e^x+x\right )\right )}{\left (e^{e^3}-e^x-x\right ) \left (4 \left (1-\frac {5 e^{10}}{4}\right )-10 e^5 \log \left (-e^{e^3}+e^x+x\right )-5 \log ^2\left (-e^{e^3}+e^x+x\right )\right )^2} \, dx\\ &=-\frac {2}{4-5 e^{10}-10 e^5 \log \left (-e^{e^3}+e^x+x\right )-5 \log ^2\left (-e^{e^3}+e^x+x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 46, normalized size = 1.59 \begin {gather*} \frac {2}{-4+5 e^{10}+10 e^5 \log \left (-e^{e^3}+e^x+x\right )+5 \log ^2\left (-e^{e^3}+e^x+x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20*E^5 + 20*E^(5 + x) + (20 + 20*E^x)*Log[-E^E^3 + E^x + x])/(E^x*(-16 + 40*E^10 - 25*E^20) + E^E^3
*(16 - 40*E^10 + 25*E^20) - 16*x + 40*E^10*x - 25*E^20*x + (E^x*(80*E^5 - 100*E^15) + E^E^3*(-80*E^5 + 100*E^1
5) + 80*E^5*x - 100*E^15*x)*Log[-E^E^3 + E^x + x] + (E^x*(40 - 150*E^10) + E^E^3*(-40 + 150*E^10) + 40*x - 150
*E^10*x)*Log[-E^E^3 + E^x + x]^2 + (100*E^(5 + E^3) - 100*E^(5 + x) - 100*E^5*x)*Log[-E^E^3 + E^x + x]^3 + (25
*E^E^3 - 25*E^x - 25*x)*Log[-E^E^3 + E^x + x]^4),x]

[Out]

2/(-4 + 5*E^10 + 10*E^5*Log[-E^E^3 + E^x + x] + 5*Log[-E^E^3 + E^x + x]^2)

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fricas [B]  time = 0.79, size = 58, normalized size = 2.00 \begin {gather*} \frac {2}{10 \, e^{5} \log \left ({\left (x e^{5} + e^{\left (x + 5\right )} - e^{\left (e^{3} + 5\right )}\right )} e^{\left (-5\right )}\right ) + 5 \, \log \left ({\left (x e^{5} + e^{\left (x + 5\right )} - e^{\left (e^{3} + 5\right )}\right )} e^{\left (-5\right )}\right )^{2} + 5 \, e^{10} - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*exp(x)+20)*log(-exp(exp(3))+exp(x)+x)+20*exp(5)*exp(x)+20*exp(5))/((25*exp(exp(3))-25*exp(x)-25
*x)*log(-exp(exp(3))+exp(x)+x)^4+(100*exp(5)*exp(exp(3))-100*exp(5)*exp(x)-100*x*exp(5))*log(-exp(exp(3))+exp(
x)+x)^3+((150*exp(5)^2-40)*exp(exp(3))+(-150*exp(5)^2+40)*exp(x)-150*x*exp(5)^2+40*x)*log(-exp(exp(3))+exp(x)+
x)^2+((100*exp(5)^3-80*exp(5))*exp(exp(3))+(-100*exp(5)^3+80*exp(5))*exp(x)-100*x*exp(5)^3+80*x*exp(5))*log(-e
xp(exp(3))+exp(x)+x)+(25*exp(5)^4-40*exp(5)^2+16)*exp(exp(3))+(-25*exp(5)^4+40*exp(5)^2-16)*exp(x)-25*x*exp(5)
^4+40*x*exp(5)^2-16*x),x, algorithm="fricas")

[Out]

2/(10*e^5*log((x*e^5 + e^(x + 5) - e^(e^3 + 5))*e^(-5)) + 5*log((x*e^5 + e^(x + 5) - e^(e^3 + 5))*e^(-5))^2 +
5*e^10 - 4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*exp(x)+20)*log(-exp(exp(3))+exp(x)+x)+20*exp(5)*exp(x)+20*exp(5))/((25*exp(exp(3))-25*exp(x)-25
*x)*log(-exp(exp(3))+exp(x)+x)^4+(100*exp(5)*exp(exp(3))-100*exp(5)*exp(x)-100*x*exp(5))*log(-exp(exp(3))+exp(
x)+x)^3+((150*exp(5)^2-40)*exp(exp(3))+(-150*exp(5)^2+40)*exp(x)-150*x*exp(5)^2+40*x)*log(-exp(exp(3))+exp(x)+
x)^2+((100*exp(5)^3-80*exp(5))*exp(exp(3))+(-100*exp(5)^3+80*exp(5))*exp(x)-100*x*exp(5)^3+80*x*exp(5))*log(-e
xp(exp(3))+exp(x)+x)+(25*exp(5)^4-40*exp(5)^2+16)*exp(exp(3))+(-25*exp(5)^4+40*exp(5)^2-16)*exp(x)-25*x*exp(5)
^4+40*x*exp(5)^2-16*x),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.85, size = 39, normalized size = 1.34

method	result	size
risch	\(\frac {2}{5 \,{\mathrm e}^{10}+10 \,{\mathrm e}^{5} \ln \left (-{\mathrm e}^{{\mathrm e}^{3}}+{\mathrm e}^{x}+x \right )+5 \ln \left (-{\mathrm e}^{{\mathrm e}^{3}}+{\mathrm e}^{x}+x \right )^{2}-4}\)	\(39\)
norman	\(\frac {2}{5 \,{\mathrm e}^{10}+10 \,{\mathrm e}^{5} \ln \left (-{\mathrm e}^{{\mathrm e}^{3}}+{\mathrm e}^{x}+x \right )+5 \ln \left (-{\mathrm e}^{{\mathrm e}^{3}}+{\mathrm e}^{x}+x \right )^{2}-4}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((20*exp(x)+20)*ln(-exp(exp(3))+exp(x)+x)+20*exp(5)*exp(x)+20*exp(5))/((25*exp(exp(3))-25*exp(x)-25*x)*ln(
-exp(exp(3))+exp(x)+x)^4+(100*exp(5)*exp(exp(3))-100*exp(5)*exp(x)-100*x*exp(5))*ln(-exp(exp(3))+exp(x)+x)^3+(
(150*exp(5)^2-40)*exp(exp(3))+(-150*exp(5)^2+40)*exp(x)-150*x*exp(5)^2+40*x)*ln(-exp(exp(3))+exp(x)+x)^2+((100
*exp(5)^3-80*exp(5))*exp(exp(3))+(-100*exp(5)^3+80*exp(5))*exp(x)-100*x*exp(5)^3+80*x*exp(5))*ln(-exp(exp(3))+
exp(x)+x)+(25*exp(5)^4-40*exp(5)^2+16)*exp(exp(3))+(-25*exp(5)^4+40*exp(5)^2-16)*exp(x)-25*x*exp(5)^4+40*x*exp
(5)^2-16*x),x,method=_RETURNVERBOSE)

[Out]

2/(5*exp(10)+10*exp(5)*ln(-exp(exp(3))+exp(x)+x)+5*ln(-exp(exp(3))+exp(x)+x)^2-4)

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maxima [A]  time = 0.76, size = 38, normalized size = 1.31 \begin {gather*} \frac {2}{10 \, e^{5} \log \left (x + e^{x} - e^{\left (e^{3}\right )}\right ) + 5 \, \log \left (x + e^{x} - e^{\left (e^{3}\right )}\right )^{2} + 5 \, e^{10} - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*exp(x)+20)*log(-exp(exp(3))+exp(x)+x)+20*exp(5)*exp(x)+20*exp(5))/((25*exp(exp(3))-25*exp(x)-25
*x)*log(-exp(exp(3))+exp(x)+x)^4+(100*exp(5)*exp(exp(3))-100*exp(5)*exp(x)-100*x*exp(5))*log(-exp(exp(3))+exp(
x)+x)^3+((150*exp(5)^2-40)*exp(exp(3))+(-150*exp(5)^2+40)*exp(x)-150*x*exp(5)^2+40*x)*log(-exp(exp(3))+exp(x)+
x)^2+((100*exp(5)^3-80*exp(5))*exp(exp(3))+(-100*exp(5)^3+80*exp(5))*exp(x)-100*x*exp(5)^3+80*x*exp(5))*log(-e
xp(exp(3))+exp(x)+x)+(25*exp(5)^4-40*exp(5)^2+16)*exp(exp(3))+(-25*exp(5)^4+40*exp(5)^2-16)*exp(x)-25*x*exp(5)
^4+40*x*exp(5)^2-16*x),x, algorithm="maxima")

[Out]

2/(10*e^5*log(x + e^x - e^(e^3)) + 5*log(x + e^x - e^(e^3))^2 + 5*e^10 - 4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {20\,{\mathrm {e}}^5+\ln \left (x-{\mathrm {e}}^{{\mathrm {e}}^3}+{\mathrm {e}}^x\right )\,\left (20\,{\mathrm {e}}^x+20\right )+20\,{\mathrm {e}}^5\,{\mathrm {e}}^x}{\left (25\,x-25\,{\mathrm {e}}^{{\mathrm {e}}^3}+25\,{\mathrm {e}}^x\right )\,{\ln \left (x-{\mathrm {e}}^{{\mathrm {e}}^3}+{\mathrm {e}}^x\right )}^4+\left (100\,x\,{\mathrm {e}}^5-100\,{\mathrm {e}}^5\,{\mathrm {e}}^{{\mathrm {e}}^3}+100\,{\mathrm {e}}^5\,{\mathrm {e}}^x\right )\,{\ln \left (x-{\mathrm {e}}^{{\mathrm {e}}^3}+{\mathrm {e}}^x\right )}^3+\left (150\,x\,{\mathrm {e}}^{10}-40\,x-{\mathrm {e}}^{{\mathrm {e}}^3}\,\left (150\,{\mathrm {e}}^{10}-40\right )+{\mathrm {e}}^x\,\left (150\,{\mathrm {e}}^{10}-40\right )\right )\,{\ln \left (x-{\mathrm {e}}^{{\mathrm {e}}^3}+{\mathrm {e}}^x\right )}^2+\left ({\mathrm {e}}^{{\mathrm {e}}^3}\,\left (80\,{\mathrm {e}}^5-100\,{\mathrm {e}}^{15}\right )-{\mathrm {e}}^x\,\left (80\,{\mathrm {e}}^5-100\,{\mathrm {e}}^{15}\right )-80\,x\,{\mathrm {e}}^5+100\,x\,{\mathrm {e}}^{15}\right )\,\ln \left (x-{\mathrm {e}}^{{\mathrm {e}}^3}+{\mathrm {e}}^x\right )+16\,x-{\mathrm {e}}^{{\mathrm {e}}^3}\,\left (25\,{\mathrm {e}}^{20}-40\,{\mathrm {e}}^{10}+16\right )+{\mathrm {e}}^x\,\left (25\,{\mathrm {e}}^{20}-40\,{\mathrm {e}}^{10}+16\right )-40\,x\,{\mathrm {e}}^{10}+25\,x\,{\mathrm {e}}^{20}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*exp(5) + log(x - exp(exp(3)) + exp(x))*(20*exp(x) + 20) + 20*exp(5)*exp(x))/(16*x - exp(exp(3))*(25*e
xp(20) - 40*exp(10) + 16) + exp(x)*(25*exp(20) - 40*exp(10) + 16) - log(x - exp(exp(3)) + exp(x))^2*(40*x - 15
0*x*exp(10) + exp(exp(3))*(150*exp(10) - 40) - exp(x)*(150*exp(10) - 40)) - 40*x*exp(10) + 25*x*exp(20) + log(
x - exp(exp(3)) + exp(x))^4*(25*x - 25*exp(exp(3)) + 25*exp(x)) + log(x - exp(exp(3)) + exp(x))*(exp(exp(3))*(
80*exp(5) - 100*exp(15)) - exp(x)*(80*exp(5) - 100*exp(15)) - 80*x*exp(5) + 100*x*exp(15)) + log(x - exp(exp(3
)) + exp(x))^3*(100*x*exp(5) - 100*exp(5)*exp(exp(3)) + 100*exp(5)*exp(x))),x)

[Out]

int(-(20*exp(5) + log(x - exp(exp(3)) + exp(x))*(20*exp(x) + 20) + 20*exp(5)*exp(x))/(16*x - exp(exp(3))*(25*e
xp(20) - 40*exp(10) + 16) + exp(x)*(25*exp(20) - 40*exp(10) + 16) - log(x - exp(exp(3)) + exp(x))^2*(40*x - 15
0*x*exp(10) + exp(exp(3))*(150*exp(10) - 40) - exp(x)*(150*exp(10) - 40)) - 40*x*exp(10) + 25*x*exp(20) + log(
x - exp(exp(3)) + exp(x))^4*(25*x - 25*exp(exp(3)) + 25*exp(x)) + log(x - exp(exp(3)) + exp(x))*(exp(exp(3))*(
80*exp(5) - 100*exp(15)) - exp(x)*(80*exp(5) - 100*exp(15)) - 80*x*exp(5) + 100*x*exp(15)) + log(x - exp(exp(3
)) + exp(x))^3*(100*x*exp(5) - 100*exp(5)*exp(exp(3)) + 100*exp(5)*exp(x))), x)

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sympy [A]  time = 0.42, size = 39, normalized size = 1.34 \begin {gather*} \frac {2}{5 \log {\left (x + e^{x} - e^{e^{3}} \right )}^{2} + 10 e^{5} \log {\left (x + e^{x} - e^{e^{3}} \right )} - 4 + 5 e^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*exp(x)+20)*ln(-exp(exp(3))+exp(x)+x)+20*exp(5)*exp(x)+20*exp(5))/((25*exp(exp(3))-25*exp(x)-25*
x)*ln(-exp(exp(3))+exp(x)+x)**4+(100*exp(5)*exp(exp(3))-100*exp(5)*exp(x)-100*x*exp(5))*ln(-exp(exp(3))+exp(x)
+x)**3+((150*exp(5)**2-40)*exp(exp(3))+(-150*exp(5)**2+40)*exp(x)-150*x*exp(5)**2+40*x)*ln(-exp(exp(3))+exp(x)
+x)**2+((100*exp(5)**3-80*exp(5))*exp(exp(3))+(-100*exp(5)**3+80*exp(5))*exp(x)-100*x*exp(5)**3+80*x*exp(5))*l
n(-exp(exp(3))+exp(x)+x)+(25*exp(5)**4-40*exp(5)**2+16)*exp(exp(3))+(-25*exp(5)**4+40*exp(5)**2-16)*exp(x)-25*
x*exp(5)**4+40*x*exp(5)**2-16*x),x)

[Out]

2/(5*log(x + exp(x) - exp(exp(3)))**2 + 10*exp(5)*log(x + exp(x) - exp(exp(3))) - 4 + 5*exp(10))

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