3.4.62 \(\int \frac {e^{1-e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}}+x} (e (75+25 x)+e (30+10 x) \log (3+x)+e (3+x) \log ^2(3+x)+e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}} (-15+e (-75-25 x)-4 x+(-3+e (-30-10 x)-x) \log (3+x)+e (-3-x) \log ^2(3+x)))}{e (150+50 x)+e (60+20 x) \log (3+x)+e (6+2 x) \log ^2(3+x)+e^{1-e^{\frac {x+5 e x+e x \log (3+x)}{5 e+e \log (3+x)}}+x} (e (75+25 x)+e (30+10 x) \log (3+x)+e (3+x) \log ^2(3+x))} \, dx\)

Optimal. Leaf size=27 \[ \log \left (2+e^{1-e^{x+\frac {x}{e (5+\log (3+x))}}+x}\right ) \]

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Rubi [F]  time = 180.00, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(1 - E^((x + 5*E*x + E*x*Log[3 + x])/(5*E + E*Log[3 + x])) + x)*(E*(75 + 25*x) + E*(30 + 10*x)*Log[3 +
x] + E*(3 + x)*Log[3 + x]^2 + E^((x + 5*E*x + E*x*Log[3 + x])/(5*E + E*Log[3 + x]))*(-15 + E*(-75 - 25*x) - 4*
x + (-3 + E*(-30 - 10*x) - x)*Log[3 + x] + E*(-3 - x)*Log[3 + x]^2)))/(E*(150 + 50*x) + E*(60 + 20*x)*Log[3 +
x] + E*(6 + 2*x)*Log[3 + x]^2 + E^(1 - E^((x + 5*E*x + E*x*Log[3 + x])/(5*E + E*Log[3 + x])) + x)*(E*(75 + 25*
x) + E*(30 + 10*x)*Log[3 + x] + E*(3 + x)*Log[3 + x]^2)),x]

[Out]

$Aborted

Rubi steps

Aborted

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Mathematica [A]  time = 1.18, size = 48, normalized size = 1.78 \begin {gather*} -e^{x+\frac {x}{e (5+\log (3+x))}}+\log \left (2 e^{e^{x+\frac {x}{e (5+\log (3+x))}}}+e^{1+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(1 - E^((x + 5*E*x + E*x*Log[3 + x])/(5*E + E*Log[3 + x])) + x)*(E*(75 + 25*x) + E*(30 + 10*x)*Lo
g[3 + x] + E*(3 + x)*Log[3 + x]^2 + E^((x + 5*E*x + E*x*Log[3 + x])/(5*E + E*Log[3 + x]))*(-15 + E*(-75 - 25*x
) - 4*x + (-3 + E*(-30 - 10*x) - x)*Log[3 + x] + E*(-3 - x)*Log[3 + x]^2)))/(E*(150 + 50*x) + E*(60 + 20*x)*Lo
g[3 + x] + E*(6 + 2*x)*Log[3 + x]^2 + E^(1 - E^((x + 5*E*x + E*x*Log[3 + x])/(5*E + E*Log[3 + x])) + x)*(E*(75
+ 25*x) + E*(30 + 10*x)*Log[3 + x] + E*(3 + x)*Log[3 + x]^2)),x]

[Out]

-E^(x + x/(E*(5 + Log[3 + x]))) + Log[2*E^E^(x + x/(E*(5 + Log[3 + x]))) + E^(1 + x)]

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fricas [A]  time = 0.65, size = 40, normalized size = 1.48 \begin {gather*} \log \left (e^{\left (x - e^{\left (\frac {x e \log \left (x + 3\right ) + 5 \, x e + x}{e \log \left (x + 3\right ) + 5 \, e}\right )} + 1\right )} + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3-x)*exp(1)*log(3+x)^2+((-10*x-30)*exp(1)-3-x)*log(3+x)+(-25*x-75)*exp(1)-4*x-15)*exp((x*exp(1)*
log(3+x)+5*x*exp(1)+x)/(exp(1)*log(3+x)+5*exp(1)))+(3+x)*exp(1)*log(3+x)^2+(10*x+30)*exp(1)*log(3+x)+(25*x+75)
*exp(1))*exp(-exp((x*exp(1)*log(3+x)+5*x*exp(1)+x)/(exp(1)*log(3+x)+5*exp(1)))+x+1)/(((3+x)*exp(1)*log(3+x)^2+
(10*x+30)*exp(1)*log(3+x)+(25*x+75)*exp(1))*exp(-exp((x*exp(1)*log(3+x)+5*x*exp(1)+x)/(exp(1)*log(3+x)+5*exp(1
)))+x+1)+(2*x+6)*exp(1)*log(3+x)^2+(20*x+60)*exp(1)*log(3+x)+(50*x+150)*exp(1)),x, algorithm="fricas")

[Out]

log(e^(x - e^((x*e*log(x + 3) + 5*x*e + x)/(e*log(x + 3) + 5*e)) + 1) + 2)

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giac [B]  time = 44.91, size = 329, normalized size = 12.19 \begin {gather*} -\frac {x e \log \left (x + 3\right ) - e \log \left (x + 3\right ) \log \left (e^{\left (\frac {2 \, x e \log \left (x + 3\right ) + 10 \, x e - e^{\left (\frac {x e \log \left (x + 3\right ) + 5 \, x e + x}{e \log \left (x + 3\right ) + 5 \, e} + 1\right )} \log \left (x + 3\right ) + x - 5 \, e^{\left (\frac {x e \log \left (x + 3\right ) + 5 \, x e + x}{e \log \left (x + 3\right ) + 5 \, e} + 1\right )}}{e \log \left (x + 3\right ) + 5 \, e} + 1\right )} + 2 \, e^{\left (\frac {x e \log \left (x + 3\right ) + 5 \, x e + x}{e \log \left (x + 3\right ) + 5 \, e}\right )}\right ) + 5 \, x e - 5 \, e \log \left (e^{\left (\frac {2 \, x e \log \left (x + 3\right ) + 10 \, x e - e^{\left (\frac {x e \log \left (x + 3\right ) + 5 \, x e + x}{e \log \left (x + 3\right ) + 5 \, e} + 1\right )} \log \left (x + 3\right ) + x - 5 \, e^{\left (\frac {x e \log \left (x + 3\right ) + 5 \, x e + x}{e \log \left (x + 3\right ) + 5 \, e} + 1\right )}}{e \log \left (x + 3\right ) + 5 \, e} + 1\right )} + 2 \, e^{\left (\frac {x e \log \left (x + 3\right ) + 5 \, x e + x}{e \log \left (x + 3\right ) + 5 \, e}\right )}\right ) + x}{e \log \left (x + 3\right ) + 5 \, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3-x)*exp(1)*log(3+x)^2+((-10*x-30)*exp(1)-3-x)*log(3+x)+(-25*x-75)*exp(1)-4*x-15)*exp((x*exp(1)*
log(3+x)+5*x*exp(1)+x)/(exp(1)*log(3+x)+5*exp(1)))+(3+x)*exp(1)*log(3+x)^2+(10*x+30)*exp(1)*log(3+x)+(25*x+75)
*exp(1))*exp(-exp((x*exp(1)*log(3+x)+5*x*exp(1)+x)/(exp(1)*log(3+x)+5*exp(1)))+x+1)/(((3+x)*exp(1)*log(3+x)^2+
(10*x+30)*exp(1)*log(3+x)+(25*x+75)*exp(1))*exp(-exp((x*exp(1)*log(3+x)+5*x*exp(1)+x)/(exp(1)*log(3+x)+5*exp(1
)))+x+1)+(2*x+6)*exp(1)*log(3+x)^2+(20*x+60)*exp(1)*log(3+x)+(50*x+150)*exp(1)),x, algorithm="giac")

[Out]

-(x*e*log(x + 3) - e*log(x + 3)*log(e^((2*x*e*log(x + 3) + 10*x*e - e^((x*e*log(x + 3) + 5*x*e + x)/(e*log(x +
3) + 5*e) + 1)*log(x + 3) + x - 5*e^((x*e*log(x + 3) + 5*x*e + x)/(e*log(x + 3) + 5*e) + 1))/(e*log(x + 3) +
5*e) + 1) + 2*e^((x*e*log(x + 3) + 5*x*e + x)/(e*log(x + 3) + 5*e))) + 5*x*e - 5*e*log(e^((2*x*e*log(x + 3) +
10*x*e - e^((x*e*log(x + 3) + 5*x*e + x)/(e*log(x + 3) + 5*e) + 1)*log(x + 3) + x - 5*e^((x*e*log(x + 3) + 5*x
*e + x)/(e*log(x + 3) + 5*e) + 1))/(e*log(x + 3) + 5*e) + 1) + 2*e^((x*e*log(x + 3) + 5*x*e + x)/(e*log(x + 3)
+ 5*e))) + x)/(e*log(x + 3) + 5*e)

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maple [A]  time = 0.21, size = 38, normalized size = 1.41

method	result	size
risch	\(-1+\ln \left ({\mathrm e}^{-{\mathrm e}^{\frac {x \left ({\mathrm e} \ln \left (3+x \right )+5 \,{\mathrm e}+1\right ) {\mathrm e}^{-1}}{\ln \left (3+x \right )+5}}+x +1}+2\right )\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-3-x)*exp(1)*ln(3+x)^2+((-10*x-30)*exp(1)-3-x)*ln(3+x)+(-25*x-75)*exp(1)-4*x-15)*exp((x*exp(1)*ln(3+x)+
5*x*exp(1)+x)/(exp(1)*ln(3+x)+5*exp(1)))+(3+x)*exp(1)*ln(3+x)^2+(10*x+30)*exp(1)*ln(3+x)+(25*x+75)*exp(1))*exp
(-exp((x*exp(1)*ln(3+x)+5*x*exp(1)+x)/(exp(1)*ln(3+x)+5*exp(1)))+x+1)/(((3+x)*exp(1)*ln(3+x)^2+(10*x+30)*exp(1
)*ln(3+x)+(25*x+75)*exp(1))*exp(-exp((x*exp(1)*ln(3+x)+5*x*exp(1)+x)/(exp(1)*ln(3+x)+5*exp(1)))+x+1)+(2*x+6)*e
xp(1)*ln(3+x)^2+(20*x+60)*exp(1)*ln(3+x)+(50*x+150)*exp(1)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.94, size = 98, normalized size = 3.63 \begin {gather*} -e^{\left (\frac {x \log \left (x + 3\right )}{\log \left (x + 3\right ) + 5} + \frac {x}{e \log \left (x + 3\right ) + 5 \, e} + \frac {5 \, x}{\log \left (x + 3\right ) + 5}\right )} + \log \left (\frac {1}{2} \, e^{\left (x + 1\right )} + e^{\left (e^{\left (\frac {x \log \left (x + 3\right )}{\log \left (x + 3\right ) + 5} + \frac {x}{e \log \left (x + 3\right ) + 5 \, e} + \frac {5 \, x}{\log \left (x + 3\right ) + 5}\right )}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3-x)*exp(1)*log(3+x)^2+((-10*x-30)*exp(1)-3-x)*log(3+x)+(-25*x-75)*exp(1)-4*x-15)*exp((x*exp(1)*
log(3+x)+5*x*exp(1)+x)/(exp(1)*log(3+x)+5*exp(1)))+(3+x)*exp(1)*log(3+x)^2+(10*x+30)*exp(1)*log(3+x)+(25*x+75)
*exp(1))*exp(-exp((x*exp(1)*log(3+x)+5*x*exp(1)+x)/(exp(1)*log(3+x)+5*exp(1)))+x+1)/(((3+x)*exp(1)*log(3+x)^2+
(10*x+30)*exp(1)*log(3+x)+(25*x+75)*exp(1))*exp(-exp((x*exp(1)*log(3+x)+5*x*exp(1)+x)/(exp(1)*log(3+x)+5*exp(1
)))+x+1)+(2*x+6)*exp(1)*log(3+x)^2+(20*x+60)*exp(1)*log(3+x)+(50*x+150)*exp(1)),x, algorithm="maxima")

[Out]

-e^(x*log(x + 3)/(log(x + 3) + 5) + x/(e*log(x + 3) + 5*e) + 5*x/(log(x + 3) + 5)) + log(1/2*e^(x + 1) + e^(e^
(x*log(x + 3)/(log(x + 3) + 5) + x/(e*log(x + 3) + 5*e) + 5*x/(log(x + 3) + 5))))

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mupad [B]  time = 1.24, size = 54, normalized size = 2.00 \begin {gather*} \ln \left (\mathrm {e}\,{\mathrm {e}}^x\,{\mathrm {e}}^{-{\mathrm {e}}^{\frac {5\,x}{\ln \left (x+3\right )+5}}\,{\mathrm {e}}^{\frac {x}{5\,\mathrm {e}+\ln \left (x+3\right )\,\mathrm {e}}}\,{\left (x+3\right )}^{\frac {x}{\ln \left (x+3\right )+5}}}+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x - exp((x + 5*x*exp(1) + x*log(x + 3)*exp(1))/(5*exp(1) + log(x + 3)*exp(1))) + 1)*(exp(1)*(25*x + 7
5) - exp((x + 5*x*exp(1) + x*log(x + 3)*exp(1))/(5*exp(1) + log(x + 3)*exp(1)))*(4*x + log(x + 3)*(x + exp(1)*
(10*x + 30) + 3) + exp(1)*(25*x + 75) + log(x + 3)^2*exp(1)*(x + 3) + 15) + log(x + 3)*exp(1)*(10*x + 30) + lo
g(x + 3)^2*exp(1)*(x + 3)))/(exp(x - exp((x + 5*x*exp(1) + x*log(x + 3)*exp(1))/(5*exp(1) + log(x + 3)*exp(1))
) + 1)*(exp(1)*(25*x + 75) + log(x + 3)*exp(1)*(10*x + 30) + log(x + 3)^2*exp(1)*(x + 3)) + exp(1)*(50*x + 150
) + log(x + 3)*exp(1)*(20*x + 60) + log(x + 3)^2*exp(1)*(2*x + 6)),x)

[Out]

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

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sympy [A]  time = 15.52, size = 41, normalized size = 1.52 \begin {gather*} \log {\left (e^{x - e^{\frac {e x \log {\left (x + 3 \right )} + x + 5 e x}{e \log {\left (x + 3 \right )} + 5 e}} + 1} + 2 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3-x)*exp(1)*ln(3+x)**2+((-10*x-30)*exp(1)-3-x)*ln(3+x)+(-25*x-75)*exp(1)-4*x-15)*exp((x*exp(1)*l
n(3+x)+5*x*exp(1)+x)/(exp(1)*ln(3+x)+5*exp(1)))+(3+x)*exp(1)*ln(3+x)**2+(10*x+30)*exp(1)*ln(3+x)+(25*x+75)*exp
(1))*exp(-exp((x*exp(1)*ln(3+x)+5*x*exp(1)+x)/(exp(1)*ln(3+x)+5*exp(1)))+x+1)/(((3+x)*exp(1)*ln(3+x)**2+(10*x+
30)*exp(1)*ln(3+x)+(25*x+75)*exp(1))*exp(-exp((x*exp(1)*ln(3+x)+5*x*exp(1)+x)/(exp(1)*ln(3+x)+5*exp(1)))+x+1)+
(2*x+6)*exp(1)*ln(3+x)**2+(20*x+60)*exp(1)*ln(3+x)+(50*x+150)*exp(1)),x)

[Out]

log(exp(x - exp((E*x*log(x + 3) + x + 5*E*x)/(E*log(x + 3) + 5*E)) + 1) + 2)

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