3.96.62 \(\int \frac {14-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (9-6 x+4 \log (3))+e^{e^x} (2+e^{4+3 x} (25+25 \log (3))+e^x (100-50 x+100 \log (3))+e^{2+x} (1-x+e^x (100-25 x+100 \log (3))))+e^{e^x} (e^{2+2 x} (-4+x-4 \log (3))+e^x (-4+2 x-4 \log (3))+e^{4+3 x} (-1-\log (3))) \log (\frac {2-x+2 \log (3)+e^{2+x} (1+\log (3))}{2+e^{2+x}})}{4-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (4-x+4 \log (3))} \, dx\)

Optimal. Leaf size=31 \[ x+\left (5+e^{e^x}\right ) \left (25-\log \left (1-\frac {x}{2+e^{2+x}}+\log (3)\right )\right ) \]

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Rubi [F]  time = 7.31, 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*} \int \frac {14-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (9-6 x+4 \log (3))+e^{e^x} \left (2+e^{4+3 x} (25+25 \log (3))+e^x (100-50 x+100 \log (3))+e^{2+x} \left (1-x+e^x (100-25 x+100 \log (3))\right )\right )+e^{e^x} \left (e^{2+2 x} (-4+x-4 \log (3))+e^x (-4+2 x-4 \log (3))+e^{4+3 x} (-1-\log (3))\right ) \log \left (\frac {2-x+2 \log (3)+e^{2+x} (1+\log (3))}{2+e^{2+x}}\right )}{4-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (4-x+4 \log (3))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(14 - 2*x + 4*Log[3] + E^(4 + 2*x)*(1 + Log[3]) + E^(2 + x)*(9 - 6*x + 4*Log[3]) + E^E^x*(2 + E^(4 + 3*x)*
(25 + 25*Log[3]) + E^x*(100 - 50*x + 100*Log[3]) + E^(2 + x)*(1 - x + E^x*(100 - 25*x + 100*Log[3]))) + E^E^x*
(E^(2 + 2*x)*(-4 + x - 4*Log[3]) + E^x*(-4 + 2*x - 4*Log[3]) + E^(4 + 3*x)*(-1 - Log[3]))*Log[(2 - x + 2*Log[3
] + E^(2 + x)*(1 + Log[3]))/(2 + E^(2 + x))])/(4 - 2*x + 4*Log[3] + E^(4 + 2*x)*(1 + Log[3]) + E^(2 + x)*(4 -
x + 4*Log[3])),x]

[Out]

25*E^E^x - 4*x + 5*Log[2 + E^(2 + x)] - E^E^x*Log[(2 - x + E^(2 + x)*(1 + Log[3]) + Log[9])/(2 + E^(2 + x))] +
(3 + Log[9])*Defer[Int][E^E^x/(x - 2*(1 + Log[3]) - E^(2 + x)*(1 + Log[3])), x] + 5*Defer[Int][x/(x - 2*(1 +
Log[3]) - E^(2 + x)*(1 + Log[3])), x] + Defer[Int][(E^E^x*x)/(x - 2*(1 + Log[3]) - E^(2 + x)*(1 + Log[3])), x]
+ 5*(3 + Log[9])*Defer[Int][(-x + 2*(1 + Log[3]) + E^(2 + x)*(1 + Log[3]))^(-1), x] + (3 + Log[9])*Defer[Int]
[E^E^x/(-x + 2*(1 + Log[3]) + E^(2 + x)*(1 + Log[3])), x] + Defer[Int][(E^E^x*x)/(-x + 2*(1 + Log[3]) + E^(2 +
x)*(1 + Log[3])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 x-14 \left (1+\frac {2 \log (3)}{7}\right )-e^{4+2 x} (1+\log (3))-e^{2+x} (9-6 x+4 \log (3))-e^{e^x} \left (2+e^{4+3 x} (25+25 \log (3))+e^x (100-50 x+100 \log (3))+e^{2+x} \left (1-x+e^x (100-25 x+100 \log (3))\right )\right )-e^{e^x} \left (e^{2+2 x} (-4+x-4 \log (3))+e^x (-4+2 x-4 \log (3))+e^{4+3 x} (-1-\log (3))\right ) \log \left (\frac {2-x+2 \log (3)+e^{2+x} (1+\log (3))}{2+e^{2+x}}\right )}{\left (2+e^{2+x}\right ) \left (x-2 (1+\log (3))-e^{2+x} (1+\log (3))\right )} \, dx\\ &=\int \left (1-\frac {2 \left (5+e^{e^x}\right )}{2+e^{2+x}}+\frac {\left (5+e^{e^x}\right ) (-3+x-\log (9))}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))}-e^{e^x+x} \left (-25+\log \left (\frac {2-x+e^{2+x} (1+\log (3))+\log (9)}{2+e^{2+x}}\right )\right )\right ) \, dx\\ &=x-2 \int \frac {5+e^{e^x}}{2+e^{2+x}} \, dx+\int \frac {\left (5+e^{e^x}\right ) (-3+x-\log (9))}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))} \, dx-\int e^{e^x+x} \left (-25+\log \left (\frac {2-x+e^{2+x} (1+\log (3))+\log (9)}{2+e^{2+x}}\right )\right ) \, dx\\ &=x-2 \operatorname {Subst}\left (\int \frac {5+e^x}{x \left (2+e^2 x\right )} \, dx,x,e^x\right )+\int \left (\frac {\left (5+e^{e^x}\right ) x}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))}+\frac {3 \left (-5-e^{e^x}\right ) \left (1+\frac {2 \log (3)}{3}\right )}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))}\right ) \, dx-\int \left (-25 e^{e^x+x}+e^{e^x+x} \log \left (\frac {-x+2 (1+\log (3))+e^{2+x} (1+\log (3))}{2+e^{2+x}}\right )\right ) \, dx\\ &=x-2 \operatorname {Subst}\left (\int \left (\frac {5}{x \left (2+e^2 x\right )}+\frac {e^x}{x \left (2+e^2 x\right )}\right ) \, dx,x,e^x\right )+25 \int e^{e^x+x} \, dx+(3+\log (9)) \int \frac {-5-e^{e^x}}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))} \, dx+\int \frac {\left (5+e^{e^x}\right ) x}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))} \, dx-\int e^{e^x+x} \log \left (\frac {-x+2 (1+\log (3))+e^{2+x} (1+\log (3))}{2+e^{2+x}}\right ) \, dx\\ &=x-e^{e^x} \log \left (\frac {2-x+e^{2+x} (1+\log (3))+\log (9)}{2+e^{2+x}}\right )-2 \operatorname {Subst}\left (\int \frac {e^x}{x \left (2+e^2 x\right )} \, dx,x,e^x\right )-10 \operatorname {Subst}\left (\int \frac {1}{x \left (2+e^2 x\right )} \, dx,x,e^x\right )+25 \operatorname {Subst}\left (\int e^x \, dx,x,e^x\right )+(3+\log (9)) \int \left (\frac {5}{-x+2 (1+\log (3))+e^{2+x} (1+\log (3))}+\frac {e^{e^x}}{-x+2 (1+\log (3))+e^{2+x} (1+\log (3))}\right ) \, dx+\int \frac {e^{e^x} \left (2-e^{2+x} (-1+x)\right )}{\left (2+e^{2+x}\right ) \left (x-2 (1+\log (3))-e^{2+x} (1+\log (3))\right )} \, dx+\int \left (\frac {5 x}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))}+\frac {e^{e^x} x}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))}\right ) \, dx\\ &=25 e^{e^x}+x-e^{e^x} \log \left (\frac {2-x+e^{2+x} (1+\log (3))+\log (9)}{2+e^{2+x}}\right )-2 \operatorname {Subst}\left (\int \left (\frac {e^x}{2 x}-\frac {e^{2+x}}{2 \left (2+e^2 x\right )}\right ) \, dx,x,e^x\right )+5 \int \frac {x}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))} \, dx-5 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )+\left (5 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{2+e^2 x} \, dx,x,e^x\right )+(3+\log (9)) \int \frac {e^{e^x}}{-x+2 (1+\log (3))+e^{2+x} (1+\log (3))} \, dx+(5 (3+\log (9))) \int \frac {1}{-x+2 (1+\log (3))+e^{2+x} (1+\log (3))} \, dx+\int \frac {e^{e^x} x}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))} \, dx+\int \left (\frac {2 e^{e^x}}{2+e^{2+x}}+\frac {e^{e^x} (3-x+\log (9))}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))}\right ) \, dx\\ &=25 e^{e^x}-4 x+5 \log \left (2+e^{2+x}\right )-e^{e^x} \log \left (\frac {2-x+e^{2+x} (1+\log (3))+\log (9)}{2+e^{2+x}}\right )+2 \int \frac {e^{e^x}}{2+e^{2+x}} \, dx+5 \int \frac {x}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))} \, dx+(3+\log (9)) \int \frac {e^{e^x}}{-x+2 (1+\log (3))+e^{2+x} (1+\log (3))} \, dx+(5 (3+\log (9))) \int \frac {1}{-x+2 (1+\log (3))+e^{2+x} (1+\log (3))} \, dx+\int \frac {e^{e^x} x}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))} \, dx+\int \frac {e^{e^x} (3-x+\log (9))}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))} \, dx-\operatorname {Subst}\left (\int \frac {e^x}{x} \, dx,x,e^x\right )+\operatorname {Subst}\left (\int \frac {e^{2+x}}{2+e^2 x} \, dx,x,e^x\right )\\ &=25 e^{e^x}-4 x-\text {Ei}\left (e^x\right )+e^{-\frac {2}{e^2}} \text {Ei}\left (\frac {2+e^{2+x}}{e^2}\right )+5 \log \left (2+e^{2+x}\right )-e^{e^x} \log \left (\frac {2-x+e^{2+x} (1+\log (3))+\log (9)}{2+e^{2+x}}\right )+2 \operatorname {Subst}\left (\int \frac {e^x}{x \left (2+e^2 x\right )} \, dx,x,e^x\right )+5 \int \frac {x}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))} \, dx+(3+\log (9)) \int \frac {e^{e^x}}{-x+2 (1+\log (3))+e^{2+x} (1+\log (3))} \, dx+(5 (3+\log (9))) \int \frac {1}{-x+2 (1+\log (3))+e^{2+x} (1+\log (3))} \, dx+\int \frac {e^{e^x} x}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))} \, dx+\int \left (\frac {3 e^{e^x} \left (1+\frac {2 \log (3)}{3}\right )}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))}+\frac {e^{e^x} x}{-x+2 (1+\log (3))+e^{2+x} (1+\log (3))}\right ) \, dx\\ &=25 e^{e^x}-4 x-\text {Ei}\left (e^x\right )+e^{-\frac {2}{e^2}} \text {Ei}\left (\frac {2+e^{2+x}}{e^2}\right )+5 \log \left (2+e^{2+x}\right )-e^{e^x} \log \left (\frac {2-x+e^{2+x} (1+\log (3))+\log (9)}{2+e^{2+x}}\right )+2 \operatorname {Subst}\left (\int \left (\frac {e^x}{2 x}-\frac {e^{2+x}}{2 \left (2+e^2 x\right )}\right ) \, dx,x,e^x\right )+5 \int \frac {x}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))} \, dx+(3+\log (9)) \int \frac {e^{e^x}}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))} \, dx+(3+\log (9)) \int \frac {e^{e^x}}{-x+2 (1+\log (3))+e^{2+x} (1+\log (3))} \, dx+(5 (3+\log (9))) \int \frac {1}{-x+2 (1+\log (3))+e^{2+x} (1+\log (3))} \, dx+\int \frac {e^{e^x} x}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))} \, dx+\int \frac {e^{e^x} x}{-x+2 (1+\log (3))+e^{2+x} (1+\log (3))} \, dx\\ &=25 e^{e^x}-4 x-\text {Ei}\left (e^x\right )+e^{-\frac {2}{e^2}} \text {Ei}\left (\frac {2+e^{2+x}}{e^2}\right )+5 \log \left (2+e^{2+x}\right )-e^{e^x} \log \left (\frac {2-x+e^{2+x} (1+\log (3))+\log (9)}{2+e^{2+x}}\right )+5 \int \frac {x}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))} \, dx+(3+\log (9)) \int \frac {e^{e^x}}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))} \, dx+(3+\log (9)) \int \frac {e^{e^x}}{-x+2 (1+\log (3))+e^{2+x} (1+\log (3))} \, dx+(5 (3+\log (9))) \int \frac {1}{-x+2 (1+\log (3))+e^{2+x} (1+\log (3))} \, dx+\int \frac {e^{e^x} x}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))} \, dx+\int \frac {e^{e^x} x}{-x+2 (1+\log (3))+e^{2+x} (1+\log (3))} \, dx+\operatorname {Subst}\left (\int \frac {e^x}{x} \, dx,x,e^x\right )-\operatorname {Subst}\left (\int \frac {e^{2+x}}{2+e^2 x} \, dx,x,e^x\right )\\ &=25 e^{e^x}-4 x+5 \log \left (2+e^{2+x}\right )-e^{e^x} \log \left (\frac {2-x+e^{2+x} (1+\log (3))+\log (9)}{2+e^{2+x}}\right )+5 \int \frac {x}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))} \, dx+(3+\log (9)) \int \frac {e^{e^x}}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))} \, dx+(3+\log (9)) \int \frac {e^{e^x}}{-x+2 (1+\log (3))+e^{2+x} (1+\log (3))} \, dx+(5 (3+\log (9))) \int \frac {1}{-x+2 (1+\log (3))+e^{2+x} (1+\log (3))} \, dx+\int \frac {e^{e^x} x}{x-2 (1+\log (3))-e^{2+x} (1+\log (3))} \, dx+\int \frac {e^{e^x} x}{-x+2 (1+\log (3))+e^{2+x} (1+\log (3))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 8.37, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {14-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (9-6 x+4 \log (3))+e^{e^x} \left (2+e^{4+3 x} (25+25 \log (3))+e^x (100-50 x+100 \log (3))+e^{2+x} \left (1-x+e^x (100-25 x+100 \log (3))\right )\right )+e^{e^x} \left (e^{2+2 x} (-4+x-4 \log (3))+e^x (-4+2 x-4 \log (3))+e^{4+3 x} (-1-\log (3))\right ) \log \left (\frac {2-x+2 \log (3)+e^{2+x} (1+\log (3))}{2+e^{2+x}}\right )}{4-2 x+4 \log (3)+e^{4+2 x} (1+\log (3))+e^{2+x} (4-x+4 \log (3))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(14 - 2*x + 4*Log[3] + E^(4 + 2*x)*(1 + Log[3]) + E^(2 + x)*(9 - 6*x + 4*Log[3]) + E^E^x*(2 + E^(4 +
3*x)*(25 + 25*Log[3]) + E^x*(100 - 50*x + 100*Log[3]) + E^(2 + x)*(1 - x + E^x*(100 - 25*x + 100*Log[3]))) +
E^E^x*(E^(2 + 2*x)*(-4 + x - 4*Log[3]) + E^x*(-4 + 2*x - 4*Log[3]) + E^(4 + 3*x)*(-1 - Log[3]))*Log[(2 - x + 2
*Log[3] + E^(2 + x)*(1 + Log[3]))/(2 + E^(2 + x))])/(4 - 2*x + 4*Log[3] + E^(4 + 2*x)*(1 + Log[3]) + E^(2 + x)
*(4 - x + 4*Log[3])),x]

[Out]

Integrate[(14 - 2*x + 4*Log[3] + E^(4 + 2*x)*(1 + Log[3]) + E^(2 + x)*(9 - 6*x + 4*Log[3]) + E^E^x*(2 + E^(4 +
3*x)*(25 + 25*Log[3]) + E^x*(100 - 50*x + 100*Log[3]) + E^(2 + x)*(1 - x + E^x*(100 - 25*x + 100*Log[3]))) +
E^E^x*(E^(2 + 2*x)*(-4 + x - 4*Log[3]) + E^x*(-4 + 2*x - 4*Log[3]) + E^(4 + 3*x)*(-1 - Log[3]))*Log[(2 - x + 2
*Log[3] + E^(2 + x)*(1 + Log[3]))/(2 + E^(2 + x))])/(4 - 2*x + 4*Log[3] + E^(4 + 2*x)*(1 + Log[3]) + E^(2 + x)
*(4 - x + 4*Log[3])), x]

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fricas [A]  time = 0.52, size = 42, normalized size = 1.35 \begin {gather*} -{\left (e^{\left (e^{x}\right )} + 5\right )} \log \left (\frac {{\left (\log \relax (3) + 1\right )} e^{\left (x + 2\right )} - x + 2 \, \log \relax (3) + 2}{e^{\left (x + 2\right )} + 2}\right ) + x + 25 \, e^{\left (e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-log(3)-1)*exp(x)*exp(2+x)^2+(-4*log(3)+x-4)*exp(x)*exp(2+x)+(-4*log(3)+2*x-4)*exp(x))*exp(exp(x)
)*log(((log(3)+1)*exp(2+x)+2*log(3)+2-x)/(exp(2+x)+2))+((25*log(3)+25)*exp(x)*exp(2+x)^2+((100*log(3)-25*x+100
)*exp(x)-x+1)*exp(2+x)+(100*log(3)-50*x+100)*exp(x)+2)*exp(exp(x))+(log(3)+1)*exp(2+x)^2+(4*log(3)-6*x+9)*exp(
2+x)+4*log(3)-2*x+14)/((log(3)+1)*exp(2+x)^2+(4*log(3)-x+4)*exp(2+x)+4*log(3)+4-2*x),x, algorithm="fricas")

[Out]

-(e^(e^x) + 5)*log(((log(3) + 1)*e^(x + 2) - x + 2*log(3) + 2)/(e^(x + 2) + 2)) + x + 25*e^(e^x)

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giac [B]  time = 0.46, size = 97, normalized size = 3.13 \begin {gather*} {\left (x e^{x} - e^{\left (x + e^{x}\right )} \log \left (e^{\left (x + 2\right )} \log \relax (3) - x + e^{\left (x + 2\right )} + 2 \, \log \relax (3) + 2\right ) - 5 \, e^{x} \log \left (-e^{\left (x + 2\right )} \log \relax (3) + x - e^{\left (x + 2\right )} - 2 \, \log \relax (3) - 2\right ) + e^{\left (x + e^{x}\right )} \log \left (e^{\left (x + 2\right )} + 2\right ) + 5 \, e^{x} \log \left (-e^{\left (x + 2\right )} - 2\right ) + 25 \, e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-log(3)-1)*exp(x)*exp(2+x)^2+(-4*log(3)+x-4)*exp(x)*exp(2+x)+(-4*log(3)+2*x-4)*exp(x))*exp(exp(x)
)*log(((log(3)+1)*exp(2+x)+2*log(3)+2-x)/(exp(2+x)+2))+((25*log(3)+25)*exp(x)*exp(2+x)^2+((100*log(3)-25*x+100
)*exp(x)-x+1)*exp(2+x)+(100*log(3)-50*x+100)*exp(x)+2)*exp(exp(x))+(log(3)+1)*exp(2+x)^2+(4*log(3)-6*x+9)*exp(
2+x)+4*log(3)-2*x+14)/((log(3)+1)*exp(2+x)^2+(4*log(3)-x+4)*exp(2+x)+4*log(3)+4-2*x),x, algorithm="giac")

[Out]

(x*e^x - e^(x + e^x)*log(e^(x + 2)*log(3) - x + e^(x + 2) + 2*log(3) + 2) - 5*e^x*log(-e^(x + 2)*log(3) + x -
e^(x + 2) - 2*log(3) - 2) + e^(x + e^x)*log(e^(x + 2) + 2) + 5*e^x*log(-e^(x + 2) - 2) + 25*e^(x + e^x))*e^(-x
)

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maple [C]  time = 0.53, size = 313, normalized size = 10.10

method	result	size
risch	\(-{\mathrm e}^{{\mathrm e}^{x}} \ln \left (\left (\ln \relax (3) {\mathrm e}^{x}+{\mathrm e}^{x}\right ) {\mathrm e}^{2}+2 \ln \relax (3)-x +2\right )+{\mathrm e}^{{\mathrm e}^{x}} \ln \left ({\mathrm e}^{2+x}+2\right )+x +5 \ln \left ({\mathrm e}^{x}+2 \,{\mathrm e}^{-2}\right )-5 \ln \left ({\mathrm e}^{x}+\frac {\left (2 \ln \relax (3)-x +2\right ) {\mathrm e}^{-2}}{\ln \relax (3)+1}\right )+\frac {i {\mathrm e}^{{\mathrm e}^{x}} \pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2+x}+2}\right ) \mathrm {csgn}\left (i \left (\left (\ln \relax (3) {\mathrm e}^{x}+{\mathrm e}^{x}\right ) {\mathrm e}^{2}+2 \ln \relax (3)-x +2\right )\right ) \mathrm {csgn}\left (\frac {i \left (\left (\ln \relax (3) {\mathrm e}^{x}+{\mathrm e}^{x}\right ) {\mathrm e}^{2}+2 \ln \relax (3)-x +2\right )}{{\mathrm e}^{2+x}+2}\right )}{2}-\frac {i {\mathrm e}^{{\mathrm e}^{x}} \pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2+x}+2}\right ) \mathrm {csgn}\left (\frac {i \left (\left (\ln \relax (3) {\mathrm e}^{x}+{\mathrm e}^{x}\right ) {\mathrm e}^{2}+2 \ln \relax (3)-x +2\right )}{{\mathrm e}^{2+x}+2}\right )^{2}}{2}-\frac {i {\mathrm e}^{{\mathrm e}^{x}} \pi \,\mathrm {csgn}\left (i \left (\left (\ln \relax (3) {\mathrm e}^{x}+{\mathrm e}^{x}\right ) {\mathrm e}^{2}+2 \ln \relax (3)-x +2\right )\right ) \mathrm {csgn}\left (\frac {i \left (\left (\ln \relax (3) {\mathrm e}^{x}+{\mathrm e}^{x}\right ) {\mathrm e}^{2}+2 \ln \relax (3)-x +2\right )}{{\mathrm e}^{2+x}+2}\right )^{2}}{2}+\frac {i {\mathrm e}^{{\mathrm e}^{x}} \pi \mathrm {csgn}\left (\frac {i \left (\left (\ln \relax (3) {\mathrm e}^{x}+{\mathrm e}^{x}\right ) {\mathrm e}^{2}+2 \ln \relax (3)-x +2\right )}{{\mathrm e}^{2+x}+2}\right )^{3}}{2}+25 \,{\mathrm e}^{{\mathrm e}^{x}}\)	\(313\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-ln(3)-1)*exp(x)*exp(2+x)^2+(-4*ln(3)+x-4)*exp(x)*exp(2+x)+(-4*ln(3)+2*x-4)*exp(x))*exp(exp(x))*ln(((ln
(3)+1)*exp(2+x)+2*ln(3)+2-x)/(exp(2+x)+2))+((25*ln(3)+25)*exp(x)*exp(2+x)^2+((100*ln(3)-25*x+100)*exp(x)-x+1)*
exp(2+x)+(100*ln(3)-50*x+100)*exp(x)+2)*exp(exp(x))+(ln(3)+1)*exp(2+x)^2+(4*ln(3)-6*x+9)*exp(2+x)+4*ln(3)-2*x+
14)/((ln(3)+1)*exp(2+x)^2+(4*ln(3)-x+4)*exp(2+x)+4*ln(3)+4-2*x),x,method=_RETURNVERBOSE)

[Out]

-exp(exp(x))*ln((ln(3)*exp(x)+exp(x))*exp(2)+2*ln(3)-x+2)+exp(exp(x))*ln(exp(2+x)+2)+x+5*ln(exp(x)+2*exp(-2))-
5*ln(exp(x)+(2*ln(3)-x+2)*exp(-2)/(ln(3)+1))+1/2*I*exp(exp(x))*Pi*csgn(I/(exp(2+x)+2))*csgn(I*((ln(3)*exp(x)+e
xp(x))*exp(2)+2*ln(3)-x+2))*csgn(I/(exp(2+x)+2)*((ln(3)*exp(x)+exp(x))*exp(2)+2*ln(3)-x+2))-1/2*I*exp(exp(x))*
Pi*csgn(I/(exp(2+x)+2))*csgn(I/(exp(2+x)+2)*((ln(3)*exp(x)+exp(x))*exp(2)+2*ln(3)-x+2))^2-1/2*I*exp(exp(x))*Pi
*csgn(I*((ln(3)*exp(x)+exp(x))*exp(2)+2*ln(3)-x+2))*csgn(I/(exp(2+x)+2)*((ln(3)*exp(x)+exp(x))*exp(2)+2*ln(3)-
x+2))^2+1/2*I*exp(exp(x))*Pi*csgn(I/(exp(2+x)+2)*((ln(3)*exp(x)+exp(x))*exp(2)+2*ln(3)-x+2))^3+25*exp(exp(x))

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maxima [A]  time = 0.54, size = 48, normalized size = 1.55 \begin {gather*} -{\left (e^{\left (e^{x}\right )} + 5\right )} \log \left ({\left (e^{2} \log \relax (3) + e^{2}\right )} e^{x} - x + 2 \, \log \relax (3) + 2\right ) + {\left (e^{\left (e^{x}\right )} + 5\right )} \log \left (e^{\left (x + 2\right )} + 2\right ) + x + 25 \, e^{\left (e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-log(3)-1)*exp(x)*exp(2+x)^2+(-4*log(3)+x-4)*exp(x)*exp(2+x)+(-4*log(3)+2*x-4)*exp(x))*exp(exp(x)
)*log(((log(3)+1)*exp(2+x)+2*log(3)+2-x)/(exp(2+x)+2))+((25*log(3)+25)*exp(x)*exp(2+x)^2+((100*log(3)-25*x+100
)*exp(x)-x+1)*exp(2+x)+(100*log(3)-50*x+100)*exp(x)+2)*exp(exp(x))+(log(3)+1)*exp(2+x)^2+(4*log(3)-6*x+9)*exp(
2+x)+4*log(3)-2*x+14)/((log(3)+1)*exp(2+x)^2+(4*log(3)-x+4)*exp(2+x)+4*log(3)+4-2*x),x, algorithm="maxima")

[Out]

-(e^(e^x) + 5)*log((e^2*log(3) + e^2)*e^x - x + 2*log(3) + 2) + (e^(e^x) + 5)*log(e^(x + 2) + 2) + x + 25*e^(e
^x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {4\,\ln \relax (3)-2\,x+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left ({\mathrm {e}}^{x+2}\,\left ({\mathrm {e}}^x\,\left (100\,\ln \relax (3)-25\,x+100\right )-x+1\right )+{\mathrm {e}}^{3\,x+4}\,\left (25\,\ln \relax (3)+25\right )+{\mathrm {e}}^x\,\left (100\,\ln \relax (3)-50\,x+100\right )+2\right )+{\mathrm {e}}^{x+2}\,\left (4\,\ln \relax (3)-6\,x+9\right )+{\mathrm {e}}^{2\,x+4}\,\left (\ln \relax (3)+1\right )-\ln \left (\frac {2\,\ln \relax (3)-x+{\mathrm {e}}^{x+2}\,\left (\ln \relax (3)+1\right )+2}{{\mathrm {e}}^{x+2}+2}\right )\,{\mathrm {e}}^{{\mathrm {e}}^x}\,\left ({\mathrm {e}}^x\,\left (4\,\ln \relax (3)-2\,x+4\right )+{\mathrm {e}}^{3\,x+4}\,\left (\ln \relax (3)+1\right )+{\mathrm {e}}^{2\,x+2}\,\left (4\,\ln \relax (3)-x+4\right )\right )+14}{4\,\ln \relax (3)-2\,x+{\mathrm {e}}^{x+2}\,\left (4\,\ln \relax (3)-x+4\right )+{\mathrm {e}}^{2\,x+4}\,\left (\ln \relax (3)+1\right )+4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*log(3) - 2*x + exp(exp(x))*(exp(x + 2)*(exp(x)*(100*log(3) - 25*x + 100) - x + 1) + exp(x)*(100*log(3)
- 50*x + 100) + exp(2*x + 4)*exp(x)*(25*log(3) + 25) + 2) + exp(x + 2)*(4*log(3) - 6*x + 9) + exp(2*x + 4)*(lo
g(3) + 1) - log((2*log(3) - x + exp(x + 2)*(log(3) + 1) + 2)/(exp(x + 2) + 2))*exp(exp(x))*(exp(x)*(4*log(3) -
2*x + 4) + exp(x + 2)*exp(x)*(4*log(3) - x + 4) + exp(2*x + 4)*exp(x)*(log(3) + 1)) + 14)/(4*log(3) - 2*x + e
xp(x + 2)*(4*log(3) - x + 4) + exp(2*x + 4)*(log(3) + 1) + 4),x)

[Out]

int((4*log(3) - 2*x + exp(exp(x))*(exp(x + 2)*(exp(x)*(100*log(3) - 25*x + 100) - x + 1) + exp(3*x + 4)*(25*lo
g(3) + 25) + exp(x)*(100*log(3) - 50*x + 100) + 2) + exp(x + 2)*(4*log(3) - 6*x + 9) + exp(2*x + 4)*(log(3) +
1) - log((2*log(3) - x + exp(x + 2)*(log(3) + 1) + 2)/(exp(x + 2) + 2))*exp(exp(x))*(exp(x)*(4*log(3) - 2*x +
4) + exp(3*x + 4)*(log(3) + 1) + exp(2*x + 2)*(4*log(3) - x + 4)) + 14)/(4*log(3) - 2*x + exp(x + 2)*(4*log(3)
- x + 4) + exp(2*x + 4)*(log(3) + 1) + 4), x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: CoercionFailed} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-ln(3)-1)*exp(x)*exp(2+x)**2+(-4*ln(3)+x-4)*exp(x)*exp(2+x)+(-4*ln(3)+2*x-4)*exp(x))*exp(exp(x))*
ln(((ln(3)+1)*exp(2+x)+2*ln(3)+2-x)/(exp(2+x)+2))+((25*ln(3)+25)*exp(x)*exp(2+x)**2+((100*ln(3)-25*x+100)*exp(
x)-x+1)*exp(2+x)+(100*ln(3)-50*x+100)*exp(x)+2)*exp(exp(x))+(ln(3)+1)*exp(2+x)**2+(4*ln(3)-6*x+9)*exp(2+x)+4*l
n(3)-2*x+14)/((ln(3)+1)*exp(2+x)**2+(4*ln(3)-x+4)*exp(2+x)+4*ln(3)+4-2*x),x)

[Out]

Exception raised: CoercionFailed

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