3.46.53 \(\int \frac {-25+4 x^2+e^2 (-625-1000 x-600 x^2-160 x^3-16 x^4)+e^2 (625+1000 x+600 x^2+160 x^3+16 x^4) \log ^2(4)+e^{2 e^x} (-e^2+e^2 \log ^2(4))+e^{e^x} (1-e^x x+e^2 (50+40 x+8 x^2)+e^2 (-50-40 x-8 x^2) \log ^2(4))}{-625-1000 x-600 x^2-160 x^3-16 x^4+(625+1000 x+600 x^2+160 x^3+16 x^4) \log ^2(4)+e^{2 e^x} (-1+\log ^2(4))+e^{e^x} (50+40 x+8 x^2+(-50-40 x-8 x^2) \log ^2(4))} \, dx\)

Optimal. Leaf size=36 \[ e^2 x-\frac {x}{\left (e^{e^x}-(5+2 x)^2\right ) \left (1-\log ^2(4)\right )} \]

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Rubi [F]  time = 3.18, 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 {-25+4 x^2+e^2 \left (-625-1000 x-600 x^2-160 x^3-16 x^4\right )+e^2 \left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-e^2+e^2 \log ^2(4)\right )+e^{e^x} \left (1-e^x x+e^2 \left (50+40 x+8 x^2\right )+e^2 \left (-50-40 x-8 x^2\right ) \log ^2(4)\right )}{-625-1000 x-600 x^2-160 x^3-16 x^4+\left (625+1000 x+600 x^2+160 x^3+16 x^4\right ) \log ^2(4)+e^{2 e^x} \left (-1+\log ^2(4)\right )+e^{e^x} \left (50+40 x+8 x^2+\left (-50-40 x-8 x^2\right ) \log ^2(4)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-25 + 4*x^2 + E^2*(-625 - 1000*x - 600*x^2 - 160*x^3 - 16*x^4) + E^2*(625 + 1000*x + 600*x^2 + 160*x^3 +
16*x^4)*Log[4]^2 + E^(2*E^x)*(-E^2 + E^2*Log[4]^2) + E^E^x*(1 - E^x*x + E^2*(50 + 40*x + 8*x^2) + E^2*(-50 - 4
0*x - 8*x^2)*Log[4]^2))/(-625 - 1000*x - 600*x^2 - 160*x^3 - 16*x^4 + (625 + 1000*x + 600*x^2 + 160*x^3 + 16*x
^4)*Log[4]^2 + E^(2*E^x)*(-1 + Log[4]^2) + E^E^x*(50 + 40*x + 8*x^2 + (-50 - 40*x - 8*x^2)*Log[4]^2)),x]

[Out]

625*E^2*Defer[Int][(-25 + E^E^x - 20*x - 4*x^2)^(-2), x] + (25*Defer[Int][(-25 + E^E^x - 20*x - 4*x^2)^(-2), x
])/(1 - Log[4]^2) - Defer[Int][E^E^x/(-25 + E^E^x - 20*x - 4*x^2)^2, x]/(1 - Log[4]^2) - 50*Defer[Int][E^(2 +
E^x)/(-25 + E^E^x - 20*x - 4*x^2)^2, x] + Defer[Int][(E^(E^x + x)*x)/(-25 + E^E^x - 20*x - 4*x^2)^2, x]/(1 - L
og[4]^2) + Defer[Int][E^(2*(1 + E^x))/(25 - E^E^x + 20*x + 4*x^2)^2, x] + 1000*E^2*Defer[Int][x/(25 - E^E^x +
20*x + 4*x^2)^2, x] - 40*Defer[Int][(E^(2 + E^x)*x)/(25 - E^E^x + 20*x + 4*x^2)^2, x] + 600*E^2*Defer[Int][x^2
/(25 - E^E^x + 20*x + 4*x^2)^2, x] - (4*Defer[Int][x^2/(25 - E^E^x + 20*x + 4*x^2)^2, x])/(1 - Log[4]^2) - 8*D
efer[Int][(E^(2 + E^x)*x^2)/(25 - E^E^x + 20*x + 4*x^2)^2, x] + 160*E^2*Defer[Int][x^3/(25 - E^E^x + 20*x + 4*
x^2)^2, x] + 16*E^2*Defer[Int][x^4/(25 - E^E^x + 20*x + 4*x^2)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25-e^{e^x}+e^{e^x+x} x-4 x^2-e^{2+2 e^x} \left (-1+\log ^2(4)\right )+2 e^{2+e^x} (5+2 x)^2 \left (-1+\log ^2(4)\right )-e^2 (5+2 x)^4 \left (-1+\log ^2(4)\right )}{\left (e^{e^x}-(5+2 x)^2\right )^2 \left (1-\log ^2(4)\right )} \, dx\\ &=\frac {\int \frac {25-e^{e^x}+e^{e^x+x} x-4 x^2-e^{2+2 e^x} \left (-1+\log ^2(4)\right )+2 e^{2+e^x} (5+2 x)^2 \left (-1+\log ^2(4)\right )-e^2 (5+2 x)^4 \left (-1+\log ^2(4)\right )}{\left (e^{e^x}-(5+2 x)^2\right )^2} \, dx}{1-\log ^2(4)}\\ &=\frac {\int \left (\frac {25}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2}-\frac {e^{e^x}}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2}+\frac {e^{e^x+x} x}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2}-\frac {4 x^2}{\left (25-e^{e^x}+20 x+4 x^2\right )^2}+\frac {e^{2 \left (1+e^x\right )} (1-\log (4)) (1+\log (4))}{\left (25-e^{e^x}+20 x+4 x^2\right )^2}+\frac {2 e^{2+e^x} (5+2 x)^2 (-1+\log (4)) (1+\log (4))}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2}-\frac {e^2 (5+2 x)^4 (-1+\log (4)) (1+\log (4))}{\left (25-e^{e^x}+20 x+4 x^2\right )^2}\right ) \, dx}{1-\log ^2(4)}\\ &=-\left (2 \int \frac {e^{2+e^x} (5+2 x)^2}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx\right )+e^2 \int \frac {(5+2 x)^4}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx-\frac {\int \frac {e^{e^x}}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx}{1-\log ^2(4)}+\frac {\int \frac {e^{e^x+x} x}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx}{1-\log ^2(4)}-\frac {4 \int \frac {x^2}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx}{1-\log ^2(4)}+\frac {25 \int \frac {1}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx}{1-\log ^2(4)}+\int \frac {e^{2 \left (1+e^x\right )}}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx\\ &=-\left (2 \int \left (\frac {25 e^{2+e^x}}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2}+\frac {20 e^{2+e^x} x}{\left (25-e^{e^x}+20 x+4 x^2\right )^2}+\frac {4 e^{2+e^x} x^2}{\left (25-e^{e^x}+20 x+4 x^2\right )^2}\right ) \, dx\right )+e^2 \int \left (\frac {625}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2}+\frac {1000 x}{\left (25-e^{e^x}+20 x+4 x^2\right )^2}+\frac {600 x^2}{\left (25-e^{e^x}+20 x+4 x^2\right )^2}+\frac {160 x^3}{\left (25-e^{e^x}+20 x+4 x^2\right )^2}+\frac {16 x^4}{\left (25-e^{e^x}+20 x+4 x^2\right )^2}\right ) \, dx-\frac {\int \frac {e^{e^x}}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx}{1-\log ^2(4)}+\frac {\int \frac {e^{e^x+x} x}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx}{1-\log ^2(4)}-\frac {4 \int \frac {x^2}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx}{1-\log ^2(4)}+\frac {25 \int \frac {1}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx}{1-\log ^2(4)}+\int \frac {e^{2 \left (1+e^x\right )}}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx\\ &=-\left (8 \int \frac {e^{2+e^x} x^2}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx\right )-40 \int \frac {e^{2+e^x} x}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx-50 \int \frac {e^{2+e^x}}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx+\left (16 e^2\right ) \int \frac {x^4}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx+\left (160 e^2\right ) \int \frac {x^3}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx+\left (600 e^2\right ) \int \frac {x^2}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx+\left (625 e^2\right ) \int \frac {1}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx+\left (1000 e^2\right ) \int \frac {x}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx-\frac {\int \frac {e^{e^x}}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx}{1-\log ^2(4)}+\frac {\int \frac {e^{e^x+x} x}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx}{1-\log ^2(4)}-\frac {4 \int \frac {x^2}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx}{1-\log ^2(4)}+\frac {25 \int \frac {1}{\left (-25+e^{e^x}-20 x-4 x^2\right )^2} \, dx}{1-\log ^2(4)}+\int \frac {e^{2 \left (1+e^x\right )}}{\left (25-e^{e^x}+20 x+4 x^2\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 40, normalized size = 1.11 \begin {gather*} \frac {\frac {x}{-25+e^{e^x}-20 x-4 x^2}+e^2 x \left (-1+\log ^2(4)\right )}{-1+\log ^2(4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25 + 4*x^2 + E^2*(-625 - 1000*x - 600*x^2 - 160*x^3 - 16*x^4) + E^2*(625 + 1000*x + 600*x^2 + 160*
x^3 + 16*x^4)*Log[4]^2 + E^(2*E^x)*(-E^2 + E^2*Log[4]^2) + E^E^x*(1 - E^x*x + E^2*(50 + 40*x + 8*x^2) + E^2*(-
50 - 40*x - 8*x^2)*Log[4]^2))/(-625 - 1000*x - 600*x^2 - 160*x^3 - 16*x^4 + (625 + 1000*x + 600*x^2 + 160*x^3
+ 16*x^4)*Log[4]^2 + E^(2*E^x)*(-1 + Log[4]^2) + E^E^x*(50 + 40*x + 8*x^2 + (-50 - 40*x - 8*x^2)*Log[4]^2)),x]

[Out]

(x/(-25 + E^E^x - 20*x - 4*x^2) + E^2*x*(-1 + Log[4]^2))/(-1 + Log[4]^2)

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fricas [B]  time = 0.63, size = 106, normalized size = 2.94 \begin {gather*} \frac {4 \, {\left (4 \, x^{3} + 20 \, x^{2} + 25 \, x\right )} e^{2} \log \relax (2)^{2} - {\left (4 \, x^{3} + 20 \, x^{2} + 25 \, x\right )} e^{2} - {\left (4 \, x e^{2} \log \relax (2)^{2} - x e^{2}\right )} e^{\left (e^{x}\right )} - x}{4 \, {\left (4 \, x^{2} + 20 \, x + 25\right )} \log \relax (2)^{2} - 4 \, x^{2} - {\left (4 \, \log \relax (2)^{2} - 1\right )} e^{\left (e^{x}\right )} - 20 \, x - 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(2)*log(2)^2-exp(2))*exp(exp(x))^2+(-exp(x)*x+4*(-8*x^2-40*x-50)*exp(2)*log(2)^2+(8*x^2+40*x+
50)*exp(2)+1)*exp(exp(x))+4*(16*x^4+160*x^3+600*x^2+1000*x+625)*exp(2)*log(2)^2+(-16*x^4-160*x^3-600*x^2-1000*
x-625)*exp(2)+4*x^2-25)/((4*log(2)^2-1)*exp(exp(x))^2+(4*(-8*x^2-40*x-50)*log(2)^2+8*x^2+40*x+50)*exp(exp(x))+
4*(16*x^4+160*x^3+600*x^2+1000*x+625)*log(2)^2-16*x^4-160*x^3-600*x^2-1000*x-625),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.25, size = 120, normalized size = 3.33 \begin {gather*} \frac {16 \, x^{3} e^{2} \log \relax (2)^{2} + 80 \, x^{2} e^{2} \log \relax (2)^{2} - 4 \, x^{3} e^{2} + 100 \, x e^{2} \log \relax (2)^{2} - 4 \, x e^{\left (e^{x} + 2\right )} \log \relax (2)^{2} - 20 \, x^{2} e^{2} - 25 \, x e^{2} + x e^{\left (e^{x} + 2\right )} - x}{16 \, x^{2} \log \relax (2)^{2} + 80 \, x \log \relax (2)^{2} - 4 \, e^{\left (e^{x}\right )} \log \relax (2)^{2} - 4 \, x^{2} + 100 \, \log \relax (2)^{2} - 20 \, x + e^{\left (e^{x}\right )} - 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(2)*log(2)^2-exp(2))*exp(exp(x))^2+(-exp(x)*x+4*(-8*x^2-40*x-50)*exp(2)*log(2)^2+(8*x^2+40*x+
50)*exp(2)+1)*exp(exp(x))+4*(16*x^4+160*x^3+600*x^2+1000*x+625)*exp(2)*log(2)^2+(-16*x^4-160*x^3-600*x^2-1000*
x-625)*exp(2)+4*x^2-25)/((4*log(2)^2-1)*exp(exp(x))^2+(4*(-8*x^2-40*x-50)*log(2)^2+8*x^2+40*x+50)*exp(exp(x))+
4*(16*x^4+160*x^3+600*x^2+1000*x+625)*log(2)^2-16*x^4-160*x^3-600*x^2-1000*x-625),x, algorithm="giac")

[Out]

(16*x^3*e^2*log(2)^2 + 80*x^2*e^2*log(2)^2 - 4*x^3*e^2 + 100*x*e^2*log(2)^2 - 4*x*e^(e^x + 2)*log(2)^2 - 20*x^
2*e^2 - 25*x*e^2 + x*e^(e^x + 2) - x)/(16*x^2*log(2)^2 + 80*x*log(2)^2 - 4*e^(e^x)*log(2)^2 - 4*x^2 + 100*log(
2)^2 - 20*x + e^(e^x) - 25)

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maple [A]  time = 0.66, size = 36, normalized size = 1.00

method	result	size
risch	\({\mathrm e}^{2} x -\frac {x}{\left (4 \ln \relax (2)^{2}-1\right ) \left (4 x^{2}+20 x -{\mathrm e}^{{\mathrm e}^{x}}+25\right )}\)	\(36\)
norman	\(\frac {\frac {\left (100 \,{\mathrm e}^{2} \ln \relax (2)^{2}-25 \,{\mathrm e}^{2}-1\right ) x}{4 \ln \relax (2)^{2}-1}+20 x^{2} {\mathrm e}^{2}+4 x^{3} {\mathrm e}^{2}-{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{x}} x}{4 x^{2}+20 x -{\mathrm e}^{{\mathrm e}^{x}}+25}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*exp(2)*ln(2)^2-exp(2))*exp(exp(x))^2+(-exp(x)*x+4*(-8*x^2-40*x-50)*exp(2)*ln(2)^2+(8*x^2+40*x+50)*exp(
2)+1)*exp(exp(x))+4*(16*x^4+160*x^3+600*x^2+1000*x+625)*exp(2)*ln(2)^2+(-16*x^4-160*x^3-600*x^2-1000*x-625)*ex
p(2)+4*x^2-25)/((4*ln(2)^2-1)*exp(exp(x))^2+(4*(-8*x^2-40*x-50)*ln(2)^2+8*x^2+40*x+50)*exp(exp(x))+4*(16*x^4+1
60*x^3+600*x^2+1000*x+625)*ln(2)^2-16*x^4-160*x^3-600*x^2-1000*x-625),x,method=_RETURNVERBOSE)

[Out]

exp(2)*x-x/(4*ln(2)^2-1)/(4*x^2+20*x-exp(exp(x))+25)

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maxima [B]  time = 0.57, size = 111, normalized size = 3.08 \begin {gather*} \frac {4 \, {\left (4 \, \log \relax (2)^{2} - 1\right )} x^{3} e^{2} + 20 \, {\left (4 \, \log \relax (2)^{2} - 1\right )} x^{2} e^{2} - {\left (4 \, \log \relax (2)^{2} - 1\right )} x e^{\left (e^{x} + 2\right )} + {\left (25 \, {\left (4 \, \log \relax (2)^{2} - 1\right )} e^{2} - 1\right )} x}{4 \, {\left (4 \, \log \relax (2)^{2} - 1\right )} x^{2} + 20 \, {\left (4 \, \log \relax (2)^{2} - 1\right )} x - {\left (4 \, \log \relax (2)^{2} - 1\right )} e^{\left (e^{x}\right )} + 100 \, \log \relax (2)^{2} - 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(2)*log(2)^2-exp(2))*exp(exp(x))^2+(-exp(x)*x+4*(-8*x^2-40*x-50)*exp(2)*log(2)^2+(8*x^2+40*x+
50)*exp(2)+1)*exp(exp(x))+4*(16*x^4+160*x^3+600*x^2+1000*x+625)*exp(2)*log(2)^2+(-16*x^4-160*x^3-600*x^2-1000*
x-625)*exp(2)+4*x^2-25)/((4*log(2)^2-1)*exp(exp(x))^2+(4*(-8*x^2-40*x-50)*log(2)^2+8*x^2+40*x+50)*exp(exp(x))+
4*(16*x^4+160*x^3+600*x^2+1000*x+625)*log(2)^2-16*x^4-160*x^3-600*x^2-1000*x-625),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int -\frac {{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left ({\mathrm {e}}^2-4\,{\mathrm {e}}^2\,{\ln \relax (2)}^2\right )+{\mathrm {e}}^2\,\left (16\,x^4+160\,x^3+600\,x^2+1000\,x+625\right )-4\,x^2-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left ({\mathrm {e}}^2\,\left (8\,x^2+40\,x+50\right )-x\,{\mathrm {e}}^x-4\,{\mathrm {e}}^2\,{\ln \relax (2)}^2\,\left (8\,x^2+40\,x+50\right )+1\right )-4\,{\mathrm {e}}^2\,{\ln \relax (2)}^2\,\left (16\,x^4+160\,x^3+600\,x^2+1000\,x+625\right )+25}{1000\,x-4\,{\ln \relax (2)}^2\,\left (16\,x^4+160\,x^3+600\,x^2+1000\,x+625\right )-{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left (4\,{\ln \relax (2)}^2-1\right )-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (40\,x-4\,{\ln \relax (2)}^2\,\left (8\,x^2+40\,x+50\right )+8\,x^2+50\right )+600\,x^2+160\,x^3+16\,x^4+625} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*exp(x))*(exp(2) - 4*exp(2)*log(2)^2) + exp(2)*(1000*x + 600*x^2 + 160*x^3 + 16*x^4 + 625) - 4*x^2 -
exp(exp(x))*(exp(2)*(40*x + 8*x^2 + 50) - x*exp(x) - 4*exp(2)*log(2)^2*(40*x + 8*x^2 + 50) + 1) - 4*exp(2)*lo
g(2)^2*(1000*x + 600*x^2 + 160*x^3 + 16*x^4 + 625) + 25)/(1000*x - 4*log(2)^2*(1000*x + 600*x^2 + 160*x^3 + 16
*x^4 + 625) - exp(2*exp(x))*(4*log(2)^2 - 1) - exp(exp(x))*(40*x - 4*log(2)^2*(40*x + 8*x^2 + 50) + 8*x^2 + 50
) + 600*x^2 + 160*x^3 + 16*x^4 + 625),x)

[Out]

-int(-(exp(2*exp(x))*(exp(2) - 4*exp(2)*log(2)^2) + exp(2)*(1000*x + 600*x^2 + 160*x^3 + 16*x^4 + 625) - 4*x^2
- exp(exp(x))*(exp(2)*(40*x + 8*x^2 + 50) - x*exp(x) - 4*exp(2)*log(2)^2*(40*x + 8*x^2 + 50) + 1) - 4*exp(2)*
log(2)^2*(1000*x + 600*x^2 + 160*x^3 + 16*x^4 + 625) + 25)/(1000*x - 4*log(2)^2*(1000*x + 600*x^2 + 160*x^3 +
16*x^4 + 625) - exp(2*exp(x))*(4*log(2)^2 - 1) - exp(exp(x))*(40*x - 4*log(2)^2*(40*x + 8*x^2 + 50) + 8*x^2 +
50) + 600*x^2 + 160*x^3 + 16*x^4 + 625), x)

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sympy [B]  time = 0.27, size = 54, normalized size = 1.50 \begin {gather*} x e^{2} + \frac {x}{- 16 x^{2} \log {\relax (2 )}^{2} + 4 x^{2} - 80 x \log {\relax (2 )}^{2} + 20 x + \left (-1 + 4 \log {\relax (2 )}^{2}\right ) e^{e^{x}} - 100 \log {\relax (2 )}^{2} + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(2)*ln(2)**2-exp(2))*exp(exp(x))**2+(-exp(x)*x+4*(-8*x**2-40*x-50)*exp(2)*ln(2)**2+(8*x**2+40
*x+50)*exp(2)+1)*exp(exp(x))+4*(16*x**4+160*x**3+600*x**2+1000*x+625)*exp(2)*ln(2)**2+(-16*x**4-160*x**3-600*x
**2-1000*x-625)*exp(2)+4*x**2-25)/((4*ln(2)**2-1)*exp(exp(x))**2+(4*(-8*x**2-40*x-50)*ln(2)**2+8*x**2+40*x+50)
*exp(exp(x))+4*(16*x**4+160*x**3+600*x**2+1000*x+625)*ln(2)**2-16*x**4-160*x**3-600*x**2-1000*x-625),x)

[Out]

x*exp(2) + x/(-16*x**2*log(2)**2 + 4*x**2 - 80*x*log(2)**2 + 20*x + (-1 + 4*log(2)**2)*exp(exp(x)) - 100*log(2
)**2 + 25)

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