3.26.61 \(\int \frac {-125 x+25 e^x x+e^{4 x} (50 x-10 e^x x)+e^{8 x} (-5 x+e^x x)+(-50 x+10 e^x x+e^{4 x} (10 x-2 e^x x)) \log (5-e^x)+(-5 x+e^x x) \log ^2(5-e^x)+x^{\frac {4}{5-e^{4 x}+\log (5-e^x)}} (-200+40 e^x+e^{4 x} (40-8 e^x)+(-40+8 e^x) \log (5-e^x)+(-8 e^x x+e^{4 x} (-160 x+32 e^x x)) \log (x))}{-125 x+25 e^x x+e^{4 x} (50 x-10 e^x x)+e^{8 x} (-5 x+e^x x)+(-50 x+10 e^x x+e^{4 x} (10 x-2 e^x x)) \log (5-e^x)+(-5 x+e^x x) \log ^2(5-e^x)} \, dx\)

Optimal. Leaf size=27 \[ x+2 x^{\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \]

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Rubi [F]  time = 26.39, 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 {-125 x+25 e^x x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (-5 x+e^x x\right )+\left (-50 x+10 e^x x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )+\left (-5 x+e^x x\right ) \log ^2\left (5-e^x\right )+x^{\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \left (-200+40 e^x+e^{4 x} \left (40-8 e^x\right )+\left (-40+8 e^x\right ) \log \left (5-e^x\right )+\left (-8 e^x x+e^{4 x} \left (-160 x+32 e^x x\right )\right ) \log (x)\right )}{-125 x+25 e^x x+e^{4 x} \left (50 x-10 e^x x\right )+e^{8 x} \left (-5 x+e^x x\right )+\left (-50 x+10 e^x x+e^{4 x} \left (10 x-2 e^x x\right )\right ) \log \left (5-e^x\right )+\left (-5 x+e^x x\right ) \log ^2\left (5-e^x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-125*x + 25*E^x*x + E^(4*x)*(50*x - 10*E^x*x) + E^(8*x)*(-5*x + E^x*x) + (-50*x + 10*E^x*x + E^(4*x)*(10*
x - 2*E^x*x))*Log[5 - E^x] + (-5*x + E^x*x)*Log[5 - E^x]^2 + x^(4/(5 - E^(4*x) + Log[5 - E^x]))*(-200 + 40*E^x
+ E^(4*x)*(40 - 8*E^x) + (-40 + 8*E^x)*Log[5 - E^x] + (-8*E^x*x + E^(4*x)*(-160*x + 32*E^x*x))*Log[x]))/(-125
*x + 25*E^x*x + E^(4*x)*(50*x - 10*E^x*x) + E^(8*x)*(-5*x + E^x*x) + (-50*x + 10*E^x*x + E^(4*x)*(10*x - 2*E^x
*x))*Log[5 - E^x] + (-5*x + E^x*x)*Log[5 - E^x]^2),x]

[Out]

x - 3075200*Defer[Int][x^(-1 + 4/(5 - E^(4*x) + Log[5 - E^x]))/((-5 + E^(4*x) - Log[5 - E^x])*(-620 + Log[5 -
E^x])^2), x] + 9920*Defer[Int][(x^(-1 + 4/(5 - E^(4*x) + Log[5 - E^x]))*Log[5 - E^x])/((-5 + E^(4*x) - Log[5 -
E^x])*(-620 + Log[5 - E^x])^2), x] - 8*Defer[Int][(x^(-1 + 4/(5 - E^(4*x) + Log[5 - E^x]))*Log[5 - E^x]^2)/((
-5 + E^(4*x) - Log[5 - E^x])*(-620 + Log[5 - E^x])^2), x] - 40*Defer[Int][(x^(4/(5 - E^(4*x) + Log[5 - E^x]))*
Log[x])/((-5 + E^x)*(-620 + Log[5 - E^x])^2), x] + 12305800*Defer[Int][(x^(4/(5 - E^(4*x) + Log[5 - E^x]))*Log
[x])/((-5 + E^(4*x) - Log[5 - E^x])*(-620 + Log[5 - E^x])^2), x] + 1000*Defer[Int][(E^x*x^(4/(5 - E^(4*x) + Lo
g[5 - E^x]))*Log[x])/((-5 + E^(4*x) - Log[5 - E^x])*(-620 + Log[5 - E^x])^2), x] + 200*Defer[Int][(E^(2*x)*x^(
4/(5 - E^(4*x) + Log[5 - E^x]))*Log[x])/((-5 + E^(4*x) - Log[5 - E^x])*(-620 + Log[5 - E^x])^2), x] + 40*Defer
[Int][(E^(3*x)*x^(4/(5 - E^(4*x) + Log[5 - E^x]))*Log[x])/((-5 + E^(4*x) - Log[5 - E^x])*(-620 + Log[5 - E^x])
^2), x] - 1000*Defer[Int][(E^x*x^(4/(5 - E^(4*x) + Log[5 - E^x]))*Log[x])/((-5 + E^(4*x) - Log[5 - E^x])^2*(-6
20 + Log[5 - E^x])), x] - 200*Defer[Int][(E^(2*x)*x^(4/(5 - E^(4*x) + Log[5 - E^x]))*Log[x])/((-5 + E^(4*x) -
Log[5 - E^x])^2*(-620 + Log[5 - E^x])), x] - 40*Defer[Int][(E^(3*x)*x^(4/(5 - E^(4*x) + Log[5 - E^x]))*Log[x])
/((-5 + E^(4*x) - Log[5 - E^x])^2*(-620 + Log[5 - E^x])), x] - 39680*Defer[Int][(x^(4/(5 - E^(4*x) + Log[5 - E
^x]))*Log[5 - E^x]*Log[x])/((-5 + E^(4*x) - Log[5 - E^x])*(-620 + Log[5 - E^x])^2), x] + 32*Defer[Int][(x^(4/(
5 - E^(4*x) + Log[5 - E^x]))*Log[5 - E^x]^2*Log[x])/((-5 + E^(4*x) - Log[5 - E^x])*(-620 + Log[5 - E^x])^2), x
] - 99240*Defer[Int][(x^(4/(5 - E^(4*x) + Log[5 - E^x]))*Log[x])/((-620 + Log[5 - E^x])*(5 - E^(4*x) + Log[5 -
E^x])^2), x] - 19688*Defer[Int][(x^(4/(5 - E^(4*x) + Log[5 - E^x]))*Log[5 - E^x]*Log[x])/((-620 + Log[5 - E^x
])*(5 - E^(4*x) + Log[5 - E^x])^2), x] + 32*Defer[Int][(x^(4/(5 - E^(4*x) + Log[5 - E^x]))*Log[5 - E^x]^2*Log[
x])/((-620 + Log[5 - E^x])*(5 - E^(4*x) + Log[5 - E^x])^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-\left (\left (25-5 e^x-5 e^{4 x}+e^{5 x}\right ) \left (\left (-5+e^{4 x}\right ) x-8 x^{\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}}\right )\right )-2 \left (-5+e^x\right ) \left (-\left (\left (-5+e^{4 x}\right ) x\right )+4 x^{\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}}\right ) \log \left (5-e^x\right )-\left (-5+e^x\right ) x \log ^2\left (5-e^x\right )-8 e^x \left (-1-20 e^{3 x}+4 e^{4 x}\right ) x^{1+\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \log (x)}{\left (5-e^x\right ) x \left (5-e^{4 x}+\log \left (5-e^x\right )\right )^2} \, dx\\ &=\int \left (1+\frac {8 x^{-1+\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \left (-25+5 e^x+5 e^{4 x}-e^{5 x}-5 \log \left (5-e^x\right )+e^x \log \left (5-e^x\right )-e^x x \log (x)-20 e^{4 x} x \log (x)+4 e^{5 x} x \log (x)\right )}{\left (-5+e^x\right ) \left (-5+e^{4 x}-\log \left (5-e^x\right )\right )^2}\right ) \, dx\\ &=x+8 \int \frac {x^{-1+\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \left (-25+5 e^x+5 e^{4 x}-e^{5 x}-5 \log \left (5-e^x\right )+e^x \log \left (5-e^x\right )-e^x x \log (x)-20 e^{4 x} x \log (x)+4 e^{5 x} x \log (x)\right )}{\left (-5+e^x\right ) \left (-5+e^{4 x}-\log \left (5-e^x\right )\right )^2} \, dx\\ &=x+8 \int \left (-\frac {5 x^{\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \log (x)}{\left (-5+e^x\right ) \left (-620+\log \left (5-e^x\right )\right )^2}+\frac {x^{\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \left (-12405-125 e^x-25 e^{2 x}-5 e^{3 x}-2461 \log \left (5-e^x\right )+4 \log ^2\left (5-e^x\right )\right ) \log (x)}{\left (-620+\log \left (5-e^x\right )\right ) \left (5-e^{4 x}+\log \left (5-e^x\right )\right )^2}+\frac {x^{-1+\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \left (-384400+1240 \log \left (5-e^x\right )-\log ^2\left (5-e^x\right )+1538225 x \log (x)+125 e^x x \log (x)+25 e^{2 x} x \log (x)+5 e^{3 x} x \log (x)-4960 x \log \left (5-e^x\right ) \log (x)+4 x \log ^2\left (5-e^x\right ) \log (x)\right )}{\left (-5+e^{4 x}-\log \left (5-e^x\right )\right ) \left (-620+\log \left (5-e^x\right )\right )^2}\right ) \, dx\\ &=x+8 \int \frac {x^{\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \left (-12405-125 e^x-25 e^{2 x}-5 e^{3 x}-2461 \log \left (5-e^x\right )+4 \log ^2\left (5-e^x\right )\right ) \log (x)}{\left (-620+\log \left (5-e^x\right )\right ) \left (5-e^{4 x}+\log \left (5-e^x\right )\right )^2} \, dx+8 \int \frac {x^{-1+\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \left (-384400+1240 \log \left (5-e^x\right )-\log ^2\left (5-e^x\right )+1538225 x \log (x)+125 e^x x \log (x)+25 e^{2 x} x \log (x)+5 e^{3 x} x \log (x)-4960 x \log \left (5-e^x\right ) \log (x)+4 x \log ^2\left (5-e^x\right ) \log (x)\right )}{\left (-5+e^{4 x}-\log \left (5-e^x\right )\right ) \left (-620+\log \left (5-e^x\right )\right )^2} \, dx-40 \int \frac {x^{\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \log (x)}{\left (-5+e^x\right ) \left (-620+\log \left (5-e^x\right )\right )^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.47, size = 27, normalized size = 1.00 \begin {gather*} x+2 x^{\frac {4}{5-e^{4 x}+\log \left (5-e^x\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-125*x + 25*E^x*x + E^(4*x)*(50*x - 10*E^x*x) + E^(8*x)*(-5*x + E^x*x) + (-50*x + 10*E^x*x + E^(4*x
)*(10*x - 2*E^x*x))*Log[5 - E^x] + (-5*x + E^x*x)*Log[5 - E^x]^2 + x^(4/(5 - E^(4*x) + Log[5 - E^x]))*(-200 +
40*E^x + E^(4*x)*(40 - 8*E^x) + (-40 + 8*E^x)*Log[5 - E^x] + (-8*E^x*x + E^(4*x)*(-160*x + 32*E^x*x))*Log[x]))
/(-125*x + 25*E^x*x + E^(4*x)*(50*x - 10*E^x*x) + E^(8*x)*(-5*x + E^x*x) + (-50*x + 10*E^x*x + E^(4*x)*(10*x -
2*E^x*x))*Log[5 - E^x] + (-5*x + E^x*x)*Log[5 - E^x]^2),x]

[Out]

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

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fricas [A]  time = 0.69, size = 49, normalized size = 1.81 \begin {gather*} \frac {x x^{\frac {4}{e^{\left (4 \, x\right )} - \log \left (-e^{x} + 5\right ) - 5}} + 2}{x^{\frac {4}{e^{\left (4 \, x\right )} - \log \left (-e^{x} + 5\right ) - 5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*exp(x)-40)*log(5-exp(x))+((32*exp(x)*x-160*x)*exp(4*x)-8*exp(x)*x)*log(x)+(-8*exp(x)+40)*exp(4*
x)+40*exp(x)-200)*exp(4*log(x)/(log(5-exp(x))-exp(4*x)+5))+(exp(x)*x-5*x)*log(5-exp(x))^2+((-2*exp(x)*x+10*x)*
exp(4*x)+10*exp(x)*x-50*x)*log(5-exp(x))+(exp(x)*x-5*x)*exp(4*x)^2+(-10*exp(x)*x+50*x)*exp(4*x)+25*exp(x)*x-12
5*x)/((exp(x)*x-5*x)*log(5-exp(x))^2+((-2*exp(x)*x+10*x)*exp(4*x)+10*exp(x)*x-50*x)*log(5-exp(x))+(exp(x)*x-5*
x)*exp(4*x)^2+(-10*exp(x)*x+50*x)*exp(4*x)+25*exp(x)*x-125*x),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x e^{x} - 5 \, x\right )} \log \left (-e^{x} + 5\right )^{2} + {\left (x e^{x} - 5 \, x\right )} e^{\left (8 \, x\right )} - 10 \, {\left (x e^{x} - 5 \, x\right )} e^{\left (4 \, x\right )} + 25 \, x e^{x} - 2 \, {\left ({\left (x e^{x} - 5 \, x\right )} e^{\left (4 \, x\right )} - 5 \, x e^{x} + 25 \, x\right )} \log \left (-e^{x} + 5\right ) - 125 \, x - \frac {8 \, {\left ({\left (e^{x} - 5\right )} e^{\left (4 \, x\right )} - {\left (4 \, {\left (x e^{x} - 5 \, x\right )} e^{\left (4 \, x\right )} - x e^{x}\right )} \log \relax (x) - {\left (e^{x} - 5\right )} \log \left (-e^{x} + 5\right ) - 5 \, e^{x} + 25\right )}}{x^{\frac {4}{e^{\left (4 \, x\right )} - \log \left (-e^{x} + 5\right ) - 5}}}}{{\left (x e^{x} - 5 \, x\right )} \log \left (-e^{x} + 5\right )^{2} + {\left (x e^{x} - 5 \, x\right )} e^{\left (8 \, x\right )} - 10 \, {\left (x e^{x} - 5 \, x\right )} e^{\left (4 \, x\right )} + 25 \, x e^{x} - 2 \, {\left ({\left (x e^{x} - 5 \, x\right )} e^{\left (4 \, x\right )} - 5 \, x e^{x} + 25 \, x\right )} \log \left (-e^{x} + 5\right ) - 125 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*exp(x)-40)*log(5-exp(x))+((32*exp(x)*x-160*x)*exp(4*x)-8*exp(x)*x)*log(x)+(-8*exp(x)+40)*exp(4*
x)+40*exp(x)-200)*exp(4*log(x)/(log(5-exp(x))-exp(4*x)+5))+(exp(x)*x-5*x)*log(5-exp(x))^2+((-2*exp(x)*x+10*x)*
exp(4*x)+10*exp(x)*x-50*x)*log(5-exp(x))+(exp(x)*x-5*x)*exp(4*x)^2+(-10*exp(x)*x+50*x)*exp(4*x)+25*exp(x)*x-12
5*x)/((exp(x)*x-5*x)*log(5-exp(x))^2+((-2*exp(x)*x+10*x)*exp(4*x)+10*exp(x)*x-50*x)*log(5-exp(x))+(exp(x)*x-5*
x)*exp(4*x)^2+(-10*exp(x)*x+50*x)*exp(4*x)+25*exp(x)*x-125*x),x, algorithm="giac")

[Out]

integrate(((x*e^x - 5*x)*log(-e^x + 5)^2 + (x*e^x - 5*x)*e^(8*x) - 10*(x*e^x - 5*x)*e^(4*x) + 25*x*e^x - 2*((x
*e^x - 5*x)*e^(4*x) - 5*x*e^x + 25*x)*log(-e^x + 5) - 125*x - 8*((e^x - 5)*e^(4*x) - (4*(x*e^x - 5*x)*e^(4*x)
- x*e^x)*log(x) - (e^x - 5)*log(-e^x + 5) - 5*e^x + 25)/x^(4/(e^(4*x) - log(-e^x + 5) - 5)))/((x*e^x - 5*x)*lo
g(-e^x + 5)^2 + (x*e^x - 5*x)*e^(8*x) - 10*(x*e^x - 5*x)*e^(4*x) + 25*x*e^x - 2*((x*e^x - 5*x)*e^(4*x) - 5*x*e
^x + 25*x)*log(-e^x + 5) - 125*x), x)

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maple [A]  time = 0.07, size = 26, normalized size = 0.96

method	result	size
risch	\(x +2 x^{\frac {4}{\ln \left (5-{\mathrm e}^{x}\right )-{\mathrm e}^{4 x}+5}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*exp(x)-40)*ln(5-exp(x))+((32*exp(x)*x-160*x)*exp(4*x)-8*exp(x)*x)*ln(x)+(-8*exp(x)+40)*exp(4*x)+40*ex
p(x)-200)*exp(4*ln(x)/(ln(5-exp(x))-exp(4*x)+5))+(exp(x)*x-5*x)*ln(5-exp(x))^2+((-2*exp(x)*x+10*x)*exp(4*x)+10
*exp(x)*x-50*x)*ln(5-exp(x))+(exp(x)*x-5*x)*exp(4*x)^2+(-10*exp(x)*x+50*x)*exp(4*x)+25*exp(x)*x-125*x)/((exp(x
)*x-5*x)*ln(5-exp(x))^2+((-2*exp(x)*x+10*x)*exp(4*x)+10*exp(x)*x-50*x)*ln(5-exp(x))+(exp(x)*x-5*x)*exp(4*x)^2+
(-10*exp(x)*x+50*x)*exp(4*x)+25*exp(x)*x-125*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} x + \int \frac {8 \, {\left ({\left (4 \, x \log \relax (x) - 1\right )} e^{\left (5 \, x\right )} - 5 \, {\left (4 \, x \log \relax (x) - 1\right )} e^{\left (4 \, x\right )} - {\left (x \log \relax (x) - 5\right )} e^{x} + {\left (e^{x} - 5\right )} \log \left (-e^{x} + 5\right ) - 25\right )}}{{\left ({\left (x e^{x} - 5 \, x\right )} \log \left (-e^{x} + 5\right )^{2} + x e^{\left (9 \, x\right )} - 5 \, x e^{\left (8 \, x\right )} - 10 \, x e^{\left (5 \, x\right )} + 50 \, x e^{\left (4 \, x\right )} + 25 \, x e^{x} - 2 \, {\left (x e^{\left (5 \, x\right )} - 5 \, x e^{\left (4 \, x\right )} - 5 \, x e^{x} + 25 \, x\right )} \log \left (-e^{x} + 5\right ) - 125 \, x\right )} x^{\frac {4}{e^{\left (4 \, x\right )} - \log \left (-e^{x} + 5\right ) - 5}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*exp(x)-40)*log(5-exp(x))+((32*exp(x)*x-160*x)*exp(4*x)-8*exp(x)*x)*log(x)+(-8*exp(x)+40)*exp(4*
x)+40*exp(x)-200)*exp(4*log(x)/(log(5-exp(x))-exp(4*x)+5))+(exp(x)*x-5*x)*log(5-exp(x))^2+((-2*exp(x)*x+10*x)*
exp(4*x)+10*exp(x)*x-50*x)*log(5-exp(x))+(exp(x)*x-5*x)*exp(4*x)^2+(-10*exp(x)*x+50*x)*exp(4*x)+25*exp(x)*x-12
5*x)/((exp(x)*x-5*x)*log(5-exp(x))^2+((-2*exp(x)*x+10*x)*exp(4*x)+10*exp(x)*x-50*x)*log(5-exp(x))+(exp(x)*x-5*
x)*exp(4*x)^2+(-10*exp(x)*x+50*x)*exp(4*x)+25*exp(x)*x-125*x),x, algorithm="maxima")

[Out]

x + integrate(8*((4*x*log(x) - 1)*e^(5*x) - 5*(4*x*log(x) - 1)*e^(4*x) - (x*log(x) - 5)*e^x + (e^x - 5)*log(-e
^x + 5) - 25)/(((x*e^x - 5*x)*log(-e^x + 5)^2 + x*e^(9*x) - 5*x*e^(8*x) - 10*x*e^(5*x) + 50*x*e^(4*x) + 25*x*e
^x - 2*(x*e^(5*x) - 5*x*e^(4*x) - 5*x*e^x + 25*x)*log(-e^x + 5) - 125*x)*x^(4/(e^(4*x) - log(-e^x + 5) - 5))),
x)

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mupad [B]  time = 1.82, size = 25, normalized size = 0.93 \begin {gather*} x+2\,x^{\frac {4}{\ln \left (5-{\mathrm {e}}^x\right )-{\mathrm {e}}^{4\,x}+5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((125*x + exp((4*log(x))/(log(5 - exp(x)) - exp(4*x) + 5))*(exp(4*x)*(8*exp(x) - 40) - log(5 - exp(x))*(8*e
xp(x) - 40) - 40*exp(x) + log(x)*(exp(4*x)*(160*x - 32*x*exp(x)) + 8*x*exp(x)) + 200) + exp(8*x)*(5*x - x*exp(
x)) - exp(4*x)*(50*x - 10*x*exp(x)) + log(5 - exp(x))^2*(5*x - x*exp(x)) - 25*x*exp(x) - log(5 - exp(x))*(exp(
4*x)*(10*x - 2*x*exp(x)) - 50*x + 10*x*exp(x)))/(125*x + exp(8*x)*(5*x - x*exp(x)) - exp(4*x)*(50*x - 10*x*exp
(x)) + log(5 - exp(x))^2*(5*x - x*exp(x)) - 25*x*exp(x) - log(5 - exp(x))*(exp(4*x)*(10*x - 2*x*exp(x)) - 50*x
+ 10*x*exp(x))),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*exp(x)-40)*ln(5-exp(x))+((32*exp(x)*x-160*x)*exp(4*x)-8*exp(x)*x)*ln(x)+(-8*exp(x)+40)*exp(4*x)
+40*exp(x)-200)*exp(4*ln(x)/(ln(5-exp(x))-exp(4*x)+5))+(exp(x)*x-5*x)*ln(5-exp(x))**2+((-2*exp(x)*x+10*x)*exp(
4*x)+10*exp(x)*x-50*x)*ln(5-exp(x))+(exp(x)*x-5*x)*exp(4*x)**2+(-10*exp(x)*x+50*x)*exp(4*x)+25*exp(x)*x-125*x)
/((exp(x)*x-5*x)*ln(5-exp(x))**2+((-2*exp(x)*x+10*x)*exp(4*x)+10*exp(x)*x-50*x)*ln(5-exp(x))+(exp(x)*x-5*x)*ex
p(4*x)**2+(-10*exp(x)*x+50*x)*exp(4*x)+25*exp(x)*x-125*x),x)

[Out]

Timed out

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