3.37.84 \(\int \frac {e^x (10+e+2 x)+(40+5 e+e^x (-8-e-2 x)+10 x) \log (5-e^x) \log (\log (5-e^x))}{(-10 e^x+2 e^{2 x}) \log (5-e^x)} \, dx\)

Optimal. Leaf size=23 \[ e^{-x} \left (5+\frac {e}{2}+x\right ) \log \left (\log \left (5-e^x\right )\right ) \]

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Rubi [F]  time = 1.44, 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 {e^x (10+e+2 x)+\left (40+5 e+e^x (-8-e-2 x)+10 x\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )}{\left (-10 e^x+2 e^{2 x}\right ) \log \left (5-e^x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(10 + E + 2*x) + (40 + 5*E + E^x*(-8 - E - 2*x) + 10*x)*Log[5 - E^x]*Log[Log[5 - E^x]])/((-10*E^x + 2
*E^(2*x))*Log[5 - E^x]),x]

[Out]

((10 + E)*Log[Log[5 - E^x]])/10 + Defer[Int][x/((-5 + E^x)*Log[5 - E^x]), x] - ((8 + E)*Defer[Int][Log[Log[5 -
E^x]]/E^x, x])/2 - Defer[Int][(x*Log[Log[5 - E^x]])/E^x, x] - ((10 + E)*Defer[Subst][Defer[Int][1/(x*Log[5 -
x]), x], x, E^x])/10

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x} \left (e^x (10+e+2 x)+\left (40+5 e+e^x (-8-e-2 x)+10 x\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )\right )}{\left (-10+2 e^x\right ) \log \left (5-e^x\right )} \, dx\\ &=\int \frac {1}{2} \left (\frac {e+2 (5+x)}{\left (-5+e^x\right ) \log \left (5-e^x\right )}-e^{-x} (8+e+2 x) \log \left (\log \left (5-e^x\right )\right )\right ) \, dx\\ &=\frac {1}{2} \int \left (\frac {e+2 (5+x)}{\left (-5+e^x\right ) \log \left (5-e^x\right )}-e^{-x} (8+e+2 x) \log \left (\log \left (5-e^x\right )\right )\right ) \, dx\\ &=\frac {1}{2} \int \frac {e+2 (5+x)}{\left (-5+e^x\right ) \log \left (5-e^x\right )} \, dx-\frac {1}{2} \int e^{-x} (8+e+2 x) \log \left (\log \left (5-e^x\right )\right ) \, dx\\ &=\frac {1}{2} \int \frac {-10-e-2 x}{\left (5-e^x\right ) \log \left (5-e^x\right )} \, dx-\frac {1}{2} \int \left (8 \left (1+\frac {e}{8}\right ) e^{-x} \log \left (\log \left (5-e^x\right )\right )+2 e^{-x} x \log \left (\log \left (5-e^x\right )\right )\right ) \, dx\\ &=\frac {1}{2} \int \left (\frac {10 \left (1+\frac {e}{10}\right )}{\left (-5+e^x\right ) \log \left (5-e^x\right )}+\frac {2 x}{\left (-5+e^x\right ) \log \left (5-e^x\right )}\right ) \, dx-\frac {1}{2} (8+e) \int e^{-x} \log \left (\log \left (5-e^x\right )\right ) \, dx-\int e^{-x} x \log \left (\log \left (5-e^x\right )\right ) \, dx\\ &=-\left (\frac {1}{2} (8+e) \int e^{-x} \log \left (\log \left (5-e^x\right )\right ) \, dx\right )+\frac {1}{2} (10+e) \int \frac {1}{\left (-5+e^x\right ) \log \left (5-e^x\right )} \, dx+\int \frac {x}{\left (-5+e^x\right ) \log \left (5-e^x\right )} \, dx-\int e^{-x} x \log \left (\log \left (5-e^x\right )\right ) \, dx\\ &=-\left (\frac {1}{2} (8+e) \int e^{-x} \log \left (\log \left (5-e^x\right )\right ) \, dx\right )+\frac {1}{2} (10+e) \operatorname {Subst}\left (\int \frac {1}{(-5+x) x \log (5-x)} \, dx,x,e^x\right )+\int \frac {x}{\left (-5+e^x\right ) \log \left (5-e^x\right )} \, dx-\int e^{-x} x \log \left (\log \left (5-e^x\right )\right ) \, dx\\ &=-\left (\frac {1}{2} (8+e) \int e^{-x} \log \left (\log \left (5-e^x\right )\right ) \, dx\right )+\frac {1}{2} (10+e) \operatorname {Subst}\left (\int \left (\frac {1}{5 (-5+x) \log (5-x)}-\frac {1}{5 x \log (5-x)}\right ) \, dx,x,e^x\right )+\int \frac {x}{\left (-5+e^x\right ) \log \left (5-e^x\right )} \, dx-\int e^{-x} x \log \left (\log \left (5-e^x\right )\right ) \, dx\\ &=\frac {1}{10} (-10-e) \operatorname {Subst}\left (\int \frac {1}{x \log (5-x)} \, dx,x,e^x\right )-\frac {1}{2} (8+e) \int e^{-x} \log \left (\log \left (5-e^x\right )\right ) \, dx+\frac {1}{10} (10+e) \operatorname {Subst}\left (\int \frac {1}{(-5+x) \log (5-x)} \, dx,x,e^x\right )+\int \frac {x}{\left (-5+e^x\right ) \log \left (5-e^x\right )} \, dx-\int e^{-x} x \log \left (\log \left (5-e^x\right )\right ) \, dx\\ &=\frac {1}{10} (-10-e) \operatorname {Subst}\left (\int \frac {1}{x \log (5-x)} \, dx,x,e^x\right )-\frac {1}{2} (8+e) \int e^{-x} \log \left (\log \left (5-e^x\right )\right ) \, dx+\frac {1}{10} (10+e) \operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,5-e^x\right )+\int \frac {x}{\left (-5+e^x\right ) \log \left (5-e^x\right )} \, dx-\int e^{-x} x \log \left (\log \left (5-e^x\right )\right ) \, dx\\ &=\frac {1}{10} (-10-e) \operatorname {Subst}\left (\int \frac {1}{x \log (5-x)} \, dx,x,e^x\right )-\frac {1}{2} (8+e) \int e^{-x} \log \left (\log \left (5-e^x\right )\right ) \, dx+\frac {1}{10} (10+e) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (5-e^x\right )\right )+\int \frac {x}{\left (-5+e^x\right ) \log \left (5-e^x\right )} \, dx-\int e^{-x} x \log \left (\log \left (5-e^x\right )\right ) \, dx\\ &=\frac {1}{10} (10+e) \log \left (\log \left (5-e^x\right )\right )+\frac {1}{10} (-10-e) \operatorname {Subst}\left (\int \frac {1}{x \log (5-x)} \, dx,x,e^x\right )-\frac {1}{2} (8+e) \int e^{-x} \log \left (\log \left (5-e^x\right )\right ) \, dx+\int \frac {x}{\left (-5+e^x\right ) \log \left (5-e^x\right )} \, dx-\int e^{-x} x \log \left (\log \left (5-e^x\right )\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 24, normalized size = 1.04 \begin {gather*} \frac {1}{2} e^{-x} (10+e+2 x) \log \left (\log \left (5-e^x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(10 + E + 2*x) + (40 + 5*E + E^x*(-8 - E - 2*x) + 10*x)*Log[5 - E^x]*Log[Log[5 - E^x]])/((-10*E
^x + 2*E^(2*x))*Log[5 - E^x]),x]

[Out]

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

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fricas [A]  time = 0.63, size = 21, normalized size = 0.91 \begin {gather*} \frac {1}{2} \, {\left (2 \, x + e + 10\right )} e^{\left (-x\right )} \log \left (\log \left (-e^{x} + 5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-exp(1)-2*x-8)*exp(x)+5*exp(1)+10*x+40)*log(5-exp(x))*log(log(5-exp(x)))+(exp(1)+2*x+10)*exp(x))/
(2*exp(x)^2-10*exp(x))/log(5-exp(x)),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left ({\left (2 \, x + e + 8\right )} e^{x} - 10 \, x - 5 \, e - 40\right )} \log \left (-e^{x} + 5\right ) \log \left (\log \left (-e^{x} + 5\right )\right ) - {\left (2 \, x + e + 10\right )} e^{x}}{2 \, {\left (e^{\left (2 \, x\right )} - 5 \, e^{x}\right )} \log \left (-e^{x} + 5\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-exp(1)-2*x-8)*exp(x)+5*exp(1)+10*x+40)*log(5-exp(x))*log(log(5-exp(x)))+(exp(1)+2*x+10)*exp(x))/
(2*exp(x)^2-10*exp(x))/log(5-exp(x)),x, algorithm="giac")

[Out]

integrate(-1/2*(((2*x + e + 8)*e^x - 10*x - 5*e - 40)*log(-e^x + 5)*log(log(-e^x + 5)) - (2*x + e + 10)*e^x)/(
(e^(2*x) - 5*e^x)*log(-e^x + 5)), x)

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maple [A]  time = 0.04, size = 22, normalized size = 0.96

method	result	size
risch	\(\frac {\left ({\mathrm e}+2 x +10\right ) {\mathrm e}^{-x} \ln \left (\ln \left (5-{\mathrm e}^{x}\right )\right )}{2}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-exp(1)-2*x-8)*exp(x)+5*exp(1)+10*x+40)*ln(5-exp(x))*ln(ln(5-exp(x)))+(exp(1)+2*x+10)*exp(x))/(2*exp(x)
^2-10*exp(x))/ln(5-exp(x)),x,method=_RETURNVERBOSE)

[Out]

1/2*(exp(1)+2*x+10)*exp(-x)*ln(ln(5-exp(x)))

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maxima [A]  time = 0.43, size = 21, normalized size = 0.91 \begin {gather*} \frac {1}{2} \, {\left (2 \, x + e + 10\right )} e^{\left (-x\right )} \log \left (\log \left (-e^{x} + 5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-exp(1)-2*x-8)*exp(x)+5*exp(1)+10*x+40)*log(5-exp(x))*log(log(5-exp(x)))+(exp(1)+2*x+10)*exp(x))/
(2*exp(x)^2-10*exp(x))/log(5-exp(x)),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^x\,\left (2\,x+\mathrm {e}+10\right )+\ln \left (5-{\mathrm {e}}^x\right )\,\ln \left (\ln \left (5-{\mathrm {e}}^x\right )\right )\,\left (10\,x+5\,\mathrm {e}-{\mathrm {e}}^x\,\left (2\,x+\mathrm {e}+8\right )+40\right )}{\ln \left (5-{\mathrm {e}}^x\right )\,\left (2\,{\mathrm {e}}^{2\,x}-10\,{\mathrm {e}}^x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(2*x + exp(1) + 10) + log(5 - exp(x))*log(log(5 - exp(x)))*(10*x + 5*exp(1) - exp(x)*(2*x + exp(1)
+ 8) + 40))/(log(5 - exp(x))*(2*exp(2*x) - 10*exp(x))),x)

[Out]

int((exp(x)*(2*x + exp(1) + 10) + log(5 - exp(x))*log(log(5 - exp(x)))*(10*x + 5*exp(1) - exp(x)*(2*x + exp(1)
+ 8) + 40))/(log(5 - exp(x))*(2*exp(2*x) - 10*exp(x))), x)

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sympy [A]  time = 0.55, size = 20, normalized size = 0.87 \begin {gather*} \frac {\left (2 x + e + 10\right ) e^{- x} \log {\left (\log {\left (5 - e^{x} \right )} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-exp(1)-2*x-8)*exp(x)+5*exp(1)+10*x+40)*ln(5-exp(x))*ln(ln(5-exp(x)))+(exp(1)+2*x+10)*exp(x))/(2*
exp(x)**2-10*exp(x))/ln(5-exp(x)),x)

[Out]

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

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