3.7.11 \(\int \frac {40+e^{-5+x} (-20-10 x)+20 x+(-40-40 x+e^{-5+x} (20+20 x+10 x^2+5 x^3)) \log (x)+\log (2-e^{-5+x}) (-20 x-10 x^2+e^{-5+x} (10 x+5 x^2)+(10 x^2-5 e^{-5+x} x^2) \log (x))}{(-8+4 e^{-5+x}) \log ^2(x)+(8 x-4 e^{-5+x} x) \log (2-e^{-5+x}) \log ^2(x)+(-2 x^2+e^{-5+x} x^2) \log ^2(2-e^{-5+x}) \log ^2(x)} \, dx\)

Optimal. Leaf size=31 \[ 4+\frac {5 (2+x)}{\left (\frac {2}{x}-\log \left (2-e^{-5+x}\right )\right ) \log (x)} \]

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Rubi [F]  time = 6.36, 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 {40+e^{-5+x} (-20-10 x)+20 x+\left (-40-40 x+e^{-5+x} \left (20+20 x+10 x^2+5 x^3\right )\right ) \log (x)+\log \left (2-e^{-5+x}\right ) \left (-20 x-10 x^2+e^{-5+x} \left (10 x+5 x^2\right )+\left (10 x^2-5 e^{-5+x} x^2\right ) \log (x)\right )}{\left (-8+4 e^{-5+x}\right ) \log ^2(x)+\left (8 x-4 e^{-5+x} x\right ) \log \left (2-e^{-5+x}\right ) \log ^2(x)+\left (-2 x^2+e^{-5+x} x^2\right ) \log ^2\left (2-e^{-5+x}\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(40 + E^(-5 + x)*(-20 - 10*x) + 20*x + (-40 - 40*x + E^(-5 + x)*(20 + 20*x + 10*x^2 + 5*x^3))*Log[x] + Log
[2 - E^(-5 + x)]*(-20*x - 10*x^2 + E^(-5 + x)*(10*x + 5*x^2) + (10*x^2 - 5*E^(-5 + x)*x^2)*Log[x]))/((-8 + 4*E
^(-5 + x))*Log[x]^2 + (8*x - 4*E^(-5 + x)*x)*Log[2 - E^(-5 + x)]*Log[x]^2 + (-2*x^2 + E^(-5 + x)*x^2)*Log[2 -
E^(-5 + x)]^2*Log[x]^2),x]

[Out]

10*Defer[Int][1/((-2 + x*Log[2 - E^(-5 + x)])*Log[x]^2), x] + 5*Defer[Int][x/((-2 + x*Log[2 - E^(-5 + x)])*Log
[x]^2), x] + 20*Defer[Int][1/((-2 + x*Log[2 - E^(-5 + x)])^2*Log[x]), x] + 20*Defer[Int][x/((-2 + x*Log[2 - E^
(-5 + x)])^2*Log[x]), x] + 10*Defer[Int][x^2/((-2 + x*Log[2 - E^(-5 + x)])^2*Log[x]), x] + 20*E^5*Defer[Int][x
^2/((-2*E^5 + E^x)*(-2 + x*Log[2 - E^(-5 + x)])^2*Log[x]), x] + 5*Defer[Int][x^3/((-2 + x*Log[2 - E^(-5 + x)])
^2*Log[x]), x] + 10*E^5*Defer[Int][x^3/((-2*E^5 + E^x)*(-2 + x*Log[2 - E^(-5 + x)])^2*Log[x]), x] - 5*Defer[In
t][(x^2*Log[2 - E^(-5 + x)])/((-2 + x*Log[2 - E^(-5 + x)])^2*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 \left (2 \left (-2 e^5+e^x\right ) (2+x)-\left (-8 e^5 (1+x)+e^x \left (4+4 x+2 x^2+x^3\right )\right ) \log (x)+\left (-2 e^5+e^x\right ) x \log \left (2-e^{-5+x}\right ) (-2-x+x \log (x))\right )}{\left (2 e^5-e^x\right ) \left (2-x \log \left (2-e^{-5+x}\right )\right )^2 \log ^2(x)} \, dx\\ &=5 \int \frac {2 \left (-2 e^5+e^x\right ) (2+x)-\left (-8 e^5 (1+x)+e^x \left (4+4 x+2 x^2+x^3\right )\right ) \log (x)+\left (-2 e^5+e^x\right ) x \log \left (2-e^{-5+x}\right ) (-2-x+x \log (x))}{\left (2 e^5-e^x\right ) \left (2-x \log \left (2-e^{-5+x}\right )\right )^2 \log ^2(x)} \, dx\\ &=5 \int \left (-\frac {2 e^5 x^2 (2+x)}{\left (2 e^5-e^x\right ) \left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)}+\frac {-4-2 x+2 x \log \left (2-e^{-5+x}\right )+x^2 \log \left (2-e^{-5+x}\right )+4 \log (x)+4 x \log (x)+2 x^2 \log (x)+x^3 \log (x)-x^2 \log \left (2-e^{-5+x}\right ) \log (x)}{\left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log ^2(x)}\right ) \, dx\\ &=5 \int \frac {-4-2 x+2 x \log \left (2-e^{-5+x}\right )+x^2 \log \left (2-e^{-5+x}\right )+4 \log (x)+4 x \log (x)+2 x^2 \log (x)+x^3 \log (x)-x^2 \log \left (2-e^{-5+x}\right ) \log (x)}{\left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log ^2(x)} \, dx-\left (10 e^5\right ) \int \frac {x^2 (2+x)}{\left (2 e^5-e^x\right ) \left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)} \, dx\\ &=5 \int \frac {-2 (2+x)+\left (4+4 x+2 x^2+x^3\right ) \log (x)+x \log \left (2-e^{-5+x}\right ) (2+x-x \log (x))}{\left (2-x \log \left (2-e^{-5+x}\right )\right )^2 \log ^2(x)} \, dx-\left (10 e^5\right ) \int \left (-\frac {2 x^2}{\left (-2 e^5+e^x\right ) \left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)}-\frac {x^3}{\left (-2 e^5+e^x\right ) \left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)}\right ) \, dx\\ &=5 \int \left (\frac {2+x}{\left (-2+x \log \left (2-e^{-5+x}\right )\right ) \log ^2(x)}+\frac {4+4 x+2 x^2+x^3-x^2 \log \left (2-e^{-5+x}\right )}{\left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)}\right ) \, dx+\left (10 e^5\right ) \int \frac {x^3}{\left (-2 e^5+e^x\right ) \left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)} \, dx+\left (20 e^5\right ) \int \frac {x^2}{\left (-2 e^5+e^x\right ) \left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)} \, dx\\ &=5 \int \frac {2+x}{\left (-2+x \log \left (2-e^{-5+x}\right )\right ) \log ^2(x)} \, dx+5 \int \frac {4+4 x+2 x^2+x^3-x^2 \log \left (2-e^{-5+x}\right )}{\left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)} \, dx+\left (10 e^5\right ) \int \frac {x^3}{\left (-2 e^5+e^x\right ) \left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)} \, dx+\left (20 e^5\right ) \int \frac {x^2}{\left (-2 e^5+e^x\right ) \left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)} \, dx\\ &=5 \int \left (\frac {2}{\left (-2+x \log \left (2-e^{-5+x}\right )\right ) \log ^2(x)}+\frac {x}{\left (-2+x \log \left (2-e^{-5+x}\right )\right ) \log ^2(x)}\right ) \, dx+5 \int \left (\frac {4}{\left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)}+\frac {4 x}{\left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)}+\frac {2 x^2}{\left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)}+\frac {x^3}{\left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)}-\frac {x^2 \log \left (2-e^{-5+x}\right )}{\left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)}\right ) \, dx+\left (10 e^5\right ) \int \frac {x^3}{\left (-2 e^5+e^x\right ) \left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)} \, dx+\left (20 e^5\right ) \int \frac {x^2}{\left (-2 e^5+e^x\right ) \left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)} \, dx\\ &=5 \int \frac {x}{\left (-2+x \log \left (2-e^{-5+x}\right )\right ) \log ^2(x)} \, dx+5 \int \frac {x^3}{\left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)} \, dx-5 \int \frac {x^2 \log \left (2-e^{-5+x}\right )}{\left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)} \, dx+10 \int \frac {1}{\left (-2+x \log \left (2-e^{-5+x}\right )\right ) \log ^2(x)} \, dx+10 \int \frac {x^2}{\left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)} \, dx+20 \int \frac {1}{\left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)} \, dx+20 \int \frac {x}{\left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)} \, dx+\left (10 e^5\right ) \int \frac {x^3}{\left (-2 e^5+e^x\right ) \left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)} \, dx+\left (20 e^5\right ) \int \frac {x^2}{\left (-2 e^5+e^x\right ) \left (-2+x \log \left (2-e^{-5+x}\right )\right )^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 26, normalized size = 0.84 \begin {gather*} -\frac {5 x (2+x)}{\left (-2+x \log \left (2-e^{-5+x}\right )\right ) \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(40 + E^(-5 + x)*(-20 - 10*x) + 20*x + (-40 - 40*x + E^(-5 + x)*(20 + 20*x + 10*x^2 + 5*x^3))*Log[x]
+ Log[2 - E^(-5 + x)]*(-20*x - 10*x^2 + E^(-5 + x)*(10*x + 5*x^2) + (10*x^2 - 5*E^(-5 + x)*x^2)*Log[x]))/((-8
+ 4*E^(-5 + x))*Log[x]^2 + (8*x - 4*E^(-5 + x)*x)*Log[2 - E^(-5 + x)]*Log[x]^2 + (-2*x^2 + E^(-5 + x)*x^2)*Lo
g[2 - E^(-5 + x)]^2*Log[x]^2),x]

[Out]

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

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fricas [A]  time = 0.65, size = 29, normalized size = 0.94 \begin {gather*} -\frac {5 \, {\left (x^{2} + 2 \, x\right )}}{x \log \relax (x) \log \left (-e^{\left (x - 5\right )} + 2\right ) - 2 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2*exp(x-5)+10*x^2)*log(x)+(5*x^2+10*x)*exp(x-5)-10*x^2-20*x)*log(-exp(x-5)+2)+((5*x^3+10*x^2
+20*x+20)*exp(x-5)-40*x-40)*log(x)+(-10*x-20)*exp(x-5)+20*x+40)/((x^2*exp(x-5)-2*x^2)*log(x)^2*log(-exp(x-5)+2
)^2+(-4*x*exp(x-5)+8*x)*log(x)^2*log(-exp(x-5)+2)+(4*exp(x-5)-8)*log(x)^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.01, size = 35, normalized size = 1.13 \begin {gather*} -\frac {5 \, {\left (x^{2} + 2 \, x\right )}}{x \log \relax (x) \log \left (2 \, e^{5} - e^{x}\right ) - 5 \, x \log \relax (x) - 2 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2*exp(x-5)+10*x^2)*log(x)+(5*x^2+10*x)*exp(x-5)-10*x^2-20*x)*log(-exp(x-5)+2)+((5*x^3+10*x^2
+20*x+20)*exp(x-5)-40*x-40)*log(x)+(-10*x-20)*exp(x-5)+20*x+40)/((x^2*exp(x-5)-2*x^2)*log(x)^2*log(-exp(x-5)+2
)^2+(-4*x*exp(x-5)+8*x)*log(x)^2*log(-exp(x-5)+2)+(4*exp(x-5)-8)*log(x)^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.08, size = 26, normalized size = 0.84

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-5*x^2*exp(x-5)+10*x^2)*ln(x)+(5*x^2+10*x)*exp(x-5)-10*x^2-20*x)*ln(-exp(x-5)+2)+((5*x^3+10*x^2+20*x+20
)*exp(x-5)-40*x-40)*ln(x)+(-10*x-20)*exp(x-5)+20*x+40)/((x^2*exp(x-5)-2*x^2)*ln(x)^2*ln(-exp(x-5)+2)^2+(-4*x*e
xp(x-5)+8*x)*ln(x)^2*ln(-exp(x-5)+2)+(4*exp(x-5)-8)*ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.51, size = 35, normalized size = 1.13 \begin {gather*} -\frac {5 \, {\left (x^{2} + 2 \, x\right )}}{x \log \relax (x) \log \left (2 \, e^{5} - e^{x}\right ) - {\left (5 \, x + 2\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2*exp(x-5)+10*x^2)*log(x)+(5*x^2+10*x)*exp(x-5)-10*x^2-20*x)*log(-exp(x-5)+2)+((5*x^3+10*x^2
+20*x+20)*exp(x-5)-40*x-40)*log(x)+(-10*x-20)*exp(x-5)+20*x+40)/((x^2*exp(x-5)-2*x^2)*log(x)^2*log(-exp(x-5)+2
)^2+(-4*x*exp(x-5)+8*x)*log(x)^2*log(-exp(x-5)+2)+(4*exp(x-5)-8)*log(x)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.91, size = 143, normalized size = 4.61 \begin {gather*} -\frac {80\,x-80\,x\,{\mathrm {e}}^{x-5}+20\,x\,{\mathrm {e}}^{2\,x-10}-40\,x^2\,{\mathrm {e}}^{x-5}-20\,x^3\,{\mathrm {e}}^{x-5}-10\,x^4\,{\mathrm {e}}^{x-5}+10\,x^2\,{\mathrm {e}}^{2\,x-10}+10\,x^3\,{\mathrm {e}}^{2\,x-10}+5\,x^4\,{\mathrm {e}}^{2\,x-10}+40\,x^2}{\ln \relax (x)\,\left (x\,\ln \left (2-{\mathrm {e}}^{-5}\,{\mathrm {e}}^x\right )-2\right )\,\left (2\,{\mathrm {e}}^{2\,x-10}-8\,{\mathrm {e}}^{x-5}-2\,x^2\,{\mathrm {e}}^{x-5}+x^2\,{\mathrm {e}}^{2\,x-10}+8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2 - exp(x - 5))*(20*x - exp(x - 5)*(10*x + 5*x^2) + log(x)*(5*x^2*exp(x - 5) - 10*x^2) + 10*x^2) - 2
0*x + log(x)*(40*x - exp(x - 5)*(20*x + 10*x^2 + 5*x^3 + 20) + 40) + exp(x - 5)*(10*x + 20) - 40)/(log(x)^2*(4
*exp(x - 5) - 8) + log(2 - exp(x - 5))*log(x)^2*(8*x - 4*x*exp(x - 5)) + log(2 - exp(x - 5))^2*log(x)^2*(x^2*e
xp(x - 5) - 2*x^2)),x)

[Out]

-(80*x - 80*x*exp(x - 5) + 20*x*exp(2*x - 10) - 40*x^2*exp(x - 5) - 20*x^3*exp(x - 5) - 10*x^4*exp(x - 5) + 10
*x^2*exp(2*x - 10) + 10*x^3*exp(2*x - 10) + 5*x^4*exp(2*x - 10) + 40*x^2)/(log(x)*(x*log(2 - exp(-5)*exp(x)) -
2)*(2*exp(2*x - 10) - 8*exp(x - 5) - 2*x^2*exp(x - 5) + x^2*exp(2*x - 10) + 8))

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sympy [A]  time = 0.45, size = 27, normalized size = 0.87 \begin {gather*} \frac {- 5 x^{2} - 10 x}{x \log {\relax (x )} \log {\left (2 - e^{x - 5} \right )} - 2 \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x**2*exp(x-5)+10*x**2)*ln(x)+(5*x**2+10*x)*exp(x-5)-10*x**2-20*x)*ln(-exp(x-5)+2)+((5*x**3+10*
x**2+20*x+20)*exp(x-5)-40*x-40)*ln(x)+(-10*x-20)*exp(x-5)+20*x+40)/((x**2*exp(x-5)-2*x**2)*ln(x)**2*ln(-exp(x-
5)+2)**2+(-4*x*exp(x-5)+8*x)*ln(x)**2*ln(-exp(x-5)+2)+(4*exp(x-5)-8)*ln(x)**2),x)

[Out]

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

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