3.60.33 \(\int \frac {e^{e^x+x} (e^x (50-50 x)-50 x)}{(-1+e^{e^x+x} (-50+50 x)) \log ^2(1+e^{e^x+x} (50-50 x))} \, dx\)

Optimal. Leaf size=23 \[ \frac {1}{\log \left (1-10 e^{e^x+x} (5-5 (2-x))\right )} \]

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Rubi [F]  time = 3.10, 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^{e^x+x} \left (e^x (50-50 x)-50 x\right )}{\left (-1+e^{e^x+x} (-50+50 x)\right ) \log ^2\left (1+e^{e^x+x} (50-50 x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(E^x + x)*(E^x*(50 - 50*x) - 50*x))/((-1 + E^(E^x + x)*(-50 + 50*x))*Log[1 + E^(E^x + x)*(50 - 50*x)]^2
),x]

[Out]

-Defer[Int][E^x/Log[1 - 50*E^(E^x + x)*(-1 + x)]^2, x] - Defer[Int][E^x/((-1 + 50*E^(E^x + x)*(-1 + x))*Log[1
- 50*E^(E^x + x)*(-1 + x)]^2), x] - 50*Defer[Int][(E^(E^x + x)*x)/((-1 + 50*E^(E^x + x)*(-1 + x))*Log[1 - 50*E
^(E^x + x)*(-1 + x)]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {50 e^{e^x+x} \left (-e^x+x+e^x x\right )}{\left (1-e^{e^x+x} (-50+50 x)\right ) \log ^2\left (1+e^{e^x+x} (50-50 x)\right )} \, dx\\ &=50 \int \frac {e^{e^x+x} \left (-e^x+x+e^x x\right )}{\left (1-e^{e^x+x} (-50+50 x)\right ) \log ^2\left (1+e^{e^x+x} (50-50 x)\right )} \, dx\\ &=50 \int \left (-\frac {e^x}{50 \log ^2\left (1-50 e^{e^x+x} (-1+x)\right )}-\frac {e^x \left (1+50 e^{e^x} x\right )}{50 \left (-1-50 e^{e^x+x}+50 e^{e^x+x} x\right ) \log ^2\left (1-50 e^{e^x+x} (-1+x)\right )}\right ) \, dx\\ &=-\int \frac {e^x}{\log ^2\left (1-50 e^{e^x+x} (-1+x)\right )} \, dx-\int \frac {e^x \left (1+50 e^{e^x} x\right )}{\left (-1-50 e^{e^x+x}+50 e^{e^x+x} x\right ) \log ^2\left (1-50 e^{e^x+x} (-1+x)\right )} \, dx\\ &=-\int \left (\frac {e^x}{\left (-1-50 e^{e^x+x}+50 e^{e^x+x} x\right ) \log ^2\left (1-50 e^{e^x+x} (-1+x)\right )}+\frac {50 e^{e^x+x} x}{\left (-1-50 e^{e^x+x}+50 e^{e^x+x} x\right ) \log ^2\left (1-50 e^{e^x+x} (-1+x)\right )}\right ) \, dx-\int \frac {e^x}{\log ^2\left (1-50 e^{e^x+x} (-1+x)\right )} \, dx\\ &=-\left (50 \int \frac {e^{e^x+x} x}{\left (-1-50 e^{e^x+x}+50 e^{e^x+x} x\right ) \log ^2\left (1-50 e^{e^x+x} (-1+x)\right )} \, dx\right )-\int \frac {e^x}{\log ^2\left (1-50 e^{e^x+x} (-1+x)\right )} \, dx-\int \frac {e^x}{\left (-1-50 e^{e^x+x}+50 e^{e^x+x} x\right ) \log ^2\left (1-50 e^{e^x+x} (-1+x)\right )} \, dx\\ &=-\left (50 \int \frac {e^{e^x+x} x}{\left (-1+50 e^{e^x+x} (-1+x)\right ) \log ^2\left (1-50 e^{e^x+x} (-1+x)\right )} \, dx\right )-\int \frac {e^x}{\log ^2\left (1-50 e^{e^x+x} (-1+x)\right )} \, dx-\int \frac {e^x}{\left (-1+50 e^{e^x+x} (-1+x)\right ) \log ^2\left (1-50 e^{e^x+x} (-1+x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.39, size = 17, normalized size = 0.74 \begin {gather*} \frac {1}{\log \left (1-50 e^{e^x+x} (-1+x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^x + x)*(E^x*(50 - 50*x) - 50*x))/((-1 + E^(E^x + x)*(-50 + 50*x))*Log[1 + E^(E^x + x)*(50 - 50
*x)]^2),x]

[Out]

Log[1 - 50*E^(E^x + x)*(-1 + x)]^(-1)

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fricas [A]  time = 0.58, size = 15, normalized size = 0.65 \begin {gather*} \frac {1}{\log \left (-50 \, {\left (x - 1\right )} e^{\left (x + e^{x}\right )} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*x+50)*exp(x)-50*x)*exp(exp(x)+x)/((50*x-50)*exp(exp(x)+x)-1)/log((-50*x+50)*exp(exp(x)+x)+1)^2
,x, algorithm="fricas")

[Out]

1/log(-50*(x - 1)*e^(x + e^x) + 1)

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giac [A]  time = 0.17, size = 20, normalized size = 0.87 \begin {gather*} \frac {1}{\log \left (-50 \, x e^{\left (x + e^{x}\right )} + 50 \, e^{\left (x + e^{x}\right )} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*x+50)*exp(x)-50*x)*exp(exp(x)+x)/((50*x-50)*exp(exp(x)+x)-1)/log((-50*x+50)*exp(exp(x)+x)+1)^2
,x, algorithm="giac")

[Out]

1/log(-50*x*e^(x + e^x) + 50*e^(x + e^x) + 1)

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maple [A]  time = 0.07, size = 17, normalized size = 0.74

method	result	size
risch	\(\frac {1}{\ln \left (\left (-50 x +50\right ) {\mathrm e}^{{\mathrm e}^{x}+x}+1\right )}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-50*x+50)*exp(x)-50*x)*exp(exp(x)+x)/((50*x-50)*exp(exp(x)+x)-1)/ln((-50*x+50)*exp(exp(x)+x)+1)^2,x,meth
od=_RETURNVERBOSE)

[Out]

1/ln((-50*x+50)*exp(exp(x)+x)+1)

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maxima [A]  time = 0.44, size = 15, normalized size = 0.65 \begin {gather*} \frac {1}{\log \left (-50 \, {\left (x - 1\right )} e^{\left (x + e^{x}\right )} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*x+50)*exp(x)-50*x)*exp(exp(x)+x)/((50*x-50)*exp(exp(x)+x)-1)/log((-50*x+50)*exp(exp(x)+x)+1)^2
,x, algorithm="maxima")

[Out]

1/log(-50*(x - 1)*e^(x + e^x) + 1)

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mupad [B]  time = 0.32, size = 17, normalized size = 0.74 \begin {gather*} \frac {1}{\ln \left (1-{\mathrm {e}}^{x+{\mathrm {e}}^x}\,\left (50\,x-50\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x + exp(x))*(50*x + exp(x)*(50*x - 50)))/(log(1 - exp(x + exp(x))*(50*x - 50))^2*(exp(x + exp(x))*(5
0*x - 50) - 1)),x)

[Out]

1/log(1 - exp(x + exp(x))*(50*x - 50))

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sympy [A]  time = 0.45, size = 15, normalized size = 0.65 \begin {gather*} \frac {1}{\log {\left (\left (50 - 50 x\right ) e^{x + e^{x}} + 1 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*x+50)*exp(x)-50*x)*exp(exp(x)+x)/((50*x-50)*exp(exp(x)+x)-1)/ln((-50*x+50)*exp(exp(x)+x)+1)**2
,x)

[Out]

1/log((50 - 50*x)*exp(x + exp(x)) + 1)

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