3.73.73 \(\int \frac {-e^{e^{\frac {1}{x}}} x^2+e^{e^{\frac {1}{x}}} (-50 x-2 x^2+e^{\frac {1}{x}} (25+x)+(e^{\frac {1}{x}} (-25-x)+50 x+2 x^2) \log (25+x)) \log (1-\log (25+x))}{(-50-2 x+(50+2 x) \log (25+x)) \log ^2(1-\log (25+x))} \, dx\)

Optimal. Leaf size=25 \[ \frac {e^{e^{\frac {1}{x}}} x^2}{2 \log (1-\log (25+x))} \]

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Rubi [F]  time = 4.94, 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^{\frac {1}{x}}} x^2+e^{e^{\frac {1}{x}}} \left (-50 x-2 x^2+e^{\frac {1}{x}} (25+x)+\left (e^{\frac {1}{x}} (-25-x)+50 x+2 x^2\right ) \log (25+x)\right ) \log (1-\log (25+x))}{(-50-2 x+(50+2 x) \log (25+x)) \log ^2(1-\log (25+x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-(E^E^x^(-1)*x^2) + E^E^x^(-1)*(-50*x - 2*x^2 + E^x^(-1)*(25 + x) + (E^x^(-1)*(-25 - x) + 50*x + 2*x^2)*L
og[25 + x])*Log[1 - Log[25 + x]])/((-50 - 2*x + (50 + 2*x)*Log[25 + x])*Log[1 - Log[25 + x]]^2),x]

[Out]

(25*Defer[Int][E^E^x^(-1)/((-1 + Log[25 + x])*Log[1 - Log[25 + x]]^2), x])/2 - Defer[Int][(E^E^x^(-1)*x)/((-1
+ Log[25 + x])*Log[1 - Log[25 + x]]^2), x]/2 - (625*Defer[Int][E^E^x^(-1)/((25 + x)*(-1 + Log[25 + x])*Log[1 -
 Log[25 + x]]^2), x])/2 - Defer[Int][E^(E^x^(-1) + x^(-1))/Log[1 - Log[25 + x]], x]/2 + Defer[Int][(E^E^x^(-1)
*x)/Log[1 - Log[25 + x]], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{e^{\frac {1}{x}}} \left (-\frac {x^2}{(25+x) (-1+\log (25+x))}-\left (e^{\frac {1}{x}}-2 x\right ) \log (1-\log (25+x))\right )}{2 \log ^2(1-\log (25+x))} \, dx\\ &=\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}} \left (-\frac {x^2}{(25+x) (-1+\log (25+x))}-\left (e^{\frac {1}{x}}-2 x\right ) \log (1-\log (25+x))\right )}{\log ^2(1-\log (25+x))} \, dx\\ &=\frac {1}{2} \int \left (-\frac {e^{e^{\frac {1}{x}}+\frac {1}{x}}}{\log (1-\log (25+x))}+\frac {e^{e^{\frac {1}{x}}} x (-x-50 \log (1-\log (25+x))-2 x \log (1-\log (25+x))+50 \log (25+x) \log (1-\log (25+x))+2 x \log (25+x) \log (1-\log (25+x)))}{(25+x) (-1+\log (25+x)) \log ^2(1-\log (25+x))}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}+\frac {1}{x}}}{\log (1-\log (25+x))} \, dx\right )+\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}} x (-x-50 \log (1-\log (25+x))-2 x \log (1-\log (25+x))+50 \log (25+x) \log (1-\log (25+x))+2 x \log (25+x) \log (1-\log (25+x)))}{(25+x) (-1+\log (25+x)) \log ^2(1-\log (25+x))} \, dx\\ &=-\left (\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}+\frac {1}{x}}}{\log (1-\log (25+x))} \, dx\right )+\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}} x (x-2 (25+x) (-1+\log (25+x)) \log (1-\log (25+x)))}{(25+x) (1-\log (25+x)) \log ^2(1-\log (25+x))} \, dx\\ &=\frac {1}{2} \int \left (-\frac {e^{e^{\frac {1}{x}}} x^2}{(25+x) (-1+\log (25+x)) \log ^2(1-\log (25+x))}+\frac {2 e^{e^{\frac {1}{x}}} x}{\log (1-\log (25+x))}\right ) \, dx-\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}+\frac {1}{x}}}{\log (1-\log (25+x))} \, dx\\ &=-\left (\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}} x^2}{(25+x) (-1+\log (25+x)) \log ^2(1-\log (25+x))} \, dx\right )-\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}+\frac {1}{x}}}{\log (1-\log (25+x))} \, dx+\int \frac {e^{e^{\frac {1}{x}}} x}{\log (1-\log (25+x))} \, dx\\ &=-\left (\frac {1}{2} \int \left (-\frac {25 e^{e^{\frac {1}{x}}}}{(-1+\log (25+x)) \log ^2(1-\log (25+x))}+\frac {e^{e^{\frac {1}{x}}} x}{(-1+\log (25+x)) \log ^2(1-\log (25+x))}+\frac {625 e^{e^{\frac {1}{x}}}}{(25+x) (-1+\log (25+x)) \log ^2(1-\log (25+x))}\right ) \, dx\right )-\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}+\frac {1}{x}}}{\log (1-\log (25+x))} \, dx+\int \frac {e^{e^{\frac {1}{x}}} x}{\log (1-\log (25+x))} \, dx\\ &=-\left (\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}} x}{(-1+\log (25+x)) \log ^2(1-\log (25+x))} \, dx\right )-\frac {1}{2} \int \frac {e^{e^{\frac {1}{x}}+\frac {1}{x}}}{\log (1-\log (25+x))} \, dx+\frac {25}{2} \int \frac {e^{e^{\frac {1}{x}}}}{(-1+\log (25+x)) \log ^2(1-\log (25+x))} \, dx-\frac {625}{2} \int \frac {e^{e^{\frac {1}{x}}}}{(25+x) (-1+\log (25+x)) \log ^2(1-\log (25+x))} \, dx+\int \frac {e^{e^{\frac {1}{x}}} x}{\log (1-\log (25+x))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.33, size = 25, normalized size = 1.00 \begin {gather*} \frac {e^{e^{\frac {1}{x}}} x^2}{2 \log (1-\log (25+x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^E^x^(-1)*x^2) + E^E^x^(-1)*(-50*x - 2*x^2 + E^x^(-1)*(25 + x) + (E^x^(-1)*(-25 - x) + 50*x + 2*
x^2)*Log[25 + x])*Log[1 - Log[25 + x]])/((-50 - 2*x + (50 + 2*x)*Log[25 + x])*Log[1 - Log[25 + x]]^2),x]

[Out]

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

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fricas [A]  time = 0.91, size = 21, normalized size = 0.84 \begin {gather*} \frac {x^{2} e^{\left (e^{\frac {1}{x}}\right )}}{2 \, \log \left (-\log \left (x + 25\right ) + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-x-25)*exp(1/x)+2*x^2+50*x)*log(x+25)+(x+25)*exp(1/x)-2*x^2-50*x)*exp(exp(1/x))*log(-log(x+25)+1
)-x^2*exp(exp(1/x)))/((2*x+50)*log(x+25)-2*x-50)/log(-log(x+25)+1)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.27, size = 21, normalized size = 0.84 \begin {gather*} \frac {x^{2} e^{\left (e^{\frac {1}{x}}\right )}}{2 \, \log \left (-\log \left (x + 25\right ) + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-x-25)*exp(1/x)+2*x^2+50*x)*log(x+25)+(x+25)*exp(1/x)-2*x^2-50*x)*exp(exp(1/x))*log(-log(x+25)+1
)-x^2*exp(exp(1/x)))/((2*x+50)*log(x+25)-2*x-50)/log(-log(x+25)+1)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.12, size = 22, normalized size = 0.88




method result size



risch \(\frac {x^{2} {\mathrm e}^{{\mathrm e}^{\frac {1}{x}}}}{2 \ln \left (-\ln \left (x +25\right )+1\right )}\) \(22\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((-x-25)*exp(1/x)+2*x^2+50*x)*ln(x+25)+(x+25)*exp(1/x)-2*x^2-50*x)*exp(exp(1/x))*ln(-ln(x+25)+1)-x^2*exp
(exp(1/x)))/((2*x+50)*ln(x+25)-2*x-50)/ln(-ln(x+25)+1)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.41, size = 21, normalized size = 0.84 \begin {gather*} \frac {x^{2} e^{\left (e^{\frac {1}{x}}\right )}}{2 \, \log \left (-\log \left (x + 25\right ) + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-x-25)*exp(1/x)+2*x^2+50*x)*log(x+25)+(x+25)*exp(1/x)-2*x^2-50*x)*exp(exp(1/x))*log(-log(x+25)+1
)-x^2*exp(exp(1/x)))/((2*x+50)*log(x+25)-2*x-50)/log(-log(x+25)+1)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.99, size = 21, normalized size = 0.84 \begin {gather*} \frac {x^2\,{\mathrm {e}}^{{\mathrm {e}}^{1/x}}}{2\,\ln \left (1-\ln \left (x+25\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*exp(exp(1/x)) + exp(exp(1/x))*log(1 - log(x + 25))*(50*x - exp(1/x)*(x + 25) - log(x + 25)*(50*x - ex
p(1/x)*(x + 25) + 2*x^2) + 2*x^2))/(log(1 - log(x + 25))^2*(2*x - log(x + 25)*(2*x + 50) + 50)),x)

[Out]

(x^2*exp(exp(1/x)))/(2*log(1 - log(x + 25)))

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sympy [A]  time = 1.47, size = 19, normalized size = 0.76 \begin {gather*} \frac {x^{2} e^{e^{\frac {1}{x}}}}{2 \log {\left (1 - \log {\left (x + 25 \right )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-x-25)*exp(1/x)+2*x**2+50*x)*ln(x+25)+(x+25)*exp(1/x)-2*x**2-50*x)*exp(exp(1/x))*ln(-ln(x+25)+1)
-x**2*exp(exp(1/x)))/((2*x+50)*ln(x+25)-2*x-50)/ln(-ln(x+25)+1)**2,x)

[Out]

x**2*exp(exp(1/x))/(2*log(1 - log(x + 25)))

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