3.79.9 \(\int \frac {e^{e^{25 x^2 \log ^2(\frac {e^3-x}{1+e^3-x})}+25 x^2 \log ^2(\frac {e^3-x}{1+e^3-x})} (-25 x^2 \log (\frac {e^3-x}{1+e^3-x})+(25 e^6 x-25 x^2+25 x^3+e^3 (25 x-50 x^2)) \log ^2(\frac {e^3-x}{1+e^3-x}))}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx\)

Optimal. Leaf size=33 \[ \frac {1}{16} e^{e^{25 x^2 \log ^2\left (\frac {x}{x+\frac {x}{e^3-x}}\right )}} \]

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Rubi [F]  time = 5.97, 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 {\exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(E^(25*x^2*Log[(E^3 - x)/(1 + E^3 - x)]^2) + 25*x^2*Log[(E^3 - x)/(1 + E^3 - x)]^2)*(-25*x^2*Log[(E^3 -
x)/(1 + E^3 - x)] + (25*E^6*x - 25*x^2 + 25*x^3 + E^3*(25*x - 50*x^2))*Log[(E^3 - x)/(1 + E^3 - x)]^2))/(8*E^
6 + E^3*(8 - 16*x) - 8*x + 8*x^2),x]

[Out]

(-25*Defer[Int][E^(E^(25*x^2*Log[(E^3 - x)/(1 + E^3 - x)]^2) + 25*x^2*Log[(E^3 - x)/(1 + E^3 - x)]^2)*Log[(E^3
- x)/(1 + E^3 - x)], x])/8 - (25*Defer[Int][(E^(6 + E^(25*x^2*Log[(E^3 - x)/(1 + E^3 - x)]^2) + 25*x^2*Log[(E
^3 - x)/(1 + E^3 - x)]^2)*Log[(E^3 - x)/(1 + E^3 - x)])/(E^3 - x), x])/8 + (25*(1 + E^3)^2*Defer[Int][(E^(E^(2
5*x^2*Log[(E^3 - x)/(1 + E^3 - x)]^2) + 25*x^2*Log[(E^3 - x)/(1 + E^3 - x)]^2)*Log[(E^3 - x)/(1 + E^3 - x)])/(
1 + E^3 - x), x])/8 + (25*Defer[Int][E^(E^(25*x^2*Log[(E^3 - x)/(1 + E^3 - x)]^2) + 25*x^2*Log[(E^3 - x)/(1 +
E^3 - x)]^2)*x*Log[(E^3 - x)/(1 + E^3 - x)]^2, x])/8

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^3 \left (1+e^3\right )-8 \left (1+2 e^3\right ) x+8 x^2} \, dx\\ &=\int \left (-\frac {25 \exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )}{8 \left (e^3-x\right ) \left (1+e^3-x\right )}+\frac {25}{8} \exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) x \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) \, dx\\ &=-\left (\frac {25}{8} \int \frac {\exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )}{\left (e^3-x\right ) \left (1+e^3-x\right )} \, dx\right )+\frac {25}{8} \int \exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) x \log ^2\left (\frac {e^3-x}{1+e^3-x}\right ) \, dx\\ &=\frac {25}{8} \int \exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) x \log ^2\left (\frac {e^3-x}{1+e^3-x}\right ) \, dx-\frac {25}{8} \int \left (\exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) \log \left (\frac {e^3-x}{1+e^3-x}\right )+\frac {\exp \left (6+e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) \log \left (\frac {e^3-x}{1+e^3-x}\right )}{e^3-x}-\frac {\exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) \left (1+e^3\right )^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )}{1+e^3-x}\right ) \, dx\\ &=-\left (\frac {25}{8} \int \exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) \log \left (\frac {e^3-x}{1+e^3-x}\right ) \, dx\right )-\frac {25}{8} \int \frac {\exp \left (6+e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) \log \left (\frac {e^3-x}{1+e^3-x}\right )}{e^3-x} \, dx+\frac {25}{8} \int \exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) x \log ^2\left (\frac {e^3-x}{1+e^3-x}\right ) \, dx+\frac {1}{8} \left (25 \left (1+e^3\right )^2\right ) \int \frac {\exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) \log \left (\frac {e^3-x}{1+e^3-x}\right )}{1+e^3-x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 34, normalized size = 1.03 \begin {gather*} \frac {1}{16} e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^(25*x^2*Log[(E^3 - x)/(1 + E^3 - x)]^2) + 25*x^2*Log[(E^3 - x)/(1 + E^3 - x)]^2)*(-25*x^2*Log[
(E^3 - x)/(1 + E^3 - x)] + (25*E^6*x - 25*x^2 + 25*x^3 + E^3*(25*x - 50*x^2))*Log[(E^3 - x)/(1 + E^3 - x)]^2))
/(8*E^6 + E^3*(8 - 16*x) - 8*x + 8*x^2),x]

[Out]

E^E^(25*x^2*Log[(E^3 - x)/(1 + E^3 - x)]^2)/16

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fricas [A]  time = 1.23, size = 28, normalized size = 0.85 \begin {gather*} \frac {1}{16} \, e^{\left (e^{\left (25 \, x^{2} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right )^{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x*exp(3)^2+(-50*x^2+25*x)*exp(3)+25*x^3-25*x^2)*log((-x+exp(3))/(exp(3)-x+1))^2-25*x^2*log((-x+
exp(3))/(exp(3)-x+1)))*exp(25*x^2*log((-x+exp(3))/(exp(3)-x+1))^2)*exp(exp(25*x^2*log((-x+exp(3))/(exp(3)-x+1)
)^2))/(8*exp(3)^2+(-16*x+8)*exp(3)+8*x^2-8*x),x, algorithm="fricas")

[Out]

1/16*e^(e^(25*x^2*log((x - e^3)/(x - e^3 - 1))^2))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {25 \, {\left (x^{2} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right ) - {\left (x^{3} - x^{2} + x e^{6} - {\left (2 \, x^{2} - x\right )} e^{3}\right )} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right )^{2}\right )} e^{\left (25 \, x^{2} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right )^{2} + e^{\left (25 \, x^{2} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right )^{2}\right )}\right )}}{8 \, {\left (x^{2} - {\left (2 \, x - 1\right )} e^{3} - x + e^{6}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x*exp(3)^2+(-50*x^2+25*x)*exp(3)+25*x^3-25*x^2)*log((-x+exp(3))/(exp(3)-x+1))^2-25*x^2*log((-x+
exp(3))/(exp(3)-x+1)))*exp(25*x^2*log((-x+exp(3))/(exp(3)-x+1))^2)*exp(exp(25*x^2*log((-x+exp(3))/(exp(3)-x+1)
)^2))/(8*exp(3)^2+(-16*x+8)*exp(3)+8*x^2-8*x),x, algorithm="giac")

[Out]

integrate(-25/8*(x^2*log((x - e^3)/(x - e^3 - 1)) - (x^3 - x^2 + x*e^6 - (2*x^2 - x)*e^3)*log((x - e^3)/(x - e
^3 - 1))^2)*e^(25*x^2*log((x - e^3)/(x - e^3 - 1))^2 + e^(25*x^2*log((x - e^3)/(x - e^3 - 1))^2))/(x^2 - (2*x
- 1)*e^3 - x + e^6), x)

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maple [A]  time = 0.31, size = 29, normalized size = 0.88

method	result	size
risch	\(\frac {{\mathrm e}^{{\mathrm e}^{25 x^{2} \ln \left (\frac {-x +{\mathrm e}^{3}}{{\mathrm e}^{3}-x +1}\right )^{2}}}}{16}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((25*x*exp(3)^2+(-50*x^2+25*x)*exp(3)+25*x^3-25*x^2)*ln((-x+exp(3))/(exp(3)-x+1))^2-25*x^2*ln((-x+exp(3))/
(exp(3)-x+1)))*exp(25*x^2*ln((-x+exp(3))/(exp(3)-x+1))^2)*exp(exp(25*x^2*ln((-x+exp(3))/(exp(3)-x+1))^2))/(8*e
xp(3)^2+(-16*x+8)*exp(3)+8*x^2-8*x),x,method=_RETURNVERBOSE)

[Out]

1/16*exp(exp(25*x^2*ln((-x+exp(3))/(exp(3)-x+1))^2))

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maxima [A]  time = 0.81, size = 54, normalized size = 1.64 \begin {gather*} \frac {1}{16} \, e^{\left (e^{\left (25 \, x^{2} \log \left (x - e^{3}\right )^{2} - 50 \, x^{2} \log \left (x - e^{3}\right ) \log \left (x - e^{3} - 1\right ) + 25 \, x^{2} \log \left (x - e^{3} - 1\right )^{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x*exp(3)^2+(-50*x^2+25*x)*exp(3)+25*x^3-25*x^2)*log((-x+exp(3))/(exp(3)-x+1))^2-25*x^2*log((-x+
exp(3))/(exp(3)-x+1)))*exp(25*x^2*log((-x+exp(3))/(exp(3)-x+1))^2)*exp(exp(25*x^2*log((-x+exp(3))/(exp(3)-x+1)
)^2))/(8*exp(3)^2+(-16*x+8)*exp(3)+8*x^2-8*x),x, algorithm="maxima")

[Out]

1/16*e^(e^(25*x^2*log(x - e^3)^2 - 50*x^2*log(x - e^3)*log(x - e^3 - 1) + 25*x^2*log(x - e^3 - 1)^2))

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mupad [B]  time = 5.87, size = 29, normalized size = 0.88 \begin {gather*} \frac {{\mathrm {e}}^{{\mathrm {e}}^{25\,x^2\,{\ln \left (-\frac {x-{\mathrm {e}}^3}{{\mathrm {e}}^3-x+1}\right )}^2}}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(25*x^2*log(-(x - exp(3))/(exp(3) - x + 1))^2)*exp(exp(25*x^2*log(-(x - exp(3))/(exp(3) - x + 1))^2))
*(log(-(x - exp(3))/(exp(3) - x + 1))^2*(exp(3)*(25*x - 50*x^2) + 25*x*exp(6) - 25*x^2 + 25*x^3) - 25*x^2*log(
-(x - exp(3))/(exp(3) - x + 1))))/(8*x - 8*exp(6) - 8*x^2 + exp(3)*(16*x - 8)),x)

[Out]

exp(exp(25*x^2*log(-(x - exp(3))/(exp(3) - x + 1))^2))/16

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sympy [A]  time = 6.78, size = 24, normalized size = 0.73 \begin {gather*} \frac {e^{e^{25 x^{2} \log {\left (\frac {- x + e^{3}}{- x + 1 + e^{3}} \right )}^{2}}}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x*exp(3)**2+(-50*x**2+25*x)*exp(3)+25*x**3-25*x**2)*ln((-x+exp(3))/(exp(3)-x+1))**2-25*x**2*ln(
(-x+exp(3))/(exp(3)-x+1)))*exp(25*x**2*ln((-x+exp(3))/(exp(3)-x+1))**2)*exp(exp(25*x**2*ln((-x+exp(3))/(exp(3)
-x+1))**2))/(8*exp(3)**2+(-16*x+8)*exp(3)+8*x**2-8*x),x)

[Out]

exp(exp(25*x**2*log((-x + exp(3))/(-x + 1 + exp(3)))**2))/16

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