3.64.80 \(\int \frac {-3 x+(e^2-3 x+e^{x^2} (-3 x^2+24 x^3-6 x^4+e^2 (x-8 x^2+2 x^3))) \log (\frac {3 e^4}{e^2-3 x})}{e^{e^{x^2} (-4+x)} (2 e^2 x^2-6 x^3)+(-e^2 x+3 x^2) \log (\frac {3 e^4}{e^2-3 x})} \, dx\)

Optimal. Leaf size=37 \[ \log \left (-2+\frac {e^{-e^{x^2} (-4+x)} \log \left (\frac {e^4}{\frac {e^2}{3}-x}\right )}{x}\right ) \]

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Rubi [F]  time = 20.40, 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 {-3 x+\left (e^2-3 x+e^{x^2} \left (-3 x^2+24 x^3-6 x^4+e^2 \left (x-8 x^2+2 x^3\right )\right )\right ) \log \left (\frac {3 e^4}{e^2-3 x}\right )}{e^{e^{x^2} (-4+x)} \left (2 e^2 x^2-6 x^3\right )+\left (-e^2 x+3 x^2\right ) \log \left (\frac {3 e^4}{e^2-3 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-3*x + (E^2 - 3*x + E^x^2*(-3*x^2 + 24*x^3 - 6*x^4 + E^2*(x - 8*x^2 + 2*x^3)))*Log[(3*E^4)/(E^2 - 3*x)])/
(E^(E^x^2*(-4 + x))*(2*E^2*x^2 - 6*x^3) + (-(E^2*x) + 3*x^2)*Log[(3*E^4)/(E^2 - 3*x)]),x]

[Out]

(4 + Log[3])*Defer[Int][E^x^2/(-4 + 2*E^(E^x^2*(-4 + x))*x - Log[3/(E^2 - 3*x)]), x] + 3*(4 + Log[3])*Defer[In
t][1/((E^2 - 3*x)*(-4 + 2*E^(E^x^2*(-4 + x))*x - Log[3/(E^2 - 3*x)])), x] - 3*(5 + Log[3])*Defer[Int][1/((E^2
- 3*x)*(-4 + 2*E^(E^x^2*(-4 + x))*x - Log[3/(E^2 - 3*x)])), x] + (4 + Log[3])*Defer[Int][1/(x*(-4 + 2*E^(E^x^2
*(-4 + x))*x - Log[3/(E^2 - 3*x)])), x] - 8*(4 + Log[3])*Defer[Int][(E^x^2*x)/(-4 + 2*E^(E^x^2*(-4 + x))*x - L
og[3/(E^2 - 3*x)]), x] + 2*(4 + Log[3])*Defer[Int][(E^x^2*x^2)/(-4 + 2*E^(E^x^2*(-4 + x))*x - Log[3/(E^2 - 3*x
)]), x] + Defer[Int][(E^x^2*Log[(E^2 - 3*x)^(-1)])/(-4 + 2*E^(E^x^2*(-4 + x))*x - Log[3/(E^2 - 3*x)]), x] + De
fer[Int][Log[(E^2 - 3*x)^(-1)]/(x*(-4 + 2*E^(E^x^2*(-4 + x))*x - Log[3/(E^2 - 3*x)])), x] - 8*Defer[Int][(E^x^
2*x*Log[(E^2 - 3*x)^(-1)])/(-4 + 2*E^(E^x^2*(-4 + x))*x - Log[3/(E^2 - 3*x)]), x] + 2*Defer[Int][(E^x^2*x^2*Lo
g[(E^2 - 3*x)^(-1)])/(-4 + 2*E^(E^x^2*(-4 + x))*x - Log[3/(E^2 - 3*x)]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 x-\left (e^2-3 x\right ) \left (1+e^{x^2} x \left (1-8 x+2 x^2\right )\right ) \left (4+\log (3)+\log \left (\frac {1}{e^2-3 x}\right )\right )}{\left (e^2-3 x\right ) x \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx\\ &=\int \left (\frac {e^{x^2} \left (1-8 x+2 x^2\right ) \left (-4 \left (1+\frac {\log (3)}{4}\right )-\log \left (\frac {1}{e^2-3 x}\right )\right )}{4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )}+\frac {15 x \left (1+\frac {\log (3)}{5}\right )-4 e^2 \left (1+\frac {\log (3)}{4}\right )-e^2 \log \left (\frac {1}{e^2-3 x}\right )+3 x \log \left (\frac {1}{e^2-3 x}\right )}{\left (e^2-3 x\right ) x \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )}\right ) \, dx\\ &=\int \frac {e^{x^2} \left (1-8 x+2 x^2\right ) \left (-4 \left (1+\frac {\log (3)}{4}\right )-\log \left (\frac {1}{e^2-3 x}\right )\right )}{4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )} \, dx+\int \frac {15 x \left (1+\frac {\log (3)}{5}\right )-4 e^2 \left (1+\frac {\log (3)}{4}\right )-e^2 \log \left (\frac {1}{e^2-3 x}\right )+3 x \log \left (\frac {1}{e^2-3 x}\right )}{\left (e^2-3 x\right ) x \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx\\ &=\int \frac {-e^2 (4+\log (3))+3 x (5+\log (3))-\left (e^2-3 x\right ) \log \left (\frac {1}{e^2-3 x}\right )}{\left (e^2-3 x\right ) x \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx+\int \left (\frac {e^{x^2} \left (-4 \left (1+\frac {\log (3)}{4}\right )-\log \left (\frac {1}{e^2-3 x}\right )\right )}{4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )}+\frac {2 e^{x^2} x^2 \left (-4 \left (1+\frac {\log (3)}{4}\right )-\log \left (\frac {1}{e^2-3 x}\right )\right )}{4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )}+\frac {8 e^{x^2} x \left (4 \left (1+\frac {\log (3)}{4}\right )+\log \left (\frac {1}{e^2-3 x}\right )\right )}{4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )}\right ) \, dx\\ &=2 \int \frac {e^{x^2} x^2 \left (-4 \left (1+\frac {\log (3)}{4}\right )-\log \left (\frac {1}{e^2-3 x}\right )\right )}{4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )} \, dx+8 \int \frac {e^{x^2} x \left (4 \left (1+\frac {\log (3)}{4}\right )+\log \left (\frac {1}{e^2-3 x}\right )\right )}{4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )} \, dx+\int \frac {e^{x^2} \left (-4 \left (1+\frac {\log (3)}{4}\right )-\log \left (\frac {1}{e^2-3 x}\right )\right )}{4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )} \, dx+\int \left (\frac {3 \left (15 x \left (1+\frac {\log (3)}{5}\right )-4 e^2 \left (1+\frac {\log (3)}{4}\right )-e^2 \log \left (\frac {1}{e^2-3 x}\right )+3 x \log \left (\frac {1}{e^2-3 x}\right )\right )}{e^2 \left (e^2-3 x\right ) \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )}+\frac {15 x \left (1+\frac {\log (3)}{5}\right )-4 e^2 \left (1+\frac {\log (3)}{4}\right )-e^2 \log \left (\frac {1}{e^2-3 x}\right )+3 x \log \left (\frac {1}{e^2-3 x}\right )}{e^2 x \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )}\right ) \, dx\\ &=2 \int \left (\frac {e^{x^2} x^2 (4+\log (3))}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )}+\frac {e^{x^2} x^2 \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )}\right ) \, dx+8 \int \left (-\frac {e^{x^2} x (4+\log (3))}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )}-\frac {e^{x^2} x \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )}\right ) \, dx+\frac {\int \frac {15 x \left (1+\frac {\log (3)}{5}\right )-4 e^2 \left (1+\frac {\log (3)}{4}\right )-e^2 \log \left (\frac {1}{e^2-3 x}\right )+3 x \log \left (\frac {1}{e^2-3 x}\right )}{x \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx}{e^2}+\frac {3 \int \frac {15 x \left (1+\frac {\log (3)}{5}\right )-4 e^2 \left (1+\frac {\log (3)}{4}\right )-e^2 \log \left (\frac {1}{e^2-3 x}\right )+3 x \log \left (\frac {1}{e^2-3 x}\right )}{\left (e^2-3 x\right ) \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx}{e^2}+\int \left (\frac {e^{x^2} (4+\log (3))}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )}+\frac {e^{x^2} \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )}\right ) \, dx\\ &=2 \int \frac {e^{x^2} x^2 \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx-8 \int \frac {e^{x^2} x \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+\frac {\int \frac {-e^2 (4+\log (3))+3 x (5+\log (3))-\left (e^2-3 x\right ) \log \left (\frac {1}{e^2-3 x}\right )}{x \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx}{e^2}+\frac {3 \int \frac {-e^2 (4+\log (3))+3 x (5+\log (3))-\left (e^2-3 x\right ) \log \left (\frac {1}{e^2-3 x}\right )}{\left (e^2-3 x\right ) \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx}{e^2}+(4+\log (3)) \int \frac {e^{x^2}}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+(2 (4+\log (3))) \int \frac {e^{x^2} x^2}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx-(8 (4+\log (3))) \int \frac {e^{x^2} x}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+\int \frac {e^{x^2} \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx\\ &=2 \int \frac {e^{x^2} x^2 \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx-8 \int \frac {e^{x^2} x \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+\frac {\int \left (\frac {e^2 (4+\log (3))}{x \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )}-\frac {3 (5+\log (3))}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )}-\frac {3 \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )}+\frac {e^2 \log \left (\frac {1}{e^2-3 x}\right )}{x \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )}\right ) \, dx}{e^2}+\frac {3 \int \left (\frac {e^2 (4+\log (3))}{\left (e^2-3 x\right ) \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )}-\frac {3 x (5+\log (3))}{\left (e^2-3 x\right ) \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )}+\frac {e^2 \log \left (\frac {1}{e^2-3 x}\right )}{\left (e^2-3 x\right ) \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )}-\frac {3 x \log \left (\frac {1}{e^2-3 x}\right )}{\left (e^2-3 x\right ) \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )}\right ) \, dx}{e^2}+(4+\log (3)) \int \frac {e^{x^2}}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+(2 (4+\log (3))) \int \frac {e^{x^2} x^2}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx-(8 (4+\log (3))) \int \frac {e^{x^2} x}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+\int \frac {e^{x^2} \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx\\ &=2 \int \frac {e^{x^2} x^2 \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+3 \int \frac {\log \left (\frac {1}{e^2-3 x}\right )}{\left (e^2-3 x\right ) \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx-8 \int \frac {e^{x^2} x \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx-\frac {3 \int \frac {\log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx}{e^2}-\frac {9 \int \frac {x \log \left (\frac {1}{e^2-3 x}\right )}{\left (e^2-3 x\right ) \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx}{e^2}+(4+\log (3)) \int \frac {e^{x^2}}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+(4+\log (3)) \int \frac {1}{x \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx+(2 (4+\log (3))) \int \frac {e^{x^2} x^2}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+(3 (4+\log (3))) \int \frac {1}{\left (e^2-3 x\right ) \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx-(8 (4+\log (3))) \int \frac {e^{x^2} x}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx-\frac {(3 (5+\log (3))) \int \frac {1}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx}{e^2}-\frac {(9 (5+\log (3))) \int \frac {x}{\left (e^2-3 x\right ) \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx}{e^2}+\int \frac {e^{x^2} \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+\int \frac {\log \left (\frac {1}{e^2-3 x}\right )}{x \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [F]  time = 20.68, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-3 x+\left (e^2-3 x+e^{x^2} \left (-3 x^2+24 x^3-6 x^4+e^2 \left (x-8 x^2+2 x^3\right )\right )\right ) \log \left (\frac {3 e^4}{e^2-3 x}\right )}{e^{e^{x^2} (-4+x)} \left (2 e^2 x^2-6 x^3\right )+\left (-e^2 x+3 x^2\right ) \log \left (\frac {3 e^4}{e^2-3 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(-3*x + (E^2 - 3*x + E^x^2*(-3*x^2 + 24*x^3 - 6*x^4 + E^2*(x - 8*x^2 + 2*x^3)))*Log[(3*E^4)/(E^2 - 3
*x)])/(E^(E^x^2*(-4 + x))*(2*E^2*x^2 - 6*x^3) + (-(E^2*x) + 3*x^2)*Log[(3*E^4)/(E^2 - 3*x)]),x]

[Out]

Integrate[(-3*x + (E^2 - 3*x + E^x^2*(-3*x^2 + 24*x^3 - 6*x^4 + E^2*(x - 8*x^2 + 2*x^3)))*Log[(3*E^4)/(E^2 - 3
*x)])/(E^(E^x^2*(-4 + x))*(2*E^2*x^2 - 6*x^3) + (-(E^2*x) + 3*x^2)*Log[(3*E^4)/(E^2 - 3*x)]), x]

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fricas [A]  time = 0.63, size = 45, normalized size = 1.22 \begin {gather*} -{\left (x - 4\right )} e^{\left (x^{2}\right )} + \log \left (\frac {2 \, x e^{\left ({\left (x - 4\right )} e^{\left (x^{2}\right )}\right )} - \log \left (-\frac {3 \, e^{4}}{3 \, x - e^{2}}\right )}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((2*x^3-8*x^2+x)*exp(2)-6*x^4+24*x^3-3*x^2)*exp(x^2)+exp(2)-3*x)*log(3*exp(4)/(exp(2)-3*x))-3*x)/(
(2*x^2*exp(2)-6*x^3)*exp((x-4)*exp(x^2))+(-exp(2)*x+3*x^2)*log(3*exp(4)/(exp(2)-3*x))),x, algorithm="fricas")

[Out]

-(x - 4)*e^(x^2) + log((2*x*e^((x - 4)*e^(x^2)) - log(-3*e^4/(3*x - e^2)))/x)

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giac [A]  time = 0.59, size = 51, normalized size = 1.38 \begin {gather*} -x e^{\left (x^{2}\right )} + 4 \, e^{\left (x^{2}\right )} + \log \left (-2 \, x e^{\left (x e^{\left (x^{2}\right )} - 4 \, e^{\left (x^{2}\right )}\right )} + \log \left (-\frac {3}{3 \, x - e^{2}}\right ) + 4\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((2*x^3-8*x^2+x)*exp(2)-6*x^4+24*x^3-3*x^2)*exp(x^2)+exp(2)-3*x)*log(3*exp(4)/(exp(2)-3*x))-3*x)/(
(2*x^2*exp(2)-6*x^3)*exp((x-4)*exp(x^2))+(-exp(2)*x+3*x^2)*log(3*exp(4)/(exp(2)-3*x))),x, algorithm="giac")

[Out]

-x*e^(x^2) + 4*e^(x^2) + log(-2*x*e^(x*e^(x^2) - 4*e^(x^2)) + log(-3/(3*x - e^2)) + 4) - log(x)

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maple [C]  time = 0.72, size = 46, normalized size = 1.24

method	result	size
risch	\(-\left (x -4\right ) {\mathrm e}^{x^{2}}+\ln \left ({\mathrm e}^{\left (x -4\right ) {\mathrm e}^{x^{2}}}+\frac {i \left (2 i \ln \relax (3)-2 i \ln \left ({\mathrm e}^{2}-3 x \right )+8 i\right )}{4 x}\right )\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((2*x^3-8*x^2+x)*exp(2)-6*x^4+24*x^3-3*x^2)*exp(x^2)+exp(2)-3*x)*ln(3*exp(4)/(exp(2)-3*x))-3*x)/((2*x^2*
exp(2)-6*x^3)*exp((x-4)*exp(x^2))+(-exp(2)*x+3*x^2)*ln(3*exp(4)/(exp(2)-3*x))),x,method=_RETURNVERBOSE)

[Out]

-(x-4)*exp(x^2)+ln(exp((x-4)*exp(x^2))+1/4*I*(2*I*ln(3)-2*I*ln(exp(2)-3*x)+8*I)/x)

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maxima [A]  time = 0.60, size = 53, normalized size = 1.43 \begin {gather*} -x e^{\left (x^{2}\right )} + \log \left (\frac {2 \, x e^{\left (x e^{\left (x^{2}\right )}\right )} - {\left (\log \relax (3) + 4\right )} e^{\left (4 \, e^{\left (x^{2}\right )}\right )} + e^{\left (4 \, e^{\left (x^{2}\right )}\right )} \log \left (-3 \, x + e^{2}\right )}{2 \, x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((2*x^3-8*x^2+x)*exp(2)-6*x^4+24*x^3-3*x^2)*exp(x^2)+exp(2)-3*x)*log(3*exp(4)/(exp(2)-3*x))-3*x)/(
(2*x^2*exp(2)-6*x^3)*exp((x-4)*exp(x^2))+(-exp(2)*x+3*x^2)*log(3*exp(4)/(exp(2)-3*x))),x, algorithm="maxima")

[Out]

-x*e^(x^2) + log(1/2*(2*x*e^(x*e^(x^2)) - (log(3) + 4)*e^(4*e^(x^2)) + e^(4*e^(x^2))*log(-3*x + e^2))/x)

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mupad [B]  time = 4.75, size = 48, normalized size = 1.30 \begin {gather*} \ln \left (\ln \left (-\frac {3\,{\mathrm {e}}^4}{3\,x-{\mathrm {e}}^2}\right )-2\,x\,{\mathrm {e}}^{-4\,{\mathrm {e}}^{x^2}}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{x^2}}\right )-\ln \relax (x)-{\mathrm {e}}^{x^2}\,\left (x-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x - log(-(3*exp(4))/(3*x - exp(2)))*(exp(2) - 3*x + exp(x^2)*(exp(2)*(x - 8*x^2 + 2*x^3) - 3*x^2 + 24*
x^3 - 6*x^4)))/(exp(exp(x^2)*(x - 4))*(2*x^2*exp(2) - 6*x^3) - log(-(3*exp(4))/(3*x - exp(2)))*(x*exp(2) - 3*x
^2)),x)

[Out]

log(log(-(3*exp(4))/(3*x - exp(2))) - 2*x*exp(-4*exp(x^2))*exp(x*exp(x^2))) - log(x) - exp(x^2)*(x - 4)

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sympy [A]  time = 1.18, size = 36, normalized size = 0.97 \begin {gather*} \left (4 - x\right ) e^{x^{2}} + \log {\left (e^{\left (x - 4\right ) e^{x^{2}}} - \frac {\log {\left (\frac {3 e^{4}}{- 3 x + e^{2}} \right )}}{2 x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((2*x**3-8*x**2+x)*exp(2)-6*x**4+24*x**3-3*x**2)*exp(x**2)+exp(2)-3*x)*ln(3*exp(4)/(exp(2)-3*x))-3
*x)/((2*x**2*exp(2)-6*x**3)*exp((x-4)*exp(x**2))+(-exp(2)*x+3*x**2)*ln(3*exp(4)/(exp(2)-3*x))),x)

[Out]

(4 - x)*exp(x**2) + log(exp((x - 4)*exp(x**2)) - log(3*exp(4)/(-3*x + exp(2)))/(2*x))

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