Optimal. Leaf size=24 \[ \frac {10 (5+\log (3))^2}{x \log ^2\left (6+e^{e^x}-x\right )} \]
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Rubi [F]
time = 2.18, 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 {500 x+200 x \log (3)+20 x \log ^2(3)+e^{e^x+x} \left (-500 x-200 x \log (3)-20 x \log ^2(3)\right )+\left (-1500+250 x+(-600+100 x) \log (3)+(-60+10 x) \log ^2(3)+e^{e^x} \left (-250-100 \log (3)-10 \log ^2(3)\right )\right ) \log \left (6+e^{e^x}-x\right )}{\left (6 x^2+e^{e^x} x^2-x^3\right ) \log ^3\left (6+e^{e^x}-x\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {20 x \log ^2(3)+x (500+200 \log (3))+e^{e^x+x} \left (-500 x-200 x \log (3)-20 x \log ^2(3)\right )+\left (-1500+250 x+(-600+100 x) \log (3)+(-60+10 x) \log ^2(3)+e^{e^x} \left (-250-100 \log (3)-10 \log ^2(3)\right )\right ) \log \left (6+e^{e^x}-x\right )}{\left (6 x^2+e^{e^x} x^2-x^3\right ) \log ^3\left (6+e^{e^x}-x\right )} \, dx\\ &=\int \frac {x \left (500+200 \log (3)+20 \log ^2(3)\right )+e^{e^x+x} \left (-500 x-200 x \log (3)-20 x \log ^2(3)\right )+\left (-1500+250 x+(-600+100 x) \log (3)+(-60+10 x) \log ^2(3)+e^{e^x} \left (-250-100 \log (3)-10 \log ^2(3)\right )\right ) \log \left (6+e^{e^x}-x\right )}{\left (6 x^2+e^{e^x} x^2-x^3\right ) \log ^3\left (6+e^{e^x}-x\right )} \, dx\\ &=\int \frac {10 (5+\log (3))^2 \left (-2 \left (-1+e^{e^x+x}\right ) x-\left (6+e^{e^x}-x\right ) \log \left (6+e^{e^x}-x\right )\right )}{\left (6+e^{e^x}-x\right ) x^2 \log ^3\left (6+e^{e^x}-x\right )} \, dx\\ &=\left (10 (5+\log (3))^2\right ) \int \frac {-2 \left (-1+e^{e^x+x}\right ) x-\left (6+e^{e^x}-x\right ) \log \left (6+e^{e^x}-x\right )}{\left (6+e^{e^x}-x\right ) x^2 \log ^3\left (6+e^{e^x}-x\right )} \, dx\\ &=\left (10 (5+\log (3))^2\right ) \int \left (-\frac {2 e^{e^x+x}}{\left (6+e^{e^x}-x\right ) x \log ^3\left (6+e^{e^x}-x\right )}-\frac {-2 x+6 \log \left (6+e^{e^x}-x\right )+e^{e^x} \log \left (6+e^{e^x}-x\right )-x \log \left (6+e^{e^x}-x\right )}{\left (6+e^{e^x}-x\right ) x^2 \log ^3\left (6+e^{e^x}-x\right )}\right ) \, dx\\ &=-\left (\left (10 (5+\log (3))^2\right ) \int \frac {-2 x+6 \log \left (6+e^{e^x}-x\right )+e^{e^x} \log \left (6+e^{e^x}-x\right )-x \log \left (6+e^{e^x}-x\right )}{\left (6+e^{e^x}-x\right ) x^2 \log ^3\left (6+e^{e^x}-x\right )} \, dx\right )-\left (20 (5+\log (3))^2\right ) \int \frac {e^{e^x+x}}{\left (6+e^{e^x}-x\right ) x \log ^3\left (6+e^{e^x}-x\right )} \, dx\\ &=-\left (\left (10 (5+\log (3))^2\right ) \int \frac {-\frac {2 x}{6+e^{e^x}-x}+\log \left (6+e^{e^x}-x\right )}{x^2 \log ^3\left (6+e^{e^x}-x\right )} \, dx\right )-\left (20 (5+\log (3))^2\right ) \int \frac {e^{e^x+x}}{\left (6+e^{e^x}-x\right ) x \log ^3\left (6+e^{e^x}-x\right )} \, dx\\ &=-\left (\left (10 (5+\log (3))^2\right ) \int \left (-\frac {2}{\left (6+e^{e^x}-x\right ) x \log ^3\left (6+e^{e^x}-x\right )}+\frac {1}{x^2 \log ^2\left (6+e^{e^x}-x\right )}\right ) \, dx\right )-\left (20 (5+\log (3))^2\right ) \int \frac {e^{e^x+x}}{\left (6+e^{e^x}-x\right ) x \log ^3\left (6+e^{e^x}-x\right )} \, dx\\ &=-\left (\left (10 (5+\log (3))^2\right ) \int \frac {1}{x^2 \log ^2\left (6+e^{e^x}-x\right )} \, dx\right )+\left (20 (5+\log (3))^2\right ) \int \frac {1}{\left (6+e^{e^x}-x\right ) x \log ^3\left (6+e^{e^x}-x\right )} \, dx-\left (20 (5+\log (3))^2\right ) \int \frac {e^{e^x+x}}{\left (6+e^{e^x}-x\right ) x \log ^3\left (6+e^{e^x}-x\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.14, size = 24, normalized size = 1.00 \begin {gather*} \frac {10 (5+\log (3))^2}{x \log ^2\left (6+e^{e^x}-x\right )} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.10, size = 27, normalized size = 1.12
method | result | size |
risch | \(\frac {10 \ln \left (3\right )^{2}+100 \ln \left (3\right )+250}{\ln \left ({\mathrm e}^{{\mathrm e}^{x}}-x +6\right )^{2} x}\) | \(27\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.54, size = 26, normalized size = 1.08 \begin {gather*} \frac {10 \, {\left (\log \left (3\right )^{2} + 10 \, \log \left (3\right ) + 25\right )}}{x \log \left (-x + e^{\left (e^{x}\right )} + 6\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.34, size = 38, normalized size = 1.58 \begin {gather*} \frac {10 \, {\left (\log \left (3\right )^{2} + 10 \, \log \left (3\right ) + 25\right )}}{x \log \left (-{\left ({\left (x - 6\right )} e^{x} - e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )}\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.19, size = 26, normalized size = 1.08 \begin {gather*} \frac {10 \log {\left (3 \right )}^{2} + 100 \log {\left (3 \right )} + 250}{x \log {\left (- x + e^{e^{x}} + 6 \right )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 4.01, size = 1482, normalized size = 61.75 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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