Integrand size = 198, antiderivative size = 26 \[ \int \frac {64^x \log ^{-1+x}\left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \left (-10 x+e^{e^x} \left (2 x+e^x \left (2 x+2 x^2\right )\right )+\left (-5-5 x+e^{e^x} (1+x)\right ) \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \log \left (64 \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right )\right )\right )}{-5-5 x+e^{e^x} (1+x)} \, dx=2+64^x \log ^x\left (\left (5-e^{e^x}\right )^2 (1+x)^2\right ) \]
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\[ \int \frac {64^x \log ^{-1+x}\left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \left (-10 x+e^{e^x} \left (2 x+e^x \left (2 x+2 x^2\right )\right )+\left (-5-5 x+e^{e^x} (1+x)\right ) \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \log \left (64 \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right )\right )\right )}{-5-5 x+e^{e^x} (1+x)} \, dx=\int \frac {64^x \log ^{-1+x}\left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \left (-10 x+e^{e^x} \left (2 x+e^x \left (2 x+2 x^2\right )\right )+\left (-5-5 x+e^{e^x} (1+x)\right ) \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \log \left (64 \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right )\right )\right )}{-5-5 x+e^{e^x} (1+x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {64^x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \left (10 x-2 e^{e^x} x \left (1+e^x (1+x)\right )-\left (-5+e^{e^x}\right ) (1+x) \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right )\right )}{\left (5-e^{e^x}\right ) (1+x)} \, dx \\ & = \int \left (\frac {2^{1+6 x} e^{e^x+x} x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{-5+e^{e^x}}+\frac {64^x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \left (2 x+\log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right )+x \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right )\right )}{1+x}\right ) \, dx \\ & = \int \frac {2^{1+6 x} e^{e^x+x} x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{-5+e^{e^x}} \, dx+\int \frac {64^x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \left (2 x+\log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right )+x \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right )\right )}{1+x} \, dx \\ & = \int \frac {2^{1+6 x} e^{e^x+x} x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{-5+e^{e^x}} \, dx+\int \frac {64^x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \left (2 x+(1+x) \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right )\right )}{1+x} \, dx \\ & = \int \frac {2^{1+6 x} e^{e^x+x} x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{-5+e^{e^x}} \, dx+\int \left (\frac {2^{1+6 x} x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{1+x}+64^x \log ^x\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right )\right ) \, dx \\ & = \int \frac {2^{1+6 x} e^{e^x+x} x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{-5+e^{e^x}} \, dx+\int \frac {2^{1+6 x} x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{1+x} \, dx+\int 64^x \log ^x\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right ) \, dx \\ & = \int \frac {2^{1+6 x} e^{e^x+x} x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{-5+e^{e^x}} \, dx+\int \left (2^{1+6 x} \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )+\frac {2^{1+6 x} \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{-1-x}\right ) \, dx+\int 64^x \log ^x\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right ) \, dx \\ & = \int 2^{1+6 x} \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \, dx+\int \frac {2^{1+6 x} \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{-1-x} \, dx+\int \frac {2^{1+6 x} e^{e^x+x} x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{-5+e^{e^x}} \, dx+\int 64^x \log ^x\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right ) \, dx \\ \end{align*}
\[ \int \frac {64^x \log ^{-1+x}\left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \left (-10 x+e^{e^x} \left (2 x+e^x \left (2 x+2 x^2\right )\right )+\left (-5-5 x+e^{e^x} (1+x)\right ) \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \log \left (64 \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right )\right )\right )}{-5-5 x+e^{e^x} (1+x)} \, dx=\int \frac {64^x \log ^{-1+x}\left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \left (-10 x+e^{e^x} \left (2 x+e^x \left (2 x+2 x^2\right )\right )+\left (-5-5 x+e^{e^x} (1+x)\right ) \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \log \left (64 \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right )\right )\right )}{-5-5 x+e^{e^x} (1+x)} \, dx \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.19 (sec) , antiderivative size = 170, normalized size of antiderivative = 6.54
\[{\left (128 \ln \left (1+x \right )+128 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )-32 i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right ) {\left (-\operatorname {csgn}\left (i \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )+\operatorname {csgn}\left (i \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )\right )\right )}^{2}-32 i \pi \,\operatorname {csgn}\left (i \left (1+x \right )^{2}\right ) {\left (-\operatorname {csgn}\left (i \left (1+x \right )^{2}\right )+\operatorname {csgn}\left (i \left (1+x \right )\right )\right )}^{2}-32 i \pi \,\operatorname {csgn}\left (i \left (1+x \right )^{2} \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right ) \left (-\operatorname {csgn}\left (i \left (1+x \right )^{2} \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )+\operatorname {csgn}\left (i \left (1+x \right )^{2}\right )\right ) \left (-\operatorname {csgn}\left (i \left (1+x \right )^{2} \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )+\operatorname {csgn}\left (i \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )\right )\right )}^{x}\]
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Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).
Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {64^x \log ^{-1+x}\left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \left (-10 x+e^{e^x} \left (2 x+e^x \left (2 x+2 x^2\right )\right )+\left (-5-5 x+e^{e^x} (1+x)\right ) \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \log \left (64 \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right )\right )\right )}{-5-5 x+e^{e^x} (1+x)} \, dx=\left (64 \, \log \left (25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, e^{x}\right )} - 10 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (e^{x}\right )} + 50 \, x + 25\right )\right )^{x} \]
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Timed out. \[ \int \frac {64^x \log ^{-1+x}\left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \left (-10 x+e^{e^x} \left (2 x+e^x \left (2 x+2 x^2\right )\right )+\left (-5-5 x+e^{e^x} (1+x)\right ) \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \log \left (64 \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right )\right )\right )}{-5-5 x+e^{e^x} (1+x)} \, dx=\text {Timed out} \]
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Time = 0.56 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {64^x \log ^{-1+x}\left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \left (-10 x+e^{e^x} \left (2 x+e^x \left (2 x+2 x^2\right )\right )+\left (-5-5 x+e^{e^x} (1+x)\right ) \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \log \left (64 \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right )\right )\right )}{-5-5 x+e^{e^x} (1+x)} \, dx=e^{\left (7 \, x \log \left (2\right ) + x \log \left (\log \left (x + 1\right ) + \log \left (e^{\left (e^{x}\right )} - 5\right )\right )\right )} \]
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\[ \int \frac {64^x \log ^{-1+x}\left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \left (-10 x+e^{e^x} \left (2 x+e^x \left (2 x+2 x^2\right )\right )+\left (-5-5 x+e^{e^x} (1+x)\right ) \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \log \left (64 \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right )\right )\right )}{-5-5 x+e^{e^x} (1+x)} \, dx=\int { \frac {{\left ({\left ({\left (x + 1\right )} e^{\left (e^{x}\right )} - 5 \, x - 5\right )} \log \left (25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, e^{x}\right )} - 10 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (e^{x}\right )} + 50 \, x + 25\right ) \log \left (64 \, \log \left (25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, e^{x}\right )} - 10 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (e^{x}\right )} + 50 \, x + 25\right )\right ) + 2 \, {\left ({\left (x^{2} + x\right )} e^{x} + x\right )} e^{\left (e^{x}\right )} - 10 \, x\right )} \left (64 \, \log \left (25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, e^{x}\right )} - 10 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (e^{x}\right )} + 50 \, x + 25\right )\right )^{x}}{{\left ({\left (x + 1\right )} e^{\left (e^{x}\right )} - 5 \, x - 5\right )} \log \left (25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, e^{x}\right )} - 10 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (e^{x}\right )} + 50 \, x + 25\right )} \,d x } \]
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Time = 11.85 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.31 \[ \int \frac {64^x \log ^{-1+x}\left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \left (-10 x+e^{e^x} \left (2 x+e^x \left (2 x+2 x^2\right )\right )+\left (-5-5 x+e^{e^x} (1+x)\right ) \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \log \left (64 \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right )\right )\right )}{-5-5 x+e^{e^x} (1+x)} \, dx=2^{6\,x}\,{\ln \left (50\,x-10\,{\mathrm {e}}^{{\mathrm {e}}^x}+{\mathrm {e}}^{2\,{\mathrm {e}}^x}-20\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}+2\,x\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}-10\,x^2\,{\mathrm {e}}^{{\mathrm {e}}^x}+25\,x^2+x^2\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}+25\right )}^x \]
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