Integrand size = 101, antiderivative size = 24 \[ \int \frac {-256+192 x-48 x^2+4 x^3+e^{2 x} (-10+2 x) \log (16)+\left (256-192 x+48 x^2-4 x^3\right ) \log (16)}{-256 x+192 x^2-48 x^3+4 x^4+e^{2 x} (-4+x) \log (16)+\left (256 x-192 x^2+48 x^3-4 x^4\right ) \log (16)} \, dx=\log \left (\frac {e^{2 x}}{(-4+x)^2}-4 \left (x-\frac {x}{\log (16)}\right )\right ) \]
[Out]
\[ \int \frac {-256+192 x-48 x^2+4 x^3+e^{2 x} (-10+2 x) \log (16)+\left (256-192 x+48 x^2-4 x^3\right ) \log (16)}{-256 x+192 x^2-48 x^3+4 x^4+e^{2 x} (-4+x) \log (16)+\left (256 x-192 x^2+48 x^3-4 x^4\right ) \log (16)} \, dx=\int \frac {-256+192 x-48 x^2+4 x^3+e^{2 x} (-10+2 x) \log (16)+\left (256-192 x+48 x^2-4 x^3\right ) \log (16)}{-256 x+192 x^2-48 x^3+4 x^4+e^{2 x} (-4+x) \log (16)+\left (256 x-192 x^2+48 x^3-4 x^4\right ) \log (16)} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \frac {256-192 x+48 x^2-4 x^3-e^{2 x} (-10+2 x) \log (16)-\left (256-192 x+48 x^2-4 x^3\right ) \log (16)}{(4-x) \left (64 x (1-\log (16))-32 x^2 (1-\log (16))+4 x^3 (1-\log (16))+e^{2 x} \log (16)\right )} \, dx \\ & = \int \left (\frac {2 (-5+x)}{-4+x}+\frac {4 \left (16-48 x+19 x^2-2 x^3\right ) (1-\log (16))}{64 x (1-\log (16))-32 x^2 (1-\log (16))+4 x^3 (1-\log (16))+e^{2 x} \log (16)}\right ) \, dx \\ & = 2 \int \frac {-5+x}{-4+x} \, dx+(4 (1-\log (16))) \int \frac {16-48 x+19 x^2-2 x^3}{64 x (1-\log (16))-32 x^2 (1-\log (16))+4 x^3 (1-\log (16))+e^{2 x} \log (16)} \, dx \\ & = 2 \int \left (1+\frac {1}{4-x}\right ) \, dx+(4 (1-\log (16))) \int \left (\frac {48 x}{-64 x (1-\log (16))+32 x^2 (1-\log (16))-4 x^3 (1-\log (16))-e^{2 x} \log (16)}+\frac {2 x^3}{-64 x (1-\log (16))+32 x^2 (1-\log (16))-4 x^3 (1-\log (16))-e^{2 x} \log (16)}+\frac {16}{64 x (1-\log (16))-32 x^2 (1-\log (16))+4 x^3 (1-\log (16))+e^{2 x} \log (16)}+\frac {19 x^2}{64 x (1-\log (16))-32 x^2 (1-\log (16))+4 x^3 (1-\log (16))+e^{2 x} \log (16)}\right ) \, dx \\ & = 2 x-2 \log (4-x)+(8 (1-\log (16))) \int \frac {x^3}{-64 x (1-\log (16))+32 x^2 (1-\log (16))-4 x^3 (1-\log (16))-e^{2 x} \log (16)} \, dx+(64 (1-\log (16))) \int \frac {1}{64 x (1-\log (16))-32 x^2 (1-\log (16))+4 x^3 (1-\log (16))+e^{2 x} \log (16)} \, dx+(76 (1-\log (16))) \int \frac {x^2}{64 x (1-\log (16))-32 x^2 (1-\log (16))+4 x^3 (1-\log (16))+e^{2 x} \log (16)} \, dx+(192 (1-\log (16))) \int \frac {x}{-64 x (1-\log (16))+32 x^2 (1-\log (16))-4 x^3 (1-\log (16))-e^{2 x} \log (16)} \, dx \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(52\) vs. \(2(24)=48\).
Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17 \[ \int \frac {-256+192 x-48 x^2+4 x^3+e^{2 x} (-10+2 x) \log (16)+\left (256-192 x+48 x^2-4 x^3\right ) \log (16)}{-256 x+192 x^2-48 x^3+4 x^4+e^{2 x} (-4+x) \log (16)+\left (256 x-192 x^2+48 x^3-4 x^4\right ) \log (16)} \, dx=-2 \log (4-x)+\log \left (-64 x+32 x^2-4 x^3-e^{2 x} \log (16)+64 x \log (16)-32 x^2 \log (16)+4 x^3 \log (16)\right ) \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. \(46\) vs. \(2(21)=42\).
Time = 0.40 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96
method | result | size |
risch | \(-2 \ln \left (x -4\right )+\ln \left ({\mathrm e}^{2 x}-\frac {x \left (4 x^{2} \ln \left (2\right )-32 x \ln \left (2\right )-x^{2}+64 \ln \left (2\right )+8 x -16\right )}{\ln \left (2\right )}\right )\) | \(47\) |
norman | \(-2 \ln \left (x -4\right )+\ln \left (4 x^{3} \ln \left (2\right )-32 x^{2} \ln \left (2\right )-\ln \left (2\right ) {\mathrm e}^{2 x}-x^{3}+64 x \ln \left (2\right )+8 x^{2}-16 x \right )\) | \(50\) |
parallelrisch | \(\ln \left (\frac {4 x^{3} \ln \left (2\right )-32 x^{2} \ln \left (2\right )-\ln \left (2\right ) {\mathrm e}^{2 x}-x^{3}+64 x \ln \left (2\right )+8 x^{2}-16 x}{4 \ln \left (2\right )-1}\right )-2 \ln \left (x -4\right )\) | \(59\) |
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (21) = 42\).
Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int \frac {-256+192 x-48 x^2+4 x^3+e^{2 x} (-10+2 x) \log (16)+\left (256-192 x+48 x^2-4 x^3\right ) \log (16)}{-256 x+192 x^2-48 x^3+4 x^4+e^{2 x} (-4+x) \log (16)+\left (256 x-192 x^2+48 x^3-4 x^4\right ) \log (16)} \, dx=\log \left (x^{3} - 8 \, x^{2} - 4 \, {\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} \log \left (2\right ) + e^{\left (2 \, x\right )} \log \left (2\right ) + 16 \, x\right ) - 2 \, \log \left (x - 4\right ) \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (19) = 38\).
Time = 0.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.12 \[ \int \frac {-256+192 x-48 x^2+4 x^3+e^{2 x} (-10+2 x) \log (16)+\left (256-192 x+48 x^2-4 x^3\right ) \log (16)}{-256 x+192 x^2-48 x^3+4 x^4+e^{2 x} (-4+x) \log (16)+\left (256 x-192 x^2+48 x^3-4 x^4\right ) \log (16)} \, dx=- 2 \log {\left (x - 4 \right )} + \log {\left (\frac {- 4 x^{3} \log {\left (2 \right )} + x^{3} - 8 x^{2} + 32 x^{2} \log {\left (2 \right )} - 64 x \log {\left (2 \right )} + 16 x}{\log {\left (2 \right )}} + e^{2 x} \right )} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (21) = 42\).
Time = 0.31 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.21 \[ \int \frac {-256+192 x-48 x^2+4 x^3+e^{2 x} (-10+2 x) \log (16)+\left (256-192 x+48 x^2-4 x^3\right ) \log (16)}{-256 x+192 x^2-48 x^3+4 x^4+e^{2 x} (-4+x) \log (16)+\left (256 x-192 x^2+48 x^3-4 x^4\right ) \log (16)} \, dx=-2 \, \log \left (x - 4\right ) + \log \left (-\frac {x^{3} {\left (4 \, \log \left (2\right ) - 1\right )} - 8 \, x^{2} {\left (4 \, \log \left (2\right ) - 1\right )} + 16 \, x {\left (4 \, \log \left (2\right ) - 1\right )} - e^{\left (2 \, x\right )} \log \left (2\right )}{\log \left (2\right )}\right ) \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (21) = 42\).
Time = 0.29 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.92 \[ \int \frac {-256+192 x-48 x^2+4 x^3+e^{2 x} (-10+2 x) \log (16)+\left (256-192 x+48 x^2-4 x^3\right ) \log (16)}{-256 x+192 x^2-48 x^3+4 x^4+e^{2 x} (-4+x) \log (16)+\left (256 x-192 x^2+48 x^3-4 x^4\right ) \log (16)} \, dx=\log \left (-4 \, x^{3} \log \left (2\right ) + x^{3} + 32 \, x^{2} \log \left (2\right ) - 8 \, x^{2} - 64 \, x \log \left (2\right ) + e^{\left (2 \, x\right )} \log \left (2\right ) + 16 \, x\right ) - 2 \, \log \left (x - 4\right ) \]
[In]
[Out]
Timed out. \[ \int \frac {-256+192 x-48 x^2+4 x^3+e^{2 x} (-10+2 x) \log (16)+\left (256-192 x+48 x^2-4 x^3\right ) \log (16)}{-256 x+192 x^2-48 x^3+4 x^4+e^{2 x} (-4+x) \log (16)+\left (256 x-192 x^2+48 x^3-4 x^4\right ) \log (16)} \, dx=\int \frac {192\,x-4\,\ln \left (2\right )\,\left (4\,x^3-48\,x^2+192\,x-256\right )-48\,x^2+4\,x^3+4\,{\mathrm {e}}^{2\,x}\,\ln \left (2\right )\,\left (2\,x-10\right )-256}{4\,\ln \left (2\right )\,\left (-4\,x^4+48\,x^3-192\,x^2+256\,x\right )-256\,x+192\,x^2-48\,x^3+4\,x^4+4\,{\mathrm {e}}^{2\,x}\,\ln \left (2\right )\,\left (x-4\right )} \,d x \]
[In]
[Out]