Integrand size = 58, antiderivative size = 19 \[ \int \frac {5 x+22 x^2+13 x^3+2 x^4+\left (5+4 x+5 x^2+4 x^3\right ) \log \left (e^x \left (1+2 x^2+x^4\right )\right )}{1+x^2} \, dx=x (5+2 x) \log \left (e^x \left (1+x^2\right )^2\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(39\) vs. \(2(19)=38\).
Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.05, number of steps used = 13, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {6857, 1816, 649, 209, 266, 2631, 1643} \[ \int \frac {5 x+22 x^2+13 x^3+2 x^4+\left (5+4 x+5 x^2+4 x^3\right ) \log \left (e^x \left (1+2 x^2+x^4\right )\right )}{1+x^2} \, dx=\frac {1}{8} (4 x+5)^2 \log \left (e^x \left (x^2+1\right )^2\right )-\frac {25}{4} \log \left (x^2+1\right )-\frac {25 x}{8} \]
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Rule 209
Rule 266
Rule 649
Rule 1643
Rule 1816
Rule 2631
Rule 6857
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {x \left (5+22 x+13 x^2+2 x^3\right )}{1+x^2}+(5+4 x) \log \left (e^x \left (1+x^2\right )^2\right )\right ) \, dx \\ & = \int \frac {x \left (5+22 x+13 x^2+2 x^3\right )}{1+x^2} \, dx+\int (5+4 x) \log \left (e^x \left (1+x^2\right )^2\right ) \, dx \\ & = \frac {1}{8} (5+4 x)^2 \log \left (e^x \left (1+x^2\right )^2\right )-\frac {1}{8} \int \frac {(5+4 x)^2 \left (1+4 x+x^2\right )}{1+x^2} \, dx+\int \left (20+13 x+2 x^2-\frac {4 (5+2 x)}{1+x^2}\right ) \, dx \\ & = 20 x+\frac {13 x^2}{2}+\frac {2 x^3}{3}+\frac {1}{8} (5+4 x)^2 \log \left (e^x \left (1+x^2\right )^2\right )-\frac {1}{8} \int \left (185+104 x+16 x^2-\frac {4 (40-9 x)}{1+x^2}\right ) \, dx-4 \int \frac {5+2 x}{1+x^2} \, dx \\ & = -\frac {25 x}{8}+\frac {1}{8} (5+4 x)^2 \log \left (e^x \left (1+x^2\right )^2\right )+\frac {1}{2} \int \frac {40-9 x}{1+x^2} \, dx-8 \int \frac {x}{1+x^2} \, dx-20 \int \frac {1}{1+x^2} \, dx \\ & = -\frac {25 x}{8}-20 \arctan (x)-4 \log \left (1+x^2\right )+\frac {1}{8} (5+4 x)^2 \log \left (e^x \left (1+x^2\right )^2\right )-\frac {9}{2} \int \frac {x}{1+x^2} \, dx+20 \int \frac {1}{1+x^2} \, dx \\ & = -\frac {25 x}{8}-\frac {25}{4} \log \left (1+x^2\right )+\frac {1}{8} (5+4 x)^2 \log \left (e^x \left (1+x^2\right )^2\right ) \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {5 x+22 x^2+13 x^3+2 x^4+\left (5+4 x+5 x^2+4 x^3\right ) \log \left (e^x \left (1+2 x^2+x^4\right )\right )}{1+x^2} \, dx=x (5+2 x) \log \left (e^x \left (1+x^2\right )^2\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. \(37\) vs. \(2(18)=36\).
Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 2.00
method | result | size |
default | \(2 \ln \left (\left (x^{4}+2 x^{2}+1\right ) {\mathrm e}^{x}\right ) x^{2}+5 \ln \left (\left (x^{4}+2 x^{2}+1\right ) {\mathrm e}^{x}\right ) x\) | \(38\) |
norman | \(2 \ln \left (\left (x^{4}+2 x^{2}+1\right ) {\mathrm e}^{x}\right ) x^{2}+5 \ln \left (\left (x^{4}+2 x^{2}+1\right ) {\mathrm e}^{x}\right ) x\) | \(38\) |
parts | \(2 \ln \left (\left (x^{4}+2 x^{2}+1\right ) {\mathrm e}^{x}\right ) x^{2}+5 \ln \left (\left (x^{4}+2 x^{2}+1\right ) {\mathrm e}^{x}\right ) x\) | \(38\) |
parallelrisch | \(2 \ln \left (\left (x^{4}+2 x^{2}+1\right ) {\mathrm e}^{x}\right ) x^{2}+5 \ln \left (\left (x^{4}+2 x^{2}+1\right ) {\mathrm e}^{x}\right ) x +2 \ln \left (x^{2}+1\right )+x -\ln \left (\left (x^{4}+2 x^{2}+1\right ) {\mathrm e}^{x}\right )\) | \(63\) |
risch | \(\left (2 x^{2}+5 x \right ) \ln \left ({\mathrm e}^{x}\right )+4 x^{2} \ln \left (x^{2}+1\right )+10 \ln \left (x^{2}+1\right ) x -i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{x} \left (x^{2}+1\right )^{2}\right )^{3}+\frac {5 i \pi x \,\operatorname {csgn}\left (i \left (x^{2}+1\right )^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (x^{2}+1\right )^{2}\right )^{2}}{2}+5 i \pi x \,\operatorname {csgn}\left (i \left (x^{2}+1\right )\right ) {\operatorname {csgn}\left (i \left (x^{2}+1\right )^{2}\right )}^{2}+\frac {5 i \pi x \,\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (x^{2}+1\right )^{2}\right )^{2}}{2}+i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (x^{2}+1\right )^{2}\right )^{2}-i \pi \,x^{2} {\operatorname {csgn}\left (i \left (x^{2}+1\right )\right )}^{2} \operatorname {csgn}\left (i \left (x^{2}+1\right )^{2}\right )-i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i \left (x^{2}+1\right )^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (x^{2}+1\right )^{2}\right )-i \pi \,x^{2} {\operatorname {csgn}\left (i \left (x^{2}+1\right )^{2}\right )}^{3}-\frac {5 i \pi x {\operatorname {csgn}\left (i \left (x^{2}+1\right )\right )}^{2} \operatorname {csgn}\left (i \left (x^{2}+1\right )^{2}\right )}{2}-\frac {5 i \pi x {\operatorname {csgn}\left (i \left (x^{2}+1\right )^{2}\right )}^{3}}{2}+2 i \pi \,x^{2} \operatorname {csgn}\left (i \left (x^{2}+1\right )\right ) {\operatorname {csgn}\left (i \left (x^{2}+1\right )^{2}\right )}^{2}+i \pi \,x^{2} \operatorname {csgn}\left (i \left (x^{2}+1\right )^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (x^{2}+1\right )^{2}\right )^{2}-\frac {5 i \pi x \operatorname {csgn}\left (i {\mathrm e}^{x} \left (x^{2}+1\right )^{2}\right )^{3}}{2}-\frac {5 i \pi x \,\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i \left (x^{2}+1\right )^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (x^{2}+1\right )^{2}\right )}{2}\) | \(417\) |
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Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {5 x+22 x^2+13 x^3+2 x^4+\left (5+4 x+5 x^2+4 x^3\right ) \log \left (e^x \left (1+2 x^2+x^4\right )\right )}{1+x^2} \, dx={\left (2 \, x^{2} + 5 \, x\right )} \log \left ({\left (x^{4} + 2 \, x^{2} + 1\right )} e^{x}\right ) \]
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Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.79 \[ \int \frac {5 x+22 x^2+13 x^3+2 x^4+\left (5+4 x+5 x^2+4 x^3\right ) \log \left (e^x \left (1+2 x^2+x^4\right )\right )}{1+x^2} \, dx=- x + \left (2 x^{2} + 5 x + 1\right ) \log {\left (\left (x^{4} + 2 x^{2} + 1\right ) e^{x} \right )} - 2 \log {\left (x^{2} + 1 \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (18) = 36\).
Time = 0.42 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.95 \[ \int \frac {5 x+22 x^2+13 x^3+2 x^4+\left (5+4 x+5 x^2+4 x^3\right ) \log \left (e^x \left (1+2 x^2+x^4\right )\right )}{1+x^2} \, dx=2 \, x^{3} + 5 \, x^{2} + 2 \, {\left (2 \, x^{2} + 5 \, x + 2\right )} \log \left (x^{2} + 1\right ) - 4 \, \log \left (x^{2} + 1\right ) \]
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Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.47 \[ \int \frac {5 x+22 x^2+13 x^3+2 x^4+\left (5+4 x+5 x^2+4 x^3\right ) \log \left (e^x \left (1+2 x^2+x^4\right )\right )}{1+x^2} \, dx=2 \, x^{3} + 5 \, x^{2} + 2 \, {\left (2 \, x^{2} + 5 \, x\right )} \log \left (x^{2} + 1\right ) \]
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Time = 9.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {5 x+22 x^2+13 x^3+2 x^4+\left (5+4 x+5 x^2+4 x^3\right ) \log \left (e^x \left (1+2 x^2+x^4\right )\right )}{1+x^2} \, dx=\left (2\,x^2+5\,x\right )\,\left (x+\ln \left (x^4+2\,x^2+1\right )\right ) \]
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