Integrand size = 74, antiderivative size = 27 \[ \int \frac {e^x \left (-25+235 x+109 x^2+13 x^3+\left (-20 x-20 x^2-4 x^3\right ) \log (2)\right )+e^x \left (-25 x-10 x^2-x^3\right ) \log (x)}{125 x+50 x^2+5 x^3} \, dx=\frac {1}{5} e^x \left (9+\frac {4 x (1-\log (2))}{5+x}-\log (x)\right ) \]
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Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 1.52 (sec) , antiderivative size = 137, normalized size of antiderivative = 5.07, number of steps used = 28, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.149, Rules used = {1608, 27, 12, 6873, 6820, 6874, 2209, 2208, 2230, 2225, 2634} \[ \int \frac {e^x \left (-25+235 x+109 x^2+13 x^3+\left (-20 x-20 x^2-4 x^3\right ) \log (2)\right )+e^x \left (-25 x-10 x^2-x^3\right ) \log (x)}{125 x+50 x^2+5 x^3} \, dx=\frac {6 \operatorname {ExpIntegralEi}(x+5)}{5 e^5}+\frac {(47-\log (16)) \operatorname {ExpIntegralEi}(x+5)}{e^5}+\frac {3 (13-\log (16)) \operatorname {ExpIntegralEi}(x+5)}{e^5}-\frac {4 (109-20 \log (2)) \operatorname {ExpIntegralEi}(x+5)}{5 e^5}-\frac {e^x}{x+5}-\frac {1}{5} e^x \log (x)-\frac {e^x (47-\log (16))}{x+5}+\frac {1}{5} e^x (13-\log (16))-\frac {5 e^x (13-\log (16))}{x+5}+\frac {e^x (109-20 \log (2))}{x+5} \]
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Rule 12
Rule 27
Rule 1608
Rule 2208
Rule 2209
Rule 2225
Rule 2230
Rule 2634
Rule 6820
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \left (-25+235 x+109 x^2+13 x^3+\left (-20 x-20 x^2-4 x^3\right ) \log (2)\right )+e^x \left (-25 x-10 x^2-x^3\right ) \log (x)}{x \left (125+50 x+5 x^2\right )} \, dx \\ & = \int \frac {e^x \left (-25+235 x+109 x^2+13 x^3+\left (-20 x-20 x^2-4 x^3\right ) \log (2)\right )+e^x \left (-25 x-10 x^2-x^3\right ) \log (x)}{5 x (5+x)^2} \, dx \\ & = \frac {1}{5} \int \frac {e^x \left (-25+235 x+109 x^2+13 x^3+\left (-20 x-20 x^2-4 x^3\right ) \log (2)\right )+e^x \left (-25 x-10 x^2-x^3\right ) \log (x)}{x (5+x)^2} \, dx \\ & = \frac {1}{5} \int \frac {e^x \left (-25+13 x^3 \left (1-\frac {4 \log (2)}{13}\right )+109 x^2 \left (1-\frac {20 \log (2)}{109}\right )+235 x \left (1-\frac {4 \log (2)}{47}\right )-25 x \log (x)-10 x^2 \log (x)-x^3 \log (x)\right )}{x (5+x)^2} \, dx \\ & = \frac {1}{5} \int \frac {e^x \left (-25-x^2 (-109+20 \log (2))-5 x (-47+\log (16))-x^3 (-13+\log (16))-x (5+x)^2 \log (x)\right )}{x (5+x)^2} \, dx \\ & = \frac {1}{5} \int \left (-\frac {25 e^x}{x (5+x)^2}-\frac {e^x x (-109+20 \log (2))}{(5+x)^2}-\frac {5 e^x (-47+\log (16))}{(5+x)^2}-\frac {e^x x^2 (-13+\log (16))}{(5+x)^2}-e^x \log (x)\right ) \, dx \\ & = -\left (\frac {1}{5} \int e^x \log (x) \, dx\right )-5 \int \frac {e^x}{x (5+x)^2} \, dx+\frac {1}{5} (109-20 \log (2)) \int \frac {e^x x}{(5+x)^2} \, dx+\frac {1}{5} (13-\log (16)) \int \frac {e^x x^2}{(5+x)^2} \, dx+(47-\log (16)) \int \frac {e^x}{(5+x)^2} \, dx \\ & = -\frac {e^x (47-\log (16))}{5+x}-\frac {1}{5} e^x \log (x)+\frac {1}{5} \int \frac {e^x}{x} \, dx-5 \int \left (\frac {e^x}{25 x}-\frac {e^x}{5 (5+x)^2}-\frac {e^x}{25 (5+x)}\right ) \, dx+\frac {1}{5} (109-20 \log (2)) \int \left (-\frac {5 e^x}{(5+x)^2}+\frac {e^x}{5+x}\right ) \, dx+\frac {1}{5} (13-\log (16)) \int \left (e^x+\frac {25 e^x}{(5+x)^2}-\frac {10 e^x}{5+x}\right ) \, dx+(47-\log (16)) \int \frac {e^x}{5+x} \, dx \\ & = \frac {\operatorname {ExpIntegralEi}(x)}{5}-\frac {e^x (47-\log (16))}{5+x}+\frac {\operatorname {ExpIntegralEi}(5+x) (47-\log (16))}{e^5}-\frac {1}{5} e^x \log (x)-\frac {1}{5} \int \frac {e^x}{x} \, dx+\frac {1}{5} \int \frac {e^x}{5+x} \, dx+\frac {1}{5} (109-20 \log (2)) \int \frac {e^x}{5+x} \, dx+(-109+20 \log (2)) \int \frac {e^x}{(5+x)^2} \, dx+\frac {1}{5} (13-\log (16)) \int e^x \, dx+(5 (13-\log (16))) \int \frac {e^x}{(5+x)^2} \, dx+(2 (-13+\log (16))) \int \frac {e^x}{5+x} \, dx+\int \frac {e^x}{(5+x)^2} \, dx \\ & = -\frac {e^x}{5+x}+\frac {\operatorname {ExpIntegralEi}(5+x)}{5 e^5}+\frac {e^x (109-20 \log (2))}{5+x}+\frac {\operatorname {ExpIntegralEi}(5+x) (109-20 \log (2))}{5 e^5}+\frac {1}{5} e^x (13-\log (16))-\frac {5 e^x (13-\log (16))}{5+x}-\frac {2 \operatorname {ExpIntegralEi}(5+x) (13-\log (16))}{e^5}-\frac {e^x (47-\log (16))}{5+x}+\frac {\operatorname {ExpIntegralEi}(5+x) (47-\log (16))}{e^5}-\frac {1}{5} e^x \log (x)+(-109+20 \log (2)) \int \frac {e^x}{5+x} \, dx+(5 (13-\log (16))) \int \frac {e^x}{5+x} \, dx+\int \frac {e^x}{5+x} \, dx \\ & = -\frac {e^x}{5+x}+\frac {6 \operatorname {ExpIntegralEi}(5+x)}{5 e^5}+\frac {e^x (109-20 \log (2))}{5+x}-\frac {4 \operatorname {ExpIntegralEi}(5+x) (109-20 \log (2))}{5 e^5}+\frac {1}{5} e^x (13-\log (16))-\frac {5 e^x (13-\log (16))}{5+x}+\frac {3 \operatorname {ExpIntegralEi}(5+x) (13-\log (16))}{e^5}-\frac {e^x (47-\log (16))}{5+x}+\frac {\operatorname {ExpIntegralEi}(5+x) (47-\log (16))}{e^5}-\frac {1}{5} e^x \log (x) \\ \end{align*}
Time = 5.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^x \left (-25+235 x+109 x^2+13 x^3+\left (-20 x-20 x^2-4 x^3\right ) \log (2)\right )+e^x \left (-25 x-10 x^2-x^3\right ) \log (x)}{125 x+50 x^2+5 x^3} \, dx=-\frac {e^x (-45+x (-13+\log (16))+(5+x) \log (x))}{5 (5+x)} \]
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Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-\frac {{\mathrm e}^{x} \ln \left (x \right )}{5}-\frac {\left (4 x \ln \left (2\right )-13 x -45\right ) {\mathrm e}^{x}}{5 \left (5+x \right )}\) | \(27\) |
norman | \(\frac {\left (-\frac {4 \ln \left (2\right )}{5}+\frac {13}{5}\right ) x \,{\mathrm e}^{x}-{\mathrm e}^{x} \ln \left (x \right )-\frac {x \,{\mathrm e}^{x} \ln \left (x \right )}{5}+9 \,{\mathrm e}^{x}}{5+x}\) | \(35\) |
parallelrisch | \(-\frac {4 x \ln \left (2\right ) {\mathrm e}^{x}+x \,{\mathrm e}^{x} \ln \left (x \right )-13 \,{\mathrm e}^{x} x +5 \,{\mathrm e}^{x} \ln \left (x \right )-45 \,{\mathrm e}^{x}}{5 \left (5+x \right )}\) | \(37\) |
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Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^x \left (-25+235 x+109 x^2+13 x^3+\left (-20 x-20 x^2-4 x^3\right ) \log (2)\right )+e^x \left (-25 x-10 x^2-x^3\right ) \log (x)}{125 x+50 x^2+5 x^3} \, dx=-\frac {{\left (x + 5\right )} e^{x} \log \left (x\right ) + {\left (4 \, x \log \left (2\right ) - 13 \, x - 45\right )} e^{x}}{5 \, {\left (x + 5\right )}} \]
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Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^x \left (-25+235 x+109 x^2+13 x^3+\left (-20 x-20 x^2-4 x^3\right ) \log (2)\right )+e^x \left (-25 x-10 x^2-x^3\right ) \log (x)}{125 x+50 x^2+5 x^3} \, dx=\frac {\left (- x \log {\left (x \right )} - 4 x \log {\left (2 \right )} + 13 x - 5 \log {\left (x \right )} + 45\right ) e^{x}}{5 x + 25} \]
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Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^x \left (-25+235 x+109 x^2+13 x^3+\left (-20 x-20 x^2-4 x^3\right ) \log (2)\right )+e^x \left (-25 x-10 x^2-x^3\right ) \log (x)}{125 x+50 x^2+5 x^3} \, dx=-\frac {{\left (x {\left (4 \, \log \left (2\right ) - 13\right )} + {\left (x + 5\right )} \log \left (x\right ) - 45\right )} e^{x}}{5 \, {\left (x + 5\right )}} \]
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Time = 0.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {e^x \left (-25+235 x+109 x^2+13 x^3+\left (-20 x-20 x^2-4 x^3\right ) \log (2)\right )+e^x \left (-25 x-10 x^2-x^3\right ) \log (x)}{125 x+50 x^2+5 x^3} \, dx=-\frac {4 \, x e^{x} \log \left (2\right ) + x e^{x} \log \left (x\right ) - 13 \, x e^{x} + 5 \, e^{x} \log \left (x\right ) - 45 \, e^{x}}{5 \, {\left (x + 5\right )}} \]
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Time = 8.57 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^x \left (-25+235 x+109 x^2+13 x^3+\left (-20 x-20 x^2-4 x^3\right ) \log (2)\right )+e^x \left (-25 x-10 x^2-x^3\right ) \log (x)}{125 x+50 x^2+5 x^3} \, dx=\frac {45\,{\mathrm {e}}^x-x\,{\mathrm {e}}^x\,\left (\ln \left (16\right )-13\right )}{5\,x+25}-\frac {{\mathrm {e}}^x\,\ln \left (x\right )}{5} \]
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