Optimal. Leaf size=32 \[ \log \left (\frac {x}{2+\frac {1}{\log (2)}+\frac {5}{x+\frac {1}{4} e^x (-x+5 \log (5))}}\right ) \]
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Rubi [F] time = 6.87, 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 {16 x^2+\left (160 x+32 x^2\right ) \log (2)+e^x \left (-8 x^2+\left (-40 x-36 x^2\right ) \log (2)+(40 x+(100+180 x) \log (2)) \log (5)\right )+e^{2 x} \left (x^2+2 x^2 \log (2)+(-10 x-20 x \log (2)) \log (5)+(25+50 \log (2)) \log ^2(5)\right )}{16 x^3+\left (80 x^2+32 x^3\right ) \log (2)+e^x \left (-8 x^3+\left (-20 x^2-16 x^3\right ) \log (2)+\left (40 x^2+\left (100 x+80 x^2\right ) \log (2)\right ) \log (5)\right )+e^{2 x} \left (x^3+2 x^3 \log (2)+\left (-10 x^2-20 x^2 \log (2)\right ) \log (5)+(25 x+50 x \log (2)) \log ^2(5)\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 {25 e^x \log (5) \left (e^x (1+\log (4)) \log (5)+\log (16)\right )+x^2 \left (16+32 \log (2)+e^{2 x} (1+\log (4))-4 e^x (2+\log (512))\right )-10 x \left (-16 \log (2)+e^{2 x} (1+\log (4)) \log (5)+e^x (\log (16)-2 \log (5) (2+\log (512)))\right )}{x \left (\left (-4+e^x\right ) x-5 e^x \log (5)\right ) \left (x \left (-4-8 \log (2)+e^x (1+\log (4))\right )-5 \left (e^x (1+\log (4)) \log (5)+\log (16)\right )\right )} \, dx\\ &=\int \left (\frac {1}{x}+\frac {4 \left (25 \log (5) \log (16)+4 x^2 \log (32)-10 x \log (25) \log (32)\right )}{(x-5 \log (5)) \left (-4 x+e^x x-5 e^x \log (5)\right ) \log (1048576)}+\frac {20 x \left (16 \log ^2(2) \log (5)+4 \log (4) \log (5)+\log (256)-4 \log (5) \log (512)+\log (16) \log (512)-8 \log (2) \left (1-\log \left (\frac {5}{4}\right )-\log (4) \log (5)+\log (5) \log (512)\right )\right )+16 x^2 \log (2) (5+\log (1024))+25 (\log (16) (\log (16)-4 \log (5) (1+\log (512)))+8 \log (2) \log (5) (4+\log (4096)))}{(x-5 \log (5)) \left (4 x (1+\log (4))-e^x x (1+\log (4))+5 e^x (1+\log (4)) \log (5)+5 \log (16)\right ) \log (1048576)}\right ) \, dx\\ &=\log (x)+\frac {\int \frac {20 x \left (16 \log ^2(2) \log (5)+4 \log (4) \log (5)+\log (256)-4 \log (5) \log (512)+\log (16) \log (512)-8 \log (2) \left (1-\log \left (\frac {5}{4}\right )-\log (4) \log (5)+\log (5) \log (512)\right )\right )+16 x^2 \log (2) (5+\log (1024))+25 (\log (16) (\log (16)-4 \log (5) (1+\log (512)))+8 \log (2) \log (5) (4+\log (4096)))}{(x-5 \log (5)) \left (4 x (1+\log (4))-e^x x (1+\log (4))+5 e^x (1+\log (4)) \log (5)+5 \log (16)\right )} \, dx}{\log (1048576)}+\frac {4 \int \frac {25 \log (5) \log (16)+4 x^2 \log (32)-10 x \log (25) \log (32)}{(x-5 \log (5)) \left (-4 x+e^x x-5 e^x \log (5)\right )} \, dx}{\log (1048576)}\\ &=\log (x)+\frac {\int \left (\frac {16 x \log (2) (5+\log (1024))}{4 x (1+\log (4))-e^x x (1+\log (4))+5 e^x (1+\log (4)) \log (5)+5 \log (16)}+\frac {20 \left (16 \log ^2(2) \log (5)+4 \log (4) \log (5)+\log (256)-4 \log (5) \log (512)+\log (16) \log (512)-4 \log (2) \left (2+\log (5) \log (16)-\log \left (\frac {78125}{16}\right )\right )\right )}{4 x (1+\log (4))-e^x x (1+\log (4))+5 e^x (1+\log (4)) \log (5)+5 \log (16)}+\frac {25 \left (32 \log \left (\frac {5}{4}\right ) \log (2) \log (5)+64 \log ^2(2) \log ^2(5)+16 \log (4) \log ^2(5)-4 \log (5) \log (16)+\log ^2(16)+4 \log (5) \log (256)-16 \log ^2(5) \log (512)+8 \log (2) \log (5) (\log (25) \log (1024)+\log (4096)+2 \log (5) (5-\log (16384)))\right )}{(x-5 \log (5)) \left (4 x (1+\log (4))-e^x x (1+\log (4))+5 e^x (1+\log (4)) \log (5)+5 \log (16)\right )}\right ) \, dx}{\log (1048576)}+\frac {4 \int \left (\frac {25 \log (5) \log (16)}{(x-5 \log (5)) \left (-4 x+e^x x-5 e^x \log (5)\right )}+\frac {4 x \log (32)}{-4 x+e^x x-5 e^x \log (5)}\right ) \, dx}{\log (1048576)}\\ &=\log (x)+\frac {(100 \log (5) \log (16)) \int \frac {1}{(x-5 \log (5)) \left (-4 x+e^x x-5 e^x \log (5)\right )} \, dx}{\log (1048576)}+\frac {(16 \log (32)) \int \frac {x}{-4 x+e^x x-5 e^x \log (5)} \, dx}{\log (1048576)}+\frac {(16 \log (2) (5+\log (1024))) \int \frac {x}{4 x (1+\log (4))-e^x x (1+\log (4))+5 e^x (1+\log (4)) \log (5)+5 \log (16)} \, dx}{\log (1048576)}+\frac {\left (20 \left (16 \log ^2(2) \log (5)+4 \log (4) \log (5)+\log (256)-4 \log (5) \log (512)+\log (16) \log (512)-4 \log (2) \left (2+\log (5) \log (16)-\log \left (\frac {78125}{16}\right )\right )\right )\right ) \int \frac {1}{4 x (1+\log (4))-e^x x (1+\log (4))+5 e^x (1+\log (4)) \log (5)+5 \log (16)} \, dx}{\log (1048576)}+\frac {\left (25 \left (32 \log \left (\frac {5}{4}\right ) \log (2) \log (5)+64 \log ^2(2) \log ^2(5)+16 \log (4) \log ^2(5)-4 \log (5) \log (16)+\log ^2(16)+4 \log (5) \log (256)-16 \log ^2(5) \log (512)+8 \log (2) \log (5) (\log (25) \log (1024)+\log (4096)+2 \log (5) (5-\log (16384)))\right )\right ) \int \frac {1}{(x-5 \log (5)) \left (4 x (1+\log (4))-e^x x (1+\log (4))+5 e^x (1+\log (4)) \log (5)+5 \log (16)\right )} \, dx}{\log (1048576)}\\ \end {aligned} \end {gather*}
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Mathematica [F] time = 2.19, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {16 x^2+\left (160 x+32 x^2\right ) \log (2)+e^x \left (-8 x^2+\left (-40 x-36 x^2\right ) \log (2)+(40 x+(100+180 x) \log (2)) \log (5)\right )+e^{2 x} \left (x^2+2 x^2 \log (2)+(-10 x-20 x \log (2)) \log (5)+(25+50 \log (2)) \log ^2(5)\right )}{16 x^3+\left (80 x^2+32 x^3\right ) \log (2)+e^x \left (-8 x^3+\left (-20 x^2-16 x^3\right ) \log (2)+\left (40 x^2+\left (100 x+80 x^2\right ) \log (2)\right ) \log (5)\right )+e^{2 x} \left (x^3+2 x^3 \log (2)+\left (-10 x^2-20 x^2 \log (2)\right ) \log (5)+(25 x+50 x \log (2)) \log ^2(5)\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [B] time = 0.73, size = 74, normalized size = 2.31 \begin {gather*} \log \relax (x) - \log \left (\frac {{\left (5 \, {\left (2 \, \log \relax (2) + 1\right )} \log \relax (5) - 2 \, x \log \relax (2) - x\right )} e^{x} + 4 \, {\left (2 \, x + 5\right )} \log \relax (2) + 4 \, x}{x - 5 \, \log \relax (5)}\right ) + \log \left (-\frac {{\left (x - 5 \, \log \relax (5)\right )} e^{x} - 4 \, x}{x - 5 \, \log \relax (5)}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.90, size = 60, normalized size = 1.88 \begin {gather*} -\log \left (-2 \, x e^{x} \log \relax (2) + 10 \, e^{x} \log \relax (5) \log \relax (2) - x e^{x} + 5 \, e^{x} \log \relax (5) + 8 \, x \log \relax (2) + 4 \, x + 20 \, \log \relax (2)\right ) + \log \left (x e^{x} - 5 \, e^{x} \log \relax (5) - 4 \, x\right ) + \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.43, size = 61, normalized size = 1.91
method | result | size |
risch | \(\ln \relax (x )+\ln \left ({\mathrm e}^{x}+\frac {4 x}{5 \ln \relax (5)-x}\right )-\ln \left ({\mathrm e}^{x}+\frac {8 x \ln \relax (2)+20 \ln \relax (2)+4 x}{10 \ln \relax (2) \ln \relax (5)-2 x \ln \relax (2)+5 \ln \relax (5)-x}\right )\) | \(61\) |
norman | \(-\ln \left (10 \ln \relax (5) \ln \relax (2) {\mathrm e}^{x}-2 x \ln \relax (2) {\mathrm e}^{x}+8 x \ln \relax (2)+5 \,{\mathrm e}^{x} \ln \relax (5)-{\mathrm e}^{x} x +20 \ln \relax (2)+4 x \right )+\ln \relax (x )+\ln \left (5 \,{\mathrm e}^{x} \ln \relax (5)-{\mathrm e}^{x} x +4 x \right )\) | \(62\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.57, size = 89, normalized size = 2.78 \begin {gather*} \log \relax (x) - \log \left (-\frac {4 \, x {\left (2 \, \log \relax (2) + 1\right )} - {\left (x {\left (2 \, \log \relax (2) + 1\right )} - 10 \, \log \relax (5) \log \relax (2) - 5 \, \log \relax (5)\right )} e^{x} + 20 \, \log \relax (2)}{x {\left (2 \, \log \relax (2) + 1\right )} - 10 \, \log \relax (5) \log \relax (2) - 5 \, \log \relax (5)}\right ) + \log \left (\frac {{\left (x - 5 \, \log \relax (5)\right )} e^{x} - 4 \, x}{x - 5 \, \log \relax (5)}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{2\,x}\,\left ({\ln \relax (5)}^2\,\left (50\,\ln \relax (2)+25\right )-\ln \relax (5)\,\left (10\,x+20\,x\,\ln \relax (2)\right )+2\,x^2\,\ln \relax (2)+x^2\right )-{\mathrm {e}}^x\,\left (\ln \relax (2)\,\left (36\,x^2+40\,x\right )-\ln \relax (5)\,\left (40\,x+\ln \relax (2)\,\left (180\,x+100\right )\right )+8\,x^2\right )+\ln \relax (2)\,\left (32\,x^2+160\,x\right )+16\,x^2}{{\mathrm {e}}^{2\,x}\,\left (2\,x^3\,\ln \relax (2)-\ln \relax (5)\,\left (20\,x^2\,\ln \relax (2)+10\,x^2\right )+{\ln \relax (5)}^2\,\left (25\,x+50\,x\,\ln \relax (2)\right )+x^3\right )-{\mathrm {e}}^x\,\left (\ln \relax (2)\,\left (16\,x^3+20\,x^2\right )-\ln \relax (5)\,\left (\ln \relax (2)\,\left (80\,x^2+100\,x\right )+40\,x^2\right )+8\,x^3\right )+\ln \relax (2)\,\left (32\,x^3+80\,x^2\right )+16\,x^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: PolynomialError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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