Optimal. Leaf size=26 \[ -\frac {10}{3 \left (x+\left (e^{e^x x}-5 x^2\right )^2\right )^2}+\log (x) \]
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Rubi [F] time = 9.75, 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 {3 e^{6 e^x x}+20 x-90 e^{5 e^x x} x^2+3 x^3+2000 x^4+225 x^6+5625 x^9+46875 x^{12}+e^{4 e^x x} \left (9 x+1125 x^4\right )+e^{3 e^x x} \left (-180 x^3-7500 x^6\right )+e^{2 e^x x} \left (9 x^2+1350 x^5+28125 x^8+e^x \left (40 x+40 x^2\right )\right )+e^{e^x x} \left (-400 x^2-90 x^4-4500 x^7-56250 x^{10}+e^x \left (-200 x^3-200 x^4\right )\right )}{3 e^{6 e^x x} x-90 e^{5 e^x x} x^3+3 x^4+225 x^7+5625 x^{10}+46875 x^{13}+e^{4 e^x x} \left (9 x^2+1125 x^5\right )+e^{3 e^x x} \left (-180 x^4-7500 x^7\right )+e^{2 e^x x} \left (9 x^3+1350 x^6+28125 x^9\right )+e^{e^x x} \left (-90 x^5-4500 x^8-56250 x^{11}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 e^{6 e^x x}-90 e^{5 e^x x} x^2+40 e^{x+2 e^x x} x (1+x)-200 e^{x+e^x x} x^3 (1+x)+9 e^{4 e^x x} x \left (1+125 x^3\right )-60 e^{3 e^x x} x^3 \left (3+125 x^3\right )+9 e^{2 e^x x} x^2 \left (1+150 x^3+3125 x^6\right )-10 e^{e^x x} x^2 \left (40+9 x^2+450 x^5+5625 x^8\right )+x \left (20+3 x^2+2000 x^3+225 x^5+5625 x^8+46875 x^{11}\right )}{3 x \left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3} \, dx\\ &=\frac {1}{3} \int \frac {3 e^{6 e^x x}-90 e^{5 e^x x} x^2+40 e^{x+2 e^x x} x (1+x)-200 e^{x+e^x x} x^3 (1+x)+9 e^{4 e^x x} x \left (1+125 x^3\right )-60 e^{3 e^x x} x^3 \left (3+125 x^3\right )+9 e^{2 e^x x} x^2 \left (1+150 x^3+3125 x^6\right )-10 e^{e^x x} x^2 \left (40+9 x^2+450 x^5+5625 x^8\right )+x \left (20+3 x^2+2000 x^3+225 x^5+5625 x^8+46875 x^{11}\right )}{x \left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3} \, dx\\ &=\frac {1}{3} \int \left (\frac {3}{x}+\frac {40 e^x (1+x)}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^2}+\frac {20 \left (1-2 e^x x-20 e^{e^x x} x-2 e^x x^2+10 e^{x+e^x x} x^2+100 x^3+10 e^{x+e^x x} x^3-50 e^x x^4-50 e^x x^5\right )}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3}\right ) \, dx\\ &=\log (x)+\frac {20}{3} \int \frac {1-2 e^x x-20 e^{e^x x} x-2 e^x x^2+10 e^{x+e^x x} x^2+100 x^3+10 e^{x+e^x x} x^3-50 e^x x^4-50 e^x x^5}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3} \, dx+\frac {40}{3} \int \frac {e^x (1+x)}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^2} \, dx\\ &=\log (x)+\frac {20}{3} \int \frac {1-20 e^{e^x x} x+100 x^3+10 e^{x+e^x x} x^2 (1+x)-2 e^x x \left (1+x+25 x^3+25 x^4\right )}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3} \, dx+\frac {40}{3} \int \left (\frac {e^x}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^2}+\frac {e^x x}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^2}\right ) \, dx\\ &=\log (x)+\frac {20}{3} \int \left (\frac {1}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3}-\frac {2 e^x x}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3}-\frac {20 e^{e^x x} x}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3}-\frac {2 e^x x^2}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3}+\frac {10 e^{x+e^x x} x^2}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3}+\frac {100 x^3}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3}+\frac {10 e^{x+e^x x} x^3}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3}-\frac {50 e^x x^4}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3}-\frac {50 e^x x^5}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3}\right ) \, dx+\frac {40}{3} \int \frac {e^x}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^2} \, dx+\frac {40}{3} \int \frac {e^x x}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^2} \, dx\\ &=\log (x)+\frac {20}{3} \int \frac {1}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3} \, dx-\frac {40}{3} \int \frac {e^x x}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3} \, dx-\frac {40}{3} \int \frac {e^x x^2}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3} \, dx+\frac {40}{3} \int \frac {e^x}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^2} \, dx+\frac {40}{3} \int \frac {e^x x}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^2} \, dx+\frac {200}{3} \int \frac {e^{x+e^x x} x^2}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3} \, dx+\frac {200}{3} \int \frac {e^{x+e^x x} x^3}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3} \, dx-\frac {400}{3} \int \frac {e^{e^x x} x}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3} \, dx-\frac {1000}{3} \int \frac {e^x x^4}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3} \, dx-\frac {1000}{3} \int \frac {e^x x^5}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3} \, dx+\frac {2000}{3} \int \frac {x^3}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^3} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.16, size = 40, normalized size = 1.54 \begin {gather*} \frac {1}{3} \left (-\frac {10}{\left (e^{2 e^x x}+x-10 e^{e^x x} x^2+25 x^4\right )^2}+3 \log (x)\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.74, size = 144, normalized size = 5.54 \begin {gather*} -\frac {60 \, x^{2} e^{\left (3 \, x e^{x}\right )} \log \relax (x) - 6 \, {\left (75 \, x^{4} + x\right )} e^{\left (2 \, x e^{x}\right )} \log \relax (x) + 60 \, {\left (25 \, x^{6} + x^{3}\right )} e^{\left (x e^{x}\right )} \log \relax (x) - 3 \, {\left (625 \, x^{8} + 50 \, x^{5} + x^{2}\right )} \log \relax (x) - 3 \, e^{\left (4 \, x e^{x}\right )} \log \relax (x) + 10}{3 \, {\left (625 \, x^{8} + 50 \, x^{5} - 20 \, x^{2} e^{\left (3 \, x e^{x}\right )} + x^{2} + 2 \, {\left (75 \, x^{4} + x\right )} e^{\left (2 \, x e^{x}\right )} - 20 \, {\left (25 \, x^{6} + x^{3}\right )} e^{\left (x e^{x}\right )} + e^{\left (4 \, x e^{x}\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 36.95, size = 169, normalized size = 6.50 \begin {gather*} \frac {1875 \, x^{8} \log \relax (x) - 1500 \, x^{6} e^{\left (x e^{x}\right )} \log \relax (x) + 150 \, x^{5} \log \relax (x) + 450 \, x^{4} e^{\left (2 \, x e^{x}\right )} \log \relax (x) - 60 \, x^{3} e^{\left (x e^{x}\right )} \log \relax (x) - 60 \, x^{2} e^{\left (3 \, x e^{x}\right )} \log \relax (x) + 3 \, x^{2} \log \relax (x) + 6 \, x e^{\left (2 \, x e^{x}\right )} \log \relax (x) + 3 \, e^{\left (4 \, x e^{x}\right )} \log \relax (x) - 20}{3 \, {\left (625 \, x^{8} - 500 \, x^{6} e^{\left (x e^{x}\right )} + 50 \, x^{5} + 150 \, x^{4} e^{\left (2 \, x e^{x}\right )} - 20 \, x^{3} e^{\left (x e^{x}\right )} - 20 \, x^{2} e^{\left (3 \, x e^{x}\right )} + x^{2} + 2 \, x e^{\left (2 \, x e^{x}\right )} + e^{\left (4 \, x e^{x}\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.09, size = 31, normalized size = 1.19
method | result | size |
risch | \(\ln \relax (x )-\frac {10}{3 \left (25 x^{4}-10 \,{\mathrm e}^{{\mathrm e}^{x} x} x^{2}+{\mathrm e}^{2 \,{\mathrm e}^{x} x}+x \right )^{2}}\) | \(31\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.96, size = 69, normalized size = 2.65 \begin {gather*} -\frac {10}{3 \, {\left (625 \, x^{8} + 50 \, x^{5} - 20 \, x^{2} e^{\left (3 \, x e^{x}\right )} + x^{2} + 2 \, {\left (75 \, x^{4} + x\right )} e^{\left (2 \, x e^{x}\right )} - 20 \, {\left (25 \, x^{6} + x^{3}\right )} e^{\left (x e^{x}\right )} + e^{\left (4 \, x e^{x}\right )}\right )}} + \log \relax (x) \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 = -0.04 \begin {gather*} \int \frac {20\,x+3\,{\mathrm {e}}^{6\,x\,{\mathrm {e}}^x}+{\mathrm {e}}^{2\,x\,{\mathrm {e}}^x}\,\left ({\mathrm {e}}^x\,\left (40\,x^2+40\,x\right )+9\,x^2+1350\,x^5+28125\,x^8\right )-{\mathrm {e}}^{3\,x\,{\mathrm {e}}^x}\,\left (7500\,x^6+180\,x^3\right )-90\,x^2\,{\mathrm {e}}^{5\,x\,{\mathrm {e}}^x}-{\mathrm {e}}^{x\,{\mathrm {e}}^x}\,\left ({\mathrm {e}}^x\,\left (200\,x^4+200\,x^3\right )+400\,x^2+90\,x^4+4500\,x^7+56250\,x^{10}\right )+3\,x^3+2000\,x^4+225\,x^6+5625\,x^9+46875\,x^{12}+{\mathrm {e}}^{4\,x\,{\mathrm {e}}^x}\,\left (1125\,x^4+9\,x\right )}{{\mathrm {e}}^{4\,x\,{\mathrm {e}}^x}\,\left (1125\,x^5+9\,x^2\right )-{\mathrm {e}}^{3\,x\,{\mathrm {e}}^x}\,\left (7500\,x^7+180\,x^4\right )-90\,x^3\,{\mathrm {e}}^{5\,x\,{\mathrm {e}}^x}+{\mathrm {e}}^{2\,x\,{\mathrm {e}}^x}\,\left (28125\,x^9+1350\,x^6+9\,x^3\right )-{\mathrm {e}}^{x\,{\mathrm {e}}^x}\,\left (56250\,x^{11}+4500\,x^8+90\,x^5\right )+3\,x^4+225\,x^7+5625\,x^{10}+46875\,x^{13}+3\,x\,{\mathrm {e}}^{6\,x\,{\mathrm {e}}^x}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.48, size = 78, normalized size = 3.00 \begin {gather*} \log {\relax (x )} - \frac {10}{1875 x^{8} + 150 x^{5} - 60 x^{2} e^{3 x e^{x}} + 3 x^{2} + \left (450 x^{4} + 6 x\right ) e^{2 x e^{x}} + \left (- 1500 x^{6} - 60 x^{3}\right ) e^{x e^{x}} + 3 e^{4 x e^{x}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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