Optimal. Leaf size=15 \[ \left (4 e^{-3 x}+e^x\right )^{20 x} \]
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Rubi [F] time = 1.58, 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 \left (4 e^{-3 x}+e^x\right )^{-1+20 x} \left (-240 e^{-3 x} x+20 e^x x+\left (80 e^{-3 x}+20 e^x\right ) \log \left (4 e^{-3 x}+e^x\right )\right ) \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} \left (-240 e^{-3 x} x+20 e^x x+\left (80 e^{-3 x}+20 e^x\right ) \log \left (4 e^{-3 x}+e^x\right )\right ) \, dx\\ &=\int \left (-240 e^{-3 x} \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} x+20 e^x \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} x+20 e^{-3 x} \left (4+e^{4 x}\right ) \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} \log \left (e^{-3 x} \left (4+e^{4 x}\right )\right )\right ) \, dx\\ &=20 \int e^x \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} x \, dx+20 \int e^{-3 x} \left (4+e^{4 x}\right ) \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} \log \left (e^{-3 x} \left (4+e^{4 x}\right )\right ) \, dx-240 \int e^{-3 x} \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} x \, dx\\ &=20 \int e^x \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} x \, dx-20 \int \frac {\left (-12+e^{4 x}\right ) \int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx}{4+e^{4 x}} \, dx-240 \int e^{-3 x} \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} x \, dx+\left (20 \log \left (e^{-3 x} \left (4+e^{4 x}\right )\right )\right ) \int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx\\ &=20 \int e^x \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} x \, dx-20 \int \left (\int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx-\frac {16 \int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx}{4+e^{4 x}}\right ) \, dx-240 \int e^{-3 x} \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} x \, dx+\left (20 \log \left (e^{-3 x} \left (4+e^{4 x}\right )\right )\right ) \int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx\\ &=20 \int e^x \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} x \, dx-20 \int \left (\int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx\right ) \, dx-240 \int e^{-3 x} \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} x \, dx+320 \int \frac {\int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx}{4+e^{4 x}} \, dx+\left (20 \log \left (e^{-3 x} \left (4+e^{4 x}\right )\right )\right ) \int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx\\ &=20 \int e^x \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} x \, dx-20 \int \left (\int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx\right ) \, dx-240 \int e^{-3 x} \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} x \, dx+320 \int \left (\frac {\left (2-e^x\right ) \int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx}{8 \left (2-2 e^x+e^{2 x}\right )}+\frac {\left (2+e^x\right ) \int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx}{8 \left (2+2 e^x+e^{2 x}\right )}\right ) \, dx+\left (20 \log \left (e^{-3 x} \left (4+e^{4 x}\right )\right )\right ) \int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx\\ &=20 \int e^x \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} x \, dx-20 \int \left (\int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx\right ) \, dx+40 \int \frac {\left (2-e^x\right ) \int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx}{2-2 e^x+e^{2 x}} \, dx+40 \int \frac {\left (2+e^x\right ) \int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx}{2+2 e^x+e^{2 x}} \, dx-240 \int e^{-3 x} \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} x \, dx+\left (20 \log \left (e^{-3 x} \left (4+e^{4 x}\right )\right )\right ) \int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx\\ &=20 \int e^x \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} x \, dx-20 \int \left (\int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx\right ) \, dx+40 \int \left (\frac {e^x \int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx}{-2+2 e^x-e^{2 x}}+\frac {2 \int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx}{2-2 e^x+e^{2 x}}\right ) \, dx+40 \int \left (\frac {2 \int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx}{2+2 e^x+e^{2 x}}+\frac {e^x \int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx}{2+2 e^x+e^{2 x}}\right ) \, dx-240 \int e^{-3 x} \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} x \, dx+\left (20 \log \left (e^{-3 x} \left (4+e^{4 x}\right )\right )\right ) \int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx\\ &=20 \int e^x \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} x \, dx-20 \int \left (\int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx\right ) \, dx+40 \int \frac {e^x \int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx}{-2+2 e^x-e^{2 x}} \, dx+40 \int \frac {e^x \int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx}{2+2 e^x+e^{2 x}} \, dx+80 \int \frac {\int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx}{2-2 e^x+e^{2 x}} \, dx+80 \int \frac {\int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx}{2+2 e^x+e^{2 x}} \, dx-240 \int e^{-3 x} \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{-1+20 x} x \, dx+\left (20 \log \left (e^{-3 x} \left (4+e^{4 x}\right )\right )\right ) \int \left (e^{-3 x} \left (4+e^{4 x}\right )\right )^{20 x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.41, size = 15, normalized size = 1.00 \begin {gather*} \left (4 e^{-3 x}+e^x\right )^{20 x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 15, normalized size = 1.00 \begin {gather*} \left ({\left (e^{\left (4 \, x\right )} + 4\right )} e^{\left (-3 \, x\right )}\right )^{20 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {20 \, {\left (12 \, x e^{\left (-3 \, x\right )} - x e^{x} - {\left (4 \, e^{\left (-3 \, x\right )} + e^{x}\right )} \log \left (4 \, e^{\left (-3 \, x\right )} + e^{x}\right )\right )} {\left (4 \, e^{\left (-3 \, x\right )} + e^{x}\right )}^{20 \, x}}{4 \, e^{\left (-3 \, x\right )} + e^{x}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.18, size = 270, normalized size = 18.00
method | result | size |
risch | \({\mathrm e}^{10 x \left (i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-2 i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2}+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{3 x}\right )-i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{3 x}\right )^{2}-i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{3 x}\right )^{2}+i \pi \mathrm {csgn}\left (i {\mathrm e}^{3 x}\right )^{3}-i \pi \mathrm {csgn}\left (i {\mathrm e}^{-3 x} \left ({\mathrm e}^{4 x}+4\right )\right )^{3}+i \pi \mathrm {csgn}\left (i {\mathrm e}^{-3 x} \left ({\mathrm e}^{4 x}+4\right )\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-3 x}\right )+i \pi \mathrm {csgn}\left (i {\mathrm e}^{-3 x} \left ({\mathrm e}^{4 x}+4\right )\right )^{2} \mathrm {csgn}\left (i \left ({\mathrm e}^{4 x}+4\right )\right )-i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-3 x} \left ({\mathrm e}^{4 x}+4\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-3 x}\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{4 x}+4\right )\right )-6 \ln \left ({\mathrm e}^{x}\right )+2 \ln \left ({\mathrm e}^{4 x}+4\right )\right )}\) | \(270\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.00, size = 35, normalized size = 2.33 \begin {gather*} e^{\left (-60 \, x^{2} + 20 \, x \log \left (e^{\left (2 \, x\right )} + 2 \, e^{x} + 2\right ) + 20 \, x \log \left (e^{\left (2 \, x\right )} - 2 \, e^{x} + 2\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.17, size = 13, normalized size = 0.87 \begin {gather*} {\left (4\,{\mathrm {e}}^{-3\,x}+{\mathrm {e}}^x\right )}^{20\,x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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