Optimal. Leaf size=30 \[ \frac {3-\frac {(-9+x) x}{3-e^{-4+x}}}{3 (1+4 x)^2} \]
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Rubi [F] time = 1.40, 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 {-189-24 e^{-8+2 x}-114 x+e^{-4+x} \left (135+47 x+35 x^2-4 x^3\right )}{27+324 x+1296 x^2+1728 x^3+e^{-4+x} \left (-18-216 x-864 x^2-1152 x^3\right )+e^{-8+2 x} \left (3+36 x+144 x^2+192 x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-24 e^{2 x}-3 e^8 (63+38 x)+e^{4+x} \left (135+47 x+35 x^2-4 x^3\right )}{3 \left (3 e^4-e^x\right )^2 (1+4 x)^3} \, dx\\ &=\frac {1}{3} \int \frac {-24 e^{2 x}-3 e^8 (63+38 x)+e^{4+x} \left (135+47 x+35 x^2-4 x^3\right )}{\left (3 e^4-e^x\right )^2 (1+4 x)^3} \, dx\\ &=\frac {1}{3} \int \left (-\frac {24}{(1+4 x)^3}-\frac {3 e^8 (-9+x) x}{\left (3 e^4-e^x\right )^2 (1+4 x)^2}+\frac {e^4 \left (9-47 x-35 x^2+4 x^3\right )}{\left (3 e^4-e^x\right ) (1+4 x)^3}\right ) \, dx\\ &=\frac {1}{(1+4 x)^2}+\frac {1}{3} e^4 \int \frac {9-47 x-35 x^2+4 x^3}{\left (3 e^4-e^x\right ) (1+4 x)^3} \, dx-e^8 \int \frac {(-9+x) x}{\left (3 e^4-e^x\right )^2 (1+4 x)^2} \, dx\\ &=\frac {1}{(1+4 x)^2}+\frac {1}{3} e^4 \int \left (-\frac {1}{16 \left (-3 e^4+e^x\right )}-\frac {37}{2 \left (-3 e^4+e^x\right ) (1+4 x)^3}+\frac {115}{16 \left (-3 e^4+e^x\right ) (1+4 x)^2}+\frac {19}{8 \left (-3 e^4+e^x\right ) (1+4 x)}\right ) \, dx-e^8 \int \left (\frac {1}{16 \left (-3 e^4+e^x\right )^2}+\frac {37}{16 \left (-3 e^4+e^x\right )^2 (1+4 x)^2}-\frac {19}{8 \left (-3 e^4+e^x\right )^2 (1+4 x)}\right ) \, dx\\ &=\frac {1}{(1+4 x)^2}-\frac {1}{48} e^4 \int \frac {1}{-3 e^4+e^x} \, dx+\frac {1}{24} \left (19 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)} \, dx+\frac {1}{48} \left (115 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)^2} \, dx-\frac {1}{6} \left (37 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)^3} \, dx-\frac {1}{16} e^8 \int \frac {1}{\left (-3 e^4+e^x\right )^2} \, dx-\frac {1}{16} \left (37 e^8\right ) \int \frac {1}{\left (-3 e^4+e^x\right )^2 (1+4 x)^2} \, dx+\frac {1}{8} \left (19 e^8\right ) \int \frac {1}{\left (-3 e^4+e^x\right )^2 (1+4 x)} \, dx\\ &=\frac {1}{(1+4 x)^2}-\frac {1}{48} e^4 \operatorname {Subst}\left (\int \frac {1}{x \left (-3 e^4+x\right )} \, dx,x,e^x\right )+\frac {1}{24} \left (19 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)} \, dx+\frac {1}{48} \left (115 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)^2} \, dx-\frac {1}{6} \left (37 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)^3} \, dx-\frac {1}{16} e^8 \operatorname {Subst}\left (\int \frac {1}{x \left (-3 e^4+x\right )^2} \, dx,x,e^x\right )-\frac {1}{16} \left (37 e^8\right ) \int \frac {1}{\left (-3 e^4+e^x\right )^2 (1+4 x)^2} \, dx+\frac {1}{8} \left (19 e^8\right ) \int \frac {1}{\left (-3 e^4+e^x\right )^2 (1+4 x)} \, dx\\ &=\frac {1}{(1+4 x)^2}+\frac {1}{144} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )-\frac {1}{144} \operatorname {Subst}\left (\int \frac {1}{-3 e^4+x} \, dx,x,e^x\right )+\frac {1}{24} \left (19 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)} \, dx+\frac {1}{48} \left (115 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)^2} \, dx-\frac {1}{6} \left (37 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)^3} \, dx-\frac {1}{16} e^8 \operatorname {Subst}\left (\int \left (\frac {1}{3 e^4 \left (3 e^4-x\right )^2}+\frac {1}{9 e^8 \left (3 e^4-x\right )}+\frac {1}{9 e^8 x}\right ) \, dx,x,e^x\right )-\frac {1}{16} \left (37 e^8\right ) \int \frac {1}{\left (-3 e^4+e^x\right )^2 (1+4 x)^2} \, dx+\frac {1}{8} \left (19 e^8\right ) \int \frac {1}{\left (-3 e^4+e^x\right )^2 (1+4 x)} \, dx\\ &=-\frac {e^4}{48 \left (3 e^4-e^x\right )}+\frac {1}{(1+4 x)^2}+\frac {1}{24} \left (19 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)} \, dx+\frac {1}{48} \left (115 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)^2} \, dx-\frac {1}{6} \left (37 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)^3} \, dx-\frac {1}{16} \left (37 e^8\right ) \int \frac {1}{\left (-3 e^4+e^x\right )^2 (1+4 x)^2} \, dx+\frac {1}{8} \left (19 e^8\right ) \int \frac {1}{\left (-3 e^4+e^x\right )^2 (1+4 x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.33, size = 43, normalized size = 1.43 \begin {gather*} \frac {1}{3} \left (\frac {3}{(1+4 x)^2}+\frac {e^4 \left (-9 x+x^2\right )}{\left (-3 e^4+e^x\right ) (1+4 x)^2}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.98, size = 44, normalized size = 1.47 \begin {gather*} -\frac {x^{2} - 9 \, x + 3 \, e^{\left (x - 4\right )} - 9}{3 \, {\left (48 \, x^{2} - {\left (16 \, x^{2} + 8 \, x + 1\right )} e^{\left (x - 4\right )} + 24 \, x + 3\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.41, size = 57, normalized size = 1.90 \begin {gather*} -\frac {x^{2} e^{4} - 9 \, x e^{4} - 9 \, e^{4} + 3 \, e^{x}}{3 \, {\left (48 \, x^{2} e^{4} - 16 \, x^{2} e^{x} + 24 \, x e^{4} - 8 \, x e^{x} + 3 \, e^{4} - e^{x}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.23, size = 31, normalized size = 1.03
method | result | size |
norman | \(\frac {-3 x +{\mathrm e}^{x -4}+\frac {x^{2}}{3}-3}{\left ({\mathrm e}^{x -4}-3\right ) \left (4 x +1\right )^{2}}\) | \(31\) |
risch | \(\frac {1}{16 x^{2}+8 x +1}+\frac {x \left (x -9\right )}{3 \left (16 x^{2}+8 x +1\right ) \left ({\mathrm e}^{x -4}-3\right )}\) | \(40\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.61, size = 55, normalized size = 1.83 \begin {gather*} -\frac {x^{2} e^{4} - 9 \, x e^{4} - 9 \, e^{4} + 3 \, e^{x}}{3 \, {\left (48 \, x^{2} e^{4} + 24 \, x e^{4} - {\left (16 \, x^{2} + 8 \, x + 1\right )} e^{x} + 3 \, e^{4}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.76, size = 34, normalized size = 1.13 \begin {gather*} \frac {1}{{\left (4\,x+1\right )}^2}-\frac {3\,x-\frac {x^2}{3}}{{\left (4\,x+1\right )}^2\,\left ({\mathrm {e}}^{x-4}-3\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.23, size = 42, normalized size = 1.40 \begin {gather*} \frac {x^{2} - 9 x}{- 144 x^{2} - 72 x + \left (48 x^{2} + 24 x + 3\right ) e^{x - 4} - 9} + \frac {8}{128 x^{2} + 64 x + 8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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