Optimal. Leaf size=23 \[ \frac {16 x^4}{\left (16-5 e^{2+2 e^{9 x}} x\right )^2} \]
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Rubi [F]
time = 1.72, 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 {-1024 x^3+e^{2+2 e^{9 x}} \left (160 x^4-2880 e^{9 x} x^5\right )}{-4096+3840 e^{2+2 e^{9 x}} x-1200 e^{4+4 e^{9 x}} x^2+125 e^{6+6 e^{9 x}} x^3} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {32 x^3 \left (32-5 e^{2+2 e^{9 x}} x+90 e^{2+2 e^{9 x}+9 x} x^2\right )}{\left (16-5 e^{2+2 e^{9 x}} x\right )^3} \, dx\\ &=32 \int \frac {x^3 \left (32-5 e^{2+2 e^{9 x}} x+90 e^{2+2 e^{9 x}+9 x} x^2\right )}{\left (16-5 e^{2+2 e^{9 x}} x\right )^3} \, dx\\ &=32 \int \left (-\frac {90 e^{2+2 e^{9 x}+9 x} x^5}{\left (-16+5 e^{2+2 e^{9 x}} x\right )^3}+\frac {x^3 \left (-32+5 e^{2+2 e^{9 x}} x\right )}{\left (-16+5 e^{2+2 e^{9 x}} x\right )^3}\right ) \, dx\\ &=32 \int \frac {x^3 \left (-32+5 e^{2+2 e^{9 x}} x\right )}{\left (-16+5 e^{2+2 e^{9 x}} x\right )^3} \, dx-2880 \int \frac {e^{2+2 e^{9 x}+9 x} x^5}{\left (-16+5 e^{2+2 e^{9 x}} x\right )^3} \, dx\\ &=32 \int \left (-\frac {16 x^3}{\left (-16+5 e^{2+2 e^{9 x}} x\right )^3}+\frac {x^3}{\left (-16+5 e^{2+2 e^{9 x}} x\right )^2}\right ) \, dx-2880 \int \frac {e^{2+2 e^{9 x}+9 x} x^5}{\left (-16+5 e^{2+2 e^{9 x}} x\right )^3} \, dx\\ &=32 \int \frac {x^3}{\left (-16+5 e^{2+2 e^{9 x}} x\right )^2} \, dx-512 \int \frac {x^3}{\left (-16+5 e^{2+2 e^{9 x}} x\right )^3} \, dx-2880 \int \frac {e^{2+2 e^{9 x}+9 x} x^5}{\left (-16+5 e^{2+2 e^{9 x}} x\right )^3} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.12, size = 23, normalized size = 1.00 \begin {gather*} \frac {16 x^4}{\left (-16+5 e^{2+2 e^{9 x}} x\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.18, size = 22, normalized size = 0.96
method | result | size |
risch | \(\frac {16 x^{4}}{\left (5 x \,{\mathrm e}^{2 \,{\mathrm e}^{9 x}+2}-16\right )^{2}}\) | \(22\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.32, size = 35, normalized size = 1.52 \begin {gather*} \frac {16 \, x^{4}}{25 \, x^{2} e^{\left (4 \, e^{\left (9 \, x\right )} + 4\right )} - 160 \, x e^{\left (2 \, e^{\left (9 \, x\right )} + 2\right )} + 256} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 35, normalized size = 1.52 \begin {gather*} \frac {16 \, x^{4}}{25 \, x^{2} e^{\left (4 \, e^{\left (9 \, x\right )} + 4\right )} - 160 \, x e^{\left (2 \, e^{\left (9 \, x\right )} + 2\right )} + 256} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.09, size = 34, normalized size = 1.48 \begin {gather*} \frac {16 x^{4}}{25 x^{2} e^{4 e^{9 x} + 4} - 160 x e^{2 e^{9 x} + 2} + 256} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.40, size = 35, normalized size = 1.52 \begin {gather*} \frac {16 \, x^{4}}{25 \, x^{2} e^{\left (4 \, e^{\left (9 \, x\right )} + 4\right )} - 160 \, x e^{\left (2 \, e^{\left (9 \, x\right )} + 2\right )} + 256} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.94, size = 35, normalized size = 1.52 \begin {gather*} \frac {16\,x^4}{25\,x^2\,{\mathrm {e}}^{4\,{\mathrm {e}}^{9\,x}}\,{\mathrm {e}}^4-160\,x\,{\mathrm {e}}^{2\,{\mathrm {e}}^{9\,x}}\,{\mathrm {e}}^2+256} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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