Optimal. Leaf size=19 \[ 30+x-\frac {\log (2)}{10+32 e^{-2 x} x} \]
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Rubi [F]
time = 0.81, 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 {25 e^{4 x}+256 x^2+e^{2 x} (160 x+(8-16 x) \log (2))}{25 e^{4 x}+160 e^{2 x} x+256 x^2} \, 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^{4 x}+256 x^2+e^{2 x} (160 x+(8-16 x) \log (2))}{\left (5 e^{2 x}+16 x\right )^2} \, dx\\ &=\int \left (1+\frac {128 x (-1+2 x) \log (2)}{5 \left (5 e^{2 x}+16 x\right )^2}-\frac {8 (-1+2 x) \log (2)}{5 \left (5 e^{2 x}+16 x\right )}\right ) \, dx\\ &=x-\frac {1}{5} (8 \log (2)) \int \frac {-1+2 x}{5 e^{2 x}+16 x} \, dx+\frac {1}{5} (128 \log (2)) \int \frac {x (-1+2 x)}{\left (5 e^{2 x}+16 x\right )^2} \, dx\\ &=x-\frac {1}{5} (4 \log (2)) \text {Subst}\left (\int \frac {-1+x}{5 e^x+8 x} \, dx,x,2 x\right )+\frac {1}{5} (128 \log (2)) \int \left (-\frac {x}{\left (5 e^{2 x}+16 x\right )^2}+\frac {2 x^2}{\left (5 e^{2 x}+16 x\right )^2}\right ) \, dx\\ &=x-\frac {1}{5} (4 \log (2)) \text {Subst}\left (\int \left (-\frac {1}{5 e^x+8 x}+\frac {x}{5 e^x+8 x}\right ) \, dx,x,2 x\right )-\frac {1}{5} (128 \log (2)) \int \frac {x}{\left (5 e^{2 x}+16 x\right )^2} \, dx+\frac {1}{5} (256 \log (2)) \int \frac {x^2}{\left (5 e^{2 x}+16 x\right )^2} \, dx\\ &=x+\frac {1}{5} (4 \log (2)) \text {Subst}\left (\int \frac {1}{5 e^x+8 x} \, dx,x,2 x\right )-\frac {1}{5} (4 \log (2)) \text {Subst}\left (\int \frac {x}{5 e^x+8 x} \, dx,x,2 x\right )-\frac {1}{5} (128 \log (2)) \int \frac {x}{\left (5 e^{2 x}+16 x\right )^2} \, dx+\frac {1}{5} (256 \log (2)) \int \frac {x^2}{\left (5 e^{2 x}+16 x\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.07, size = 22, normalized size = 1.16 \begin {gather*} x+\frac {8 x \log (2)}{5 \left (5 e^{2 x}+16 x\right )} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.37, size = 20, normalized size = 1.05
method | result | size |
risch | \(x +\frac {8 x \ln \left (2\right )}{5 \left (5 \,{\mathrm e}^{2 x}+16 x \right )}\) | \(20\) |
norman | \(\frac {-\frac {\ln \left (2\right ) {\mathrm e}^{2 x}}{2}+16 x^{2}+5 x \,{\mathrm e}^{2 x}}{5 \,{\mathrm e}^{2 x}+16 x}\) | \(35\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.52, size = 32, normalized size = 1.68 \begin {gather*} \frac {80 \, x^{2} + 25 \, x e^{\left (2 \, x\right )} + 8 \, x \log \left (2\right )}{5 \, {\left (16 \, x + 5 \, e^{\left (2 \, x\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 32, normalized size = 1.68 \begin {gather*} \frac {80 \, x^{2} + 25 \, x e^{\left (2 \, x\right )} + 8 \, x \log \left (2\right )}{5 \, {\left (16 \, x + 5 \, e^{\left (2 \, x\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.05, size = 17, normalized size = 0.89 \begin {gather*} x + \frac {8 x \log {\left (2 \right )}}{80 x + 25 e^{2 x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.43, size = 32, normalized size = 1.68 \begin {gather*} \frac {80 \, x^{2} + 25 \, x e^{\left (2 \, x\right )} + 8 \, x \log \left (2\right )}{5 \, {\left (16 \, x + 5 \, e^{\left (2 \, x\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 3.24, size = 19, normalized size = 1.00 \begin {gather*} x+\frac {8\,x\,\ln \left (2\right )}{5\,\left (16\,x+5\,{\mathrm {e}}^{2\,x}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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