Optimal. Leaf size=20 \[ 5+e^{e^x}+e^{\frac {144 x}{5 (-4+\log (x))^2}} \]
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Rubi [A]
time = 1.35, antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps
used = 8, number of rules used = 7, integrand size = 85, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {6873, 12, 6820,
6874, 2320, 2225, 6838} \begin {gather*} e^{e^x}+e^{\frac {144 x}{5 (4-\log (x))^2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2225
Rule 2320
Rule 6820
Rule 6838
Rule 6873
Rule 6874
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e^{\frac {144 x}{80-40 \log (x)+5 \log ^2(x)}} (-864+144 \log (x))-e^{e^x} \left (-320 e^x+240 e^x \log (x)-60 e^x \log ^2(x)+5 e^x \log ^3(x)\right )}{5 (4-\log (x))^3} \, dx\\ &=\frac {1}{5} \int \frac {-e^{\frac {144 x}{80-40 \log (x)+5 \log ^2(x)}} (-864+144 \log (x))-e^{e^x} \left (-320 e^x+240 e^x \log (x)-60 e^x \log ^2(x)+5 e^x \log ^3(x)\right )}{(4-\log (x))^3} \, dx\\ &=\frac {1}{5} \int \frac {-144 e^{\frac {144 x}{5 (-4+\log (x))^2}} (-6+\log (x))-5 e^{e^x+x} (-4+\log (x))^3}{(4-\log (x))^3} \, dx\\ &=\frac {1}{5} \int \left (5 e^{e^x+x}+\frac {144 e^{\frac {144 x}{5 (-4+\log (x))^2}} (-6+\log (x))}{(-4+\log (x))^3}\right ) \, dx\\ &=\frac {144}{5} \int \frac {e^{\frac {144 x}{5 (-4+\log (x))^2}} (-6+\log (x))}{(-4+\log (x))^3} \, dx+\int e^{e^x+x} \, dx\\ &=e^{\frac {144 x}{5 (4-\log (x))^2}}+\text {Subst}\left (\int e^x \, dx,x,e^x\right )\\ &=e^{e^x}+e^{\frac {144 x}{5 (4-\log (x))^2}}\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.16, size = 27, normalized size = 1.35 \begin {gather*} \frac {1}{5} \left (5 e^{e^x}+5 e^{\frac {144 x}{5 (-4+\log (x))^2}}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(79\) vs.
\(2(15)=30\).
time = 0.32, size = 80, normalized size = 4.00
method | result | size |
risch | \({\mathrm e}^{{\mathrm e}^{x}}+{\mathrm e}^{\frac {144 x}{5 \left (\ln \left (x \right )-4\right )^{2}}}\) | \(15\) |
default | \(\frac {-40 \ln \left (x \right ) {\mathrm e}^{\frac {144 x}{5 \ln \left (x \right )^{2}-40 \ln \left (x \right )+80}}+5 \ln \left (x \right )^{2} {\mathrm e}^{\frac {144 x}{5 \ln \left (x \right )^{2}-40 \ln \left (x \right )+80}}+80 \,{\mathrm e}^{\frac {144 x}{5 \ln \left (x \right )^{2}-40 \ln \left (x \right )+80}}}{5 \left (\ln \left (x \right )-4\right )^{2}}+{\mathrm e}^{{\mathrm e}^{x}}\) | \(80\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.32, size = 20, normalized size = 1.00 \begin {gather*} e^{\left (\frac {144 \, x}{5 \, {\left (\log \left (x\right )^{2} - 8 \, \log \left (x\right ) + 16\right )}}\right )} + e^{\left (e^{x}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 20, normalized size = 1.00 \begin {gather*} e^{\left (\frac {144 \, x}{5 \, {\left (\log \left (x\right )^{2} - 8 \, \log \left (x\right ) + 16\right )}}\right )} + e^{\left (e^{x}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.66, size = 22, normalized size = 1.10 \begin {gather*} e^{\frac {144 x}{5 \log {\left (x \right )}^{2} - 40 \log {\left (x \right )} + 80}} + e^{e^{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.45, size = 20, normalized size = 1.00 \begin {gather*} e^{\left (\frac {144 \, x}{5 \, {\left (\log \left (x\right )^{2} - 8 \, \log \left (x\right ) + 16\right )}}\right )} + e^{\left (e^{x}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 7.23, size = 22, normalized size = 1.10 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^x}+{\mathrm {e}}^{\frac {144\,x}{5\,{\ln \left (x\right )}^2-40\,\ln \left (x\right )+80}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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