Optimal. Leaf size=20 \[ \frac {15}{5+e+\frac {3 e^{53}}{2 x}+2 x} \]
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Rubi [A] time = 0.17, antiderivative size = 22, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 5, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {6, 1680, 12, 1814, 8} \begin {gather*} \frac {30 x}{4 x^2+2 (5+e) x+3 e^{53}} \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 8
Rule 12
Rule 1680
Rule 1814
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {90 e^{53}-120 x^2}{9 e^{106}+\left (100+4 e^2\right ) x^2+80 x^3+16 x^4+e^{53} \left (60 x+12 e x+24 x^2\right )+e \left (40 x^2+16 x^3\right )} \, dx\\ &=\operatorname {Subst}\left (\int \frac {120 \left (-25-10 e-e^2+12 e^{53}+8 (5+e) x-16 x^2\right )}{\left (25+10 e+e^2-12 e^{53}-16 x^2\right )^2} \, dx,x,\frac {1}{64} (80+16 e)+x\right )\\ &=120 \operatorname {Subst}\left (\int \frac {-25-10 e-e^2+12 e^{53}+8 (5+e) x-16 x^2}{\left (25+10 e+e^2-12 e^{53}-16 x^2\right )^2} \, dx,x,\frac {1}{64} (80+16 e)+x\right )\\ &=\frac {30 x}{3 e^{53}+2 (5+e) x+4 x^2}-\frac {60 \operatorname {Subst}\left (\int 0 \, dx,x,\frac {1}{64} (80+16 e)+x\right )}{25+10 e+e^2-12 e^{53}}\\ &=\frac {30 x}{3 e^{53}+2 (5+e) x+4 x^2}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.02, size = 23, normalized size = 1.15 \begin {gather*} \frac {30 x}{3 e^{53}+10 x+2 e x+4 x^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.68, size = 23, normalized size = 1.15 \begin {gather*} \frac {30 \, x}{4 \, x^{2} + 2 \, x e + 10 \, x + 3 \, e^{53}} \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 {30 \, {\left (4 \, x^{2} - 3 \, e^{53}\right )}}{16 \, x^{4} + 80 \, x^{3} + 4 \, x^{2} e^{2} + 100 \, x^{2} + 12 \, {\left (2 \, x^{2} + x e + 5 \, x\right )} e^{53} + 8 \, {\left (2 \, x^{3} + 5 \, x^{2}\right )} e + 9 \, e^{106}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.18, size = 22, normalized size = 1.10
method | result | size |
risch | \(\frac {10 x}{{\mathrm e}^{53}+\frac {2 x \,{\mathrm e}}{3}+\frac {4 x^{2}}{3}+\frac {10 x}{3}}\) | \(22\) |
gosper | \(\frac {30 x}{3 \,{\mathrm e}^{3} {\mathrm e}^{50}+2 x \,{\mathrm e}+4 x^{2}+10 x}\) | \(28\) |
norman | \(\frac {30 x}{3 \,{\mathrm e}^{3} {\mathrm e}^{50}+2 x \,{\mathrm e}+4 x^{2}+10 x}\) | \(28\) |
default | \(\frac {15 \left (\munderset {\textit {\_R} =\RootOf \left (16 \textit {\_Z}^{4}+\left (16 \,{\mathrm e}+80\right ) \textit {\_Z}^{3}+\left (24 \,{\mathrm e}^{53}+4 \,{\mathrm e}^{2}+40 \,{\mathrm e}+100\right ) \textit {\_Z}^{2}+\left (12 \,{\mathrm e}^{54}+60 \,{\mathrm e}^{53}\right ) \textit {\_Z} +9 \,{\mathrm e}^{106}\right )}{\sum }\frac {\left (3 \,{\mathrm e}^{53}-4 \textit {\_R}^{2}\right ) \ln \left (x -\textit {\_R} \right )}{3 \,{\mathrm e}^{54}+12 \textit {\_R} \,{\mathrm e}^{53}+2 \,{\mathrm e}^{2} \textit {\_R} +12 \textit {\_R}^{2} {\mathrm e}+15 \,{\mathrm e}^{53}+16 \textit {\_R}^{3}+20 \textit {\_R} \,{\mathrm e}+60 \textit {\_R}^{2}+50 \textit {\_R}}\right )}{2}\) | \(119\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.36, size = 22, normalized size = 1.10 \begin {gather*} \frac {30 \, x}{4 \, x^{2} + 2 \, x {\left (e + 5\right )} + 3 \, e^{53}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.12, size = 23, normalized size = 1.15 \begin {gather*} \frac {30\,x}{4\,x^2+\left (2\,\mathrm {e}+10\right )\,x+3\,{\mathrm {e}}^{53}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.62, size = 20, normalized size = 1.00 \begin {gather*} \frac {30 x}{4 x^{2} + x \left (2 e + 10\right ) + 3 e^{53}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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