Optimal. Leaf size=26 \[ 2 x+\frac {4 e^{10} x^2}{\left (4-x \left (5-x^2\right )\right )^2} \]
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Rubi [B] time = 0.31, antiderivative size = 96, normalized size of antiderivative = 3.69, number of steps used = 11, number of rules used = 5, integrand size = 94, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2074, 638, 614, 618, 206} \begin {gather*} \frac {21 e^{10} (2 x+1)}{17 \left (-x^2-x+4\right )}+\frac {e^{10} (43 x+166)}{17 \left (-x^2-x+4\right )}+\frac {e^{10} (7 x+20)}{\left (-x^2-x+4\right )^2}+2 x-\frac {5 e^{10}}{1-x}+\frac {e^{10}}{(1-x)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 206
Rule 614
Rule 618
Rule 638
Rule 2074
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2-\frac {2 e^{10}}{(-1+x)^3}-\frac {5 e^{10}}{(-1+x)^2}-\frac {2 e^{10} (76+33 x)}{\left (-4+x+x^2\right )^3}+\frac {e^{10} (30+17 x)}{\left (-4+x+x^2\right )^2}+\frac {5 e^{10}}{-4+x+x^2}\right ) \, dx\\ &=\frac {e^{10}}{(1-x)^2}-\frac {5 e^{10}}{1-x}+2 x+e^{10} \int \frac {30+17 x}{\left (-4+x+x^2\right )^2} \, dx-\left (2 e^{10}\right ) \int \frac {76+33 x}{\left (-4+x+x^2\right )^3} \, dx+\left (5 e^{10}\right ) \int \frac {1}{-4+x+x^2} \, dx\\ &=\frac {e^{10}}{(1-x)^2}-\frac {5 e^{10}}{1-x}+2 x+\frac {e^{10} (20+7 x)}{\left (4-x-x^2\right )^2}+\frac {e^{10} (166+43 x)}{17 \left (4-x-x^2\right )}-\frac {1}{17} \left (43 e^{10}\right ) \int \frac {1}{-4+x+x^2} \, dx-\left (10 e^{10}\right ) \operatorname {Subst}\left (\int \frac {1}{17-x^2} \, dx,x,1+2 x\right )+\left (21 e^{10}\right ) \int \frac {1}{\left (-4+x+x^2\right )^2} \, dx\\ &=\frac {e^{10}}{(1-x)^2}-\frac {5 e^{10}}{1-x}+2 x+\frac {e^{10} (20+7 x)}{\left (4-x-x^2\right )^2}+\frac {21 e^{10} (1+2 x)}{17 \left (4-x-x^2\right )}+\frac {e^{10} (166+43 x)}{17 \left (4-x-x^2\right )}-\frac {10 e^{10} \tanh ^{-1}\left (\frac {1+2 x}{\sqrt {17}}\right )}{\sqrt {17}}-\frac {1}{17} \left (42 e^{10}\right ) \int \frac {1}{-4+x+x^2} \, dx+\frac {1}{17} \left (86 e^{10}\right ) \operatorname {Subst}\left (\int \frac {1}{17-x^2} \, dx,x,1+2 x\right )\\ &=\frac {e^{10}}{(1-x)^2}-\frac {5 e^{10}}{1-x}+2 x+\frac {e^{10} (20+7 x)}{\left (4-x-x^2\right )^2}+\frac {21 e^{10} (1+2 x)}{17 \left (4-x-x^2\right )}+\frac {e^{10} (166+43 x)}{17 \left (4-x-x^2\right )}-\frac {84 e^{10} \tanh ^{-1}\left (\frac {1+2 x}{\sqrt {17}}\right )}{17 \sqrt {17}}+\frac {1}{17} \left (84 e^{10}\right ) \operatorname {Subst}\left (\int \frac {1}{17-x^2} \, dx,x,1+2 x\right )\\ &=\frac {e^{10}}{(1-x)^2}-\frac {5 e^{10}}{1-x}+2 x+\frac {e^{10} (20+7 x)}{\left (4-x-x^2\right )^2}+\frac {21 e^{10} (1+2 x)}{17 \left (4-x-x^2\right )}+\frac {e^{10} (166+43 x)}{17 \left (4-x-x^2\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.02, size = 22, normalized size = 0.85 \begin {gather*} 2 \left (x+\frac {2 e^{10} x^2}{\left (4-5 x+x^3\right )^2}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.61, size = 61, normalized size = 2.35 \begin {gather*} \frac {2 \, {\left (x^{7} - 10 \, x^{5} + 8 \, x^{4} + 25 \, x^{3} + 2 \, x^{2} e^{10} - 40 \, x^{2} + 16 \, x\right )}}{x^{6} - 10 \, x^{4} + 8 \, x^{3} + 25 \, x^{2} - 40 \, x + 16} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.15, size = 21, normalized size = 0.81 \begin {gather*} 2 \, x + \frac {4 \, x^{2} e^{10}}{{\left (x^{3} - 5 \, x + 4\right )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 37, normalized size = 1.42
method | result | size |
risch | \(2 x +\frac {4 \,{\mathrm e}^{10} x^{2}}{x^{6}-10 x^{4}+8 x^{3}+25 x^{2}-40 x +16}\) | \(37\) |
default | \(2 x +\frac {{\mathrm e}^{10} \left (-5 x^{3}-16 x^{2}+16 x +64\right )}{\left (x^{2}+x -4\right )^{2}}+\frac {{\mathrm e}^{10}}{\left (x -1\right )^{2}}+\frac {5 \,{\mathrm e}^{10}}{x -1}\) | \(48\) |
norman | \(\frac {16 x^{4}+32 x +50 x^{3}+\left (4 \,{\mathrm e}^{10}-80\right ) x^{2}-20 x^{5}+2 x^{7}}{\left (x^{3}-5 x +4\right )^{2}}\) | \(48\) |
gosper | \(\frac {2 x \left (x^{6}-10 x^{4}+2 x \,{\mathrm e}^{10}+8 x^{3}+25 x^{2}-40 x +16\right )}{x^{6}-10 x^{4}+8 x^{3}+25 x^{2}-40 x +16}\) | \(59\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.35, size = 36, normalized size = 1.38 \begin {gather*} \frac {4 \, x^{2} e^{10}}{x^{6} - 10 \, x^{4} + 8 \, x^{3} + 25 \, x^{2} - 40 \, x + 16} + 2 \, x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.09, size = 21, normalized size = 0.81 \begin {gather*} 2\,x+\frac {4\,x^2\,{\mathrm {e}}^{10}}{{\left (x^3-5\,x+4\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.46, size = 34, normalized size = 1.31 \begin {gather*} \frac {4 x^{2} e^{10}}{x^{6} - 10 x^{4} + 8 x^{3} + 25 x^{2} - 40 x + 16} + 2 x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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