Optimal. Leaf size=25 \[ 25 \left (4-e^{x+4 x^2}+\frac {25 x}{-25+x}\right )^2 \]
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Rubi [B] time = 0.51, antiderivative size = 59, normalized size of antiderivative = 2.36, number of steps used = 15, number of rules used = 8, integrand size = 83, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.096, Rules used = {6742, 2244, 2236, 37, 2234, 2204, 2242, 2240} \begin {gather*} -1450 e^{4 x^2+x}+25 e^{8 x^2+2 x}+\frac {31250 e^{4 x^2+x}}{25-x}+\frac {25 (100-29 x)^2}{(25-x)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 37
Rule 2204
Rule 2234
Rule 2236
Rule 2240
Rule 2242
Rule 2244
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (50 e^{2 x (1+4 x)} (1+8 x)-\frac {31250 (-100+29 x)}{(-25+x)^3}-\frac {50 e^{x+4 x^2} \left (1875+19175 x-6571 x^2+232 x^3\right )}{(-25+x)^2}\right ) \, dx\\ &=50 \int e^{2 x (1+4 x)} (1+8 x) \, dx-50 \int \frac {e^{x+4 x^2} \left (1875+19175 x-6571 x^2+232 x^3\right )}{(-25+x)^2} \, dx-31250 \int \frac {-100+29 x}{(-25+x)^3} \, dx\\ &=\frac {25 (100-29 x)^2}{(25-x)^2}+50 \int e^{2 x+8 x^2} (1+8 x) \, dx-50 \int \left (5029 e^{x+4 x^2}-\frac {625 e^{x+4 x^2}}{(-25+x)^2}+\frac {125625 e^{x+4 x^2}}{-25+x}+232 e^{x+4 x^2} x\right ) \, dx\\ &=25 e^{2 x+8 x^2}+\frac {25 (100-29 x)^2}{(25-x)^2}-11600 \int e^{x+4 x^2} x \, dx+31250 \int \frac {e^{x+4 x^2}}{(-25+x)^2} \, dx-251450 \int e^{x+4 x^2} \, dx-6281250 \int \frac {e^{x+4 x^2}}{-25+x} \, dx\\ &=-1450 e^{x+4 x^2}+25 e^{2 x+8 x^2}+\frac {25 (100-29 x)^2}{(25-x)^2}+\frac {31250 e^{x+4 x^2}}{25-x}+1450 \int e^{x+4 x^2} \, dx+250000 \int e^{x+4 x^2} \, dx-\frac {251450 \int e^{\frac {1}{16} (1+8 x)^2} \, dx}{\sqrt [16]{e}}\\ &=-1450 e^{x+4 x^2}+25 e^{2 x+8 x^2}+\frac {25 (100-29 x)^2}{(25-x)^2}+\frac {31250 e^{x+4 x^2}}{25-x}-\frac {125725 \sqrt {\pi } \text {erfi}\left (\frac {1}{4} (1+8 x)\right )}{2 \sqrt [16]{e}}+\frac {1450 \int e^{\frac {1}{16} (1+8 x)^2} \, dx}{\sqrt [16]{e}}+\frac {250000 \int e^{\frac {1}{16} (1+8 x)^2} \, dx}{\sqrt [16]{e}}\\ &=-1450 e^{x+4 x^2}+25 e^{2 x+8 x^2}+\frac {25 (100-29 x)^2}{(25-x)^2}+\frac {31250 e^{x+4 x^2}}{25-x}\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.64, size = 52, normalized size = 2.08 \begin {gather*} \frac {25 \left (e^{2 x (1+4 x)} (-25+x)^2+625 (-825+58 x)-2 e^{x+4 x^2} \left (2500-825 x+29 x^2\right )\right )}{(-25+x)^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.71, size = 56, normalized size = 2.24 \begin {gather*} \frac {25 \, {\left ({\left (x^{2} - 50 \, x + 625\right )} e^{\left (8 \, x^{2} + 2 \, x\right )} - 2 \, {\left (29 \, x^{2} - 825 \, x + 2500\right )} e^{\left (4 \, x^{2} + x\right )} + 36250 \, x - 515625\right )}}{x^{2} - 50 \, x + 625} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.20, size = 90, normalized size = 3.60 \begin {gather*} \frac {25 \, {\left (x^{2} e^{\left (8 \, x^{2} + 2 \, x\right )} - 58 \, x^{2} e^{\left (4 \, x^{2} + x\right )} - 50 \, x e^{\left (8 \, x^{2} + 2 \, x\right )} + 1650 \, x e^{\left (4 \, x^{2} + x\right )} + 36250 \, x + 625 \, e^{\left (8 \, x^{2} + 2 \, x\right )} - 5000 \, e^{\left (4 \, x^{2} + x\right )} - 515625\right )}}{x^{2} - 50 \, x + 625} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.09, size = 49, normalized size = 1.96
method | result | size |
risch | \(\frac {906250 x -12890625}{x^{2}-50 x +625}+25 \,{\mathrm e}^{2 x \left (4 x +1\right )}-\frac {50 \left (29 x -100\right ) {\mathrm e}^{x \left (4 x +1\right )}}{x -25}\) | \(49\) |
norman | \(\frac {906250 x +15625 \,{\mathrm e}^{8 x^{2}+2 x}+41250 \,{\mathrm e}^{4 x^{2}+x} x -1450 \,{\mathrm e}^{4 x^{2}+x} x^{2}-1250 \,{\mathrm e}^{8 x^{2}+2 x} x +25 \,{\mathrm e}^{8 x^{2}+2 x} x^{2}-125000 \,{\mathrm e}^{4 x^{2}+x}-12890625}{\left (x -25\right )^{2}}\) | \(86\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.44, size = 67, normalized size = 2.68 \begin {gather*} \frac {453125 \, {\left (2 \, x - 25\right )}}{x^{2} - 50 \, x + 625} + \frac {25 \, {\left ({\left (x - 25\right )} e^{\left (8 \, x^{2} + 2 \, x\right )} - 2 \, {\left (29 \, x - 100\right )} e^{\left (4 \, x^{2} + x\right )}\right )}}{x - 25} - \frac {1562500}{x^{2} - 50 \, x + 625} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.35, size = 56, normalized size = 2.24 \begin {gather*} 25\,{\mathrm {e}}^{8\,x^2+2\,x}-1450\,{\mathrm {e}}^{4\,x^2+x}-\frac {x\,\left (31250\,{\mathrm {e}}^{4\,x^2+x}-906250\right )-781250\,{\mathrm {e}}^{4\,x^2+x}+12890625}{{\left (x-25\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.20, size = 44, normalized size = 1.76 \begin {gather*} - \frac {12890625 - 906250 x}{x^{2} - 50 x + 625} + \frac {\left (5000 - 1450 x\right ) e^{4 x^{2} + x} + \left (25 x - 625\right ) e^{8 x^{2} + 2 x}}{x - 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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