Integrand size = 20, antiderivative size = 45 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^8} \, dx=\frac {7}{81 (2+3 x)^7}-\frac {91}{162 (2+3 x)^6}+\frac {16}{45 (2+3 x)^5}-\frac {5}{81 (2+3 x)^4} \]
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Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^8} \, dx=-\frac {5}{81 (3 x+2)^4}+\frac {16}{45 (3 x+2)^5}-\frac {91}{162 (3 x+2)^6}+\frac {7}{81 (3 x+2)^7} \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {49}{27 (2+3 x)^8}+\frac {91}{9 (2+3 x)^7}-\frac {16}{3 (2+3 x)^6}+\frac {20}{27 (2+3 x)^5}\right ) \, dx \\ & = \frac {7}{81 (2+3 x)^7}-\frac {91}{162 (2+3 x)^6}+\frac {16}{45 (2+3 x)^5}-\frac {5}{81 (2+3 x)^4} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.58 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^8} \, dx=-\frac {88-291 x+108 x^2+1350 x^3}{810 (2+3 x)^7} \]
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Time = 2.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.53
method | result | size |
norman | \(\frac {-\frac {5}{3} x^{3}-\frac {2}{15} x^{2}+\frac {97}{270} x -\frac {44}{405}}{\left (2+3 x \right )^{7}}\) | \(24\) |
gosper | \(-\frac {1350 x^{3}+108 x^{2}-291 x +88}{810 \left (2+3 x \right )^{7}}\) | \(25\) |
risch | \(\frac {-\frac {5}{3} x^{3}-\frac {2}{15} x^{2}+\frac {97}{270} x -\frac {44}{405}}{\left (2+3 x \right )^{7}}\) | \(25\) |
default | \(\frac {7}{81 \left (2+3 x \right )^{7}}-\frac {91}{162 \left (2+3 x \right )^{6}}+\frac {16}{45 \left (2+3 x \right )^{5}}-\frac {5}{81 \left (2+3 x \right )^{4}}\) | \(38\) |
parallelrisch | \(\frac {3564 x^{7}+16632 x^{6}+33264 x^{5}+36960 x^{4}+21440 x^{3}+9600 x^{2}+2880 x}{1920 \left (2+3 x \right )^{7}}\) | \(44\) |
meijerg | \(\frac {3 x \left (\frac {729}{64} x^{6}+\frac {1701}{32} x^{5}+\frac {1701}{16} x^{4}+\frac {945}{8} x^{3}+\frac {315}{4} x^{2}+\frac {63}{2} x +7\right )}{1792 \left (1+\frac {3 x}{2}\right )^{7}}-\frac {x^{2} \left (\frac {243}{32} x^{5}+\frac {567}{16} x^{4}+\frac {567}{8} x^{3}+\frac {315}{4} x^{2}+\frac {105}{2} x +21\right )}{1536 \left (1+\frac {3 x}{2}\right )^{7}}-\frac {x^{3} \left (\frac {81}{16} x^{4}+\frac {189}{8} x^{3}+\frac {189}{4} x^{2}+\frac {105}{2} x +35\right )}{3360 \left (1+\frac {3 x}{2}\right )^{7}}+\frac {x^{4} \left (\frac {27}{8} x^{3}+\frac {63}{4} x^{2}+\frac {63}{2} x +35\right )}{1792 \left (1+\frac {3 x}{2}\right )^{7}}\) | \(138\) |
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Time = 0.21 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.20 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^8} \, dx=-\frac {1350 \, x^{3} + 108 \, x^{2} - 291 \, x + 88}{810 \, {\left (2187 \, x^{7} + 10206 \, x^{6} + 20412 \, x^{5} + 22680 \, x^{4} + 15120 \, x^{3} + 6048 \, x^{2} + 1344 \, x + 128\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.09 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^8} \, dx=\frac {- 1350 x^{3} - 108 x^{2} + 291 x - 88}{1771470 x^{7} + 8266860 x^{6} + 16533720 x^{5} + 18370800 x^{4} + 12247200 x^{3} + 4898880 x^{2} + 1088640 x + 103680} \]
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Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.20 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^8} \, dx=-\frac {1350 \, x^{3} + 108 \, x^{2} - 291 \, x + 88}{810 \, {\left (2187 \, x^{7} + 10206 \, x^{6} + 20412 \, x^{5} + 22680 \, x^{4} + 15120 \, x^{3} + 6048 \, x^{2} + 1344 \, x + 128\right )}} \]
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Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.53 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^8} \, dx=-\frac {1350 \, x^{3} + 108 \, x^{2} - 291 \, x + 88}{810 \, {\left (3 \, x + 2\right )}^{7}} \]
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Time = 1.15 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^8} \, dx=\frac {16}{45\,{\left (3\,x+2\right )}^5}-\frac {5}{81\,{\left (3\,x+2\right )}^4}-\frac {91}{162\,{\left (3\,x+2\right )}^6}+\frac {7}{81\,{\left (3\,x+2\right )}^7} \]
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