Integrand size = 20, antiderivative size = 37 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^6} \, dx=\frac {(1-2 x)^4}{105 (2+3 x)^5}-\frac {173 (1-2 x)^4}{2940 (2+3 x)^4} \]
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Time = 0.00 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {79, 37} \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^6} \, dx=\frac {(1-2 x)^4}{105 (3 x+2)^5}-\frac {173 (1-2 x)^4}{2940 (3 x+2)^4} \]
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Rule 37
Rule 79
Rubi steps \begin{align*} \text {integral}& = \frac {(1-2 x)^4}{105 (2+3 x)^5}+\frac {173}{105} \int \frac {(1-2 x)^3}{(2+3 x)^5} \, dx \\ & = \frac {(1-2 x)^4}{105 (2+3 x)^5}-\frac {173 (1-2 x)^4}{2940 (2+3 x)^4} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^6} \, dx=\frac {1282+16905 x+34920 x^2+57240 x^3+64800 x^4}{4860 (2+3 x)^5} \]
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Time = 2.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78
method | result | size |
norman | \(\frac {\frac {40}{3} x^{4}+\frac {106}{9} x^{3}+\frac {194}{27} x^{2}+\frac {1127}{324} x +\frac {641}{2430}}{\left (2+3 x \right )^{5}}\) | \(29\) |
gosper | \(\frac {64800 x^{4}+57240 x^{3}+34920 x^{2}+16905 x +1282}{4860 \left (2+3 x \right )^{5}}\) | \(30\) |
risch | \(\frac {\frac {40}{3} x^{4}+\frac {106}{9} x^{3}+\frac {194}{27} x^{2}+\frac {1127}{324} x +\frac {641}{2430}}{\left (2+3 x \right )^{5}}\) | \(30\) |
parallelrisch | \(\frac {-641 x^{5}+2130 x^{4}+920 x^{3}+400 x^{2}+480 x}{320 \left (2+3 x \right )^{5}}\) | \(34\) |
default | \(\frac {40}{243 \left (2+3 x \right )}-\frac {214}{243 \left (2+3 x \right )^{2}}-\frac {2009}{972 \left (2+3 x \right )^{4}}+\frac {343}{1215 \left (2+3 x \right )^{5}}+\frac {518}{243 \left (2+3 x \right )^{3}}\) | \(47\) |
meijerg | \(\frac {3 x \left (\frac {81}{16} x^{4}+\frac {135}{8} x^{3}+\frac {45}{2} x^{2}+15 x +5\right )}{320 \left (1+\frac {3 x}{2}\right )^{5}}-\frac {13 x^{2} \left (\frac {27}{8} x^{3}+\frac {45}{4} x^{2}+15 x +10\right )}{1280 \left (1+\frac {3 x}{2}\right )^{5}}+\frac {x^{3} \left (\frac {9}{4} x^{2}+\frac {15}{2} x +10\right )}{320 \left (1+\frac {3 x}{2}\right )^{5}}+\frac {9 x^{4} \left (\frac {3 x}{2}+5\right )}{320 \left (1+\frac {3 x}{2}\right )^{5}}-\frac {x^{5}}{8 \left (1+\frac {3 x}{2}\right )^{5}}\) | \(110\) |
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Time = 0.22 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.32 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^6} \, dx=\frac {64800 \, x^{4} + 57240 \, x^{3} + 34920 \, x^{2} + 16905 \, x + 1282}{4860 \, {\left (243 \, x^{5} + 810 \, x^{4} + 1080 \, x^{3} + 720 \, x^{2} + 240 \, x + 32\right )}} \]
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Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.30 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^6} \, dx=- \frac {- 64800 x^{4} - 57240 x^{3} - 34920 x^{2} - 16905 x - 1282}{1180980 x^{5} + 3936600 x^{4} + 5248800 x^{3} + 3499200 x^{2} + 1166400 x + 155520} \]
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Time = 0.20 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.32 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^6} \, dx=\frac {64800 \, x^{4} + 57240 \, x^{3} + 34920 \, x^{2} + 16905 \, x + 1282}{4860 \, {\left (243 \, x^{5} + 810 \, x^{4} + 1080 \, x^{3} + 720 \, x^{2} + 240 \, x + 32\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^6} \, dx=\frac {64800 \, x^{4} + 57240 \, x^{3} + 34920 \, x^{2} + 16905 \, x + 1282}{4860 \, {\left (3 \, x + 2\right )}^{5}} \]
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Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.24 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^6} \, dx=\frac {40}{243\,\left (3\,x+2\right )}-\frac {214}{243\,{\left (3\,x+2\right )}^2}+\frac {518}{243\,{\left (3\,x+2\right )}^3}-\frac {2009}{972\,{\left (3\,x+2\right )}^4}+\frac {343}{1215\,{\left (3\,x+2\right )}^5} \]
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