Integrand size = 20, antiderivative size = 40 \[ \int \sqrt {1-2 x} (2+3 x) (3+5 x) \, dx=-\frac {77}{12} (1-2 x)^{3/2}+\frac {17}{5} (1-2 x)^{5/2}-\frac {15}{28} (1-2 x)^{7/2} \]
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Time = 0.01 (sec) , antiderivative size = 40, 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} \[ \int \sqrt {1-2 x} (2+3 x) (3+5 x) \, dx=-\frac {15}{28} (1-2 x)^{7/2}+\frac {17}{5} (1-2 x)^{5/2}-\frac {77}{12} (1-2 x)^{3/2} \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {77}{4} \sqrt {1-2 x}-17 (1-2 x)^{3/2}+\frac {15}{4} (1-2 x)^{5/2}\right ) \, dx \\ & = -\frac {77}{12} (1-2 x)^{3/2}+\frac {17}{5} (1-2 x)^{5/2}-\frac {15}{28} (1-2 x)^{7/2} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.58 \[ \int \sqrt {1-2 x} (2+3 x) (3+5 x) \, dx=-\frac {1}{105} (1-2 x)^{3/2} \left (373+489 x+225 x^2\right ) \]
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Time = 0.92 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.50
method | result | size |
gosper | \(-\frac {\left (1-2 x \right )^{\frac {3}{2}} \left (225 x^{2}+489 x +373\right )}{105}\) | \(20\) |
trager | \(\left (\frac {30}{7} x^{3}+\frac {251}{35} x^{2}+\frac {257}{105} x -\frac {373}{105}\right ) \sqrt {1-2 x}\) | \(24\) |
pseudoelliptic | \(\frac {\sqrt {1-2 x}\, \left (450 x^{3}+753 x^{2}+257 x -373\right )}{105}\) | \(25\) |
derivativedivides | \(-\frac {77 \left (1-2 x \right )^{\frac {3}{2}}}{12}+\frac {17 \left (1-2 x \right )^{\frac {5}{2}}}{5}-\frac {15 \left (1-2 x \right )^{\frac {7}{2}}}{28}\) | \(29\) |
default | \(-\frac {77 \left (1-2 x \right )^{\frac {3}{2}}}{12}+\frac {17 \left (1-2 x \right )^{\frac {5}{2}}}{5}-\frac {15 \left (1-2 x \right )^{\frac {7}{2}}}{28}\) | \(29\) |
risch | \(-\frac {\left (450 x^{3}+753 x^{2}+257 x -373\right ) \left (-1+2 x \right )}{105 \sqrt {1-2 x}}\) | \(30\) |
meijerg | \(\frac {2 \sqrt {\pi }-\sqrt {\pi }\, \left (2-4 x \right ) \sqrt {1-2 x}}{\sqrt {\pi }}-\frac {19 \left (-\frac {8 \sqrt {\pi }}{15}+\frac {4 \sqrt {\pi }\, \left (1-2 x \right )^{\frac {3}{2}} \left (6 x +2\right )}{15}\right )}{8 \sqrt {\pi }}+\frac {\frac {2 \sqrt {\pi }}{7}-\frac {\sqrt {\pi }\, \left (1-2 x \right )^{\frac {3}{2}} \left (60 x^{2}+24 x +8\right )}{28}}{\sqrt {\pi }}\) | \(91\) |
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Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.60 \[ \int \sqrt {1-2 x} (2+3 x) (3+5 x) \, dx=\frac {1}{105} \, {\left (450 \, x^{3} + 753 \, x^{2} + 257 \, x - 373\right )} \sqrt {-2 \, x + 1} \]
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Time = 0.59 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85 \[ \int \sqrt {1-2 x} (2+3 x) (3+5 x) \, dx=- \frac {15 \left (1 - 2 x\right )^{\frac {7}{2}}}{28} + \frac {17 \left (1 - 2 x\right )^{\frac {5}{2}}}{5} - \frac {77 \left (1 - 2 x\right )^{\frac {3}{2}}}{12} \]
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Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.70 \[ \int \sqrt {1-2 x} (2+3 x) (3+5 x) \, dx=-\frac {15}{28} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} + \frac {17}{5} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - \frac {77}{12} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} \]
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Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.05 \[ \int \sqrt {1-2 x} (2+3 x) (3+5 x) \, dx=\frac {15}{28} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} + \frac {17}{5} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - \frac {77}{12} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} \]
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Time = 0.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.58 \[ \int \sqrt {1-2 x} (2+3 x) (3+5 x) \, dx=-\frac {{\left (1-2\,x\right )}^{3/2}\,\left (2856\,x+225\,{\left (2\,x-1\right )}^2+1267\right )}{420} \]
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