Integrand size = 27, antiderivative size = 79 \[ \int (5-x) (3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^2 \, dx=\frac {325}{224} (3+2 x)^{7/2}-\frac {355}{96} (3+2 x)^{9/2}+\frac {651}{176} (3+2 x)^{11/2}-\frac {359}{208} (3+2 x)^{13/2}+\frac {11}{32} (3+2 x)^{15/2}-\frac {9}{544} (3+2 x)^{17/2} \]
325/224*(3+2*x)^(7/2)-355/96*(3+2*x)^(9/2)+651/176*(3+2*x)^(11/2)-359/208* (3+2*x)^(13/2)+11/32*(3+2*x)^(15/2)-9/544*(3+2*x)^(17/2)
Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.48 \[ \int (5-x) (3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^2 \, dx=-\frac {(3+2 x)^{7/2} \left (-44388-236768 x-461664 x^2-371679 x^3-78078 x^4+27027 x^5\right )}{51051} \]
-1/51051*((3 + 2*x)^(7/2)*(-44388 - 236768*x - 461664*x^2 - 371679*x^3 - 7 8078*x^4 + 27027*x^5))
Time = 0.21 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1195, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (5-x) (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )^2 \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (-\frac {9}{32} (2 x+3)^{15/2}+\frac {165}{32} (2 x+3)^{13/2}-\frac {359}{16} (2 x+3)^{11/2}+\frac {651}{16} (2 x+3)^{9/2}-\frac {1065}{32} (2 x+3)^{7/2}+\frac {325}{32} (2 x+3)^{5/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {9}{544} (2 x+3)^{17/2}+\frac {11}{32} (2 x+3)^{15/2}-\frac {359}{208} (2 x+3)^{13/2}+\frac {651}{176} (2 x+3)^{11/2}-\frac {355}{96} (2 x+3)^{9/2}+\frac {325}{224} (2 x+3)^{7/2}\) |
(325*(3 + 2*x)^(7/2))/224 - (355*(3 + 2*x)^(9/2))/96 + (651*(3 + 2*x)^(11/ 2))/176 - (359*(3 + 2*x)^(13/2))/208 + (11*(3 + 2*x)^(15/2))/32 - (9*(3 + 2*x)^(17/2))/544
3.26.35.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Time = 0.36 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.44
method | result | size |
gosper | \(-\frac {\left (27027 x^{5}-78078 x^{4}-371679 x^{3}-461664 x^{2}-236768 x -44388\right ) \left (3+2 x \right )^{\frac {7}{2}}}{51051}\) | \(35\) |
pseudoelliptic | \(-\frac {\left (27027 x^{5}-78078 x^{4}-371679 x^{3}-461664 x^{2}-236768 x -44388\right ) \left (3+2 x \right )^{\frac {7}{2}}}{51051}\) | \(35\) |
trager | \(\left (-\frac {72}{17} x^{8}-\frac {116}{17} x^{7}+\frac {18722}{221} x^{6}+\frac {979059}{2431} x^{5}+\frac {5813260}{7293} x^{4}+\frac {14614647}{17017} x^{3}+\frac {8949456}{17017} x^{2}+\frac {2929896}{17017} x +\frac {399492}{17017}\right ) \sqrt {3+2 x}\) | \(49\) |
risch | \(-\frac {\left (216216 x^{8}+348348 x^{7}-4324782 x^{6}-20560239 x^{5}-40692820 x^{4}-43843941 x^{3}-26848368 x^{2}-8789688 x -1198476\right ) \sqrt {3+2 x}}{51051}\) | \(50\) |
derivativedivides | \(\frac {325 \left (3+2 x \right )^{\frac {7}{2}}}{224}-\frac {355 \left (3+2 x \right )^{\frac {9}{2}}}{96}+\frac {651 \left (3+2 x \right )^{\frac {11}{2}}}{176}-\frac {359 \left (3+2 x \right )^{\frac {13}{2}}}{208}+\frac {11 \left (3+2 x \right )^{\frac {15}{2}}}{32}-\frac {9 \left (3+2 x \right )^{\frac {17}{2}}}{544}\) | \(56\) |
default | \(\frac {325 \left (3+2 x \right )^{\frac {7}{2}}}{224}-\frac {355 \left (3+2 x \right )^{\frac {9}{2}}}{96}+\frac {651 \left (3+2 x \right )^{\frac {11}{2}}}{176}-\frac {359 \left (3+2 x \right )^{\frac {13}{2}}}{208}+\frac {11 \left (3+2 x \right )^{\frac {15}{2}}}{32}-\frac {9 \left (3+2 x \right )^{\frac {17}{2}}}{544}\) | \(56\) |
meijerg | \(-\frac {601425 \sqrt {3}\, \left (\frac {128 \sqrt {\pi }}{10395}-\frac {8 \sqrt {\pi }\, \left (\frac {448}{27} x^{5}+\frac {5152}{81} x^{4}+\frac {1808}{27} x^{3}+\frac {8}{3} x^{2}-\frac {16}{3} x +16\right ) \sqrt {1+\frac {2 x}{3}}}{10395}\right )}{64 \sqrt {\pi }}-\frac {1235655 \sqrt {3}\, \left (-\frac {256 \sqrt {\pi }}{45045}+\frac {2 \sqrt {\pi }\, \left (-\frac {39424}{243} x^{6}-\frac {1792}{3} x^{5}-\frac {47488}{81} x^{4}-\frac {320}{27} x^{3}+\frac {64}{3} x^{2}-\frac {128}{3} x +128\right ) \sqrt {1+\frac {2 x}{3}}}{45045}\right )}{128 \sqrt {\pi }}-\frac {492075 \sqrt {3}\, \left (\frac {2048 \sqrt {\pi }}{675675}-\frac {8 \sqrt {\pi }\, \left (\frac {256256}{729} x^{7}+\frac {305536}{243} x^{6}+\frac {31808}{27} x^{5}+\frac {1120}{81} x^{4}-\frac {640}{27} x^{3}+\frac {128}{3} x^{2}-\frac {256}{3} x +256\right ) \sqrt {1+\frac {2 x}{3}}}{675675}\right )}{256 \sqrt {\pi }}-\frac {3645 \sqrt {3}\, \left (-\frac {32 \sqrt {\pi }}{945}+\frac {4 \sqrt {\pi }\, \left (-\frac {448}{81} x^{4}-\frac {608}{27} x^{3}-\frac {80}{3} x^{2}-\frac {8}{3} x +8\right ) \sqrt {1+\frac {2 x}{3}}}{945}\right )}{\sqrt {\pi }}-\frac {2025 \sqrt {3}\, \left (\frac {16 \sqrt {\pi }}{105}-\frac {8 \sqrt {\pi }\, \left (\frac {16}{27} x^{3}+\frac {8}{3} x^{2}+4 x +2\right ) \sqrt {1+\frac {2 x}{3}}}{105}\right )}{4 \sqrt {\pi }}+\frac {885735 \sqrt {3}\, \left (-\frac {4096 \sqrt {\pi }}{2297295}+\frac {4 \sqrt {\pi }\, \left (-\frac {1025024}{729} x^{8}-\frac {3587584}{729} x^{7}-\frac {1084160}{243} x^{6}-\frac {896}{27} x^{5}+\frac {4480}{81} x^{4}-\frac {2560}{27} x^{3}+\frac {512}{3} x^{2}-\frac {1024}{3} x +1024\right ) \sqrt {1+\frac {2 x}{3}}}{2297295}\right )}{512 \sqrt {\pi }}\) | \(323\) |
Time = 0.36 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.62 \[ \int (5-x) (3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^2 \, dx=-\frac {1}{51051} \, {\left (216216 \, x^{8} + 348348 \, x^{7} - 4324782 \, x^{6} - 20560239 \, x^{5} - 40692820 \, x^{4} - 43843941 \, x^{3} - 26848368 \, x^{2} - 8789688 \, x - 1198476\right )} \sqrt {2 \, x + 3} \]
-1/51051*(216216*x^8 + 348348*x^7 - 4324782*x^6 - 20560239*x^5 - 40692820* x^4 - 43843941*x^3 - 26848368*x^2 - 8789688*x - 1198476)*sqrt(2*x + 3)
Time = 1.16 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.89 \[ \int (5-x) (3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^2 \, dx=- \frac {9 \left (2 x + 3\right )^{\frac {17}{2}}}{544} + \frac {11 \left (2 x + 3\right )^{\frac {15}{2}}}{32} - \frac {359 \left (2 x + 3\right )^{\frac {13}{2}}}{208} + \frac {651 \left (2 x + 3\right )^{\frac {11}{2}}}{176} - \frac {355 \left (2 x + 3\right )^{\frac {9}{2}}}{96} + \frac {325 \left (2 x + 3\right )^{\frac {7}{2}}}{224} \]
-9*(2*x + 3)**(17/2)/544 + 11*(2*x + 3)**(15/2)/32 - 359*(2*x + 3)**(13/2) /208 + 651*(2*x + 3)**(11/2)/176 - 355*(2*x + 3)**(9/2)/96 + 325*(2*x + 3) **(7/2)/224
Time = 0.19 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.70 \[ \int (5-x) (3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^2 \, dx=-\frac {9}{544} \, {\left (2 \, x + 3\right )}^{\frac {17}{2}} + \frac {11}{32} \, {\left (2 \, x + 3\right )}^{\frac {15}{2}} - \frac {359}{208} \, {\left (2 \, x + 3\right )}^{\frac {13}{2}} + \frac {651}{176} \, {\left (2 \, x + 3\right )}^{\frac {11}{2}} - \frac {355}{96} \, {\left (2 \, x + 3\right )}^{\frac {9}{2}} + \frac {325}{224} \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} \]
-9/544*(2*x + 3)^(17/2) + 11/32*(2*x + 3)^(15/2) - 359/208*(2*x + 3)^(13/2 ) + 651/176*(2*x + 3)^(11/2) - 355/96*(2*x + 3)^(9/2) + 325/224*(2*x + 3)^ (7/2)
Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.70 \[ \int (5-x) (3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^2 \, dx=-\frac {9}{544} \, {\left (2 \, x + 3\right )}^{\frac {17}{2}} + \frac {11}{32} \, {\left (2 \, x + 3\right )}^{\frac {15}{2}} - \frac {359}{208} \, {\left (2 \, x + 3\right )}^{\frac {13}{2}} + \frac {651}{176} \, {\left (2 \, x + 3\right )}^{\frac {11}{2}} - \frac {355}{96} \, {\left (2 \, x + 3\right )}^{\frac {9}{2}} + \frac {325}{224} \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} \]
-9/544*(2*x + 3)^(17/2) + 11/32*(2*x + 3)^(15/2) - 359/208*(2*x + 3)^(13/2 ) + 651/176*(2*x + 3)^(11/2) - 355/96*(2*x + 3)^(9/2) + 325/224*(2*x + 3)^ (7/2)
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.70 \[ \int (5-x) (3+2 x)^{5/2} \left (2+5 x+3 x^2\right )^2 \, dx=\frac {325\,{\left (2\,x+3\right )}^{7/2}}{224}-\frac {355\,{\left (2\,x+3\right )}^{9/2}}{96}+\frac {651\,{\left (2\,x+3\right )}^{11/2}}{176}-\frac {359\,{\left (2\,x+3\right )}^{13/2}}{208}+\frac {11\,{\left (2\,x+3\right )}^{15/2}}{32}-\frac {9\,{\left (2\,x+3\right )}^{17/2}}{544} \]