Integrand size = 24, antiderivative size = 79 \[ \int (1-2 x)^{5/2} (2+3 x)^2 (3+5 x)^3 \, dx=-\frac {9317}{32} (1-2 x)^{7/2}+\frac {16093}{32} (1-2 x)^{9/2}-\frac {5847}{16} (1-2 x)^{11/2}+\frac {28555}{208} (1-2 x)^{13/2}-\frac {845}{32} (1-2 x)^{15/2}+\frac {1125}{544} (1-2 x)^{17/2} \]
-9317/32*(1-2*x)^(7/2)+16093/32*(1-2*x)^(9/2)-5847/16*(1-2*x)^(11/2)+28555 /208*(1-2*x)^(13/2)-845/32*(1-2*x)^(15/2)+1125/544*(1-2*x)^(17/2)
Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.48 \[ \int (1-2 x)^{5/2} (2+3 x)^2 (3+5 x)^3 \, dx=-\frac {1}{221} (1-2 x)^{7/2} \left (9004+39160 x+80748 x^2+92535 x^3+56810 x^4+14625 x^5\right ) \]
Time = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {99, 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.
\(\displaystyle \int (1-2 x)^{5/2} (3 x+2)^2 (5 x+3)^3 \, dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \int \left (-\frac {1125}{32} (1-2 x)^{15/2}+\frac {12675}{32} (1-2 x)^{13/2}-\frac {28555}{16} (1-2 x)^{11/2}+\frac {64317}{16} (1-2 x)^{9/2}-\frac {144837}{32} (1-2 x)^{7/2}+\frac {65219}{32} (1-2 x)^{5/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1125}{544} (1-2 x)^{17/2}-\frac {845}{32} (1-2 x)^{15/2}+\frac {28555}{208} (1-2 x)^{13/2}-\frac {5847}{16} (1-2 x)^{11/2}+\frac {16093}{32} (1-2 x)^{9/2}-\frac {9317}{32} (1-2 x)^{7/2}\) |
(-9317*(1 - 2*x)^(7/2))/32 + (16093*(1 - 2*x)^(9/2))/32 - (5847*(1 - 2*x)^ (11/2))/16 + (28555*(1 - 2*x)^(13/2))/208 - (845*(1 - 2*x)^(15/2))/32 + (1 125*(1 - 2*x)^(17/2))/544
3.20.57.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Time = 1.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.44
method | result | size |
gosper | \(-\frac {\left (1-2 x \right )^{\frac {7}{2}} \left (14625 x^{5}+56810 x^{4}+92535 x^{3}+80748 x^{2}+39160 x +9004\right )}{221}\) | \(35\) |
pseudoelliptic | \(\frac {\left (14625 x^{5}+56810 x^{4}+92535 x^{3}+80748 x^{2}+39160 x +9004\right ) \sqrt {1-2 x}\, \left (-1+2 x \right )^{3}}{221}\) | \(42\) |
trager | \(\left (\frac {9000}{17} x^{8}+\frac {21460}{17} x^{7}+\frac {146310}{221} x^{6}-\frac {138201}{221} x^{5}-\frac {157296}{221} x^{4}-\frac {5935}{221} x^{3}+\frac {46164}{221} x^{2}+\frac {14864}{221} x -\frac {9004}{221}\right ) \sqrt {1-2 x}\) | \(49\) |
risch | \(-\frac {\left (117000 x^{8}+278980 x^{7}+146310 x^{6}-138201 x^{5}-157296 x^{4}-5935 x^{3}+46164 x^{2}+14864 x -9004\right ) \left (-1+2 x \right )}{221 \sqrt {1-2 x}}\) | \(55\) |
derivativedivides | \(-\frac {9317 \left (1-2 x \right )^{\frac {7}{2}}}{32}+\frac {16093 \left (1-2 x \right )^{\frac {9}{2}}}{32}-\frac {5847 \left (1-2 x \right )^{\frac {11}{2}}}{16}+\frac {28555 \left (1-2 x \right )^{\frac {13}{2}}}{208}-\frac {845 \left (1-2 x \right )^{\frac {15}{2}}}{32}+\frac {1125 \left (1-2 x \right )^{\frac {17}{2}}}{544}\) | \(56\) |
default | \(-\frac {9317 \left (1-2 x \right )^{\frac {7}{2}}}{32}+\frac {16093 \left (1-2 x \right )^{\frac {9}{2}}}{32}-\frac {5847 \left (1-2 x \right )^{\frac {11}{2}}}{16}+\frac {28555 \left (1-2 x \right )^{\frac {13}{2}}}{208}-\frac {845 \left (1-2 x \right )^{\frac {15}{2}}}{32}+\frac {1125 \left (1-2 x \right )^{\frac {17}{2}}}{544}\) | \(56\) |
meijerg | \(\frac {\frac {108 \sqrt {\pi }}{7}-\frac {54 \sqrt {\pi }\, \left (-16 x^{3}+24 x^{2}-12 x +2\right ) \sqrt {1-2 x}}{7}}{\sqrt {\pi }}-\frac {405 \left (-\frac {32 \sqrt {\pi }}{945}+\frac {4 \sqrt {\pi }\, \left (-448 x^{4}+608 x^{3}-240 x^{2}+8 x +8\right ) \sqrt {1-2 x}}{945}\right )}{\sqrt {\pi }}+\frac {\frac {614 \sqrt {\pi }}{77}-\frac {307 \sqrt {\pi }\, \left (-4032 x^{5}+5152 x^{4}-1808 x^{3}+24 x^{2}+16 x +16\right ) \sqrt {1-2 x}}{616}}{\sqrt {\pi }}-\frac {66225 \left (-\frac {256 \sqrt {\pi }}{45045}+\frac {2 \sqrt {\pi }\, \left (-118272 x^{6}+145152 x^{5}-47488 x^{4}+320 x^{3}+192 x^{2}+128 x +128\right ) \sqrt {1-2 x}}{45045}\right )}{128 \sqrt {\pi }}+\frac {\frac {1880 \sqrt {\pi }}{3003}-\frac {235 \sqrt {\pi }\, \left (-768768 x^{7}+916608 x^{6}-286272 x^{5}+1120 x^{4}+640 x^{3}+384 x^{2}+256 x +256\right ) \sqrt {1-2 x}}{96096}}{\sqrt {\pi }}-\frac {16875 \left (-\frac {4096 \sqrt {\pi }}{2297295}+\frac {4 \sqrt {\pi }\, \left (-9225216 x^{8}+10762752 x^{7}-3252480 x^{6}+8064 x^{5}+4480 x^{4}+2560 x^{3}+1536 x^{2}+1024 x +1024\right ) \sqrt {1-2 x}}{2297295}\right )}{512 \sqrt {\pi }}\) | \(305\) |
Time = 0.21 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.62 \[ \int (1-2 x)^{5/2} (2+3 x)^2 (3+5 x)^3 \, dx=\frac {1}{221} \, {\left (117000 \, x^{8} + 278980 \, x^{7} + 146310 \, x^{6} - 138201 \, x^{5} - 157296 \, x^{4} - 5935 \, x^{3} + 46164 \, x^{2} + 14864 \, x - 9004\right )} \sqrt {-2 \, x + 1} \]
1/221*(117000*x^8 + 278980*x^7 + 146310*x^6 - 138201*x^5 - 157296*x^4 - 59 35*x^3 + 46164*x^2 + 14864*x - 9004)*sqrt(-2*x + 1)
Time = 0.92 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.89 \[ \int (1-2 x)^{5/2} (2+3 x)^2 (3+5 x)^3 \, dx=\frac {1125 \left (1 - 2 x\right )^{\frac {17}{2}}}{544} - \frac {845 \left (1 - 2 x\right )^{\frac {15}{2}}}{32} + \frac {28555 \left (1 - 2 x\right )^{\frac {13}{2}}}{208} - \frac {5847 \left (1 - 2 x\right )^{\frac {11}{2}}}{16} + \frac {16093 \left (1 - 2 x\right )^{\frac {9}{2}}}{32} - \frac {9317 \left (1 - 2 x\right )^{\frac {7}{2}}}{32} \]
1125*(1 - 2*x)**(17/2)/544 - 845*(1 - 2*x)**(15/2)/32 + 28555*(1 - 2*x)**( 13/2)/208 - 5847*(1 - 2*x)**(11/2)/16 + 16093*(1 - 2*x)**(9/2)/32 - 9317*( 1 - 2*x)**(7/2)/32
Time = 0.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.70 \[ \int (1-2 x)^{5/2} (2+3 x)^2 (3+5 x)^3 \, dx=\frac {1125}{544} \, {\left (-2 \, x + 1\right )}^{\frac {17}{2}} - \frac {845}{32} \, {\left (-2 \, x + 1\right )}^{\frac {15}{2}} + \frac {28555}{208} \, {\left (-2 \, x + 1\right )}^{\frac {13}{2}} - \frac {5847}{16} \, {\left (-2 \, x + 1\right )}^{\frac {11}{2}} + \frac {16093}{32} \, {\left (-2 \, x + 1\right )}^{\frac {9}{2}} - \frac {9317}{32} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} \]
1125/544*(-2*x + 1)^(17/2) - 845/32*(-2*x + 1)^(15/2) + 28555/208*(-2*x + 1)^(13/2) - 5847/16*(-2*x + 1)^(11/2) + 16093/32*(-2*x + 1)^(9/2) - 9317/3 2*(-2*x + 1)^(7/2)
Time = 0.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.23 \[ \int (1-2 x)^{5/2} (2+3 x)^2 (3+5 x)^3 \, dx=\frac {1125}{544} \, {\left (2 \, x - 1\right )}^{8} \sqrt {-2 \, x + 1} + \frac {845}{32} \, {\left (2 \, x - 1\right )}^{7} \sqrt {-2 \, x + 1} + \frac {28555}{208} \, {\left (2 \, x - 1\right )}^{6} \sqrt {-2 \, x + 1} + \frac {5847}{16} \, {\left (2 \, x - 1\right )}^{5} \sqrt {-2 \, x + 1} + \frac {16093}{32} \, {\left (2 \, x - 1\right )}^{4} \sqrt {-2 \, x + 1} + \frac {9317}{32} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} \]
1125/544*(2*x - 1)^8*sqrt(-2*x + 1) + 845/32*(2*x - 1)^7*sqrt(-2*x + 1) + 28555/208*(2*x - 1)^6*sqrt(-2*x + 1) + 5847/16*(2*x - 1)^5*sqrt(-2*x + 1) + 16093/32*(2*x - 1)^4*sqrt(-2*x + 1) + 9317/32*(2*x - 1)^3*sqrt(-2*x + 1)
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.70 \[ \int (1-2 x)^{5/2} (2+3 x)^2 (3+5 x)^3 \, dx=\frac {16093\,{\left (1-2\,x\right )}^{9/2}}{32}-\frac {9317\,{\left (1-2\,x\right )}^{7/2}}{32}-\frac {5847\,{\left (1-2\,x\right )}^{11/2}}{16}+\frac {28555\,{\left (1-2\,x\right )}^{13/2}}{208}-\frac {845\,{\left (1-2\,x\right )}^{15/2}}{32}+\frac {1125\,{\left (1-2\,x\right )}^{17/2}}{544} \]