Integrand size = 22, antiderivative size = 77 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)^3} \, dx=\frac {49}{4 (2+3 x)^4}+\frac {707}{3 (2+3 x)^3}+\frac {3467}{(2+3 x)^2}+\frac {57110}{2+3 x}-\frac {3025}{2 (3+5 x)^2}+\frac {46475}{3+5 x}-424975 \log (2+3 x)+424975 \log (3+5 x) \]
49/4/(2+3*x)^4+707/3/(2+3*x)^3+3467/(2+3*x)^2+57110/(2+3*x)-3025/2/(3+5*x) ^2+46475/(3+5*x)-424975*ln(2+3*x)+424975*ln(3+5*x)
Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.03 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)^3} \, dx=\frac {49}{4 (2+3 x)^4}+\frac {707}{3 (2+3 x)^3}+\frac {3467}{(2+3 x)^2}+\frac {57110}{2+3 x}-\frac {3025}{2 (3+5 x)^2}+\frac {46475}{3+5 x}-424975 \log (5 (2+3 x))+424975 \log (3+5 x) \]
49/(4*(2 + 3*x)^4) + 707/(3*(2 + 3*x)^3) + 3467/(2 + 3*x)^2 + 57110/(2 + 3 *x) - 3025/(2*(3 + 5*x)^2) + 46475/(3 + 5*x) - 424975*Log[5*(2 + 3*x)] + 4 24975*Log[3 + 5*x]
Time = 0.21 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, 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.
\(\displaystyle \int \frac {(1-2 x)^2}{(3 x+2)^5 (5 x+3)^3} \, dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \int \left (\frac {2124875}{5 x+3}-\frac {232375}{(5 x+3)^2}+\frac {15125}{(5 x+3)^3}-\frac {1274925}{3 x+2}-\frac {171330}{(3 x+2)^2}-\frac {20802}{(3 x+2)^3}-\frac {2121}{(3 x+2)^4}-\frac {147}{(3 x+2)^5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {57110}{3 x+2}+\frac {46475}{5 x+3}+\frac {3467}{(3 x+2)^2}-\frac {3025}{2 (5 x+3)^2}+\frac {707}{3 (3 x+2)^3}+\frac {49}{4 (3 x+2)^4}-424975 \log (3 x+2)+424975 \log (5 x+3)\) |
49/(4*(2 + 3*x)^4) + 707/(3*(2 + 3*x)^3) + 3467/(2 + 3*x)^2 + 57110/(2 + 3 *x) - 3025/(2*(3 + 5*x)^2) + 46475/(3 + 5*x) - 424975*Log[2 + 3*x] + 42497 5*Log[3 + 5*x]
3.14.34.3.1 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 = 2.37 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.75
method | result | size |
norman | \(\frac {57371625 x^{5}+239770895 x^{3}+\frac {99957973}{2} x +\frac {371003175}{2} x^{4}+\frac {1858347679}{12} x^{2}+\frac {25790737}{4}}{\left (2+3 x \right )^{4} \left (3+5 x \right )^{2}}-424975 \ln \left (2+3 x \right )+424975 \ln \left (3+5 x \right )\) | \(58\) |
risch | \(\frac {57371625 x^{5}+239770895 x^{3}+\frac {99957973}{2} x +\frac {371003175}{2} x^{4}+\frac {1858347679}{12} x^{2}+\frac {25790737}{4}}{\left (2+3 x \right )^{4} \left (3+5 x \right )^{2}}-424975 \ln \left (2+3 x \right )+424975 \ln \left (3+5 x \right )\) | \(59\) |
default | \(\frac {49}{4 \left (2+3 x \right )^{4}}+\frac {707}{3 \left (2+3 x \right )^{3}}+\frac {3467}{\left (2+3 x \right )^{2}}+\frac {57110}{2+3 x}-\frac {3025}{2 \left (3+5 x \right )^{2}}+\frac {46475}{3+5 x}-424975 \ln \left (2+3 x \right )+424975 \ln \left (3+5 x \right )\) | \(72\) |
parallelrisch | \(-\frac {5874854304 x -1278759974400 \ln \left (x +\frac {3}{5}\right ) x^{2}+2649559334400 \ln \left (\frac {2}{3}+x \right ) x^{3}-328991846400 \ln \left (x +\frac {3}{5}\right ) x +1278759974400 \ln \left (\frac {2}{3}+x \right ) x^{2}+328991846400 \ln \left (\frac {2}{3}+x \right ) x +168895414710 x^{5}+52226242425 x^{6}+141050901768 x^{3}+218346488433 x^{4}+45530121496 x^{2}+3086501630400 \ln \left (\frac {2}{3}+x \right ) x^{4}+35249126400 \ln \left (\frac {2}{3}+x \right )-35249126400 \ln \left (x +\frac {3}{5}\right )+1916671248000 \ln \left (\frac {2}{3}+x \right ) x^{5}-2649559334400 \ln \left (x +\frac {3}{5}\right ) x^{3}-1916671248000 \ln \left (x +\frac {3}{5}\right ) x^{5}-3086501630400 \ln \left (x +\frac {3}{5}\right ) x^{4}+495690840000 \ln \left (\frac {2}{3}+x \right ) x^{6}-495690840000 \ln \left (x +\frac {3}{5}\right ) x^{6}}{576 \left (2+3 x \right )^{4} \left (3+5 x \right )^{2}}\) | \(162\) |
(57371625*x^5+239770895*x^3+99957973/2*x+371003175/2*x^4+1858347679/12*x^2 +25790737/4)/(2+3*x)^4/(3+5*x)^2-424975*ln(2+3*x)+424975*ln(3+5*x)
Time = 0.22 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.75 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)^3} \, dx=\frac {688459500 \, x^{5} + 2226019050 \, x^{4} + 2877250740 \, x^{3} + 1858347679 \, x^{2} + 5099700 \, {\left (2025 \, x^{6} + 7830 \, x^{5} + 12609 \, x^{4} + 10824 \, x^{3} + 5224 \, x^{2} + 1344 \, x + 144\right )} \log \left (5 \, x + 3\right ) - 5099700 \, {\left (2025 \, x^{6} + 7830 \, x^{5} + 12609 \, x^{4} + 10824 \, x^{3} + 5224 \, x^{2} + 1344 \, x + 144\right )} \log \left (3 \, x + 2\right ) + 599747838 \, x + 77372211}{12 \, {\left (2025 \, x^{6} + 7830 \, x^{5} + 12609 \, x^{4} + 10824 \, x^{3} + 5224 \, x^{2} + 1344 \, x + 144\right )}} \]
1/12*(688459500*x^5 + 2226019050*x^4 + 2877250740*x^3 + 1858347679*x^2 + 5 099700*(2025*x^6 + 7830*x^5 + 12609*x^4 + 10824*x^3 + 5224*x^2 + 1344*x + 144)*log(5*x + 3) - 5099700*(2025*x^6 + 7830*x^5 + 12609*x^4 + 10824*x^3 + 5224*x^2 + 1344*x + 144)*log(3*x + 2) + 599747838*x + 77372211)/(2025*x^6 + 7830*x^5 + 12609*x^4 + 10824*x^3 + 5224*x^2 + 1344*x + 144)
Time = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.92 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)^3} \, dx=\frac {688459500 x^{5} + 2226019050 x^{4} + 2877250740 x^{3} + 1858347679 x^{2} + 599747838 x + 77372211}{24300 x^{6} + 93960 x^{5} + 151308 x^{4} + 129888 x^{3} + 62688 x^{2} + 16128 x + 1728} + 424975 \log {\left (x + \frac {3}{5} \right )} - 424975 \log {\left (x + \frac {2}{3} \right )} \]
(688459500*x**5 + 2226019050*x**4 + 2877250740*x**3 + 1858347679*x**2 + 59 9747838*x + 77372211)/(24300*x**6 + 93960*x**5 + 151308*x**4 + 129888*x**3 + 62688*x**2 + 16128*x + 1728) + 424975*log(x + 3/5) - 424975*log(x + 2/3 )
Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)^3} \, dx=\frac {688459500 \, x^{5} + 2226019050 \, x^{4} + 2877250740 \, x^{3} + 1858347679 \, x^{2} + 599747838 \, x + 77372211}{12 \, {\left (2025 \, x^{6} + 7830 \, x^{5} + 12609 \, x^{4} + 10824 \, x^{3} + 5224 \, x^{2} + 1344 \, x + 144\right )}} + 424975 \, \log \left (5 \, x + 3\right ) - 424975 \, \log \left (3 \, x + 2\right ) \]
1/12*(688459500*x^5 + 2226019050*x^4 + 2877250740*x^3 + 1858347679*x^2 + 5 99747838*x + 77372211)/(2025*x^6 + 7830*x^5 + 12609*x^4 + 10824*x^3 + 5224 *x^2 + 1344*x + 144) + 424975*log(5*x + 3) - 424975*log(3*x + 2)
Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)^3} \, dx=\frac {57110}{3 \, x + 2} - \frac {4125 \, {\left (\frac {404}{3 \, x + 2} - 1855\right )}}{2 \, {\left (\frac {1}{3 \, x + 2} - 5\right )}^{2}} + \frac {3467}{{\left (3 \, x + 2\right )}^{2}} + \frac {707}{3 \, {\left (3 \, x + 2\right )}^{3}} + \frac {49}{4 \, {\left (3 \, x + 2\right )}^{4}} + 424975 \, \log \left ({\left | -\frac {1}{3 \, x + 2} + 5 \right |}\right ) \]
57110/(3*x + 2) - 4125/2*(404/(3*x + 2) - 1855)/(1/(3*x + 2) - 5)^2 + 3467 /(3*x + 2)^2 + 707/3/(3*x + 2)^3 + 49/4/(3*x + 2)^4 + 424975*log(abs(-1/(3 *x + 2) + 5))
Time = 1.49 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)^3} \, dx=\frac {\frac {84995\,x^5}{3}+\frac {1648903\,x^4}{18}+\frac {47954179\,x^3}{405}+\frac {1858347679\,x^2}{24300}+\frac {99957973\,x}{4050}+\frac {25790737}{8100}}{x^6+\frac {58\,x^5}{15}+\frac {467\,x^4}{75}+\frac {3608\,x^3}{675}+\frac {5224\,x^2}{2025}+\frac {448\,x}{675}+\frac {16}{225}}-849950\,\mathrm {atanh}\left (30\,x+19\right ) \]