Integrand size = 20, antiderivative size = 65 \[ \int \frac {3+5 x}{(1-2 x)^3 (2+3 x)^3} \, dx=\frac {11}{343 (1-2 x)^2}+\frac {128}{2401 (1-2 x)}+\frac {3}{686 (2+3 x)^2}-\frac {87}{2401 (2+3 x)}-\frac {558 \log (1-2 x)}{16807}+\frac {558 \log (2+3 x)}{16807} \] Output:
11/343/(1-2*x)^2+128/(2401-4802*x)+3/686/(2+3*x)^2-87/(4802+7203*x)-558/16 807*ln(1-2*x)+558/16807*ln(2+3*x)
Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74 \[ \int \frac {3+5 x}{(1-2 x)^3 (2+3 x)^3} \, dx=\frac {\frac {7 \left (1313+3658 x-1674 x^2-6696 x^3\right )}{\left (-2+x+6 x^2\right )^2}-1116 \log (1-2 x)+1116 \log (2+3 x)}{33614} \] Input:
Integrate[(3 + 5*x)/((1 - 2*x)^3*(2 + 3*x)^3),x]
Output:
((7*(1313 + 3658*x - 1674*x^2 - 6696*x^3))/(-2 + x + 6*x^2)^2 - 1116*Log[1 - 2*x] + 1116*Log[2 + 3*x])/33614
Time = 0.22 (sec) , antiderivative size = 65, 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 = {86, 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.
\(\displaystyle \int \frac {5 x+3}{(1-2 x)^3 (3 x+2)^3} \, dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {1674}{16807 (3 x+2)}+\frac {261}{2401 (3 x+2)^2}-\frac {9}{343 (3 x+2)^3}-\frac {1116}{16807 (2 x-1)}+\frac {256}{2401 (2 x-1)^2}-\frac {44}{343 (2 x-1)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {128}{2401 (1-2 x)}-\frac {87}{2401 (3 x+2)}+\frac {11}{343 (1-2 x)^2}+\frac {3}{686 (3 x+2)^2}-\frac {558 \log (1-2 x)}{16807}+\frac {558 \log (3 x+2)}{16807}\) |
Input:
Int[(3 + 5*x)/((1 - 2*x)^3*(2 + 3*x)^3),x]
Output:
11/(343*(1 - 2*x)^2) + 128/(2401*(1 - 2*x)) + 3/(686*(2 + 3*x)^2) - 87/(24 01*(2 + 3*x)) - (558*Log[1 - 2*x])/16807 + (558*Log[2 + 3*x])/16807
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Time = 0.21 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74
method | result | size |
norman | \(\frac {-\frac {3348}{2401} x^{3}-\frac {837}{2401} x^{2}+\frac {1829}{2401} x +\frac {1313}{4802}}{\left (-1+2 x \right )^{2} \left (2+3 x \right )^{2}}-\frac {558 \ln \left (-1+2 x \right )}{16807}+\frac {558 \ln \left (2+3 x \right )}{16807}\) | \(48\) |
risch | \(\frac {-\frac {3348}{2401} x^{3}-\frac {837}{2401} x^{2}+\frac {1829}{2401} x +\frac {1313}{4802}}{\left (-1+2 x \right )^{2} \left (2+3 x \right )^{2}}-\frac {558 \ln \left (-1+2 x \right )}{16807}+\frac {558 \ln \left (2+3 x \right )}{16807}\) | \(49\) |
default | \(\frac {3}{686 \left (2+3 x \right )^{2}}-\frac {87}{2401 \left (2+3 x \right )}+\frac {558 \ln \left (2+3 x \right )}{16807}+\frac {11}{343 \left (-1+2 x \right )^{2}}-\frac {128}{2401 \left (-1+2 x \right )}-\frac {558 \ln \left (-1+2 x \right )}{16807}\) | \(54\) |
parallelrisch | \(\frac {80352 \ln \left (\frac {2}{3}+x \right ) x^{4}-80352 \ln \left (x -\frac {1}{2}\right ) x^{4}+69594+26784 \ln \left (\frac {2}{3}+x \right ) x^{3}-26784 \ln \left (x -\frac {1}{2}\right ) x^{3}+460908 x^{4}-51336 \ln \left (\frac {2}{3}+x \right ) x^{2}+51336 \ln \left (x -\frac {1}{2}\right ) x^{2}+59892 x^{3}-8928 \ln \left (\frac {2}{3}+x \right ) x +8928 \ln \left (x -\frac {1}{2}\right ) x -317905 x^{2}+8928 \ln \left (\frac {2}{3}+x \right )-8928 \ln \left (x -\frac {1}{2}\right )}{67228 \left (-1+2 x \right )^{2} \left (2+3 x \right )^{2}}\) | \(114\) |
Input:
int((3+5*x)/(1-2*x)^3/(2+3*x)^3,x,method=_RETURNVERBOSE)
Output:
(-3348/2401*x^3-837/2401*x^2+1829/2401*x+1313/4802)/(-1+2*x)^2/(2+3*x)^2-5 58/16807*ln(-1+2*x)+558/16807*ln(2+3*x)
Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.46 \[ \int \frac {3+5 x}{(1-2 x)^3 (2+3 x)^3} \, dx=-\frac {46872 \, x^{3} + 11718 \, x^{2} - 1116 \, {\left (36 \, x^{4} + 12 \, x^{3} - 23 \, x^{2} - 4 \, x + 4\right )} \log \left (3 \, x + 2\right ) + 1116 \, {\left (36 \, x^{4} + 12 \, x^{3} - 23 \, x^{2} - 4 \, x + 4\right )} \log \left (2 \, x - 1\right ) - 25606 \, x - 9191}{33614 \, {\left (36 \, x^{4} + 12 \, x^{3} - 23 \, x^{2} - 4 \, x + 4\right )}} \] Input:
integrate((3+5*x)/(1-2*x)^3/(2+3*x)^3,x, algorithm="fricas")
Output:
-1/33614*(46872*x^3 + 11718*x^2 - 1116*(36*x^4 + 12*x^3 - 23*x^2 - 4*x + 4 )*log(3*x + 2) + 1116*(36*x^4 + 12*x^3 - 23*x^2 - 4*x + 4)*log(2*x - 1) - 25606*x - 9191)/(36*x^4 + 12*x^3 - 23*x^2 - 4*x + 4)
Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.83 \[ \int \frac {3+5 x}{(1-2 x)^3 (2+3 x)^3} \, dx=- \frac {6696 x^{3} + 1674 x^{2} - 3658 x - 1313}{172872 x^{4} + 57624 x^{3} - 110446 x^{2} - 19208 x + 19208} - \frac {558 \log {\left (x - \frac {1}{2} \right )}}{16807} + \frac {558 \log {\left (x + \frac {2}{3} \right )}}{16807} \] Input:
integrate((3+5*x)/(1-2*x)**3/(2+3*x)**3,x)
Output:
-(6696*x**3 + 1674*x**2 - 3658*x - 1313)/(172872*x**4 + 57624*x**3 - 11044 6*x**2 - 19208*x + 19208) - 558*log(x - 1/2)/16807 + 558*log(x + 2/3)/1680 7
Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int \frac {3+5 x}{(1-2 x)^3 (2+3 x)^3} \, dx=-\frac {6696 \, x^{3} + 1674 \, x^{2} - 3658 \, x - 1313}{4802 \, {\left (36 \, x^{4} + 12 \, x^{3} - 23 \, x^{2} - 4 \, x + 4\right )}} + \frac {558}{16807} \, \log \left (3 \, x + 2\right ) - \frac {558}{16807} \, \log \left (2 \, x - 1\right ) \] Input:
integrate((3+5*x)/(1-2*x)^3/(2+3*x)^3,x, algorithm="maxima")
Output:
-1/4802*(6696*x^3 + 1674*x^2 - 3658*x - 1313)/(36*x^4 + 12*x^3 - 23*x^2 - 4*x + 4) + 558/16807*log(3*x + 2) - 558/16807*log(2*x - 1)
Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.71 \[ \int \frac {3+5 x}{(1-2 x)^3 (2+3 x)^3} \, dx=-\frac {6696 \, x^{3} + 1674 \, x^{2} - 3658 \, x - 1313}{4802 \, {\left (6 \, x^{2} + x - 2\right )}^{2}} + \frac {558}{16807} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {558}{16807} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \] Input:
integrate((3+5*x)/(1-2*x)^3/(2+3*x)^3,x, algorithm="giac")
Output:
-1/4802*(6696*x^3 + 1674*x^2 - 3658*x - 1313)/(6*x^2 + x - 2)^2 + 558/1680 7*log(abs(3*x + 2)) - 558/16807*log(abs(2*x - 1))
Time = 1.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.69 \[ \int \frac {3+5 x}{(1-2 x)^3 (2+3 x)^3} \, dx=\frac {1116\,\mathrm {atanh}\left (\frac {12\,x}{7}+\frac {1}{7}\right )}{16807}+\frac {-\frac {93\,x^3}{2401}-\frac {93\,x^2}{9604}+\frac {1829\,x}{86436}+\frac {1313}{172872}}{x^4+\frac {x^3}{3}-\frac {23\,x^2}{36}-\frac {x}{9}+\frac {1}{9}} \] Input:
int(-(5*x + 3)/((2*x - 1)^3*(3*x + 2)^3),x)
Output:
(1116*atanh((12*x)/7 + 1/7))/16807 + ((1829*x)/86436 - (93*x^2)/9604 - (93 *x^3)/2401 + 1313/172872)/(x^3/3 - (23*x^2)/36 - x/9 + x^4 + 1/9)
Time = 0.17 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.12 \[ \int \frac {3+5 x}{(1-2 x)^3 (2+3 x)^3} \, dx=\frac {40176 \,\mathrm {log}\left (3 x +2\right ) x^{4}+13392 \,\mathrm {log}\left (3 x +2\right ) x^{3}-25668 \,\mathrm {log}\left (3 x +2\right ) x^{2}-4464 \,\mathrm {log}\left (3 x +2\right ) x +4464 \,\mathrm {log}\left (3 x +2\right )-40176 \,\mathrm {log}\left (2 x -1\right ) x^{4}-13392 \,\mathrm {log}\left (2 x -1\right ) x^{3}+25668 \,\mathrm {log}\left (2 x -1\right ) x^{2}+4464 \,\mathrm {log}\left (2 x -1\right ) x -4464 \,\mathrm {log}\left (2 x -1\right )+140616 x^{4}-101556 x^{2}+9982 x +24815}{1210104 x^{4}+403368 x^{3}-773122 x^{2}-134456 x +134456} \] Input:
int((3+5*x)/(1-2*x)^3/(2+3*x)^3,x)
Output:
(40176*log(3*x + 2)*x**4 + 13392*log(3*x + 2)*x**3 - 25668*log(3*x + 2)*x* *2 - 4464*log(3*x + 2)*x + 4464*log(3*x + 2) - 40176*log(2*x - 1)*x**4 - 1 3392*log(2*x - 1)*x**3 + 25668*log(2*x - 1)*x**2 + 4464*log(2*x - 1)*x - 4 464*log(2*x - 1) + 140616*x**4 - 101556*x**2 + 9982*x + 24815)/(33614*(36* x**4 + 12*x**3 - 23*x**2 - 4*x + 4))