\(\int \frac {(3+5 x)^2}{(1-2 x)^3 (2+3 x)^3} \, dx\) [1655]

Optimal result

Integrand size = 22, antiderivative size = 65 \[ \int \frac {(3+5 x)^2}{(1-2 x)^3 (2+3 x)^3} \, dx=\frac {121}{686 (1-2 x)^2}+\frac {319}{2401 (1-2 x)}-\frac {1}{686 (2+3 x)^2}+\frac {64}{2401 (2+3 x)}-\frac {829 \log (1-2 x)}{16807}+\frac {829 \log (2+3 x)}{16807} \]

[Out]

121/686/(1-2*x)^2+319/2401/(1-2*x)-1/686/(2+3*x)^2+64/2401/(2+3*x)-829/16807*ln(1-2*x)+829/16807*ln(2+3*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {90} \[ \int \frac {(3+5 x)^2}{(1-2 x)^3 (2+3 x)^3} \, dx=\frac {319}{2401 (1-2 x)}+\frac {64}{2401 (3 x+2)}+\frac {121}{686 (1-2 x)^2}-\frac {1}{686 (3 x+2)^2}-\frac {829 \log (1-2 x)}{16807}+\frac {829 \log (3 x+2)}{16807} \]

[In]

Int[(3 + 5*x)^2/((1 - 2*x)^3*(2 + 3*x)^3),x]

[Out]

121/(686*(1 - 2*x)^2) + 319/(2401*(1 - 2*x)) - 1/(686*(2 + 3*x)^2) + 64/(2401*(2 + 3*x)) - (829*Log[1 - 2*x])/
16807 + (829*Log[2 + 3*x])/16807

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(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]))

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {242}{343 (-1+2 x)^3}+\frac {638}{2401 (-1+2 x)^2}-\frac {1658}{16807 (-1+2 x)}+\frac {3}{343 (2+3 x)^3}-\frac {192}{2401 (2+3 x)^2}+\frac {2487}{16807 (2+3 x)}\right ) \, dx \\ & = \frac {121}{686 (1-2 x)^2}+\frac {319}{2401 (1-2 x)}-\frac {1}{686 (2+3 x)^2}+\frac {64}{2401 (2+3 x)}-\frac {829 \log (1-2 x)}{16807}+\frac {829 \log (2+3 x)}{16807} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74 \[ \int \frac {(3+5 x)^2}{(1-2 x)^3 (2+3 x)^3} \, dx=\frac {-\frac {7 \left (-6189-12104 x+2487 x^2+9948 x^3\right )}{\left (-2+x+6 x^2\right )^2}-1658 \log (1-2 x)+1658 \log (2+3 x)}{33614} \]

[In]

Integrate[(3 + 5*x)^2/((1 - 2*x)^3*(2 + 3*x)^3),x]

[Out]

((-7*(-6189 - 12104*x + 2487*x^2 + 9948*x^3))/(-2 + x + 6*x^2)^2 - 1658*Log[1 - 2*x] + 1658*Log[2 + 3*x])/3361
4

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74

method	result	size
norman	\(\frac {-\frac {4974}{2401} x^{3}-\frac {2487}{4802} x^{2}+\frac {6052}{2401} x +\frac {6189}{4802}}{\left (-1+2 x \right )^{2} \left (2+3 x \right )^{2}}-\frac {829 \ln \left (-1+2 x \right )}{16807}+\frac {829 \ln \left (2+3 x \right )}{16807}\)	\(48\)
risch	\(\frac {-\frac {4974}{2401} x^{3}-\frac {2487}{4802} x^{2}+\frac {6052}{2401} x +\frac {6189}{4802}}{\left (-1+2 x \right )^{2} \left (2+3 x \right )^{2}}-\frac {829 \ln \left (-1+2 x \right )}{16807}+\frac {829 \ln \left (2+3 x \right )}{16807}\)	\(49\)
default	\(\frac {121}{686 \left (-1+2 x \right )^{2}}-\frac {319}{2401 \left (-1+2 x \right )}-\frac {829 \ln \left (-1+2 x \right )}{16807}-\frac {1}{686 \left (2+3 x \right )^{2}}+\frac {64}{2401 \left (2+3 x \right )}+\frac {829 \ln \left (2+3 x \right )}{16807}\)	\(54\)
parallelrisch	\(\frac {119376 \ln \left (\frac {2}{3}+x \right ) x^{4}-119376 \ln \left (x -\frac {1}{2}\right ) x^{4}+256102+39792 \ln \left (\frac {2}{3}+x \right ) x^{3}-39792 \ln \left (x -\frac {1}{2}\right ) x^{3}+1525104 x^{4}-76268 \ln \left (\frac {2}{3}+x \right ) x^{2}+76268 \ln \left (x -\frac {1}{2}\right ) x^{2}+369096 x^{3}-13264 \ln \left (\frac {2}{3}+x \right ) x +13264 \ln \left (x -\frac {1}{2}\right ) x -1009190 x^{2}+13264 \ln \left (\frac {2}{3}+x \right )-13264 \ln \left (x -\frac {1}{2}\right )}{67228 \left (-1+2 x \right )^{2} \left (2+3 x \right )^{2}}\)	\(114\)

[In]

int((3+5*x)^2/(1-2*x)^3/(2+3*x)^3,x,method=_RETURNVERBOSE)

[Out]

(-4974/2401*x^3-2487/4802*x^2+6052/2401*x+6189/4802)/(-1+2*x)^2/(2+3*x)^2-829/16807*ln(-1+2*x)+829/16807*ln(2+
3*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.46 \[ \int \frac {(3+5 x)^2}{(1-2 x)^3 (2+3 x)^3} \, dx=-\frac {69636 \, x^{3} + 17409 \, x^{2} - 1658 \, {\left (36 \, x^{4} + 12 \, x^{3} - 23 \, x^{2} - 4 \, x + 4\right )} \log \left (3 \, x + 2\right ) + 1658 \, {\left (36 \, x^{4} + 12 \, x^{3} - 23 \, x^{2} - 4 \, x + 4\right )} \log \left (2 \, x - 1\right ) - 84728 \, x - 43323}{33614 \, {\left (36 \, x^{4} + 12 \, x^{3} - 23 \, x^{2} - 4 \, x + 4\right )}} \]

[In]

integrate((3+5*x)^2/(1-2*x)^3/(2+3*x)^3,x, algorithm="fricas")

[Out]

-1/33614*(69636*x^3 + 17409*x^2 - 1658*(36*x^4 + 12*x^3 - 23*x^2 - 4*x + 4)*log(3*x + 2) + 1658*(36*x^4 + 12*x
^3 - 23*x^2 - 4*x + 4)*log(2*x - 1) - 84728*x - 43323)/(36*x^4 + 12*x^3 - 23*x^2 - 4*x + 4)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.83 \[ \int \frac {(3+5 x)^2}{(1-2 x)^3 (2+3 x)^3} \, dx=- \frac {9948 x^{3} + 2487 x^{2} - 12104 x - 6189}{172872 x^{4} + 57624 x^{3} - 110446 x^{2} - 19208 x + 19208} - \frac {829 \log {\left (x - \frac {1}{2} \right )}}{16807} + \frac {829 \log {\left (x + \frac {2}{3} \right )}}{16807} \]

[In]

integrate((3+5*x)**2/(1-2*x)**3/(2+3*x)**3,x)

[Out]

-(9948*x**3 + 2487*x**2 - 12104*x - 6189)/(172872*x**4 + 57624*x**3 - 110446*x**2 - 19208*x + 19208) - 829*log
(x - 1/2)/16807 + 829*log(x + 2/3)/16807

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int \frac {(3+5 x)^2}{(1-2 x)^3 (2+3 x)^3} \, dx=-\frac {9948 \, x^{3} + 2487 \, x^{2} - 12104 \, x - 6189}{4802 \, {\left (36 \, x^{4} + 12 \, x^{3} - 23 \, x^{2} - 4 \, x + 4\right )}} + \frac {829}{16807} \, \log \left (3 \, x + 2\right ) - \frac {829}{16807} \, \log \left (2 \, x - 1\right ) \]

[In]

integrate((3+5*x)^2/(1-2*x)^3/(2+3*x)^3,x, algorithm="maxima")

[Out]

-1/4802*(9948*x^3 + 2487*x^2 - 12104*x - 6189)/(36*x^4 + 12*x^3 - 23*x^2 - 4*x + 4) + 829/16807*log(3*x + 2) -
829/16807*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.71 \[ \int \frac {(3+5 x)^2}{(1-2 x)^3 (2+3 x)^3} \, dx=-\frac {9948 \, x^{3} + 2487 \, x^{2} - 12104 \, x - 6189}{4802 \, {\left (6 \, x^{2} + x - 2\right )}^{2}} + \frac {829}{16807} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {829}{16807} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

integrate((3+5*x)^2/(1-2*x)^3/(2+3*x)^3,x, algorithm="giac")

[Out]

-1/4802*(9948*x^3 + 2487*x^2 - 12104*x - 6189)/(6*x^2 + x - 2)^2 + 829/16807*log(abs(3*x + 2)) - 829/16807*log
(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 1.34 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.69 \[ \int \frac {(3+5 x)^2}{(1-2 x)^3 (2+3 x)^3} \, dx=\frac {1658\,\mathrm {atanh}\left (\frac {12\,x}{7}+\frac {1}{7}\right )}{16807}+\frac {-\frac {829\,x^3}{14406}-\frac {829\,x^2}{57624}+\frac {1513\,x}{21609}+\frac {2063}{57624}}{x^4+\frac {x^3}{3}-\frac {23\,x^2}{36}-\frac {x}{9}+\frac {1}{9}} \]

[In]

int(-(5*x + 3)^2/((2*x - 1)^3*(3*x + 2)^3),x)

[Out]

(1658*atanh((12*x)/7 + 1/7))/16807 + ((1513*x)/21609 - (829*x^2)/57624 - (829*x^3)/14406 + 2063/57624)/(x^3/3
- (23*x^2)/36 - x/9 + x^4 + 1/9)