\(\int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)^2} \, dx\) [1317]

Optimal result

Integrand size = 22, antiderivative size = 68 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)^2} \, dx=-\frac {49}{12 (2+3 x)^4}-\frac {154}{3 (2+3 x)^3}-\frac {1133}{2 (2+3 x)^2}-\frac {7480}{2+3 x}-\frac {3025}{3+5 x}+46475 \log (2+3 x)-46475 \log (3+5 x) \]

[Out]

-49/12/(2+3*x)^4-154/3/(2+3*x)^3-1133/2/(2+3*x)^2-7480/(2+3*x)-3025/(3+5*x)+46475*ln(2+3*x)-46475*ln(3+5*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 68, 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} \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)^2} \, dx=-\frac {7480}{3 x+2}-\frac {3025}{5 x+3}-\frac {1133}{2 (3 x+2)^2}-\frac {154}{3 (3 x+2)^3}-\frac {49}{12 (3 x+2)^4}+46475 \log (3 x+2)-46475 \log (5 x+3) \]

[In]

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

[Out]

-49/(12*(2 + 3*x)^4) - 154/(3*(2 + 3*x)^3) - 1133/(2*(2 + 3*x)^2) - 7480/(2 + 3*x) - 3025/(3 + 5*x) + 46475*Lo
g[2 + 3*x] - 46475*Log[3 + 5*x]

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 {49}{(2+3 x)^5}+\frac {462}{(2+3 x)^4}+\frac {3399}{(2+3 x)^3}+\frac {22440}{(2+3 x)^2}+\frac {139425}{2+3 x}+\frac {15125}{(3+5 x)^2}-\frac {232375}{3+5 x}\right ) \, dx \\ & = -\frac {49}{12 (2+3 x)^4}-\frac {154}{3 (2+3 x)^3}-\frac {1133}{2 (2+3 x)^2}-\frac {7480}{2+3 x}-\frac {3025}{3+5 x}+46475 \log (2+3 x)-46475 \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)^2} \, dx=-\frac {940153+5720639 x+13046462 x^2+13217490 x^3+5019300 x^4}{4 (2+3 x)^4 (3+5 x)}+46475 \log (5 (2+3 x))-46475 \log (3+5 x) \]

[In]

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

[Out]

-1/4*(940153 + 5720639*x + 13046462*x^2 + 13217490*x^3 + 5019300*x^4)/((2 + 3*x)^4*(3 + 5*x)) + 46475*Log[5*(2
+ 3*x)] - 46475*Log[3 + 5*x]

Maple [A] (verified)

Time = 2.33 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.78

method	result	size
norman	\(\frac {-1254825 x^{4}-\frac {6608745}{2} x^{3}-\frac {6523231}{2} x^{2}-\frac {5720639}{4} x -\frac {940153}{4}}{\left (2+3 x \right )^{4} \left (3+5 x \right )}+46475 \ln \left (2+3 x \right )-46475 \ln \left (3+5 x \right )\)	\(53\)
risch	\(\frac {-1254825 x^{4}-\frac {6608745}{2} x^{3}-\frac {6523231}{2} x^{2}-\frac {5720639}{4} x -\frac {940153}{4}}{\left (2+3 x \right )^{4} \left (3+5 x \right )}+46475 \ln \left (2+3 x \right )-46475 \ln \left (3+5 x \right )\)	\(54\)
default	\(-\frac {49}{12 \left (2+3 x \right )^{4}}-\frac {154}{3 \left (2+3 x \right )^{3}}-\frac {1133}{2 \left (2+3 x \right )^{2}}-\frac {7480}{2+3 x}-\frac {3025}{3+5 x}+46475 \ln \left (2+3 x \right )-46475 \ln \left (3+5 x \right )\)	\(63\)
parallelrisch	\(\frac {3613896000 \ln \left (\frac {2}{3}+x \right ) x^{5}-3613896000 \ln \left (x +\frac {3}{5}\right ) x^{5}[phone] \ln \left (\frac {2}{3}+x \right ) x^{4}-11805393600 \ln \left (x +\frac {3}{5}\right ) x^{4}[phone] x^{5}[phone] \ln \left (\frac {2}{3}+x \right ) x^{3}-15419289600 \ln \left (x +\frac {3}{5}\right ) x^{3}[phone] x^{4}[phone] \ln \left (\frac {2}{3}+x \right ) x^{2}-10065369600 \ln \left (x +\frac {3}{5}\right ) x^{2}[phone] x^{3}[phone] \ln \left (\frac {2}{3}+x \right ) x -3283737600 \ln \left (x +\frac {3}{5}\right ) x [phone] x^{2}[phone] \ln \left (\frac {2}{3}+x \right )-428313600 \ln \left (x +\frac {3}{5}\right )[phone] x}{192 \left (2+3 x \right )^{4} \left (3+5 x \right )}\)	\(139\)

[In]

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

[Out]

(-1254825*x^4-6608745/2*x^3-6523231/2*x^2-5720639/4*x-940153/4)/(2+3*x)^4/(3+5*x)+46475*ln(2+3*x)-46475*ln(3+5
*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.69 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)^2} \, dx=-\frac {5019300 \, x^{4} + 13217490 \, x^{3} + 13046462 \, x^{2} + 185900 \, {\left (405 \, x^{5} + 1323 \, x^{4} + 1728 \, x^{3} + 1128 \, x^{2} + 368 \, x + 48\right )} \log \left (5 \, x + 3\right ) - 185900 \, {\left (405 \, x^{5} + 1323 \, x^{4} + 1728 \, x^{3} + 1128 \, x^{2} + 368 \, x + 48\right )} \log \left (3 \, x + 2\right ) + 5720639 \, x + 940153}{4 \, {\left (405 \, x^{5} + 1323 \, x^{4} + 1728 \, x^{3} + 1128 \, x^{2} + 368 \, x + 48\right )}} \]

[In]

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

[Out]

-1/4*(5019300*x^4 + 13217490*x^3 + 13046462*x^2 + 185900*(405*x^5 + 1323*x^4 + 1728*x^3 + 1128*x^2 + 368*x + 4
8)*log(5*x + 3) - 185900*(405*x^5 + 1323*x^4 + 1728*x^3 + 1128*x^2 + 368*x + 48)*log(3*x + 2) + 5720639*x + 94
0153)/(405*x^5 + 1323*x^4 + 1728*x^3 + 1128*x^2 + 368*x + 48)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.93 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)^2} \, dx=\frac {- 5019300 x^{4} - 13217490 x^{3} - 13046462 x^{2} - 5720639 x - 940153}{1620 x^{5} + 5292 x^{4} + 6912 x^{3} + 4512 x^{2} + 1472 x + 192} - 46475 \log {\left (x + \frac {3}{5} \right )} + 46475 \log {\left (x + \frac {2}{3} \right )} \]

[In]

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

[Out]

(-5019300*x**4 - 13217490*x**3 - 13046462*x**2 - 5720639*x - 940153)/(1620*x**5 + 5292*x**4 + 6912*x**3 + 4512
*x**2 + 1472*x + 192) - 46475*log(x + 3/5) + 46475*log(x + 2/3)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)^2} \, dx=-\frac {5019300 \, x^{4} + 13217490 \, x^{3} + 13046462 \, x^{2} + 5720639 \, x + 940153}{4 \, {\left (405 \, x^{5} + 1323 \, x^{4} + 1728 \, x^{3} + 1128 \, x^{2} + 368 \, x + 48\right )}} - 46475 \, \log \left (5 \, x + 3\right ) + 46475 \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

-1/4*(5019300*x^4 + 13217490*x^3 + 13046462*x^2 + 5720639*x + 940153)/(405*x^5 + 1323*x^4 + 1728*x^3 + 1128*x^
2 + 368*x + 48) - 46475*log(5*x + 3) + 46475*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.99 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)^2} \, dx=-\frac {3025}{5 \, x + 3} + \frac {25 \, {\left (\frac {884412}{5 \, x + 3} + \frac {341028}{{\left (5 \, x + 3\right )}^{2}} + \frac {45688}{{\left (5 \, x + 3\right )}^{3}} + 784485\right )}}{4 \, {\left (\frac {1}{5 \, x + 3} + 3\right )}^{4}} + 46475 \, \log \left ({\left | -\frac {1}{5 \, x + 3} - 3 \right |}\right ) \]

[In]

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

[Out]

-3025/(5*x + 3) + 25/4*(884412/(5*x + 3) + 341028/(5*x + 3)^2 + 45688/(5*x + 3)^3 + 784485)/(1/(5*x + 3) + 3)^
4 + 46475*log(abs(-1/(5*x + 3) - 3))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)^2} \, dx=92950\,\mathrm {atanh}\left (30\,x+19\right )-\frac {\frac {9295\,x^4}{3}+\frac {146861\,x^3}{18}+\frac {6523231\,x^2}{810}+\frac {5720639\,x}{1620}+\frac {940153}{1620}}{x^5+\frac {49\,x^4}{15}+\frac {64\,x^3}{15}+\frac {376\,x^2}{135}+\frac {368\,x}{405}+\frac {16}{135}} \]

[In]

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

[Out]

92950*atanh(30*x + 19) - ((5720639*x)/1620 + (6523231*x^2)/810 + (146861*x^3)/18 + (9295*x^4)/3 + 940153/1620)
/((368*x)/405 + (376*x^2)/135 + (64*x^3)/15 + (49*x^4)/15 + x^5 + 16/135)