\(\int \frac {(1-2 x)^2}{(2+3 x)^3 (3+5 x)^3} \, dx\) [1332]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 22, antiderivative size = 57 \[ \int \frac {(1-2 x)^2}{(2+3 x)^3 (3+5 x)^3} \, dx=\frac {49}{2 (2+3 x)^2}+\frac {707}{2+3 x}-\frac {121}{2 (3+5 x)^2}+\frac {1133}{3+5 x}-6934 \log (2+3 x)+6934 \log (3+5 x) \]

[Out]

49/2/(2+3*x)^2+707/(2+3*x)-121/2/(3+5*x)^2+1133/(3+5*x)-6934*ln(2+3*x)+6934*ln(3+5*x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 57, 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} \[ \int \frac {(1-2 x)^2}{(2+3 x)^3 (3+5 x)^3} \, dx=\frac {707}{3 x+2}+\frac {1133}{5 x+3}+\frac {49}{2 (3 x+2)^2}-\frac {121}{2 (5 x+3)^2}-6934 \log (3 x+2)+6934 \log (5 x+3) \]

[In]

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

[Out]

49/(2*(2 + 3*x)^2) + 707/(2 + 3*x) - 121/(2*(3 + 5*x)^2) + 1133/(3 + 5*x) - 6934*Log[2 + 3*x] + 6934*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 {147}{(2+3 x)^3}-\frac {2121}{(2+3 x)^2}-\frac {20802}{2+3 x}+\frac {605}{(3+5 x)^3}-\frac {5665}{(3+5 x)^2}+\frac {34670}{3+5 x}\right ) \, dx \\ & = \frac {49}{2 (2+3 x)^2}+\frac {707}{2+3 x}-\frac {121}{2 (3+5 x)^2}+\frac {1133}{3+5 x}-6934 \log (2+3 x)+6934 \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.04 \[ \int \frac {(1-2 x)^2}{(2+3 x)^3 (3+5 x)^3} \, dx=\frac {49}{2 (2+3 x)^2}+\frac {707}{2+3 x}-\frac {121}{2 (3+5 x)^2}+\frac {1133}{3+5 x}-6934 \log (5 (2+3 x))+6934 \log (3+5 x) \]

[In]

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

[Out]

49/(2*(2 + 3*x)^2) + 707/(2 + 3*x) - 121/(2*(3 + 5*x)^2) + 1133/(3 + 5*x) - 6934*Log[5*(2 + 3*x)] + 6934*Log[3
 + 5*x]

Maple [A] (verified)

Time = 2.37 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.84

method result size
norman \(\frac {104010 x^{3}+124966 x +197619 x^{2}+\frac {52601}{2}}{\left (2+3 x \right )^{2} \left (3+5 x \right )^{2}}-6934 \ln \left (2+3 x \right )+6934 \ln \left (3+5 x \right )\) \(48\)
risch \(\frac {104010 x^{3}+124966 x +197619 x^{2}+\frac {52601}{2}}{\left (2+3 x \right )^{2} \left (3+5 x \right )^{2}}-6934 \ln \left (2+3 x \right )+6934 \ln \left (3+5 x \right )\) \(49\)
default \(\frac {49}{2 \left (2+3 x \right )^{2}}+\frac {707}{2+3 x}-\frac {121}{2 \left (3+5 x \right )^{2}}+\frac {1133}{3+5 x}-6934 \ln \left (2+3 x \right )+6934 \ln \left (3+5 x \right )\) \(54\)
parallelrisch \(-\frac {112330800 \ln \left (\frac {2}{3}+x \right ) x^{4}-112330800 \ln \left (x +\frac {3}{5}\right ) x^{4}+284571360 \ln \left (\frac {2}{3}+x \right ) x^{3}-284571360 \ln \left (x +\frac {3}{5}\right ) x^{3}+11835225 x^{4}+270093168 \ln \left (\frac {2}{3}+x \right ) x^{2}-270093168 \ln \left (x +\frac {3}{5}\right ) x^{2}+22493850 x^{3}+113828544 \ln \left (\frac {2}{3}+x \right ) x -113828544 \ln \left (x +\frac {3}{5}\right ) x +14228573 x^{2}+17972928 \ln \left (\frac {2}{3}+x \right )-17972928 \ln \left (x +\frac {3}{5}\right )+2995476 x}{72 \left (2+3 x \right )^{2} \left (3+5 x \right )^{2}}\) \(116\)

[In]

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

[Out]

(104010*x^3+124966*x+197619*x^2+52601/2)/(2+3*x)^2/(3+5*x)^2-6934*ln(2+3*x)+6934*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.67 \[ \int \frac {(1-2 x)^2}{(2+3 x)^3 (3+5 x)^3} \, dx=\frac {208020 \, x^{3} + 395238 \, x^{2} + 13868 \, {\left (225 \, x^{4} + 570 \, x^{3} + 541 \, x^{2} + 228 \, x + 36\right )} \log \left (5 \, x + 3\right ) - 13868 \, {\left (225 \, x^{4} + 570 \, x^{3} + 541 \, x^{2} + 228 \, x + 36\right )} \log \left (3 \, x + 2\right ) + 249932 \, x + 52601}{2 \, {\left (225 \, x^{4} + 570 \, x^{3} + 541 \, x^{2} + 228 \, x + 36\right )}} \]

[In]

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

[Out]

1/2*(208020*x^3 + 395238*x^2 + 13868*(225*x^4 + 570*x^3 + 541*x^2 + 228*x + 36)*log(5*x + 3) - 13868*(225*x^4
+ 570*x^3 + 541*x^2 + 228*x + 36)*log(3*x + 2) + 249932*x + 52601)/(225*x^4 + 570*x^3 + 541*x^2 + 228*x + 36)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.89 \[ \int \frac {(1-2 x)^2}{(2+3 x)^3 (3+5 x)^3} \, dx=\frac {208020 x^{3} + 395238 x^{2} + 249932 x + 52601}{450 x^{4} + 1140 x^{3} + 1082 x^{2} + 456 x + 72} + 6934 \log {\left (x + \frac {3}{5} \right )} - 6934 \log {\left (x + \frac {2}{3} \right )} \]

[In]

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

[Out]

(208020*x**3 + 395238*x**2 + 249932*x + 52601)/(450*x**4 + 1140*x**3 + 1082*x**2 + 456*x + 72) + 6934*log(x +
3/5) - 6934*log(x + 2/3)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.98 \[ \int \frac {(1-2 x)^2}{(2+3 x)^3 (3+5 x)^3} \, dx=\frac {208020 \, x^{3} + 395238 \, x^{2} + 249932 \, x + 52601}{2 \, {\left (225 \, x^{4} + 570 \, x^{3} + 541 \, x^{2} + 228 \, x + 36\right )}} + 6934 \, \log \left (5 \, x + 3\right ) - 6934 \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

1/2*(208020*x^3 + 395238*x^2 + 249932*x + 52601)/(225*x^4 + 570*x^3 + 541*x^2 + 228*x + 36) + 6934*log(5*x + 3
) - 6934*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^2}{(2+3 x)^3 (3+5 x)^3} \, dx=\frac {208020 \, x^{3} + 395238 \, x^{2} + 249932 \, x + 52601}{2 \, {\left (15 \, x^{2} + 19 \, x + 6\right )}^{2}} + 6934 \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - 6934 \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

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

[Out]

1/2*(208020*x^3 + 395238*x^2 + 249932*x + 52601)/(15*x^2 + 19*x + 6)^2 + 6934*log(abs(5*x + 3)) - 6934*log(abs
(3*x + 2))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.79 \[ \int \frac {(1-2 x)^2}{(2+3 x)^3 (3+5 x)^3} \, dx=\frac {\frac {6934\,x^3}{15}+\frac {65873\,x^2}{75}+\frac {124966\,x}{225}+\frac {52601}{450}}{x^4+\frac {38\,x^3}{15}+\frac {541\,x^2}{225}+\frac {76\,x}{75}+\frac {4}{25}}-13868\,\mathrm {atanh}\left (30\,x+19\right ) \]

[In]

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

[Out]

((124966*x)/225 + (65873*x^2)/75 + (6934*x^3)/15 + 52601/450)/((76*x)/75 + (541*x^2)/225 + (38*x^3)/15 + x^4 +
 4/25) - 13868*atanh(30*x + 19)