\(\int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)} \, dx\) [1301]

Optimal result

Integrand size = 22, antiderivative size = 59 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)} \, dx=\frac {49}{36 (2+3 x)^4}+\frac {217}{27 (2+3 x)^3}+\frac {121}{2 (2+3 x)^2}+\frac {605}{2+3 x}-3025 \log (2+3 x)+3025 \log (3+5 x) \]

[Out]

49/36/(2+3*x)^4+217/27/(2+3*x)^3+121/2/(2+3*x)^2+605/(2+3*x)-3025*ln(2+3*x)+3025*ln(3+5*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 59, 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} \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)} \, dx=\frac {605}{3 x+2}+\frac {121}{2 (3 x+2)^2}+\frac {217}{27 (3 x+2)^3}+\frac {49}{36 (3 x+2)^4}-3025 \log (3 x+2)+3025 \log (5 x+3) \]

[In]

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

[Out]

49/(36*(2 + 3*x)^4) + 217/(27*(2 + 3*x)^3) + 121/(2*(2 + 3*x)^2) + 605/(2 + 3*x) - 3025*Log[2 + 3*x] + 3025*Lo
g[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}{3 (2+3 x)^5}-\frac {217}{3 (2+3 x)^4}-\frac {363}{(2+3 x)^3}-\frac {1815}{(2+3 x)^2}-\frac {9075}{2+3 x}+\frac {15125}{3+5 x}\right ) \, dx \\ & = \frac {49}{36 (2+3 x)^4}+\frac {217}{27 (2+3 x)^3}+\frac {121}{2 (2+3 x)^2}+\frac {605}{2+3 x}-3025 \log (2+3 x)+3025 \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)} \, dx=\frac {550739+2433252 x+3587166 x^2+1764180 x^3}{108 (2+3 x)^4}-3025 \log (5 (2+3 x))+3025 \log (3+5 x) \]

[In]

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

[Out]

(550739 + 2433252*x + 3587166*x^2 + 1764180*x^3)/(108*(2 + 3*x)^4) - 3025*Log[5*(2 + 3*x)] + 3025*Log[3 + 5*x]

Maple [A] (verified)

Time = 2.32 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.69

method	result	size
norman	\(\frac {16335 x^{3}+\frac {66429}{2} x^{2}+\frac {202771}{9} x +\frac {550739}{108}}{\left (2+3 x \right )^{4}}-3025 \ln \left (2+3 x \right )+3025 \ln \left (3+5 x \right )\)	\(41\)
risch	\(\frac {16335 x^{3}+\frac {66429}{2} x^{2}+\frac {202771}{9} x +\frac {550739}{108}}{\left (2+3 x \right )^{4}}-3025 \ln \left (2+3 x \right )+3025 \ln \left (3+5 x \right )\)	\(42\)
default	\(\frac {49}{36 \left (2+3 x \right )^{4}}+\frac {217}{27 \left (2+3 x \right )^{3}}+\frac {121}{2 \left (2+3 x \right )^{2}}+\frac {605}{2+3 x}-3025 \ln \left (2+3 x \right )+3025 \ln \left (3+5 x \right )\)	\(54\)
parallelrisch	\(-\frac {15681600 \ln \left (\frac {2}{3}+x \right ) x^{4}-15681600 \ln \left (x +\frac {3}{5}\right ) x^{4}[phone] \ln \left (\frac {2}{3}+x \right ) x^{3}-41817600 \ln \left (x +\frac {3}{5}\right ) x^{3}+1652217 x^{4}[phone] \ln \left (\frac {2}{3}+x \right ) x^{2}-41817600 \ln \left (x +\frac {3}{5}\right ) x^{2}+3360472 x^{3}[phone] \ln \left (\frac {2}{3}+x \right ) x -18585600 \ln \left (x +\frac {3}{5}\right ) x +2280184 x^{2}+3097600 \ln \left (\frac {2}{3}+x \right )-3097600 \ln \left (x +\frac {3}{5}\right )+516256 x}{64 \left (2+3 x \right )^{4}}\)	\(109\)

[In]

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

[Out]

(16335*x^3+66429/2*x^2+202771/9*x+550739/108)/(2+3*x)^4-3025*ln(2+3*x)+3025*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.61 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)} \, dx=\frac {1764180 \, x^{3} + 3587166 \, x^{2} + 326700 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )} \log \left (5 \, x + 3\right ) - 326700 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )} \log \left (3 \, x + 2\right ) + 2433252 \, x + 550739}{108 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} \]

[In]

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

[Out]

1/108*(1764180*x^3 + 3587166*x^2 + 326700*(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)*log(5*x + 3) - 326700*(81*x
^4 + 216*x^3 + 216*x^2 + 96*x + 16)*log(3*x + 2) + 2433252*x + 550739)/(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16
)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.86 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)} \, dx=\frac {1764180 x^{3} + 3587166 x^{2} + 2433252 x + 550739}{8748 x^{4} + 23328 x^{3} + 23328 x^{2} + 10368 x + 1728} + 3025 \log {\left (x + \frac {3}{5} \right )} - 3025 \log {\left (x + \frac {2}{3} \right )} \]

[In]

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

[Out]

(1764180*x**3 + 3587166*x**2 + 2433252*x + 550739)/(8748*x**4 + 23328*x**3 + 23328*x**2 + 10368*x + 1728) + 30
25*log(x + 3/5) - 3025*log(x + 2/3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.95 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)} \, dx=\frac {1764180 \, x^{3} + 3587166 \, x^{2} + 2433252 \, x + 550739}{108 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} + 3025 \, \log \left (5 \, x + 3\right ) - 3025 \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

1/108*(1764180*x^3 + 3587166*x^2 + 2433252*x + 550739)/(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16) + 3025*log(5*x
+ 3) - 3025*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.88 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)} \, dx=\frac {605}{3 \, x + 2} + \frac {121}{2 \, {\left (3 \, x + 2\right )}^{2}} + \frac {217}{27 \, {\left (3 \, x + 2\right )}^{3}} + \frac {49}{36 \, {\left (3 \, x + 2\right )}^{4}} + 3025 \, \log \left ({\left | -\frac {1}{3 \, x + 2} + 5 \right |}\right ) \]

[In]

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

[Out]

605/(3*x + 2) + 121/2/(3*x + 2)^2 + 217/27/(3*x + 2)^3 + 49/36/(3*x + 2)^4 + 3025*log(abs(-1/(3*x + 2) + 5))

Mupad [B] (verification not implemented)

Time = 1.20 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x)^2}{(2+3 x)^5 (3+5 x)} \, dx=\frac {\frac {605\,x^3}{3}+\frac {7381\,x^2}{18}+\frac {202771\,x}{729}+\frac {550739}{8748}}{x^4+\frac {8\,x^3}{3}+\frac {8\,x^2}{3}+\frac {32\,x}{27}+\frac {16}{81}}-6050\,\mathrm {atanh}\left (30\,x+19\right ) \]

[In]

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

[Out]

((202771*x)/729 + (7381*x^2)/18 + (605*x^3)/3 + 550739/8748)/((32*x)/27 + (8*x^2)/3 + (8*x^3)/3 + x^4 + 16/81)
- 6050*atanh(30*x + 19)