\(\int \frac {(1-2 x)^2}{(2+3 x)^3 (3+5 x)} \, dx\) [1299]

Optimal result

Integrand size = 22, antiderivative size = 39 \[ \int \frac {(1-2 x)^2}{(2+3 x)^3 (3+5 x)} \, dx=\frac {49}{18 (2+3 x)^2}+\frac {217}{9 (2+3 x)}-121 \log (2+3 x)+121 \log (3+5 x) \]

[Out]

49/18/(2+3*x)^2+217/9/(2+3*x)-121*ln(2+3*x)+121*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, 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} \[ \int \frac {(1-2 x)^2}{(2+3 x)^3 (3+5 x)} \, dx=\frac {217}{9 (3 x+2)}+\frac {49}{18 (3 x+2)^2}-121 \log (3 x+2)+121 \log (5 x+3) \]

[In]

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

[Out]

49/(18*(2 + 3*x)^2) + 217/(9*(2 + 3*x)) - 121*Log[2 + 3*x] + 121*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}{3 (2+3 x)^3}-\frac {217}{3 (2+3 x)^2}-\frac {363}{2+3 x}+\frac {605}{3+5 x}\right ) \, dx \\ & = \frac {49}{18 (2+3 x)^2}+\frac {217}{9 (2+3 x)}-121 \log (2+3 x)+121 \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.23 \[ \int \frac {(1-2 x)^2}{(2+3 x)^3 (3+5 x)} \, dx=\frac {917+1302 x-2178 (2+3 x)^2 \log (5 (2+3 x))+2178 (2+3 x)^2 \log (3+5 x)}{18 (2+3 x)^2} \]

[In]

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

[Out]

(917 + 1302*x - 2178*(2 + 3*x)^2*Log[5*(2 + 3*x)] + 2178*(2 + 3*x)^2*Log[3 + 5*x])/(18*(2 + 3*x)^2)

Maple [A] (verified)

Time = 2.33 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.82

method	result	size
risch	\(\frac {\frac {217 x}{3}+\frac {917}{18}}{\left (2+3 x \right )^{2}}-121 \ln \left (2+3 x \right )+121 \ln \left (3+5 x \right )\)	\(32\)
norman	\(\frac {-\frac {161}{2} x -\frac {917}{8} x^{2}}{\left (2+3 x \right )^{2}}-121 \ln \left (2+3 x \right )+121 \ln \left (3+5 x \right )\)	\(35\)
default	\(\frac {49}{18 \left (2+3 x \right )^{2}}+\frac {217}{9 \left (2+3 x \right )}-121 \ln \left (2+3 x \right )+121 \ln \left (3+5 x \right )\)	\(36\)
parallelrisch	\(-\frac {8712 \ln \left (\frac {2}{3}+x \right ) x^{2}-8712 \ln \left (x +\frac {3}{5}\right ) x^{2}+11616 \ln \left (\frac {2}{3}+x \right ) x -11616 \ln \left (x +\frac {3}{5}\right ) x +917 x^{2}+3872 \ln \left (\frac {2}{3}+x \right )-3872 \ln \left (x +\frac {3}{5}\right )+644 x}{8 \left (2+3 x \right )^{2}}\)	\(63\)

[In]

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

[Out]

9*(217/27*x+917/162)/(2+3*x)^2-121*ln(2+3*x)+121*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.41 \[ \int \frac {(1-2 x)^2}{(2+3 x)^3 (3+5 x)} \, dx=\frac {2178 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (5 \, x + 3\right ) - 2178 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (3 \, x + 2\right ) + 1302 \, x + 917}{18 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \]

[In]

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

[Out]

1/18*(2178*(9*x^2 + 12*x + 4)*log(5*x + 3) - 2178*(9*x^2 + 12*x + 4)*log(3*x + 2) + 1302*x + 917)/(9*x^2 + 12*
x + 4)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.79 \[ \int \frac {(1-2 x)^2}{(2+3 x)^3 (3+5 x)} \, dx=\frac {1302 x + 917}{162 x^{2} + 216 x + 72} + 121 \log {\left (x + \frac {3}{5} \right )} - 121 \log {\left (x + \frac {2}{3} \right )} \]

[In]

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

[Out]

(1302*x + 917)/(162*x**2 + 216*x + 72) + 121*log(x + 3/5) - 121*log(x + 2/3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.92 \[ \int \frac {(1-2 x)^2}{(2+3 x)^3 (3+5 x)} \, dx=\frac {7 \, {\left (186 \, x + 131\right )}}{18 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} + 121 \, \log \left (5 \, x + 3\right ) - 121 \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

7/18*(186*x + 131)/(9*x^2 + 12*x + 4) + 121*log(5*x + 3) - 121*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.85 \[ \int \frac {(1-2 x)^2}{(2+3 x)^3 (3+5 x)} \, dx=\frac {7 \, {\left (186 \, x + 131\right )}}{18 \, {\left (3 \, x + 2\right )}^{2}} + 121 \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - 121 \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

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

[Out]

7/18*(186*x + 131)/(3*x + 2)^2 + 121*log(abs(5*x + 3)) - 121*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.64 \[ \int \frac {(1-2 x)^2}{(2+3 x)^3 (3+5 x)} \, dx=\frac {\frac {217\,x}{27}+\frac {917}{162}}{x^2+\frac {4\,x}{3}+\frac {4}{9}}-242\,\mathrm {atanh}\left (30\,x+19\right ) \]

[In]

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

[Out]

((217*x)/27 + 917/162)/((4*x)/3 + x^2 + 4/9) - 242*atanh(30*x + 19)