\(\int \frac {(1-2 x)^2}{(2+3 x)^2 (3+5 x)} \, dx\) [1298]

Optimal result

Integrand size = 22, antiderivative size = 32 \[ \int \frac {(1-2 x)^2}{(2+3 x)^2 (3+5 x)} \, dx=\frac {49}{9 (2+3 x)}-\frac {217}{9} \log (2+3 x)+\frac {121}{5} \log (3+5 x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, 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} \[ \int \frac {(1-2 x)^2}{(2+3 x)^2 (3+5 x)} \, dx=\frac {49}{9 (3 x+2)}-\frac {217}{9} \log (3 x+2)+\frac {121}{5} \log (5 x+3) \]

[In]

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

[Out]

49/(9*(2 + 3*x)) - (217*Log[2 + 3*x])/9 + (121*Log[3 + 5*x])/5

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)^2}-\frac {217}{3 (2+3 x)}+\frac {121}{3+5 x}\right ) \, dx \\ & = \frac {49}{9 (2+3 x)}-\frac {217}{9} \log (2+3 x)+\frac {121}{5} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {(1-2 x)^2}{(2+3 x)^2 (3+5 x)} \, dx=\frac {49}{18+27 x}-\frac {217}{9} \log (5 (2+3 x))+\frac {121}{5} \log (3+5 x) \]

[In]

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

[Out]

49/(18 + 27*x) - (217*Log[5*(2 + 3*x)])/9 + (121*Log[3 + 5*x])/5

Maple [A] (verified)

Time = 1.91 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {49}{27 \left (\frac {2}{3}+x \right )}-\frac {217 \ln \left (2+3 x \right )}{9}+\frac {121 \ln \left (3+5 x \right )}{5}\)	\(25\)
default	\(\frac {49}{9 \left (2+3 x \right )}-\frac {217 \ln \left (2+3 x \right )}{9}+\frac {121 \ln \left (3+5 x \right )}{5}\)	\(27\)
norman	\(-\frac {49 x}{6 \left (2+3 x \right )}-\frac {217 \ln \left (2+3 x \right )}{9}+\frac {121 \ln \left (3+5 x \right )}{5}\)	\(28\)
parallelrisch	\(-\frac {6510 \ln \left (\frac {2}{3}+x \right ) x -6534 \ln \left (x +\frac {3}{5}\right ) x +4340 \ln \left (\frac {2}{3}+x \right )-4356 \ln \left (x +\frac {3}{5}\right )+735 x}{90 \left (2+3 x \right )}\)	\(40\)

[In]

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

[Out]

49/27/(2/3+x)-217/9*ln(2+3*x)+121/5*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {(1-2 x)^2}{(2+3 x)^2 (3+5 x)} \, dx=\frac {1089 \, {\left (3 \, x + 2\right )} \log \left (5 \, x + 3\right ) - 1085 \, {\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) + 245}{45 \, {\left (3 \, x + 2\right )}} \]

[In]

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

[Out]

1/45*(1089*(3*x + 2)*log(5*x + 3) - 1085*(3*x + 2)*log(3*x + 2) + 245)/(3*x + 2)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {(1-2 x)^2}{(2+3 x)^2 (3+5 x)} \, dx=\frac {121 \log {\left (x + \frac {3}{5} \right )}}{5} - \frac {217 \log {\left (x + \frac {2}{3} \right )}}{9} + \frac {49}{27 x + 18} \]

[In]

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

[Out]

121*log(x + 3/5)/5 - 217*log(x + 2/3)/9 + 49/(27*x + 18)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {(1-2 x)^2}{(2+3 x)^2 (3+5 x)} \, dx=\frac {49}{9 \, {\left (3 \, x + 2\right )}} + \frac {121}{5} \, \log \left (5 \, x + 3\right ) - \frac {217}{9} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

49/9/(3*x + 2) + 121/5*log(5*x + 3) - 217/9*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.34 \[ \int \frac {(1-2 x)^2}{(2+3 x)^2 (3+5 x)} \, dx=\frac {49}{9 \, {\left (3 \, x + 2\right )}} - \frac {4}{45} \, \log \left (\frac {{\left | 3 \, x + 2 \right |}}{3 \, {\left (3 \, x + 2\right )}^{2}}\right ) + \frac {121}{5} \, \log \left ({\left | -\frac {1}{3 \, x + 2} + 5 \right |}\right ) \]

[In]

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

[Out]

49/9/(3*x + 2) - 4/45*log(1/3*abs(3*x + 2)/(3*x + 2)^2) + 121/5*log(abs(-1/(3*x + 2) + 5))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {(1-2 x)^2}{(2+3 x)^2 (3+5 x)} \, dx=\frac {121\,\ln \left (x+\frac {3}{5}\right )}{5}-\frac {217\,\ln \left (x+\frac {2}{3}\right )}{9}+\frac {49}{27\,\left (x+\frac {2}{3}\right )} \]

[In]

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

[Out]

(121*log(x + 3/5))/5 - (217*log(x + 2/3))/9 + 49/(27*(x + 2/3))