\(\int \frac {(1-2 x)^2}{(2+3 x)^4 (3+5 x)} \, dx\) [1300]

Optimal result

Integrand size = 22, antiderivative size = 48 \[ \int \frac {(1-2 x)^2}{(2+3 x)^4 (3+5 x)} \, dx=\frac {49}{27 (2+3 x)^3}+\frac {217}{18 (2+3 x)^2}+\frac {121}{2+3 x}-605 \log (2+3 x)+605 \log (3+5 x) \]

[Out]

49/27/(2+3*x)^3+217/18/(2+3*x)^2+121/(2+3*x)-605*ln(2+3*x)+605*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, 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} \[ \int \frac {(1-2 x)^2}{(2+3 x)^4 (3+5 x)} \, dx=\frac {121}{3 x+2}+\frac {217}{18 (3 x+2)^2}+\frac {49}{27 (3 x+2)^3}-605 \log (3 x+2)+605 \log (5 x+3) \]

[In]

Int[(1 - 2*x)^2/((2 + 3*x)^4*(3 + 5*x)),x]

[Out]

49/(27*(2 + 3*x)^3) + 217/(18*(2 + 3*x)^2) + 121/(2 + 3*x) - 605*Log[2 + 3*x] + 605*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)^4}-\frac {217}{3 (2+3 x)^3}-\frac {363}{(2+3 x)^2}-\frac {1815}{2+3 x}+\frac {3025}{3+5 x}\right ) \, dx \\ & = \frac {49}{27 (2+3 x)^3}+\frac {217}{18 (2+3 x)^2}+\frac {121}{2+3 x}-605 \log (2+3 x)+605 \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.83 \[ \int \frac {(1-2 x)^2}{(2+3 x)^4 (3+5 x)} \, dx=\frac {27536+80361 x+58806 x^2}{54 (2+3 x)^3}-605 \log (5 (2+3 x))+605 \log (3+5 x) \]

[In]

Integrate[(1 - 2*x)^2/((2 + 3*x)^4*(3 + 5*x)),x]

[Out]

(27536 + 80361*x + 58806*x^2)/(54*(2 + 3*x)^3) - 605*Log[5*(2 + 3*x)] + 605*Log[3 + 5*x]

Maple [A] (verified)

Time = 2.32 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.75

method	result	size
norman	\(\frac {1089 x^{2}+\frac {8929}{6} x +\frac {13768}{27}}{\left (2+3 x \right )^{3}}-605 \ln \left (2+3 x \right )+605 \ln \left (3+5 x \right )\)	\(36\)
risch	\(\frac {1089 x^{2}+\frac {8929}{6} x +\frac {13768}{27}}{\left (2+3 x \right )^{3}}-605 \ln \left (2+3 x \right )+605 \ln \left (3+5 x \right )\)	\(37\)
default	\(\frac {49}{27 \left (2+3 x \right )^{3}}+\frac {217}{18 \left (2+3 x \right )^{2}}+\frac {121}{2+3 x}-605 \ln \left (2+3 x \right )+605 \ln \left (3+5 x \right )\)	\(45\)
parallelrisch	\(-\frac {130680 \ln \left (\frac {2}{3}+x \right ) x^{3}-130680 \ln \left (x +\frac {3}{5}\right ) x^{3}+261360 \ln \left (\frac {2}{3}+x \right ) x^{2}-261360 \ln \left (x +\frac {3}{5}\right ) x^{2}+13768 x^{3}+174240 \ln \left (\frac {2}{3}+x \right ) x -174240 \ln \left (x +\frac {3}{5}\right ) x +18824 x^{2}+38720 \ln \left (\frac {2}{3}+x \right )-38720 \ln \left (x +\frac {3}{5}\right )+6452 x}{8 \left (2+3 x \right )^{3}}\)	\(86\)

[In]

int((1-2*x)^2/(2+3*x)^4/(3+5*x),x,method=_RETURNVERBOSE)

[Out]

(1089*x^2+8929/6*x+13768/27)/(2+3*x)^3-605*ln(2+3*x)+605*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.56 \[ \int \frac {(1-2 x)^2}{(2+3 x)^4 (3+5 x)} \, dx=\frac {58806 \, x^{2} + 32670 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (5 \, x + 3\right ) - 32670 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (3 \, x + 2\right ) + 80361 \, x + 27536}{54 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \]

[In]

integrate((1-2*x)^2/(2+3*x)^4/(3+5*x),x, algorithm="fricas")

[Out]

1/54*(58806*x^2 + 32670*(27*x^3 + 54*x^2 + 36*x + 8)*log(5*x + 3) - 32670*(27*x^3 + 54*x^2 + 36*x + 8)*log(3*x
+ 2) + 80361*x + 27536)/(27*x^3 + 54*x^2 + 36*x + 8)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.85 \[ \int \frac {(1-2 x)^2}{(2+3 x)^4 (3+5 x)} \, dx=\frac {58806 x^{2} + 80361 x + 27536}{1458 x^{3} + 2916 x^{2} + 1944 x + 432} + 605 \log {\left (x + \frac {3}{5} \right )} - 605 \log {\left (x + \frac {2}{3} \right )} \]

[In]

integrate((1-2*x)**2/(2+3*x)**4/(3+5*x),x)

[Out]

(58806*x**2 + 80361*x + 27536)/(1458*x**3 + 2916*x**2 + 1944*x + 432) + 605*log(x + 3/5) - 605*log(x + 2/3)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.96 \[ \int \frac {(1-2 x)^2}{(2+3 x)^4 (3+5 x)} \, dx=\frac {58806 \, x^{2} + 80361 \, x + 27536}{54 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} + 605 \, \log \left (5 \, x + 3\right ) - 605 \, \log \left (3 \, x + 2\right ) \]

[In]

integrate((1-2*x)^2/(2+3*x)^4/(3+5*x),x, algorithm="maxima")

[Out]

1/54*(58806*x^2 + 80361*x + 27536)/(27*x^3 + 54*x^2 + 36*x + 8) + 605*log(5*x + 3) - 605*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.79 \[ \int \frac {(1-2 x)^2}{(2+3 x)^4 (3+5 x)} \, dx=\frac {58806 \, x^{2} + 80361 \, x + 27536}{54 \, {\left (3 \, x + 2\right )}^{3}} + 605 \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - 605 \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

integrate((1-2*x)^2/(2+3*x)^4/(3+5*x),x, algorithm="giac")

[Out]

1/54*(58806*x^2 + 80361*x + 27536)/(3*x + 2)^3 + 605*log(abs(5*x + 3)) - 605*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x)^2}{(2+3 x)^4 (3+5 x)} \, dx=\frac {\frac {121\,x^2}{3}+\frac {8929\,x}{162}+\frac {13768}{729}}{x^3+2\,x^2+\frac {4\,x}{3}+\frac {8}{27}}-1210\,\mathrm {atanh}\left (30\,x+19\right ) \]

[In]

int((2*x - 1)^2/((3*x + 2)^4*(5*x + 3)),x)

[Out]

((8929*x)/162 + (121*x^2)/3 + 13768/729)/((4*x)/3 + 2*x^2 + x^3 + 8/27) - 1210*atanh(30*x + 19)