\(\int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^4} \, dx\) [1248]

Optimal result

Integrand size = 20, antiderivative size = 44 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^4} \, dx=\frac {49}{243 (2+3 x)^3}-\frac {91}{54 (2+3 x)^2}+\frac {16}{9 (2+3 x)}+\frac {20}{81} \log (2+3 x) \]

[Out]

49/243/(2+3*x)^3-91/54/(2+3*x)^2+16/9/(2+3*x)+20/81*ln(2+3*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^4} \, dx=\frac {16}{9 (3 x+2)}-\frac {91}{54 (3 x+2)^2}+\frac {49}{243 (3 x+2)^3}+\frac {20}{81} \log (3 x+2) \]

[In]

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

[Out]

49/(243*(2 + 3*x)^3) - 91/(54*(2 + 3*x)^2) + 16/(9*(2 + 3*x)) + (20*Log[2 + 3*x])/81

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {49}{27 (2+3 x)^4}+\frac {91}{9 (2+3 x)^3}-\frac {16}{3 (2+3 x)^2}+\frac {20}{27 (2+3 x)}\right ) \, dx \\ & = \frac {49}{243 (2+3 x)^3}-\frac {91}{54 (2+3 x)^2}+\frac {16}{9 (2+3 x)}+\frac {20}{81} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^4} \, dx=\frac {1916+7911 x+7776 x^2+120 (2+3 x)^3 \log (2+3 x)}{486 (2+3 x)^3} \]

[In]

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

[Out]

(1916 + 7911*x + 7776*x^2 + 120*(2 + 3*x)^3*Log[2 + 3*x])/(486*(2 + 3*x)^3)

Maple [A] (verified)

Time = 2.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.66

method	result	size
risch	\(\frac {16 x^{2}+\frac {293}{18} x +\frac {958}{243}}{\left (2+3 x \right )^{3}}+\frac {20 \ln \left (2+3 x \right )}{81}\)	\(29\)
norman	\(\frac {-\frac {191}{18} x^{2}-\frac {79}{54} x -\frac {479}{36} x^{3}}{\left (2+3 x \right )^{3}}+\frac {20 \ln \left (2+3 x \right )}{81}\)	\(32\)
default	\(\frac {49}{243 \left (2+3 x \right )^{3}}-\frac {91}{54 \left (2+3 x \right )^{2}}+\frac {16}{9 \left (2+3 x \right )}+\frac {20 \ln \left (2+3 x \right )}{81}\)	\(37\)
parallelrisch	\(\frac {4320 \ln \left (\frac {2}{3}+x \right ) x^{3}+8640 \ln \left (\frac {2}{3}+x \right ) x^{2}-8622 x^{3}+5760 \ln \left (\frac {2}{3}+x \right ) x -6876 x^{2}+1280 \ln \left (\frac {2}{3}+x \right )-948 x}{648 \left (2+3 x \right )^{3}}\)	\(55\)
meijerg	\(\frac {x \left (\frac {9}{4} x^{2}+\frac {9}{2} x +3\right )}{16 \left (1+\frac {3 x}{2}\right )^{3}}-\frac {7 x^{2} \left (3+\frac {3 x}{2}\right )}{96 \left (1+\frac {3 x}{2}\right )^{3}}-\frac {x^{3}}{6 \left (1+\frac {3 x}{2}\right )^{3}}-\frac {5 x \left (\frac {99}{2} x^{2}+45 x +12\right )}{162 \left (1+\frac {3 x}{2}\right )^{3}}+\frac {20 \ln \left (1+\frac {3 x}{2}\right )}{81}\)	\(79\)

[In]

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

[Out]

27*(16/27*x^2+293/486*x+958/6561)/(2+3*x)^3+20/81*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.18 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^4} \, dx=\frac {7776 \, x^{2} + 120 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (3 \, x + 2\right ) + 7911 \, x + 1916}{486 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \]

[In]

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

[Out]

1/486*(7776*x^2 + 120*(27*x^3 + 54*x^2 + 36*x + 8)*log(3*x + 2) + 7911*x + 1916)/(27*x^3 + 54*x^2 + 36*x + 8)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.77 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^4} \, dx=\frac {7776 x^{2} + 7911 x + 1916}{13122 x^{3} + 26244 x^{2} + 17496 x + 3888} + \frac {20 \log {\left (3 x + 2 \right )}}{81} \]

[In]

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

[Out]

(7776*x**2 + 7911*x + 1916)/(13122*x**3 + 26244*x**2 + 17496*x + 3888) + 20*log(3*x + 2)/81

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^4} \, dx=\frac {7776 \, x^{2} + 7911 \, x + 1916}{486 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} + \frac {20}{81} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

1/486*(7776*x^2 + 7911*x + 1916)/(27*x^3 + 54*x^2 + 36*x + 8) + 20/81*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.66 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^4} \, dx=\frac {7776 \, x^{2} + 7911 \, x + 1916}{486 \, {\left (3 \, x + 2\right )}^{3}} + \frac {20}{81} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

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

[Out]

1/486*(7776*x^2 + 7911*x + 1916)/(3*x + 2)^3 + 20/81*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.75 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^4} \, dx=\frac {20\,\ln \left (x+\frac {2}{3}\right )}{81}+\frac {\frac {16\,x^2}{27}+\frac {293\,x}{486}+\frac {958}{6561}}{x^3+2\,x^2+\frac {4\,x}{3}+\frac {8}{27}} \]

[In]

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

[Out]

(20*log(x + 2/3))/81 + ((293*x)/486 + (16*x^2)/27 + 958/6561)/((4*x)/3 + 2*x^2 + x^3 + 8/27)