\(\int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^3} \, dx\) [1247]

Optimal result

Integrand size = 20, antiderivative size = 38 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^3} \, dx=\frac {20 x}{27}+\frac {49}{162 (2+3 x)^2}-\frac {91}{27 (2+3 x)}-\frac {16}{9} \log (2+3 x) \]

[Out]

20/27*x+49/162/(2+3*x)^2-91/27/(2+3*x)-16/9*ln(2+3*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, 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} \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^3} \, dx=\frac {20 x}{27}-\frac {91}{27 (3 x+2)}+\frac {49}{162 (3 x+2)^2}-\frac {16}{9} \log (3 x+2) \]

[In]

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

[Out]

(20*x)/27 + 49/(162*(2 + 3*x)^2) - 91/(27*(2 + 3*x)) - (16*Log[2 + 3*x])/9

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 {20}{27}-\frac {49}{27 (2+3 x)^3}+\frac {91}{9 (2+3 x)^2}-\frac {16}{3 (2+3 x)}\right ) \, dx \\ & = \frac {20 x}{27}+\frac {49}{162 (2+3 x)^2}-\frac {91}{27 (2+3 x)}-\frac {16}{9} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.08 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^3} \, dx=\frac {-1283-1878 x+900 x^2+1080 x^3-288 (2+3 x)^2 \log (4+6 x)}{162 (2+3 x)^2} \]

[In]

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

[Out]

(-1283 - 1878*x + 900*x^2 + 1080*x^3 - 288*(2 + 3*x)^2*Log[4 + 6*x])/(162*(2 + 3*x)^2)

Maple [A] (verified)

Time = 2.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71

method	result	size
risch	\(\frac {20 x}{27}+\frac {-\frac {91 x}{9}-\frac {1043}{162}}{\left (2+3 x \right )^{2}}-\frac {16 \ln \left (2+3 x \right )}{9}\)	\(27\)
default	\(\frac {20 x}{27}+\frac {49}{162 \left (2+3 x \right )^{2}}-\frac {91}{27 \left (2+3 x \right )}-\frac {16 \ln \left (2+3 x \right )}{9}\)	\(31\)
norman	\(\frac {\frac {73}{6} x +\frac {187}{8} x^{2}+\frac {20}{3} x^{3}}{\left (2+3 x \right )^{2}}-\frac {16 \ln \left (2+3 x \right )}{9}\)	\(32\)
parallelrisch	\(-\frac {1152 \ln \left (\frac {2}{3}+x \right ) x^{2}-480 x^{3}+1536 \ln \left (\frac {2}{3}+x \right ) x -1683 x^{2}+512 \ln \left (\frac {2}{3}+x \right )-876 x}{72 \left (2+3 x \right )^{2}}\)	\(46\)
meijerg	\(\frac {3 x \left (\frac {3 x}{2}+2\right )}{16 \left (1+\frac {3 x}{2}\right )^{2}}-\frac {7 x^{2}}{16 \left (1+\frac {3 x}{2}\right )^{2}}+\frac {2 x \left (\frac {27 x}{2}+6\right )}{27 \left (1+\frac {3 x}{2}\right )^{2}}-\frac {16 \ln \left (1+\frac {3 x}{2}\right )}{9}+\frac {5 x \left (9 x^{2}+27 x +12\right )}{27 \left (1+\frac {3 x}{2}\right )^{2}}\)	\(72\)

[In]

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

[Out]

20/27*x+9*(-91/81*x-1043/1458)/(2+3*x)^2-16/9*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^3} \, dx=\frac {1080 \, x^{3} + 1440 \, x^{2} - 288 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (3 \, x + 2\right ) - 1158 \, x - 1043}{162 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \]

[In]

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

[Out]

1/162*(1080*x^3 + 1440*x^2 - 288*(9*x^2 + 12*x + 4)*log(3*x + 2) - 1158*x - 1043)/(9*x^2 + 12*x + 4)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^3} \, dx=\frac {20 x}{27} + \frac {- 1638 x - 1043}{1458 x^{2} + 1944 x + 648} - \frac {16 \log {\left (3 x + 2 \right )}}{9} \]

[In]

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

[Out]

20*x/27 + (-1638*x - 1043)/(1458*x**2 + 1944*x + 648) - 16*log(3*x + 2)/9

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^3} \, dx=\frac {20}{27} \, x - \frac {7 \, {\left (234 \, x + 149\right )}}{162 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} - \frac {16}{9} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

20/27*x - 7/162*(234*x + 149)/(9*x^2 + 12*x + 4) - 16/9*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^3} \, dx=\frac {20}{27} \, x - \frac {7 \, {\left (234 \, x + 149\right )}}{162 \, {\left (3 \, x + 2\right )}^{2}} - \frac {16}{9} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

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

[Out]

20/27*x - 7/162*(234*x + 149)/(3*x + 2)^2 - 16/9*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 1.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^3} \, dx=\frac {20\,x}{27}-\frac {16\,\ln \left (x+\frac {2}{3}\right )}{9}-\frac {\frac {91\,x}{81}+\frac {1043}{1458}}{x^2+\frac {4\,x}{3}+\frac {4}{9}} \]

[In]

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

[Out]

(20*x)/27 - (16*log(x + 2/3))/9 - ((91*x)/81 + 1043/1458)/((4*x)/3 + x^2 + 4/9)