\(\int \frac {(1-2 x)^2 (2+3 x)}{(3+5 x)^2} \, dx\) [1311]

Optimal result

Integrand size = 20, antiderivative size = 34 \[ \int \frac {(1-2 x)^2 (2+3 x)}{(3+5 x)^2} \, dx=-\frac {92 x}{125}+\frac {6 x^2}{25}-\frac {121}{625 (3+5 x)}+\frac {319}{625} \log (3+5 x) \]

[Out]

-92/125*x+6/25*x^2-121/625/(3+5*x)+319/625*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, 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} \[ \int \frac {(1-2 x)^2 (2+3 x)}{(3+5 x)^2} \, dx=\frac {6 x^2}{25}-\frac {92 x}{125}-\frac {121}{625 (5 x+3)}+\frac {319}{625} \log (5 x+3) \]

[In]

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

[Out]

(-92*x)/125 + (6*x^2)/25 - 121/(625*(3 + 5*x)) + (319*Log[3 + 5*x])/625

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 {92}{125}+\frac {12 x}{25}+\frac {121}{125 (3+5 x)^2}+\frac {319}{125 (3+5 x)}\right ) \, dx \\ & = -\frac {92 x}{125}+\frac {6 x^2}{25}-\frac {121}{625 (3+5 x)}+\frac {319}{625} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {(1-2 x)^2 (2+3 x)}{(3+5 x)^2} \, dx=\frac {913-835 x-3700 x^2+1500 x^3+638 (3+5 x) \log (6+10 x)}{1250 (3+5 x)} \]

[In]

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

[Out]

(913 - 835*x - 3700*x^2 + 1500*x^3 + 638*(3 + 5*x)*Log[6 + 10*x])/(1250*(3 + 5*x))

Maple [A] (verified)

Time = 2.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {6 x^{2}}{25}-\frac {92 x}{125}-\frac {121}{3125 \left (x +\frac {3}{5}\right )}+\frac {319 \ln \left (3+5 x \right )}{625}\)	\(25\)
default	\(-\frac {92 x}{125}+\frac {6 x^{2}}{25}-\frac {121}{625 \left (3+5 x \right )}+\frac {319 \ln \left (3+5 x \right )}{625}\)	\(27\)
norman	\(\frac {-\frac {707}{375} x -\frac {74}{25} x^{2}+\frac {6}{5} x^{3}}{3+5 x}+\frac {319 \ln \left (3+5 x \right )}{625}\)	\(32\)
parallelrisch	\(\frac {2250 x^{3}+4785 \ln \left (x +\frac {3}{5}\right ) x -5550 x^{2}+2871 \ln \left (x +\frac {3}{5}\right )-3535 x}{5625+9375 x}\)	\(37\)
meijerg	\(\frac {5 x}{9 \left (1+\frac {5 x}{3}\right )}+\frac {319 \ln \left (1+\frac {5 x}{3}\right )}{625}-\frac {4 x \left (5 x +6\right )}{75 \left (1+\frac {5 x}{3}\right )}-\frac {9 x \left (-\frac {50}{9} x^{2}+10 x +12\right )}{125 \left (1+\frac {5 x}{3}\right )}\)	\(55\)

[In]

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

[Out]

6/25*x^2-92/125*x-121/3125/(x+3/5)+319/625*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {(1-2 x)^2 (2+3 x)}{(3+5 x)^2} \, dx=\frac {750 \, x^{3} - 1850 \, x^{2} + 319 \, {\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 1380 \, x - 121}{625 \, {\left (5 \, x + 3\right )}} \]

[In]

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

[Out]

1/625*(750*x^3 - 1850*x^2 + 319*(5*x + 3)*log(5*x + 3) - 1380*x - 121)/(5*x + 3)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {(1-2 x)^2 (2+3 x)}{(3+5 x)^2} \, dx=\frac {6 x^{2}}{25} - \frac {92 x}{125} + \frac {319 \log {\left (5 x + 3 \right )}}{625} - \frac {121}{3125 x + 1875} \]

[In]

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

[Out]

6*x**2/25 - 92*x/125 + 319*log(5*x + 3)/625 - 121/(3125*x + 1875)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x)^2 (2+3 x)}{(3+5 x)^2} \, dx=\frac {6}{25} \, x^{2} - \frac {92}{125} \, x - \frac {121}{625 \, {\left (5 \, x + 3\right )}} + \frac {319}{625} \, \log \left (5 \, x + 3\right ) \]

[In]

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

[Out]

6/25*x^2 - 92/125*x - 121/625/(5*x + 3) + 319/625*log(5*x + 3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41 \[ \int \frac {(1-2 x)^2 (2+3 x)}{(3+5 x)^2} \, dx=-\frac {2}{625} \, {\left (5 \, x + 3\right )}^{2} {\left (\frac {64}{5 \, x + 3} - 3\right )} - \frac {121}{625 \, {\left (5 \, x + 3\right )}} - \frac {319}{625} \, \log \left (\frac {{\left | 5 \, x + 3 \right |}}{5 \, {\left (5 \, x + 3\right )}^{2}}\right ) \]

[In]

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

[Out]

-2/625*(5*x + 3)^2*(64/(5*x + 3) - 3) - 121/625/(5*x + 3) - 319/625*log(1/5*abs(5*x + 3)/(5*x + 3)^2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x)^2 (2+3 x)}{(3+5 x)^2} \, dx=\frac {319\,\ln \left (x+\frac {3}{5}\right )}{625}-\frac {92\,x}{125}-\frac {121}{3125\,\left (x+\frac {3}{5}\right )}+\frac {6\,x^2}{25} \]

[In]

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

[Out]

(319*log(x + 3/5))/625 - (92*x)/125 - 121/(3125*(x + 3/5)) + (6*x^2)/25