\(\int \frac {(1-2 x)^2 (2+3 x)}{(3+5 x)^3} \, dx\) [1328]

Optimal result

Integrand size = 20, antiderivative size = 38 \[ \int \frac {(1-2 x)^2 (2+3 x)}{(3+5 x)^3} \, dx=\frac {12 x}{125}-\frac {121}{1250 (3+5 x)^2}-\frac {319}{625 (3+5 x)}-\frac {128}{625} \log (3+5 x) \]

[Out]

12/125*x-121/1250/(3+5*x)^2-319/625/(3+5*x)-128/625*ln(3+5*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} \[ \int \frac {(1-2 x)^2 (2+3 x)}{(3+5 x)^3} \, dx=\frac {12 x}{125}-\frac {319}{625 (5 x+3)}-\frac {121}{1250 (5 x+3)^2}-\frac {128}{625} \log (5 x+3) \]

[In]

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

[Out]

(12*x)/125 - 121/(1250*(3 + 5*x)^2) - 319/(625*(3 + 5*x)) - (128*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 {12}{125}+\frac {121}{125 (3+5 x)^3}+\frac {319}{125 (3+5 x)^2}-\frac {128}{125 (3+5 x)}\right ) \, dx \\ & = \frac {12 x}{125}-\frac {121}{1250 (3+5 x)^2}-\frac {319}{625 (3+5 x)}-\frac {128}{625} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97 \[ \int \frac {(1-2 x)^2 (2+3 x)}{(3+5 x)^3} \, dx=\frac {\frac {5 \left (-515-782 x+420 x^2+600 x^3\right )}{(3+5 x)^2}-256 \log (6+10 x)}{1250} \]

[In]

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

[Out]

((5*(-515 - 782*x + 420*x^2 + 600*x^3))/(3 + 5*x)^2 - 256*Log[6 + 10*x])/1250

Maple [A] (verified)

Time = 2.36 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71

method	result	size
risch	\(\frac {12 x}{125}+\frac {-\frac {319 x}{125}-\frac {407}{250}}{\left (3+5 x \right )^{2}}-\frac {128 \ln \left (3+5 x \right )}{625}\)	\(27\)
default	\(\frac {12 x}{125}-\frac {121}{1250 \left (3+5 x \right )^{2}}-\frac {319}{625 \left (3+5 x \right )}-\frac {128 \ln \left (3+5 x \right )}{625}\)	\(31\)
norman	\(\frac {\frac {1402}{375} x +\frac {3331}{450} x^{2}+\frac {12}{5} x^{3}}{\left (3+5 x \right )^{2}}-\frac {128 \ln \left (3+5 x \right )}{625}\)	\(32\)
parallelrisch	\(-\frac {57600 \ln \left (x +\frac {3}{5}\right ) x^{2}-27000 x^{3}+69120 \ln \left (x +\frac {3}{5}\right ) x -83275 x^{2}+20736 \ln \left (x +\frac {3}{5}\right )-42060 x}{11250 \left (3+5 x \right )^{2}}\)	\(46\)
meijerg	\(\frac {x \left (\frac {5 x}{3}+2\right )}{27 \left (1+\frac {5 x}{3}\right )^{2}}-\frac {5 x^{2}}{54 \left (1+\frac {5 x}{3}\right )^{2}}+\frac {2 x \left (15 x +6\right )}{225 \left (1+\frac {5 x}{3}\right )^{2}}-\frac {128 \ln \left (1+\frac {5 x}{3}\right )}{625}+\frac {3 x \left (\frac {100}{9} x^{2}+30 x +12\right )}{125 \left (1+\frac {5 x}{3}\right )^{2}}\)	\(72\)

[In]

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

[Out]

12/125*x+25*(-319/3125*x-407/6250)/(3+5*x)^2-128/625*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int \frac {(1-2 x)^2 (2+3 x)}{(3+5 x)^3} \, dx=\frac {3000 \, x^{3} + 3600 \, x^{2} - 256 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (5 \, x + 3\right ) - 2110 \, x - 2035}{1250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

[In]

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

[Out]

1/1250*(3000*x^3 + 3600*x^2 - 256*(25*x^2 + 30*x + 9)*log(5*x + 3) - 2110*x - 2035)/(25*x^2 + 30*x + 9)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^2 (2+3 x)}{(3+5 x)^3} \, dx=\frac {12 x}{125} + \frac {- 638 x - 407}{6250 x^{2} + 7500 x + 2250} - \frac {128 \log {\left (5 x + 3 \right )}}{625} \]

[In]

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

[Out]

12*x/125 + (-638*x - 407)/(6250*x**2 + 7500*x + 2250) - 128*log(5*x + 3)/625

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^2 (2+3 x)}{(3+5 x)^3} \, dx=\frac {12}{125} \, x - \frac {11 \, {\left (58 \, x + 37\right )}}{250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} - \frac {128}{625} \, \log \left (5 \, x + 3\right ) \]

[In]

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

[Out]

12/125*x - 11/250*(58*x + 37)/(25*x^2 + 30*x + 9) - 128/625*log(5*x + 3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x)^2 (2+3 x)}{(3+5 x)^3} \, dx=\frac {12}{125} \, x - \frac {11 \, {\left (58 \, x + 37\right )}}{250 \, {\left (5 \, x + 3\right )}^{2}} - \frac {128}{625} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \]

[In]

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

[Out]

12/125*x - 11/250*(58*x + 37)/(5*x + 3)^2 - 128/625*log(abs(5*x + 3))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x)^2 (2+3 x)}{(3+5 x)^3} \, dx=\frac {12\,x}{125}-\frac {128\,\ln \left (x+\frac {3}{5}\right )}{625}-\frac {\frac {319\,x}{3125}+\frac {407}{6250}}{x^2+\frac {6\,x}{5}+\frac {9}{25}} \]

[In]

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

[Out]

(12*x)/125 - (128*log(x + 3/5))/625 - ((319*x)/3125 + 407/6250)/((6*x)/5 + x^2 + 9/25)