\(\int \frac {(1-2 x)^2 (2+3 x)^2}{(3+5 x)^2} \, dx\) [1310]

Optimal result

Integrand size = 22, antiderivative size = 41 \[ \int \frac {(1-2 x)^2 (2+3 x)^2}{(3+5 x)^2} \, dx=\frac {37 x}{625}-\frac {78 x^2}{125}+\frac {12 x^3}{25}-\frac {121}{3125 (3+5 x)}+\frac {682 \log (3+5 x)}{3125} \]

[Out]

37/625*x-78/125*x^2+12/25*x^3-121/3125/(3+5*x)+682/3125*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, 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} \[ \int \frac {(1-2 x)^2 (2+3 x)^2}{(3+5 x)^2} \, dx=\frac {12 x^3}{25}-\frac {78 x^2}{125}+\frac {37 x}{625}-\frac {121}{3125 (5 x+3)}+\frac {682 \log (5 x+3)}{3125} \]

[In]

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

[Out]

(37*x)/625 - (78*x^2)/125 + (12*x^3)/25 - 121/(3125*(3 + 5*x)) + (682*Log[3 + 5*x])/3125

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 {37}{625}-\frac {156 x}{125}+\frac {36 x^2}{25}+\frac {121}{625 (3+5 x)^2}+\frac {682}{625 (3+5 x)}\right ) \, dx \\ & = \frac {37 x}{625}-\frac {78 x^2}{125}+\frac {12 x^3}{25}-\frac {121}{3125 (3+5 x)}+\frac {682 \log (3+5 x)}{3125} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.02 \[ \int \frac {(1-2 x)^2 (2+3 x)^2}{(3+5 x)^2} \, dx=\frac {\frac {5 \left (658+1248 x-985 x^2-1050 x^3+1500 x^4\right )}{3+5 x}+682 \log (3+5 x)}{3125} \]

[In]

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

[Out]

((5*(658 + 1248*x - 985*x^2 - 1050*x^3 + 1500*x^4))/(3 + 5*x) + 682*Log[3 + 5*x])/3125

Maple [A] (verified)

Time = 2.34 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.73

method	result	size
risch	\(\frac {12 x^{3}}{25}-\frac {78 x^{2}}{125}+\frac {37 x}{625}-\frac {121}{15625 \left (x +\frac {3}{5}\right )}+\frac {682 \ln \left (3+5 x \right )}{3125}\)	\(30\)
default	\(\frac {37 x}{625}-\frac {78 x^{2}}{125}+\frac {12 x^{3}}{25}-\frac {121}{3125 \left (3+5 x \right )}+\frac {682 \ln \left (3+5 x \right )}{3125}\)	\(32\)
norman	\(\frac {\frac {454}{1875} x -\frac {197}{125} x^{2}-\frac {42}{25} x^{3}+\frac {12}{5} x^{4}}{3+5 x}+\frac {682 \ln \left (3+5 x \right )}{3125}\)	\(37\)
parallelrisch	\(\frac {22500 x^{4}-15750 x^{3}+10230 \ln \left (x +\frac {3}{5}\right ) x -14775 x^{2}+6138 \ln \left (x +\frac {3}{5}\right )+2270 x}{28125+46875 x}\)	\(42\)
meijerg	\(\frac {32 x}{45 \left (1+\frac {5 x}{3}\right )}+\frac {682 \ln \left (1+\frac {5 x}{3}\right )}{3125}-\frac {23 x \left (5 x +6\right )}{75 \left (1+\frac {5 x}{3}\right )}-\frac {9 x \left (-\frac {50}{9} x^{2}+10 x +12\right )}{125 \left (1+\frac {5 x}{3}\right )}+\frac {108 x \left (\frac {625}{27} x^{3}-\frac {250}{9} x^{2}+50 x +60\right )}{3125 \left (1+\frac {5 x}{3}\right )}\)	\(80\)

[In]

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

[Out]

12/25*x^3-78/125*x^2+37/625*x-121/15625/(x+3/5)+682/3125*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.02 \[ \int \frac {(1-2 x)^2 (2+3 x)^2}{(3+5 x)^2} \, dx=\frac {7500 \, x^{4} - 5250 \, x^{3} - 4925 \, x^{2} + 682 \, {\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) + 555 \, x - 121}{3125 \, {\left (5 \, x + 3\right )}} \]

[In]

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

[Out]

1/3125*(7500*x^4 - 5250*x^3 - 4925*x^2 + 682*(5*x + 3)*log(5*x + 3) + 555*x - 121)/(5*x + 3)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {(1-2 x)^2 (2+3 x)^2}{(3+5 x)^2} \, dx=\frac {12 x^{3}}{25} - \frac {78 x^{2}}{125} + \frac {37 x}{625} + \frac {682 \log {\left (5 x + 3 \right )}}{3125} - \frac {121}{15625 x + 9375} \]

[In]

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

[Out]

12*x**3/25 - 78*x**2/125 + 37*x/625 + 682*log(5*x + 3)/3125 - 121/(15625*x + 9375)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x)^2 (2+3 x)^2}{(3+5 x)^2} \, dx=\frac {12}{25} \, x^{3} - \frac {78}{125} \, x^{2} + \frac {37}{625} \, x - \frac {121}{3125 \, {\left (5 \, x + 3\right )}} + \frac {682}{3125} \, \log \left (5 \, x + 3\right ) \]

[In]

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

[Out]

12/25*x^3 - 78/125*x^2 + 37/625*x - 121/3125/(5*x + 3) + 682/3125*log(5*x + 3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.39 \[ \int \frac {(1-2 x)^2 (2+3 x)^2}{(3+5 x)^2} \, dx=-\frac {1}{3125} \, {\left (5 \, x + 3\right )}^{3} {\left (\frac {186}{5 \, x + 3} - \frac {829}{{\left (5 \, x + 3\right )}^{2}} - 12\right )} - \frac {121}{3125 \, {\left (5 \, x + 3\right )}} - \frac {682}{3125} \, \log \left (\frac {{\left | 5 \, x + 3 \right |}}{5 \, {\left (5 \, x + 3\right )}^{2}}\right ) \]

[In]

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

[Out]

-1/3125*(5*x + 3)^3*(186/(5*x + 3) - 829/(5*x + 3)^2 - 12) - 121/3125/(5*x + 3) - 682/3125*log(1/5*abs(5*x + 3
)/(5*x + 3)^2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x)^2 (2+3 x)^2}{(3+5 x)^2} \, dx=\frac {37\,x}{625}+\frac {682\,\ln \left (x+\frac {3}{5}\right )}{3125}-\frac {121}{15625\,\left (x+\frac {3}{5}\right )}-\frac {78\,x^2}{125}+\frac {12\,x^3}{25} \]

[In]

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

[Out]

(37*x)/625 + (682*log(x + 3/5))/3125 - 121/(15625*(x + 3/5)) - (78*x^2)/125 + (12*x^3)/25