\(\int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^2} \, dx\) [1263]

Optimal result

Integrand size = 22, antiderivative size = 41 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^2} \, dx=\frac {143 x}{27}-\frac {170 x^2}{27}+\frac {100 x^3}{27}-\frac {49}{243 (2+3 x)}-\frac {518}{243} \log (2+3 x) \]

[Out]

143/27*x-170/27*x^2+100/27*x^3-49/243/(2+3*x)-518/243*ln(2+3*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} \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^2} \, dx=\frac {100 x^3}{27}-\frac {170 x^2}{27}+\frac {143 x}{27}-\frac {49}{243 (3 x+2)}-\frac {518}{243} \log (3 x+2) \]

[In]

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

[Out]

(143*x)/27 - (170*x^2)/27 + (100*x^3)/27 - 49/(243*(2 + 3*x)) - (518*Log[2 + 3*x])/243

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 {143}{27}-\frac {340 x}{27}+\frac {100 x^2}{9}+\frac {49}{81 (2+3 x)^2}-\frac {518}{81 (2+3 x)}\right ) \, dx \\ & = \frac {143 x}{27}-\frac {170 x^2}{27}+\frac {100 x^3}{27}-\frac {49}{243 (2+3 x)}-\frac {518}{243} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.07 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^2} \, dx=\frac {10681+23964 x+2403 x^2-8370 x^3+8100 x^4-1554 (2+3 x) \log (2+3 x)}{729 (2+3 x)} \]

[In]

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

[Out]

(10681 + 23964*x + 2403*x^2 - 8370*x^3 + 8100*x^4 - 1554*(2 + 3*x)*Log[2 + 3*x])/(729*(2 + 3*x))

Maple [A] (verified)

Time = 2.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.73

method	result	size
risch	\(\frac {100 x^{3}}{27}-\frac {170 x^{2}}{27}+\frac {143 x}{27}-\frac {49}{729 \left (\frac {2}{3}+x \right )}-\frac {518 \ln \left (2+3 x \right )}{243}\)	\(30\)
default	\(\frac {143 x}{27}-\frac {170 x^{2}}{27}+\frac {100 x^{3}}{27}-\frac {49}{243 \left (2+3 x \right )}-\frac {518 \ln \left (2+3 x \right )}{243}\)	\(32\)
norman	\(\frac {\frac {1765}{162} x +\frac {89}{27} x^{2}-\frac {310}{27} x^{3}+\frac {100}{9} x^{4}}{2+3 x}-\frac {518 \ln \left (2+3 x \right )}{243}\)	\(37\)
parallelrisch	\(-\frac {-5400 x^{4}+5580 x^{3}+3108 \ln \left (\frac {2}{3}+x \right ) x -1602 x^{2}+2072 \ln \left (\frac {2}{3}+x \right )-5295 x}{486 \left (2+3 x \right )}\)	\(42\)
meijerg	\(\frac {13 x}{4 \left (1+\frac {3 x}{2}\right )}-\frac {518 \ln \left (1+\frac {3 x}{2}\right )}{243}-\frac {59 x \left (\frac {9 x}{2}+6\right )}{27 \left (1+\frac {3 x}{2}\right )}-\frac {10 x \left (-\frac {9}{2} x^{2}+9 x +12\right )}{27 \left (1+\frac {3 x}{2}\right )}+\frac {80 x \left (\frac {135}{8} x^{3}-\frac {45}{2} x^{2}+45 x +60\right )}{243 \left (1+\frac {3 x}{2}\right )}\)	\(80\)

[In]

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

[Out]

100/27*x^3-170/27*x^2+143/27*x-49/729/(2/3+x)-518/243*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.02 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^2} \, dx=\frac {2700 \, x^{4} - 2790 \, x^{3} + 801 \, x^{2} - 518 \, {\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) + 2574 \, x - 49}{243 \, {\left (3 \, x + 2\right )}} \]

[In]

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

[Out]

1/243*(2700*x^4 - 2790*x^3 + 801*x^2 - 518*(3*x + 2)*log(3*x + 2) + 2574*x - 49)/(3*x + 2)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^2} \, dx=\frac {100 x^{3}}{27} - \frac {170 x^{2}}{27} + \frac {143 x}{27} - \frac {518 \log {\left (3 x + 2 \right )}}{243} - \frac {49}{729 x + 486} \]

[In]

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

[Out]

100*x**3/27 - 170*x**2/27 + 143*x/27 - 518*log(3*x + 2)/243 - 49/(729*x + 486)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^2} \, dx=\frac {100}{27} \, x^{3} - \frac {170}{27} \, x^{2} + \frac {143}{27} \, x - \frac {49}{243 \, {\left (3 \, x + 2\right )}} - \frac {518}{243} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

100/27*x^3 - 170/27*x^2 + 143/27*x - 49/243/(3*x + 2) - 518/243*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.39 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^2} \, dx=-\frac {1}{729} \, {\left (3 \, x + 2\right )}^{3} {\left (\frac {1110}{3 \, x + 2} - \frac {4527}{{\left (3 \, x + 2\right )}^{2}} - 100\right )} - \frac {49}{243 \, {\left (3 \, x + 2\right )}} + \frac {518}{243} \, \log \left (\frac {{\left | 3 \, x + 2 \right |}}{3 \, {\left (3 \, x + 2\right )}^{2}}\right ) \]

[In]

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

[Out]

-1/729*(3*x + 2)^3*(1110/(3*x + 2) - 4527/(3*x + 2)^2 - 100) - 49/243/(3*x + 2) + 518/243*log(1/3*abs(3*x + 2)
/(3*x + 2)^2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^2} \, dx=\frac {143\,x}{27}-\frac {518\,\ln \left (x+\frac {2}{3}\right )}{243}-\frac {49}{729\,\left (x+\frac {2}{3}\right )}-\frac {170\,x^2}{27}+\frac {100\,x^3}{27} \]

[In]

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

[Out]

(143*x)/27 - (518*log(x + 2/3))/243 - 49/(729*(x + 2/3)) - (170*x^2)/27 + (100*x^3)/27