\(\int \frac {(1-2 x)^2 (3+5 x)^3}{(2+3 x)^4} \, dx\) [1284]

Optimal result

Integrand size = 22, antiderivative size = 56 \[ \int \frac {(1-2 x)^2 (3+5 x)^3}{(2+3 x)^4} \, dx=-\frac {2800 x}{243}+\frac {250 x^2}{81}+\frac {49}{2187 (2+3 x)^3}-\frac {763}{1458 (2+3 x)^2}+\frac {4099}{729 (2+3 x)}+\frac {8285}{729} \log (2+3 x) \]

[Out]

-2800/243*x+250/81*x^2+49/2187/(2+3*x)^3-763/1458/(2+3*x)^2+4099/729/(2+3*x)+8285/729*ln(2+3*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, 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} \[ \int \frac {(1-2 x)^2 (3+5 x)^3}{(2+3 x)^4} \, dx=\frac {250 x^2}{81}-\frac {2800 x}{243}+\frac {4099}{729 (3 x+2)}-\frac {763}{1458 (3 x+2)^2}+\frac {49}{2187 (3 x+2)^3}+\frac {8285}{729} \log (3 x+2) \]

[In]

Int[((1 - 2*x)^2*(3 + 5*x)^3)/(2 + 3*x)^4,x]

[Out]

(-2800*x)/243 + (250*x^2)/81 + 49/(2187*(2 + 3*x)^3) - 763/(1458*(2 + 3*x)^2) + 4099/(729*(2 + 3*x)) + (8285*L
og[2 + 3*x])/729

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 {2800}{243}+\frac {500 x}{81}-\frac {49}{243 (2+3 x)^4}+\frac {763}{243 (2+3 x)^3}-\frac {4099}{243 (2+3 x)^2}+\frac {8285}{243 (2+3 x)}\right ) \, dx \\ & = -\frac {2800 x}{243}+\frac {250 x^2}{81}+\frac {49}{2187 (2+3 x)^3}-\frac {763}{1458 (2+3 x)^2}+\frac {4099}{729 (2+3 x)}+\frac {8285}{729} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.91 \[ \int \frac {(1-2 x)^2 (3+5 x)^3}{(2+3 x)^4} \, dx=-\frac {222904+1540539 x+3623454 x^2+3304800 x^3+631800 x^4-364500 x^5-49710 (2+3 x)^3 \log (20+30 x)}{4374 (2+3 x)^3} \]

[In]

Integrate[((1 - 2*x)^2*(3 + 5*x)^3)/(2 + 3*x)^4,x]

[Out]

-1/4374*(222904 + 1540539*x + 3623454*x^2 + 3304800*x^3 + 631800*x^4 - 364500*x^5 - 49710*(2 + 3*x)^3*Log[20 +
30*x])/(2 + 3*x)^3

Maple [A] (verified)

Time = 2.33 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.66

method	result	size
risch	\(\frac {250 x^{2}}{81}-\frac {2800 x}{243}+\frac {\frac {4099}{81} x^{2}+\frac {32029}{486} x +\frac {46948}{2187}}{\left (2+3 x \right )^{3}}+\frac {8285 \ln \left (2+3 x \right )}{729}\)	\(37\)
norman	\(\frac {-\frac {59719}{486} x -\frac {39238}{81} x^{2}-\frac {94537}{162} x^{3}-\frac {1300}{9} x^{4}+\frac {250}{3} x^{5}}{\left (2+3 x \right )^{3}}+\frac {8285 \ln \left (2+3 x \right )}{729}\)	\(42\)
default	\(-\frac {2800 x}{243}+\frac {250 x^{2}}{81}+\frac {49}{2187 \left (2+3 x \right )^{3}}-\frac {763}{1458 \left (2+3 x \right )^{2}}+\frac {4099}{729 \left (2+3 x \right )}+\frac {8285 \ln \left (2+3 x \right )}{729}\)	\(45\)
parallelrisch	\(\frac {486000 x^{5}+1789560 \ln \left (\frac {2}{3}+x \right ) x^{3}-842400 x^{4}+3579120 \ln \left (\frac {2}{3}+x \right ) x^{2}-3403332 x^{3}+2386080 \ln \left (\frac {2}{3}+x \right ) x -2825136 x^{2}+530240 \ln \left (\frac {2}{3}+x \right )-716628 x}{5832 \left (2+3 x \right )^{3}}\)	\(65\)
meijerg	\(\frac {9 x \left (\frac {9}{4} x^{2}+\frac {9}{2} x +3\right )}{16 \left (1+\frac {3 x}{2}\right )^{3}}+\frac {9 x^{2} \left (3+\frac {3 x}{2}\right )}{32 \left (1+\frac {3 x}{2}\right )^{3}}-\frac {69 x^{3}}{16 \left (1+\frac {3 x}{2}\right )^{3}}+\frac {235 x \left (\frac {99}{2} x^{2}+45 x +12\right )}{648 \left (1+\frac {3 x}{2}\right )^{3}}+\frac {8285 \ln \left (1+\frac {3 x}{2}\right )}{729}+\frac {80 x \left (\frac {405}{8} x^{3}+\frac {495}{2} x^{2}+225 x +60\right )}{243 \left (1+\frac {3 x}{2}\right )^{3}}-\frac {500 x \left (-\frac {243}{16} x^{4}+\frac {405}{8} x^{3}+\frac {495}{2} x^{2}+225 x +60\right )}{729 \left (1+\frac {3 x}{2}\right )^{3}}\)	\(134\)

[In]

int((1-2*x)^2*(3+5*x)^3/(2+3*x)^4,x,method=_RETURNVERBOSE)

[Out]

250/81*x^2-2800/243*x+27*(4099/2187*x^2+32029/13122*x+46948/59049)/(2+3*x)^3+8285/729*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.20 \[ \int \frac {(1-2 x)^2 (3+5 x)^3}{(2+3 x)^4} \, dx=\frac {364500 \, x^{5} - 631800 \, x^{4} - 2235600 \, x^{3} - 1485054 \, x^{2} + 49710 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (3 \, x + 2\right ) - 114939 \, x + 93896}{4374 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \]

[In]

integrate((1-2*x)^2*(3+5*x)^3/(2+3*x)^4,x, algorithm="fricas")

[Out]

1/4374*(364500*x^5 - 631800*x^4 - 2235600*x^3 - 1485054*x^2 + 49710*(27*x^3 + 54*x^2 + 36*x + 8)*log(3*x + 2)
- 114939*x + 93896)/(27*x^3 + 54*x^2 + 36*x + 8)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^2 (3+5 x)^3}{(2+3 x)^4} \, dx=\frac {250 x^{2}}{81} - \frac {2800 x}{243} + \frac {221346 x^{2} + 288261 x + 93896}{118098 x^{3} + 236196 x^{2} + 157464 x + 34992} + \frac {8285 \log {\left (3 x + 2 \right )}}{729} \]

[In]

integrate((1-2*x)**2*(3+5*x)**3/(2+3*x)**4,x)

[Out]

250*x**2/81 - 2800*x/243 + (221346*x**2 + 288261*x + 93896)/(118098*x**3 + 236196*x**2 + 157464*x + 34992) + 8
285*log(3*x + 2)/729

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^2 (3+5 x)^3}{(2+3 x)^4} \, dx=\frac {250}{81} \, x^{2} - \frac {2800}{243} \, x + \frac {221346 \, x^{2} + 288261 \, x + 93896}{4374 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} + \frac {8285}{729} \, \log \left (3 \, x + 2\right ) \]

[In]

integrate((1-2*x)^2*(3+5*x)^3/(2+3*x)^4,x, algorithm="maxima")

[Out]

250/81*x^2 - 2800/243*x + 1/4374*(221346*x^2 + 288261*x + 93896)/(27*x^3 + 54*x^2 + 36*x + 8) + 8285/729*log(3
*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.66 \[ \int \frac {(1-2 x)^2 (3+5 x)^3}{(2+3 x)^4} \, dx=\frac {250}{81} \, x^{2} - \frac {2800}{243} \, x + \frac {221346 \, x^{2} + 288261 \, x + 93896}{4374 \, {\left (3 \, x + 2\right )}^{3}} + \frac {8285}{729} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

integrate((1-2*x)^2*(3+5*x)^3/(2+3*x)^4,x, algorithm="giac")

[Out]

250/81*x^2 - 2800/243*x + 1/4374*(221346*x^2 + 288261*x + 93896)/(3*x + 2)^3 + 8285/729*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x)^2 (3+5 x)^3}{(2+3 x)^4} \, dx=\frac {8285\,\ln \left (x+\frac {2}{3}\right )}{729}-\frac {2800\,x}{243}+\frac {\frac {4099\,x^2}{2187}+\frac {32029\,x}{13122}+\frac {46948}{59049}}{x^3+2\,x^2+\frac {4\,x}{3}+\frac {8}{27}}+\frac {250\,x^2}{81} \]

[In]

int(((2*x - 1)^2*(5*x + 3)^3)/(3*x + 2)^4,x)

[Out]

(8285*log(x + 2/3))/729 - (2800*x)/243 + ((32029*x)/13122 + (4099*x^2)/2187 + 46948/59049)/((4*x)/3 + 2*x^2 +
x^3 + 8/27) + (250*x^2)/81