\(\int \frac {(1-2 x)^2 (3+5 x)^3}{2+3 x} \, dx\) [1281]

Optimal result

Integrand size = 22, antiderivative size = 44 \[ \int \frac {(1-2 x)^2 (3+5 x)^3}{2+3 x} \, dx=\frac {3305 x}{243}-\frac {559 x^2}{162}-\frac {2515 x^3}{81}+\frac {50 x^4}{9}+\frac {100 x^5}{3}-\frac {49}{729} \log (2+3 x) \]

[Out]

3305/243*x-559/162*x^2-2515/81*x^3+50/9*x^4+100/3*x^5-49/729*ln(2+3*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, 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} \[ \int \frac {(1-2 x)^2 (3+5 x)^3}{2+3 x} \, dx=\frac {100 x^5}{3}+\frac {50 x^4}{9}-\frac {2515 x^3}{81}-\frac {559 x^2}{162}+\frac {3305 x}{243}-\frac {49}{729} \log (3 x+2) \]

[In]

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

[Out]

(3305*x)/243 - (559*x^2)/162 - (2515*x^3)/81 + (50*x^4)/9 + (100*x^5)/3 - (49*Log[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 {3305}{243}-\frac {559 x}{81}-\frac {2515 x^2}{27}+\frac {200 x^3}{9}+\frac {500 x^4}{3}-\frac {49}{243 (2+3 x)}\right ) \, dx \\ & = \frac {3305 x}{243}-\frac {559 x^2}{162}-\frac {2515 x^3}{81}+\frac {50 x^4}{9}+\frac {100 x^5}{3}-\frac {49}{729} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^2 (3+5 x)^3}{2+3 x} \, dx=\frac {20528+59490 x-15093 x^2-135810 x^3+24300 x^4+145800 x^5-294 \log (2+3 x)}{4374} \]

[In]

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

[Out]

(20528 + 59490*x - 15093*x^2 - 135810*x^3 + 24300*x^4 + 145800*x^5 - 294*Log[2 + 3*x])/4374

Maple [A] (verified)

Time = 1.90 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(\frac {100 x^{5}}{3}+\frac {50 x^{4}}{9}-\frac {2515 x^{3}}{81}-\frac {559 x^{2}}{162}+\frac {3305 x}{243}-\frac {49 \ln \left (\frac {2}{3}+x \right )}{729}\)	\(31\)
default	\(\frac {3305 x}{243}-\frac {559 x^{2}}{162}-\frac {2515 x^{3}}{81}+\frac {50 x^{4}}{9}+\frac {100 x^{5}}{3}-\frac {49 \ln \left (2+3 x \right )}{729}\)	\(33\)
norman	\(\frac {3305 x}{243}-\frac {559 x^{2}}{162}-\frac {2515 x^{3}}{81}+\frac {50 x^{4}}{9}+\frac {100 x^{5}}{3}-\frac {49 \ln \left (2+3 x \right )}{729}\)	\(33\)
risch	\(\frac {3305 x}{243}-\frac {559 x^{2}}{162}-\frac {2515 x^{3}}{81}+\frac {50 x^{4}}{9}+\frac {100 x^{5}}{3}-\frac {49 \ln \left (2+3 x \right )}{729}\)	\(33\)
meijerg	\(-\frac {49 \ln \left (1+\frac {3 x}{2}\right )}{729}+9 x +\frac {23 x \left (-\frac {9 x}{2}+6\right )}{3}-\frac {235 x \left (9 x^{2}-9 x +12\right )}{81}-\frac {160 x \left (-\frac {405}{8} x^{3}+45 x^{2}-45 x +60\right )}{243}+\frac {400 x \left (\frac {243}{4} x^{4}-\frac {405}{8} x^{3}+45 x^{2}-45 x +60\right )}{729}\)	\(75\)

[In]

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

[Out]

100/3*x^5+50/9*x^4-2515/81*x^3-559/162*x^2+3305/243*x-49/729*ln(2/3+x)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x)^2 (3+5 x)^3}{2+3 x} \, dx=\frac {100}{3} \, x^{5} + \frac {50}{9} \, x^{4} - \frac {2515}{81} \, x^{3} - \frac {559}{162} \, x^{2} + \frac {3305}{243} \, x - \frac {49}{729} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

100/3*x^5 + 50/9*x^4 - 2515/81*x^3 - 559/162*x^2 + 3305/243*x - 49/729*log(3*x + 2)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93 \[ \int \frac {(1-2 x)^2 (3+5 x)^3}{2+3 x} \, dx=\frac {100 x^{5}}{3} + \frac {50 x^{4}}{9} - \frac {2515 x^{3}}{81} - \frac {559 x^{2}}{162} + \frac {3305 x}{243} - \frac {49 \log {\left (3 x + 2 \right )}}{729} \]

[In]

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

[Out]

100*x**5/3 + 50*x**4/9 - 2515*x**3/81 - 559*x**2/162 + 3305*x/243 - 49*log(3*x + 2)/729

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x)^2 (3+5 x)^3}{2+3 x} \, dx=\frac {100}{3} \, x^{5} + \frac {50}{9} \, x^{4} - \frac {2515}{81} \, x^{3} - \frac {559}{162} \, x^{2} + \frac {3305}{243} \, x - \frac {49}{729} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

100/3*x^5 + 50/9*x^4 - 2515/81*x^3 - 559/162*x^2 + 3305/243*x - 49/729*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.75 \[ \int \frac {(1-2 x)^2 (3+5 x)^3}{2+3 x} \, dx=\frac {100}{3} \, x^{5} + \frac {50}{9} \, x^{4} - \frac {2515}{81} \, x^{3} - \frac {559}{162} \, x^{2} + \frac {3305}{243} \, x - \frac {49}{729} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

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

[Out]

100/3*x^5 + 50/9*x^4 - 2515/81*x^3 - 559/162*x^2 + 3305/243*x - 49/729*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.68 \[ \int \frac {(1-2 x)^2 (3+5 x)^3}{2+3 x} \, dx=\frac {3305\,x}{243}-\frac {49\,\ln \left (x+\frac {2}{3}\right )}{729}-\frac {559\,x^2}{162}-\frac {2515\,x^3}{81}+\frac {50\,x^4}{9}+\frac {100\,x^5}{3} \]

[In]

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

[Out]

(3305*x)/243 - (49*log(x + 2/3))/729 - (559*x^2)/162 - (2515*x^3)/81 + (50*x^4)/9 + (100*x^5)/3