\(\int \frac {(1-2 x)^2 (2+3 x)^3}{3+5 x} \, dx\) [1293]

Optimal result

Integrand size = 22, antiderivative size = 44 \[ \int \frac {(1-2 x)^2 (2+3 x)^3}{3+5 x} \, dx=\frac {8293 x}{3125}-\frac {1931 x^2}{1250}-\frac {591 x^3}{125}+\frac {54 x^4}{25}+\frac {108 x^5}{25}+\frac {121 \log (3+5 x)}{15625} \]

[Out]

8293/3125*x-1931/1250*x^2-591/125*x^3+54/25*x^4+108/25*x^5+121/15625*ln(3+5*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} \[ \int \frac {(1-2 x)^2 (2+3 x)^3}{3+5 x} \, dx=\frac {108 x^5}{25}+\frac {54 x^4}{25}-\frac {591 x^3}{125}-\frac {1931 x^2}{1250}+\frac {8293 x}{3125}+\frac {121 \log (5 x+3)}{15625} \]

[In]

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

[Out]

(8293*x)/3125 - (1931*x^2)/1250 - (591*x^3)/125 + (54*x^4)/25 + (108*x^5)/25 + (121*Log[3 + 5*x])/15625

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 {8293}{3125}-\frac {1931 x}{625}-\frac {1773 x^2}{125}+\frac {216 x^3}{25}+\frac {108 x^4}{5}+\frac {121}{3125 (3+5 x)}\right ) \, dx \\ & = \frac {8293 x}{3125}-\frac {1931 x^2}{1250}-\frac {591 x^3}{125}+\frac {54 x^4}{25}+\frac {108 x^5}{25}+\frac {121 \log (3+5 x)}{15625} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^2 (2+3 x)^3}{3+5 x} \, dx=\frac {184863+414650 x-241375 x^2-738750 x^3+337500 x^4+675000 x^5+1210 \log (3+5 x)}{156250} \]

[In]

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

[Out]

(184863 + 414650*x - 241375*x^2 - 738750*x^3 + 337500*x^4 + 675000*x^5 + 1210*Log[3 + 5*x])/156250

Maple [A] (verified)

Time = 2.33 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(\frac {108 x^{5}}{25}+\frac {54 x^{4}}{25}-\frac {591 x^{3}}{125}-\frac {1931 x^{2}}{1250}+\frac {8293 x}{3125}+\frac {121 \ln \left (x +\frac {3}{5}\right )}{15625}\)	\(31\)
default	\(\frac {8293 x}{3125}-\frac {1931 x^{2}}{1250}-\frac {591 x^{3}}{125}+\frac {54 x^{4}}{25}+\frac {108 x^{5}}{25}+\frac {121 \ln \left (3+5 x \right )}{15625}\)	\(33\)
norman	\(\frac {8293 x}{3125}-\frac {1931 x^{2}}{1250}-\frac {591 x^{3}}{125}+\frac {54 x^{4}}{25}+\frac {108 x^{5}}{25}+\frac {121 \ln \left (3+5 x \right )}{15625}\)	\(33\)
risch	\(\frac {8293 x}{3125}-\frac {1931 x^{2}}{1250}-\frac {591 x^{3}}{125}+\frac {54 x^{4}}{25}+\frac {108 x^{5}}{25}+\frac {121 \ln \left (3+5 x \right )}{15625}\)	\(33\)
meijerg	\(\frac {121 \ln \left (1+\frac {5 x}{3}\right )}{15625}+\frac {4 x}{5}+\frac {29 x \left (-5 x +6\right )}{25}-\frac {27 x \left (\frac {100}{9} x^{2}-10 x +12\right )}{100}-\frac {243 x \left (-\frac {625}{9} x^{3}+\frac {500}{9} x^{2}-50 x +60\right )}{3125}+\frac {729 x \left (\frac {2500}{27} x^{4}-\frac {625}{9} x^{3}+\frac {500}{9} x^{2}-50 x +60\right )}{15625}\)	\(75\)

[In]

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

[Out]

108/25*x^5+54/25*x^4-591/125*x^3-1931/1250*x^2+8293/3125*x+121/15625*ln(x+3/5)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x)^2 (2+3 x)^3}{3+5 x} \, dx=\frac {108}{25} \, x^{5} + \frac {54}{25} \, x^{4} - \frac {591}{125} \, x^{3} - \frac {1931}{1250} \, x^{2} + \frac {8293}{3125} \, x + \frac {121}{15625} \, \log \left (5 \, x + 3\right ) \]

[In]

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

[Out]

108/25*x^5 + 54/25*x^4 - 591/125*x^3 - 1931/1250*x^2 + 8293/3125*x + 121/15625*log(5*x + 3)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93 \[ \int \frac {(1-2 x)^2 (2+3 x)^3}{3+5 x} \, dx=\frac {108 x^{5}}{25} + \frac {54 x^{4}}{25} - \frac {591 x^{3}}{125} - \frac {1931 x^{2}}{1250} + \frac {8293 x}{3125} + \frac {121 \log {\left (5 x + 3 \right )}}{15625} \]

[In]

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

[Out]

108*x**5/25 + 54*x**4/25 - 591*x**3/125 - 1931*x**2/1250 + 8293*x/3125 + 121*log(5*x + 3)/15625

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x)^2 (2+3 x)^3}{3+5 x} \, dx=\frac {108}{25} \, x^{5} + \frac {54}{25} \, x^{4} - \frac {591}{125} \, x^{3} - \frac {1931}{1250} \, x^{2} + \frac {8293}{3125} \, x + \frac {121}{15625} \, \log \left (5 \, x + 3\right ) \]

[In]

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

[Out]

108/25*x^5 + 54/25*x^4 - 591/125*x^3 - 1931/1250*x^2 + 8293/3125*x + 121/15625*log(5*x + 3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.75 \[ \int \frac {(1-2 x)^2 (2+3 x)^3}{3+5 x} \, dx=\frac {108}{25} \, x^{5} + \frac {54}{25} \, x^{4} - \frac {591}{125} \, x^{3} - \frac {1931}{1250} \, x^{2} + \frac {8293}{3125} \, x + \frac {121}{15625} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \]

[In]

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

[Out]

108/25*x^5 + 54/25*x^4 - 591/125*x^3 - 1931/1250*x^2 + 8293/3125*x + 121/15625*log(abs(5*x + 3))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.68 \[ \int \frac {(1-2 x)^2 (2+3 x)^3}{3+5 x} \, dx=\frac {8293\,x}{3125}+\frac {121\,\ln \left (x+\frac {3}{5}\right )}{15625}-\frac {1931\,x^2}{1250}-\frac {591\,x^3}{125}+\frac {54\,x^4}{25}+\frac {108\,x^5}{25} \]

[In]

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

[Out]

(8293*x)/3125 + (121*log(x + 3/5))/15625 - (1931*x^2)/1250 - (591*x^3)/125 + (54*x^4)/25 + (108*x^5)/25