\(\int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^2} \, dx\) [1185]

Optimal result

Integrand size = 20, antiderivative size = 41 \[ \int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^2} \, dx=\frac {55 x}{9}+\frac {25 x^2}{54}-\frac {250 x^3}{27}+\frac {7}{243 (2+3 x)}+\frac {107}{243} \log (2+3 x) \]

[Out]

55/9*x+25/54*x^2-250/27*x^3+7/243/(2+3*x)+107/243*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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \[ \int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^2} \, dx=-\frac {250 x^3}{27}+\frac {25 x^2}{54}+\frac {55 x}{9}+\frac {7}{243 (3 x+2)}+\frac {107}{243} \log (3 x+2) \]

[In]

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

[Out]

(55*x)/9 + (25*x^2)/54 - (250*x^3)/27 + 7/(243*(2 + 3*x)) + (107*Log[2 + 3*x])/243

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {55}{9}+\frac {25 x}{27}-\frac {250 x^2}{9}-\frac {7}{81 (2+3 x)^2}+\frac {107}{81 (2+3 x)}\right ) \, dx \\ & = \frac {55 x}{9}+\frac {25 x^2}{54}-\frac {250 x^3}{27}+\frac {7}{243 (2+3 x)}+\frac {107}{243} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.07 \[ \int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^2} \, dx=\frac {3322+22740 x+28080 x^2-24975 x^3-40500 x^4+642 (2+3 x) \log (2+3 x)}{1458 (2+3 x)} \]

[In]

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

[Out]

(3322 + 22740*x + 28080*x^2 - 24975*x^3 - 40500*x^4 + 642*(2 + 3*x)*Log[2 + 3*x])/(1458*(2 + 3*x))

Maple [A] (verified)

Time = 2.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.73

method	result	size
risch	\(-\frac {250 x^{3}}{27}+\frac {25 x^{2}}{54}+\frac {55 x}{9}+\frac {7}{729 \left (\frac {2}{3}+x \right )}+\frac {107 \ln \left (2+3 x \right )}{243}\)	\(30\)
default	\(\frac {55 x}{9}+\frac {25 x^{2}}{54}-\frac {250 x^{3}}{27}+\frac {7}{243 \left (2+3 x \right )}+\frac {107 \ln \left (2+3 x \right )}{243}\)	\(32\)
norman	\(\frac {\frac {1973}{162} x +\frac {520}{27} x^{2}-\frac {925}{54} x^{3}-\frac {250}{9} x^{4}}{2+3 x}+\frac {107 \ln \left (2+3 x \right )}{243}\)	\(37\)
parallelrisch	\(\frac {-13500 x^{4}-8325 x^{3}+642 \ln \left (\frac {2}{3}+x \right ) x +9360 x^{2}+428 \ln \left (\frac {2}{3}+x \right )+5919 x}{972+1458 x}\)	\(42\)
meijerg	\(-\frac {27 x}{4 \left (1+\frac {3 x}{2}\right )}+\frac {107 \ln \left (1+\frac {3 x}{2}\right )}{243}-\frac {5 x \left (\frac {9 x}{2}+6\right )}{3 \left (1+\frac {3 x}{2}\right )}+\frac {325 x \left (-\frac {9}{2} x^{2}+9 x +12\right )}{54 \left (1+\frac {3 x}{2}\right )}-\frac {200 x \left (\frac {135}{8} x^{3}-\frac {45}{2} x^{2}+45 x +60\right )}{243 \left (1+\frac {3 x}{2}\right )}\)	\(80\)

[In]

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

[Out]

-250/27*x^3+25/54*x^2+55/9*x+7/729/(2/3+x)+107/243*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.02 \[ \int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^2} \, dx=-\frac {13500 \, x^{4} + 8325 \, x^{3} - 9360 \, x^{2} - 214 \, {\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) - 5940 \, x - 14}{486 \, {\left (3 \, x + 2\right )}} \]

[In]

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

[Out]

-1/486*(13500*x^4 + 8325*x^3 - 9360*x^2 - 214*(3*x + 2)*log(3*x + 2) - 5940*x - 14)/(3*x + 2)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^2} \, dx=- \frac {250 x^{3}}{27} + \frac {25 x^{2}}{54} + \frac {55 x}{9} + \frac {107 \log {\left (3 x + 2 \right )}}{243} + \frac {7}{729 x + 486} \]

[In]

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

[Out]

-250*x**3/27 + 25*x**2/54 + 55*x/9 + 107*log(3*x + 2)/243 + 7/(729*x + 486)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^2} \, dx=-\frac {250}{27} \, x^{3} + \frac {25}{54} \, x^{2} + \frac {55}{9} \, x + \frac {7}{243 \, {\left (3 \, x + 2\right )}} + \frac {107}{243} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

-250/27*x^3 + 25/54*x^2 + 55/9*x + 7/243/(3*x + 2) + 107/243*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.39 \[ \int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^2} \, dx=\frac {5}{1458} \, {\left (3 \, x + 2\right )}^{3} {\left (\frac {615}{3 \, x + 2} - \frac {666}{{\left (3 \, x + 2\right )}^{2}} - 100\right )} + \frac {7}{243 \, {\left (3 \, x + 2\right )}} - \frac {107}{243} \, \log \left (\frac {{\left | 3 \, x + 2 \right |}}{3 \, {\left (3 \, x + 2\right )}^{2}}\right ) \]

[In]

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

[Out]

5/1458*(3*x + 2)^3*(615/(3*x + 2) - 666/(3*x + 2)^2 - 100) + 7/243/(3*x + 2) - 107/243*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 \[ \int \frac {(1-2 x) (3+5 x)^3}{(2+3 x)^2} \, dx=\frac {55\,x}{9}+\frac {107\,\ln \left (x+\frac {2}{3}\right )}{243}+\frac {7}{729\,\left (x+\frac {2}{3}\right )}+\frac {25\,x^2}{54}-\frac {250\,x^3}{27} \]

[In]

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

[Out]

(55*x)/9 + (107*log(x + 2/3))/243 + 7/(729*(x + 2/3)) + (25*x^2)/54 - (250*x^3)/27