\(\int \frac {(2+3 x)^2 (3+5 x)}{(1-2 x)^3} \, dx\) [1636]

Optimal result

Integrand size = 20, antiderivative size = 38 \[ \int \frac {(2+3 x)^2 (3+5 x)}{(1-2 x)^3} \, dx=\frac {539}{32 (1-2 x)^2}-\frac {707}{16 (1-2 x)}-\frac {45 x}{8}-\frac {309}{16} \log (1-2 x) \]

[Out]

539/32/(1-2*x)^2-707/16/(1-2*x)-45/8*x-309/16*ln(1-2*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, 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} \[ \int \frac {(2+3 x)^2 (3+5 x)}{(1-2 x)^3} \, dx=-\frac {45 x}{8}-\frac {707}{16 (1-2 x)}+\frac {539}{32 (1-2 x)^2}-\frac {309}{16} \log (1-2 x) \]

[In]

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

[Out]

539/(32*(1 - 2*x)^2) - 707/(16*(1 - 2*x)) - (45*x)/8 - (309*Log[1 - 2*x])/16

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 {45}{8}-\frac {539}{8 (-1+2 x)^3}-\frac {707}{8 (-1+2 x)^2}-\frac {309}{8 (-1+2 x)}\right ) \, dx \\ & = \frac {539}{32 (1-2 x)^2}-\frac {707}{16 (1-2 x)}-\frac {45 x}{8}-\frac {309}{16} \log (1-2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {(2+3 x)^2 (3+5 x)}{(1-2 x)^3} \, dx=\frac {1}{32} \left (-180 x+\frac {-785+2468 x+360 x^2}{(1-2 x)^2}-618 \log (1-2 x)\right ) \]

[In]

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

[Out]

(-180*x + (-785 + 2468*x + 360*x^2)/(1 - 2*x)^2 - 618*Log[1 - 2*x])/32

Maple [A] (verified)

Time = 2.72 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71

method	result	size
risch	\(-\frac {45 x}{8}+\frac {\frac {707 x}{8}-\frac {875}{32}}{\left (-1+2 x \right )^{2}}-\frac {309 \ln \left (-1+2 x \right )}{16}\)	\(27\)
default	\(-\frac {45 x}{8}-\frac {309 \ln \left (-1+2 x \right )}{16}+\frac {707}{16 \left (-1+2 x \right )}+\frac {539}{32 \left (-1+2 x \right )^{2}}\)	\(31\)
norman	\(\frac {-\frac {213}{8} x +\frac {1055}{8} x^{2}-\frac {45}{2} x^{3}}{\left (-1+2 x \right )^{2}}-\frac {309 \ln \left (-1+2 x \right )}{16}\)	\(32\)
parallelrisch	\(-\frac {1236 \ln \left (x -\frac {1}{2}\right ) x^{2}+360 x^{3}-1236 \ln \left (x -\frac {1}{2}\right ) x -2110 x^{2}+309 \ln \left (x -\frac {1}{2}\right )+426 x}{16 \left (-1+2 x \right )^{2}}\)	\(46\)
meijerg	\(\frac {6 x \left (2-2 x \right )}{\left (1-2 x \right )^{2}}+\frac {28 x^{2}}{\left (1-2 x \right )^{2}}-\frac {29 x \left (-18 x +6\right )}{8 \left (1-2 x \right )^{2}}-\frac {309 \ln \left (1-2 x \right )}{16}-\frac {45 x \left (16 x^{2}-36 x +12\right )}{32 \left (1-2 x \right )^{2}}\)	\(72\)

[In]

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

[Out]

-45/8*x+4*(707/32*x-875/128)/(-1+2*x)^2-309/16*ln(-1+2*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int \frac {(2+3 x)^2 (3+5 x)}{(1-2 x)^3} \, dx=-\frac {720 \, x^{3} - 720 \, x^{2} + 618 \, {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (2 \, x - 1\right ) - 2648 \, x + 875}{32 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \]

[In]

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

[Out]

-1/32*(720*x^3 - 720*x^2 + 618*(4*x^2 - 4*x + 1)*log(2*x - 1) - 2648*x + 875)/(4*x^2 - 4*x + 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {(2+3 x)^2 (3+5 x)}{(1-2 x)^3} \, dx=- \frac {45 x}{8} - \frac {875 - 2828 x}{128 x^{2} - 128 x + 32} - \frac {309 \log {\left (2 x - 1 \right )}}{16} \]

[In]

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

[Out]

-45*x/8 - (875 - 2828*x)/(128*x**2 - 128*x + 32) - 309*log(2*x - 1)/16

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {(2+3 x)^2 (3+5 x)}{(1-2 x)^3} \, dx=-\frac {45}{8} \, x + \frac {7 \, {\left (404 \, x - 125\right )}}{32 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} - \frac {309}{16} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-45/8*x + 7/32*(404*x - 125)/(4*x^2 - 4*x + 1) - 309/16*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {(2+3 x)^2 (3+5 x)}{(1-2 x)^3} \, dx=-\frac {45}{8} \, x + \frac {7 \, {\left (404 \, x - 125\right )}}{32 \, {\left (2 \, x - 1\right )}^{2}} - \frac {309}{16} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

-45/8*x + 7/32*(404*x - 125)/(2*x - 1)^2 - 309/16*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.68 \[ \int \frac {(2+3 x)^2 (3+5 x)}{(1-2 x)^3} \, dx=\frac {\frac {707\,x}{32}-\frac {875}{128}}{x^2-x+\frac {1}{4}}-\frac {309\,\ln \left (x-\frac {1}{2}\right )}{16}-\frac {45\,x}{8} \]

[In]

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

[Out]

((707*x)/32 - 875/128)/(x^2 - x + 1/4) - (309*log(x - 1/2))/16 - (45*x)/8