\(\int \frac {(2+3 x)^3 (3+5 x)^2}{1-2 x} \, dx\) [1456]

Optimal result

Integrand size = 22, antiderivative size = 44 \[ \int \frac {(2+3 x)^3 (3+5 x)^2}{1-2 x} \, dx=-\frac {39199 x}{32}-\frac {30175 x^2}{32}-\frac {5349 x^3}{8}-\frac {4995 x^4}{16}-\frac {135 x^5}{2}-\frac {41503}{64} \log (1-2 x) \]

[Out]

-39199/32*x-30175/32*x^2-5349/8*x^3-4995/16*x^4-135/2*x^5-41503/64*ln(1-2*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} \[ \int \frac {(2+3 x)^3 (3+5 x)^2}{1-2 x} \, dx=-\frac {135 x^5}{2}-\frac {4995 x^4}{16}-\frac {5349 x^3}{8}-\frac {30175 x^2}{32}-\frac {39199 x}{32}-\frac {41503}{64} \log (1-2 x) \]

[In]

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

[Out]

(-39199*x)/32 - (30175*x^2)/32 - (5349*x^3)/8 - (4995*x^4)/16 - (135*x^5)/2 - (41503*Log[1 - 2*x])/64

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 {39199}{32}-\frac {30175 x}{16}-\frac {16047 x^2}{8}-\frac {4995 x^3}{4}-\frac {675 x^4}{2}-\frac {41503}{32 (-1+2 x)}\right ) \, dx \\ & = -\frac {39199 x}{32}-\frac {30175 x^2}{32}-\frac {5349 x^3}{8}-\frac {4995 x^4}{16}-\frac {135 x^5}{2}-\frac {41503}{64} \log (1-2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.84 \[ \int \frac {(2+3 x)^3 (3+5 x)^2}{1-2 x} \, dx=\frac {1}{256} \left (244077-313592 x-241400 x^2-171168 x^3-79920 x^4-17280 x^5-166012 \log (1-2 x)\right ) \]

[In]

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

[Out]

(244077 - 313592*x - 241400*x^2 - 171168*x^3 - 79920*x^4 - 17280*x^5 - 166012*Log[1 - 2*x])/256

Maple [A] (verified)

Time = 2.49 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(-\frac {135 x^{5}}{2}-\frac {4995 x^{4}}{16}-\frac {5349 x^{3}}{8}-\frac {30175 x^{2}}{32}-\frac {39199 x}{32}-\frac {41503 \ln \left (x -\frac {1}{2}\right )}{64}\)	\(31\)
default	\(-\frac {135 x^{5}}{2}-\frac {4995 x^{4}}{16}-\frac {5349 x^{3}}{8}-\frac {30175 x^{2}}{32}-\frac {39199 x}{32}-\frac {41503 \ln \left (-1+2 x \right )}{64}\)	\(33\)
norman	\(-\frac {135 x^{5}}{2}-\frac {4995 x^{4}}{16}-\frac {5349 x^{3}}{8}-\frac {30175 x^{2}}{32}-\frac {39199 x}{32}-\frac {41503 \ln \left (-1+2 x \right )}{64}\)	\(33\)
risch	\(-\frac {135 x^{5}}{2}-\frac {4995 x^{4}}{16}-\frac {5349 x^{3}}{8}-\frac {30175 x^{2}}{32}-\frac {39199 x}{32}-\frac {41503 \ln \left (-1+2 x \right )}{64}\)	\(33\)
meijerg	\(-\frac {41503 \ln \left (1-2 x \right )}{64}-282 x -\frac {883 x \left (6 x +6\right )}{12}-\frac {921 x \left (16 x^{2}+12 x +12\right )}{32}-\frac {9 x \left (120 x^{3}+80 x^{2}+60 x +60\right )}{4}-\frac {45 x \left (192 x^{4}+120 x^{3}+80 x^{2}+60 x +60\right )}{128}\)	\(75\)

[In]

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

[Out]

-135/2*x^5-4995/16*x^4-5349/8*x^3-30175/32*x^2-39199/32*x-41503/64*ln(x-1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int \frac {(2+3 x)^3 (3+5 x)^2}{1-2 x} \, dx=-\frac {135}{2} \, x^{5} - \frac {4995}{16} \, x^{4} - \frac {5349}{8} \, x^{3} - \frac {30175}{32} \, x^{2} - \frac {39199}{32} \, x - \frac {41503}{64} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-135/2*x^5 - 4995/16*x^4 - 5349/8*x^3 - 30175/32*x^2 - 39199/32*x - 41503/64*log(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95 \[ \int \frac {(2+3 x)^3 (3+5 x)^2}{1-2 x} \, dx=- \frac {135 x^{5}}{2} - \frac {4995 x^{4}}{16} - \frac {5349 x^{3}}{8} - \frac {30175 x^{2}}{32} - \frac {39199 x}{32} - \frac {41503 \log {\left (2 x - 1 \right )}}{64} \]

[In]

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

[Out]

-135*x**5/2 - 4995*x**4/16 - 5349*x**3/8 - 30175*x**2/32 - 39199*x/32 - 41503*log(2*x - 1)/64

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int \frac {(2+3 x)^3 (3+5 x)^2}{1-2 x} \, dx=-\frac {135}{2} \, x^{5} - \frac {4995}{16} \, x^{4} - \frac {5349}{8} \, x^{3} - \frac {30175}{32} \, x^{2} - \frac {39199}{32} \, x - \frac {41503}{64} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-135/2*x^5 - 4995/16*x^4 - 5349/8*x^3 - 30175/32*x^2 - 39199/32*x - 41503/64*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.75 \[ \int \frac {(2+3 x)^3 (3+5 x)^2}{1-2 x} \, dx=-\frac {135}{2} \, x^{5} - \frac {4995}{16} \, x^{4} - \frac {5349}{8} \, x^{3} - \frac {30175}{32} \, x^{2} - \frac {39199}{32} \, x - \frac {41503}{64} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

-135/2*x^5 - 4995/16*x^4 - 5349/8*x^3 - 30175/32*x^2 - 39199/32*x - 41503/64*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.68 \[ \int \frac {(2+3 x)^3 (3+5 x)^2}{1-2 x} \, dx=-\frac {39199\,x}{32}-\frac {41503\,\ln \left (x-\frac {1}{2}\right )}{64}-\frac {30175\,x^2}{32}-\frac {5349\,x^3}{8}-\frac {4995\,x^4}{16}-\frac {135\,x^5}{2} \]

[In]

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

[Out]

- (39199*x)/32 - (41503*log(x - 1/2))/64 - (30175*x^2)/32 - (5349*x^3)/8 - (4995*x^4)/16 - (135*x^5)/2