\(\int \frac {(2+3 x)^2 (3+5 x)^3}{1-2 x} \, dx\) [1473]

Optimal result

Integrand size = 22, antiderivative size = 44 \[ \int \frac {(2+3 x)^2 (3+5 x)^3}{1-2 x} \, dx=-\frac {61763 x}{32}-\frac {47939 x^2}{32}-\frac {25835 x^3}{24}-\frac {8175 x^4}{16}-\frac {225 x^5}{2}-\frac {65219}{64} \log (1-2 x) \]

[Out]

-61763/32*x-47939/32*x^2-25835/24*x^3-8175/16*x^4-225/2*x^5-65219/64*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} \[ \int \frac {(2+3 x)^2 (3+5 x)^3}{1-2 x} \, dx=-\frac {225 x^5}{2}-\frac {8175 x^4}{16}-\frac {25835 x^3}{24}-\frac {47939 x^2}{32}-\frac {61763 x}{32}-\frac {65219}{64} \log (1-2 x) \]

[In]

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

[Out]

(-61763*x)/32 - (47939*x^2)/32 - (25835*x^3)/24 - (8175*x^4)/16 - (225*x^5)/2 - (65219*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 {61763}{32}-\frac {47939 x}{16}-\frac {25835 x^2}{8}-\frac {8175 x^3}{4}-\frac {1125 x^4}{2}-\frac {65219}{32 (-1+2 x)}\right ) \, dx \\ & = -\frac {61763 x}{32}-\frac {47939 x^2}{32}-\frac {25835 x^3}{24}-\frac {8175 x^4}{16}-\frac {225 x^5}{2}-\frac {65219}{64} \log (1-2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.84 \[ \int \frac {(2+3 x)^2 (3+5 x)^3}{1-2 x} \, dx=\frac {1}{768} \left (1159355-1482312 x-1150536 x^2-826720 x^3-392400 x^4-86400 x^5-782628 \log (1-2 x)\right ) \]

[In]

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

[Out]

(1159355 - 1482312*x - 1150536*x^2 - 826720*x^3 - 392400*x^4 - 86400*x^5 - 782628*Log[1 - 2*x])/768

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(-\frac {225 x^{5}}{2}-\frac {8175 x^{4}}{16}-\frac {25835 x^{3}}{24}-\frac {47939 x^{2}}{32}-\frac {61763 x}{32}-\frac {65219 \ln \left (x -\frac {1}{2}\right )}{64}\)	\(31\)
default	\(-\frac {225 x^{5}}{2}-\frac {8175 x^{4}}{16}-\frac {25835 x^{3}}{24}-\frac {47939 x^{2}}{32}-\frac {61763 x}{32}-\frac {65219 \ln \left (-1+2 x \right )}{64}\)	\(33\)
norman	\(-\frac {225 x^{5}}{2}-\frac {8175 x^{4}}{16}-\frac {25835 x^{3}}{24}-\frac {47939 x^{2}}{32}-\frac {61763 x}{32}-\frac {65219 \ln \left (-1+2 x \right )}{64}\)	\(33\)
risch	\(-\frac {225 x^{5}}{2}-\frac {8175 x^{4}}{16}-\frac {25835 x^{3}}{24}-\frac {47939 x^{2}}{32}-\frac {61763 x}{32}-\frac {65219 \ln \left (-1+2 x \right )}{64}\)	\(33\)
meijerg	\(-\frac {65219 \ln \left (1-2 x \right )}{64}-432 x -\frac {921 x \left (6 x +6\right )}{8}-\frac {4415 x \left (16 x^{2}+12 x +12\right )}{96}-\frac {235 x \left (120 x^{3}+80 x^{2}+60 x +60\right )}{64}-\frac {75 x \left (192 x^{4}+120 x^{3}+80 x^{2}+60 x +60\right )}{128}\)	\(75\)

[In]

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

[Out]

-225/2*x^5-8175/16*x^4-25835/24*x^3-47939/32*x^2-61763/32*x-65219/64*ln(x-1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int \frac {(2+3 x)^2 (3+5 x)^3}{1-2 x} \, dx=-\frac {225}{2} \, x^{5} - \frac {8175}{16} \, x^{4} - \frac {25835}{24} \, x^{3} - \frac {47939}{32} \, x^{2} - \frac {61763}{32} \, x - \frac {65219}{64} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-225/2*x^5 - 8175/16*x^4 - 25835/24*x^3 - 47939/32*x^2 - 61763/32*x - 65219/64*log(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95 \[ \int \frac {(2+3 x)^2 (3+5 x)^3}{1-2 x} \, dx=- \frac {225 x^{5}}{2} - \frac {8175 x^{4}}{16} - \frac {25835 x^{3}}{24} - \frac {47939 x^{2}}{32} - \frac {61763 x}{32} - \frac {65219 \log {\left (2 x - 1 \right )}}{64} \]

[In]

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

[Out]

-225*x**5/2 - 8175*x**4/16 - 25835*x**3/24 - 47939*x**2/32 - 61763*x/32 - 65219*log(2*x - 1)/64

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int \frac {(2+3 x)^2 (3+5 x)^3}{1-2 x} \, dx=-\frac {225}{2} \, x^{5} - \frac {8175}{16} \, x^{4} - \frac {25835}{24} \, x^{3} - \frac {47939}{32} \, x^{2} - \frac {61763}{32} \, x - \frac {65219}{64} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-225/2*x^5 - 8175/16*x^4 - 25835/24*x^3 - 47939/32*x^2 - 61763/32*x - 65219/64*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.75 \[ \int \frac {(2+3 x)^2 (3+5 x)^3}{1-2 x} \, dx=-\frac {225}{2} \, x^{5} - \frac {8175}{16} \, x^{4} - \frac {25835}{24} \, x^{3} - \frac {47939}{32} \, x^{2} - \frac {61763}{32} \, x - \frac {65219}{64} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

-225/2*x^5 - 8175/16*x^4 - 25835/24*x^3 - 47939/32*x^2 - 61763/32*x - 65219/64*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.68 \[ \int \frac {(2+3 x)^2 (3+5 x)^3}{1-2 x} \, dx=-\frac {61763\,x}{32}-\frac {65219\,\ln \left (x-\frac {1}{2}\right )}{64}-\frac {47939\,x^2}{32}-\frac {25835\,x^3}{24}-\frac {8175\,x^4}{16}-\frac {225\,x^5}{2} \]

[In]

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

[Out]

- (61763*x)/32 - (65219*log(x - 1/2))/64 - (47939*x^2)/32 - (25835*x^3)/24 - (8175*x^4)/16 - (225*x^5)/2