\(\int \frac {(2+3 x)^3 (3+5 x)^3}{1-2 x} \, dx\) [1472]

Optimal result

Integrand size = 22, antiderivative size = 51 \[ \int \frac {(2+3 x)^3 (3+5 x)^3}{1-2 x} \, dx=-\frac {442709 x}{64}-\frac {377045 x^2}{64}-\frac {247157 x^3}{48}-\frac {110205 x^4}{32}-\frac {5805 x^5}{4}-\frac {1125 x^6}{4}-\frac {456533}{128} \log (1-2 x) \]

[Out]

-442709/64*x-377045/64*x^2-247157/48*x^3-110205/32*x^4-5805/4*x^5-1125/4*x^6-456533/128*ln(1-2*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 51, 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} \[ \int \frac {(2+3 x)^3 (3+5 x)^3}{1-2 x} \, dx=-\frac {1125 x^6}{4}-\frac {5805 x^5}{4}-\frac {110205 x^4}{32}-\frac {247157 x^3}{48}-\frac {377045 x^2}{64}-\frac {442709 x}{64}-\frac {456533}{128} \log (1-2 x) \]

[In]

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

[Out]

(-442709*x)/64 - (377045*x^2)/64 - (247157*x^3)/48 - (110205*x^4)/32 - (5805*x^5)/4 - (1125*x^6)/4 - (456533*L
og[1 - 2*x])/128

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 {442709}{64}-\frac {377045 x}{32}-\frac {247157 x^2}{16}-\frac {110205 x^3}{8}-\frac {29025 x^4}{4}-\frac {3375 x^5}{2}-\frac {456533}{64 (-1+2 x)}\right ) \, dx \\ & = -\frac {442709 x}{64}-\frac {377045 x^2}{64}-\frac {247157 x^3}{48}-\frac {110205 x^4}{32}-\frac {5805 x^5}{4}-\frac {1125 x^6}{4}-\frac {456533}{128} \log (1-2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.82 \[ \int \frac {(2+3 x)^3 (3+5 x)^3}{1-2 x} \, dx=\frac {8970431-10625016 x-9049080 x^2-7909024 x^3-5289840 x^4-2229120 x^5-432000 x^6-5478396 \log (1-2 x)}{1536} \]

[In]

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

[Out]

(8970431 - 10625016*x - 9049080*x^2 - 7909024*x^3 - 5289840*x^4 - 2229120*x^5 - 432000*x^6 - 5478396*Log[1 - 2
*x])/1536

Maple [A] (verified)

Time = 3.34 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.71

method	result	size
parallelrisch	\(-\frac {1125 x^{6}}{4}-\frac {5805 x^{5}}{4}-\frac {110205 x^{4}}{32}-\frac {247157 x^{3}}{48}-\frac {377045 x^{2}}{64}-\frac {442709 x}{64}-\frac {456533 \ln \left (x -\frac {1}{2}\right )}{128}\)	\(36\)
default	\(-\frac {1125 x^{6}}{4}-\frac {5805 x^{5}}{4}-\frac {110205 x^{4}}{32}-\frac {247157 x^{3}}{48}-\frac {377045 x^{2}}{64}-\frac {442709 x}{64}-\frac {456533 \ln \left (-1+2 x \right )}{128}\)	\(38\)
norman	\(-\frac {1125 x^{6}}{4}-\frac {5805 x^{5}}{4}-\frac {110205 x^{4}}{32}-\frac {247157 x^{3}}{48}-\frac {377045 x^{2}}{64}-\frac {442709 x}{64}-\frac {456533 \ln \left (-1+2 x \right )}{128}\)	\(38\)
risch	\(-\frac {1125 x^{6}}{4}-\frac {5805 x^{5}}{4}-\frac {110205 x^{4}}{32}-\frac {247157 x^{3}}{48}-\frac {377045 x^{2}}{64}-\frac {442709 x}{64}-\frac {456533 \ln \left (-1+2 x \right )}{128}\)	\(38\)
meijerg	\(-\frac {456533 \ln \left (1-2 x \right )}{128}-1026 x -\frac {1353 x \left (6 x +6\right )}{4}-\frac {17119 x \left (16 x^{2}+12 x +12\right )}{96}-\frac {1353 x \left (120 x^{3}+80 x^{2}+60 x +60\right )}{64}-\frac {855 x \left (192 x^{4}+120 x^{3}+80 x^{2}+60 x +60\right )}{128}-\frac {225 x \left (2240 x^{5}+1344 x^{4}+840 x^{3}+560 x^{2}+420 x +420\right )}{1792}\)	\(103\)

[In]

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

[Out]

-1125/4*x^6-5805/4*x^5-110205/32*x^4-247157/48*x^3-377045/64*x^2-442709/64*x-456533/128*ln(x-1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.73 \[ \int \frac {(2+3 x)^3 (3+5 x)^3}{1-2 x} \, dx=-\frac {1125}{4} \, x^{6} - \frac {5805}{4} \, x^{5} - \frac {110205}{32} \, x^{4} - \frac {247157}{48} \, x^{3} - \frac {377045}{64} \, x^{2} - \frac {442709}{64} \, x - \frac {456533}{128} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-1125/4*x^6 - 5805/4*x^5 - 110205/32*x^4 - 247157/48*x^3 - 377045/64*x^2 - 442709/64*x - 456533/128*log(2*x -
1)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96 \[ \int \frac {(2+3 x)^3 (3+5 x)^3}{1-2 x} \, dx=- \frac {1125 x^{6}}{4} - \frac {5805 x^{5}}{4} - \frac {110205 x^{4}}{32} - \frac {247157 x^{3}}{48} - \frac {377045 x^{2}}{64} - \frac {442709 x}{64} - \frac {456533 \log {\left (2 x - 1 \right )}}{128} \]

[In]

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

[Out]

-1125*x**6/4 - 5805*x**5/4 - 110205*x**4/32 - 247157*x**3/48 - 377045*x**2/64 - 442709*x/64 - 456533*log(2*x -
1)/128

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.73 \[ \int \frac {(2+3 x)^3 (3+5 x)^3}{1-2 x} \, dx=-\frac {1125}{4} \, x^{6} - \frac {5805}{4} \, x^{5} - \frac {110205}{32} \, x^{4} - \frac {247157}{48} \, x^{3} - \frac {377045}{64} \, x^{2} - \frac {442709}{64} \, x - \frac {456533}{128} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-1125/4*x^6 - 5805/4*x^5 - 110205/32*x^4 - 247157/48*x^3 - 377045/64*x^2 - 442709/64*x - 456533/128*log(2*x -
1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.75 \[ \int \frac {(2+3 x)^3 (3+5 x)^3}{1-2 x} \, dx=-\frac {1125}{4} \, x^{6} - \frac {5805}{4} \, x^{5} - \frac {110205}{32} \, x^{4} - \frac {247157}{48} \, x^{3} - \frac {377045}{64} \, x^{2} - \frac {442709}{64} \, x - \frac {456533}{128} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

-1125/4*x^6 - 5805/4*x^5 - 110205/32*x^4 - 247157/48*x^3 - 377045/64*x^2 - 442709/64*x - 456533/128*log(abs(2*
x - 1))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int \frac {(2+3 x)^3 (3+5 x)^3}{1-2 x} \, dx=-\frac {442709\,x}{64}-\frac {456533\,\ln \left (x-\frac {1}{2}\right )}{128}-\frac {377045\,x^2}{64}-\frac {247157\,x^3}{48}-\frac {110205\,x^4}{32}-\frac {5805\,x^5}{4}-\frac {1125\,x^6}{4} \]

[In]

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

[Out]

- (442709*x)/64 - (456533*log(x - 1/2))/128 - (377045*x^2)/64 - (247157*x^3)/48 - (110205*x^4)/32 - (5805*x^5)
/4 - (1125*x^6)/4