3.1472 \(\int \frac {(2+3 x)^3 (3+5 x)^3}{1-2 x} \, dx\)

Optimal. Leaf size=51 \[ -\frac {1125 x^6}{4}-\frac {5805 x^5}{4}-\frac {110205 x^4}{32}-\frac {247157 x^3}{48}-\frac {377045 x^2}{64}-\frac {442709 x}{64}-\frac {456533}{128} \log (1-2 x) \]

[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]  time = 0.02, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {88} \[ -\frac {1125 x^6}{4}-\frac {5805 x^5}{4}-\frac {110205 x^4}{32}-\frac {247157 x^3}{48}-\frac {377045 x^2}{64}-\frac {442709 x}{64}-\frac {456533}{128} \log (1-2 x) \]

Antiderivative was successfully verified.

[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 88

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

Antiderivative was successfully verified.

[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

________________________________________________________________________________________

fricas [A]  time = 0.93, size = 37, normalized size = 0.73 \[ -\frac {1125}{4} \, x^{6} - \frac {5805}{4} \, x^{5} - \frac {110205}{32} \, x^{4} - \frac {247157}{48} \, x^{3} - \frac {377045}{64} \, x^{2} - \frac {442709}{64} \, x - \frac {456533}{128} \, \log \left (2 \, x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [A]  time = 0.95, size = 38, normalized size = 0.75 \[ -\frac {1125}{4} \, x^{6} - \frac {5805}{4} \, x^{5} - \frac {110205}{32} \, x^{4} - \frac {247157}{48} \, x^{3} - \frac {377045}{64} \, x^{2} - \frac {442709}{64} \, x - \frac {456533}{128} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

maple [A]  time = 0.00, size = 38, normalized size = 0.75 \[ -\frac {1125 x^{6}}{4}-\frac {5805 x^{5}}{4}-\frac {110205 x^{4}}{32}-\frac {247157 x^{3}}{48}-\frac {377045 x^{2}}{64}-\frac {442709 x}{64}-\frac {456533 \ln \left (2 x -1\right )}{128} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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(2*x-1)

________________________________________________________________________________________

maxima [A]  time = 0.50, size = 37, normalized size = 0.73 \[ -\frac {1125}{4} \, x^{6} - \frac {5805}{4} \, x^{5} - \frac {110205}{32} \, x^{4} - \frac {247157}{48} \, x^{3} - \frac {377045}{64} \, x^{2} - \frac {442709}{64} \, x - \frac {456533}{128} \, \log \left (2 \, x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 0.03, size = 35, normalized size = 0.69 \[ -\frac {442709\,x}{64}-\frac {456533\,\ln \left (x-\frac {1}{2}\right )}{128}-\frac {377045\,x^2}{64}-\frac {247157\,x^3}{48}-\frac {110205\,x^4}{32}-\frac {5805\,x^5}{4}-\frac {1125\,x^6}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

sympy [A]  time = 0.11, size = 49, normalized size = 0.96 \[ - \frac {1125 x^{6}}{4} - \frac {5805 x^{5}}{4} - \frac {110205 x^{4}}{32} - \frac {247157 x^{3}}{48} - \frac {377045 x^{2}}{64} - \frac {442709 x}{64} - \frac {456533 \log {\left (2 x - 1 \right )}}{128} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________