\(\int \frac {(2+3 x)^4 (3+5 x)^2}{1-2 x} \, dx\) [1455]

Optimal result

Integrand size = 22, antiderivative size = 51 \[ \int \frac {(2+3 x)^4 (3+5 x)^2}{1-2 x} \, dx=-\frac {281305 x}{64}-\frac {238297 x^2}{64}-\frac {51571 x^3}{16}-\frac {68121 x^4}{32}-\frac {3537 x^5}{4}-\frac {675 x^6}{4}-\frac {290521}{128} \log (1-2 x) \]

[Out]

-281305/64*x-238297/64*x^2-51571/16*x^3-68121/32*x^4-3537/4*x^5-675/4*x^6-290521/128*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} \[ \int \frac {(2+3 x)^4 (3+5 x)^2}{1-2 x} \, dx=-\frac {675 x^6}{4}-\frac {3537 x^5}{4}-\frac {68121 x^4}{32}-\frac {51571 x^3}{16}-\frac {238297 x^2}{64}-\frac {281305 x}{64}-\frac {290521}{128} \log (1-2 x) \]

[In]

Int[((2 + 3*x)^4*(3 + 5*x)^2)/(1 - 2*x),x]

[Out]

(-281305*x)/64 - (238297*x^2)/64 - (51571*x^3)/16 - (68121*x^4)/32 - (3537*x^5)/4 - (675*x^6)/4 - (290521*Log[
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 {281305}{64}-\frac {238297 x}{32}-\frac {154713 x^2}{16}-\frac {68121 x^3}{8}-\frac {17685 x^4}{4}-\frac {2025 x^5}{2}-\frac {290521}{64 (-1+2 x)}\right ) \, dx \\ & = -\frac {281305 x}{64}-\frac {238297 x^2}{64}-\frac {51571 x^3}{16}-\frac {68121 x^4}{32}-\frac {3537 x^5}{4}-\frac {675 x^6}{4}-\frac {290521}{128} \log (1-2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.82 \[ \int \frac {(2+3 x)^4 (3+5 x)^2}{1-2 x} \, dx=\frac {1}{512} \left (1891717-2250440 x-1906376 x^2-1650272 x^3-1089936 x^4-452736 x^5-86400 x^6-1162084 \log (1-2 x)\right ) \]

[In]

Integrate[((2 + 3*x)^4*(3 + 5*x)^2)/(1 - 2*x),x]

[Out]

(1891717 - 2250440*x - 1906376*x^2 - 1650272*x^3 - 1089936*x^4 - 452736*x^5 - 86400*x^6 - 1162084*Log[1 - 2*x]
)/512

Maple [A] (verified)

Time = 2.52 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.71

method	result	size
parallelrisch	\(-\frac {675 x^{6}}{4}-\frac {3537 x^{5}}{4}-\frac {68121 x^{4}}{32}-\frac {51571 x^{3}}{16}-\frac {238297 x^{2}}{64}-\frac {281305 x}{64}-\frac {290521 \ln \left (x -\frac {1}{2}\right )}{128}\)	\(36\)
default	\(-\frac {675 x^{6}}{4}-\frac {3537 x^{5}}{4}-\frac {68121 x^{4}}{32}-\frac {51571 x^{3}}{16}-\frac {238297 x^{2}}{64}-\frac {281305 x}{64}-\frac {290521 \ln \left (-1+2 x \right )}{128}\)	\(38\)
norman	\(-\frac {675 x^{6}}{4}-\frac {3537 x^{5}}{4}-\frac {68121 x^{4}}{32}-\frac {51571 x^{3}}{16}-\frac {238297 x^{2}}{64}-\frac {281305 x}{64}-\frac {290521 \ln \left (-1+2 x \right )}{128}\)	\(38\)
risch	\(-\frac {675 x^{6}}{4}-\frac {3537 x^{5}}{4}-\frac {68121 x^{4}}{32}-\frac {51571 x^{3}}{16}-\frac {238297 x^{2}}{64}-\frac {281305 x}{64}-\frac {290521 \ln \left (-1+2 x \right )}{128}\)	\(38\)
meijerg	\(-\frac {290521 \ln \left (1-2 x \right )}{128}-672 x -\frac {653 x \left (6 x +6\right )}{3}-\frac {451 x \left (16 x^{2}+12 x +12\right )}{4}-\frac {4203 x \left (120 x^{3}+80 x^{2}+60 x +60\right )}{320}-\frac {261 x \left (192 x^{4}+120 x^{3}+80 x^{2}+60 x +60\right )}{64}-\frac {135 x \left (2240 x^{5}+1344 x^{4}+840 x^{3}+560 x^{2}+420 x +420\right )}{1792}\)	\(103\)

[In]

int((2+3*x)^4*(3+5*x)^2/(1-2*x),x,method=_RETURNVERBOSE)

[Out]

-675/4*x^6-3537/4*x^5-68121/32*x^4-51571/16*x^3-238297/64*x^2-281305/64*x-290521/128*ln(x-1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.73 \[ \int \frac {(2+3 x)^4 (3+5 x)^2}{1-2 x} \, dx=-\frac {675}{4} \, x^{6} - \frac {3537}{4} \, x^{5} - \frac {68121}{32} \, x^{4} - \frac {51571}{16} \, x^{3} - \frac {238297}{64} \, x^{2} - \frac {281305}{64} \, x - \frac {290521}{128} \, \log \left (2 \, x - 1\right ) \]

[In]

integrate((2+3*x)^4*(3+5*x)^2/(1-2*x),x, algorithm="fricas")

[Out]

-675/4*x^6 - 3537/4*x^5 - 68121/32*x^4 - 51571/16*x^3 - 238297/64*x^2 - 281305/64*x - 290521/128*log(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96 \[ \int \frac {(2+3 x)^4 (3+5 x)^2}{1-2 x} \, dx=- \frac {675 x^{6}}{4} - \frac {3537 x^{5}}{4} - \frac {68121 x^{4}}{32} - \frac {51571 x^{3}}{16} - \frac {238297 x^{2}}{64} - \frac {281305 x}{64} - \frac {290521 \log {\left (2 x - 1 \right )}}{128} \]

[In]

integrate((2+3*x)**4*(3+5*x)**2/(1-2*x),x)

[Out]

-675*x**6/4 - 3537*x**5/4 - 68121*x**4/32 - 51571*x**3/16 - 238297*x**2/64 - 281305*x/64 - 290521*log(2*x - 1)
/128

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.73 \[ \int \frac {(2+3 x)^4 (3+5 x)^2}{1-2 x} \, dx=-\frac {675}{4} \, x^{6} - \frac {3537}{4} \, x^{5} - \frac {68121}{32} \, x^{4} - \frac {51571}{16} \, x^{3} - \frac {238297}{64} \, x^{2} - \frac {281305}{64} \, x - \frac {290521}{128} \, \log \left (2 \, x - 1\right ) \]

[In]

integrate((2+3*x)^4*(3+5*x)^2/(1-2*x),x, algorithm="maxima")

[Out]

-675/4*x^6 - 3537/4*x^5 - 68121/32*x^4 - 51571/16*x^3 - 238297/64*x^2 - 281305/64*x - 290521/128*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.75 \[ \int \frac {(2+3 x)^4 (3+5 x)^2}{1-2 x} \, dx=-\frac {675}{4} \, x^{6} - \frac {3537}{4} \, x^{5} - \frac {68121}{32} \, x^{4} - \frac {51571}{16} \, x^{3} - \frac {238297}{64} \, x^{2} - \frac {281305}{64} \, x - \frac {290521}{128} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

integrate((2+3*x)^4*(3+5*x)^2/(1-2*x),x, algorithm="giac")

[Out]

-675/4*x^6 - 3537/4*x^5 - 68121/32*x^4 - 51571/16*x^3 - 238297/64*x^2 - 281305/64*x - 290521/128*log(abs(2*x -
1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int \frac {(2+3 x)^4 (3+5 x)^2}{1-2 x} \, dx=-\frac {281305\,x}{64}-\frac {290521\,\ln \left (x-\frac {1}{2}\right )}{128}-\frac {238297\,x^2}{64}-\frac {51571\,x^3}{16}-\frac {68121\,x^4}{32}-\frac {3537\,x^5}{4}-\frac {675\,x^6}{4} \]

[In]

int(-((3*x + 2)^4*(5*x + 3)^2)/(2*x - 1),x)

[Out]

- (281305*x)/64 - (290521*log(x - 1/2))/128 - (238297*x^2)/64 - (51571*x^3)/16 - (68121*x^4)/32 - (3537*x^5)/4
- (675*x^6)/4