3.16.90 \(\int \frac {(2+3 x)^6}{(1-2 x)^2 (3+5 x)} \, dx\) [1590]

3.16.90.1 Optimal result

Integrand size = 22, antiderivative size = 58 \[ \int \frac {(2+3 x)^6}{(1-2 x)^2 (3+5 x)} \, dx=\frac {117649}{704 (1-2 x)}+\frac {3946293 x}{10000}+\frac {639819 x^2}{4000}+\frac {2673 x^3}{50}+\frac {729 x^4}{80}+\frac {2739541 \log (1-2 x)}{7744}+\frac {\log (3+5 x)}{378125} \]

output

117649/704/(1-2*x)+3946293/10000*x+639819/4000*x^2+2673/50*x^3+729/80*x^4+ 
2739541/7744*ln(1-2*x)+1/378125*ln(3+5*x)
 

3.16.90.2 Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.95 \[ \int \frac {(2+3 x)^6}{(1-2 x)^2 (3+5 x)} \, dx=\frac {\frac {55 \left (-156937135-6823872 x+276893694 x^2+117237780 x^3+43035300 x^4+8019000 x^5\right )}{-1+2 x}[phone] \log (5-10 x)+64 \log (3+5 x)}{24200000} \]

input

Integrate[(2 + 3*x)^6/((1 - 2*x)^2*(3 + 5*x)),x]
 

output

((55*(-156937135 - 6823872*x + 276893694*x^2 + 117237780*x^3 + 43035300*x^ 
4 + 8019000*x^5))/(-1 + 2*x) + 8561065625*Log[5 - 10*x] + 64*Log[3 + 5*x]) 
/24200000
 

3.16.90.3 Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {99, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(3 x+2)^6}{(1-2 x)^2 (5 x+3)} \, dx\)

\(\Big \downarrow \) 99

\(\displaystyle \int \left (\frac {729 x^3}{20}+\frac {8019 x^2}{50}+\frac {639819 x}{2000}+\frac {2739541}{3872 (2 x-1)}+\frac {1}{75625 (5 x+3)}+\frac {117649}{352 (2 x-1)^2}+\frac {3946293}{10000}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {729 x^4}{80}+\frac {2673 x^3}{50}+\frac {639819 x^2}{4000}+\frac {3946293 x}{10000}+\frac {117649}{704 (1-2 x)}+\frac {2739541 \log (1-2 x)}{7744}+\frac {\log (5 x+3)}{378125}\)

input

Int[(2 + 3*x)^6/((1 - 2*x)^2*(3 + 5*x)),x]
 

output

117649/(704*(1 - 2*x)) + (3946293*x)/10000 + (639819*x^2)/4000 + (2673*x^3 
)/50 + (729*x^4)/80 + (2739541*Log[1 - 2*x])/7744 + Log[3 + 5*x]/378125
 

3.16.90.3.1 Defintions of rubi rules used

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(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]))
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

3.16.90.4 Maple [A] (verified)

Time = 2.91 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {729 x^{4}}{80}+\frac {2673 x^{3}}{50}+\frac {639819 x^{2}}{4000}+\frac {3946293 x}{10000}-\frac {117649}{1408 \left (x -\frac {1}{2}\right )}+\frac {2739541 \ln \left (-1+2 x \right )}{7744}+\frac {\ln \left (3+5 x \right )}{378125}\)	\(43\)
default	\(\frac {729 x^{4}}{80}+\frac {2673 x^{3}}{50}+\frac {639819 x^{2}}{4000}+\frac {3946293 x}{10000}+\frac {\ln \left (3+5 x \right )}{378125}-\frac {117649}{704 \left (-1+2 x \right )}+\frac {2739541 \ln \left (-1+2 x \right )}{7744}\)	\(45\)
norman	\(\frac {-\frac {160349071}{220000} x +\frac {12586077}{20000} x^{2}+\frac {532899}{2000} x^{3}+\frac {39123}{400} x^{4}+\frac {729}{40} x^{5}}{-1+2 x}+\frac {2739541 \ln \left (-1+2 x \right )}{7744}+\frac {\ln \left (3+5 x \right )}{378125}\)	\(50\)
parallelrisch	\(\frac {441045000 x^{5}[phone] x^{4}[phone] x^{3}+128 \ln \left (x +\frac {3}{5}\right ) x [phone] \ln \left (x -\frac {1}{2}\right ) x [phone] x^{2}-64 \ln \left (x +\frac {3}{5}\right )-8561065625 \ln \left (x -\frac {1}{2}\right )-17638397810 x}{-24200000+48400000 x}\)	\(60\)

input

int((2+3*x)^6/(1-2*x)^2/(3+5*x),x,method=_RETURNVERBOSE)
 

output

729/80*x^4+2673/50*x^3+639819/4000*x^2+3946293/10000*x-117649/1408/(x-1/2) 
+2739541/7744*ln(-1+2*x)+1/378125*ln(3+5*x)
 

3.16.90.5 Fricas [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.03 \[ \int \frac {(2+3 x)^6}{(1-2 x)^2 (3+5 x)} \, dx=\frac {441045000 \, x^{5} + 2366941500 \, x^{4} + 6448077900 \, x^{3} + 15229153170 \, x^{2} + 64 \, {\left (2 \, x - 1\right )} \log \left (5 \, x + 3\right ) + 8561065625 \, {\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 9550029060 \, x - 4044184375}{24200000 \, {\left (2 \, x - 1\right )}} \]

input

integrate((2+3*x)^6/(1-2*x)^2/(3+5*x),x, algorithm="fricas")
 

output

1/24200000*(441045000*x^5 + 2366941500*x^4 + 6448077900*x^3 + 15229153170* 
x^2 + 64*(2*x - 1)*log(5*x + 3) + 8561065625*(2*x - 1)*log(2*x - 1) - 9550 
029060*x - 4044184375)/(2*x - 1)
 

3.16.90.6 Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.84 \[ \int \frac {(2+3 x)^6}{(1-2 x)^2 (3+5 x)} \, dx=\frac {729 x^{4}}{80} + \frac {2673 x^{3}}{50} + \frac {639819 x^{2}}{4000} + \frac {3946293 x}{10000} + \frac {2739541 \log {\left (x - \frac {1}{2} \right )}}{7744} + \frac {\log {\left (x + \frac {3}{5} \right )}}{378125} - \frac {117649}{1408 x - 704} \]

input

integrate((2+3*x)**6/(1-2*x)**2/(3+5*x),x)
 

output

729*x**4/80 + 2673*x**3/50 + 639819*x**2/4000 + 3946293*x/10000 + 2739541* 
log(x - 1/2)/7744 + log(x + 3/5)/378125 - 117649/(1408*x - 704)
 

3.16.90.7 Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.76 \[ \int \frac {(2+3 x)^6}{(1-2 x)^2 (3+5 x)} \, dx=\frac {729}{80} \, x^{4} + \frac {2673}{50} \, x^{3} + \frac {639819}{4000} \, x^{2} + \frac {3946293}{10000} \, x - \frac {117649}{704 \, {\left (2 \, x - 1\right )}} + \frac {1}{378125} \, \log \left (5 \, x + 3\right ) + \frac {2739541}{7744} \, \log \left (2 \, x - 1\right ) \]

input

integrate((2+3*x)^6/(1-2*x)^2/(3+5*x),x, algorithm="maxima")
 

output

729/80*x^4 + 2673/50*x^3 + 639819/4000*x^2 + 3946293/10000*x - 117649/704/ 
(2*x - 1) + 1/378125*log(5*x + 3) + 2739541/7744*log(2*x - 1)
 

3.16.90.8 Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.40 \[ \int \frac {(2+3 x)^6}{(1-2 x)^2 (3+5 x)} \, dx=\frac {27}{160000} \, {\left (2 \, x - 1\right )}^{4} {\left (\frac {53100}{2 \, x - 1} + \frac {376020}{{\left (2 \, x - 1\right )}^{2}} + \frac {1775512}{{\left (2 \, x - 1\right )}^{3}} + 3375\right )} - \frac {117649}{704 \, {\left (2 \, x - 1\right )}} - \frac {70752609}{200000} \, \log \left (\frac {{\left | 2 \, x - 1 \right |}}{2 \, {\left (2 \, x - 1\right )}^{2}}\right ) + \frac {1}{378125} \, \log \left ({\left | -\frac {11}{2 \, x - 1} - 5 \right |}\right ) \]

input

integrate((2+3*x)^6/(1-2*x)^2/(3+5*x),x, algorithm="giac")
 

output

27/160000*(2*x - 1)^4*(53100/(2*x - 1) + 376020/(2*x - 1)^2 + 1775512/(2*x 
 - 1)^3 + 3375) - 117649/704/(2*x - 1) - 70752609/200000*log(1/2*abs(2*x - 
 1)/(2*x - 1)^2) + 1/378125*log(abs(-11/(2*x - 1) - 5))
 

3.16.90.9 Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.69 \[ \int \frac {(2+3 x)^6}{(1-2 x)^2 (3+5 x)} \, dx=\frac {3946293\,x}{10000}+\frac {2739541\,\ln \left (x-\frac {1}{2}\right )}{7744}+\frac {\ln \left (x+\frac {3}{5}\right )}{378125}-\frac {117649}{1408\,\left (x-\frac {1}{2}\right )}+\frac {639819\,x^2}{4000}+\frac {2673\,x^3}{50}+\frac {729\,x^4}{80} \]

input

int((3*x + 2)^6/((2*x - 1)^2*(5*x + 3)),x)
 

output

(3946293*x)/10000 + (2739541*log(x - 1/2))/7744 + log(x + 3/5)/378125 - 11 
7649/(1408*(x - 1/2)) + (639819*x^2)/4000 + (2673*x^3)/50 + (729*x^4)/80