\(\int \frac {(3+5 x)^3}{(1-2 x)^2 (2+3 x)} \, dx\) [1580]

Optimal result

Integrand size = 22, antiderivative size = 37 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2 (2+3 x)} \, dx=\frac {1331}{56 (1-2 x)}+\frac {125 x}{12}+\frac {1089}{49} \log (1-2 x)-\frac {1}{441} \log (2+3 x) \]

[Out]

1331/56/(1-2*x)+125/12*x+1089/49*ln(1-2*x)-1/441*ln(2+3*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, 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} \[ \int \frac {(3+5 x)^3}{(1-2 x)^2 (2+3 x)} \, dx=\frac {125 x}{12}+\frac {1331}{56 (1-2 x)}+\frac {1089}{49} \log (1-2 x)-\frac {1}{441} \log (3 x+2) \]

[In]

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

[Out]

1331/(56*(1 - 2*x)) + (125*x)/12 + (1089*Log[1 - 2*x])/49 - Log[2 + 3*x]/441

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 {125}{12}+\frac {1331}{28 (-1+2 x)^2}+\frac {2178}{49 (-1+2 x)}-\frac {1}{147 (2+3 x)}\right ) \, dx \\ & = \frac {1331}{56 (1-2 x)}+\frac {125 x}{12}+\frac {1089}{49} \log (1-2 x)-\frac {1}{441} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2 (2+3 x)} \, dx=\frac {\frac {83853}{1-2 x}+12250 (2+3 x)+78408 \log (3-6 x)-8 \log (2+3 x)}{3528} \]

[In]

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

[Out]

(83853/(1 - 2*x) + 12250*(2 + 3*x) + 78408*Log[3 - 6*x] - 8*Log[2 + 3*x])/3528

Maple [A] (verified)

Time = 2.66 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76

method	result	size
risch	\(\frac {125 x}{12}-\frac {1331}{112 \left (x -\frac {1}{2}\right )}+\frac {1089 \ln \left (-1+2 x \right )}{49}-\frac {\ln \left (2+3 x \right )}{441}\)	\(28\)
default	\(\frac {125 x}{12}-\frac {1331}{56 \left (-1+2 x \right )}+\frac {1089 \ln \left (-1+2 x \right )}{49}-\frac {\ln \left (2+3 x \right )}{441}\)	\(30\)
norman	\(\frac {-\frac {1217}{21} x +\frac {125}{6} x^{2}}{-1+2 x}+\frac {1089 \ln \left (-1+2 x \right )}{49}-\frac {\ln \left (2+3 x \right )}{441}\)	\(35\)
parallelrisch	\(-\frac {4 \ln \left (\frac {2}{3}+x \right ) x -39204 \ln \left (x -\frac {1}{2}\right ) x -18375 x^{2}-2 \ln \left (\frac {2}{3}+x \right )+19602 \ln \left (x -\frac {1}{2}\right )+51114 x}{882 \left (-1+2 x \right )}\)	\(45\)

[In]

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

[Out]

125/12*x-1331/112/(x-1/2)+1089/49*ln(-1+2*x)-1/441*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.22 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2 (2+3 x)} \, dx=\frac {73500 \, x^{2} - 8 \, {\left (2 \, x - 1\right )} \log \left (3 \, x + 2\right ) + 78408 \, {\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 36750 \, x - 83853}{3528 \, {\left (2 \, x - 1\right )}} \]

[In]

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

[Out]

1/3528*(73500*x^2 - 8*(2*x - 1)*log(3*x + 2) + 78408*(2*x - 1)*log(2*x - 1) - 36750*x - 83853)/(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2 (2+3 x)} \, dx=\frac {125 x}{12} + \frac {1089 \log {\left (x - \frac {1}{2} \right )}}{49} - \frac {\log {\left (x + \frac {2}{3} \right )}}{441} - \frac {1331}{112 x - 56} \]

[In]

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

[Out]

125*x/12 + 1089*log(x - 1/2)/49 - log(x + 2/3)/441 - 1331/(112*x - 56)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2 (2+3 x)} \, dx=\frac {125}{12} \, x - \frac {1331}{56 \, {\left (2 \, x - 1\right )}} - \frac {1}{441} \, \log \left (3 \, x + 2\right ) + \frac {1089}{49} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

125/12*x - 1331/56/(2*x - 1) - 1/441*log(3*x + 2) + 1089/49*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.27 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2 (2+3 x)} \, dx=\frac {125}{12} \, x - \frac {1331}{56 \, {\left (2 \, x - 1\right )}} - \frac {200}{9} \, \log \left (\frac {{\left | 2 \, x - 1 \right |}}{2 \, {\left (2 \, x - 1\right )}^{2}}\right ) - \frac {1}{441} \, \log \left ({\left | -\frac {7}{2 \, x - 1} - 3 \right |}\right ) - \frac {125}{24} \]

[In]

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

[Out]

125/12*x - 1331/56/(2*x - 1) - 200/9*log(1/2*abs(2*x - 1)/(2*x - 1)^2) - 1/441*log(abs(-7/(2*x - 1) - 3)) - 12
5/24

Mupad [B] (verification not implemented)

Time = 1.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.68 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2 (2+3 x)} \, dx=\frac {125\,x}{12}+\frac {1089\,\ln \left (x-\frac {1}{2}\right )}{49}-\frac {\ln \left (x+\frac {2}{3}\right )}{441}-\frac {1331}{112\,\left (x-\frac {1}{2}\right )} \]

[In]

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

[Out]

(125*x)/12 + (1089*log(x - 1/2))/49 - log(x + 2/3)/441 - 1331/(112*(x - 1/2))