3.1477 \(\int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)^2} \, dx\)

Optimal. Leaf size=37 \[ -\frac {125 x}{18}+\frac {1}{189 (3 x+2)}-\frac {1331}{196} \log (1-2 x)+\frac {103 \log (3 x+2)}{1323} \]

[Out]

-125/18*x+1/189/(2+3*x)-1331/196*ln(1-2*x)+103/1323*ln(2+3*x)

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 37, 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} \[ -\frac {125 x}{18}+\frac {1}{189 (3 x+2)}-\frac {1331}{196} \log (1-2 x)+\frac {103 \log (3 x+2)}{1323} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-125*x)/18 + 1/(189*(2 + 3*x)) - (1331*Log[1 - 2*x])/196 + (103*Log[2 + 3*x])/1323

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 {(3+5 x)^3}{(1-2 x) (2+3 x)^2} \, dx &=\int \left (-\frac {125}{18}-\frac {1331}{98 (-1+2 x)}-\frac {1}{63 (2+3 x)^2}+\frac {103}{441 (2+3 x)}\right ) \, dx\\ &=-\frac {125 x}{18}+\frac {1}{189 (2+3 x)}-\frac {1331}{196} \log (1-2 x)+\frac {103 \log (2+3 x)}{1323}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 37, normalized size = 1.00 \[ \frac {18375 (1-2 x)+\frac {28}{3 x+2}-35937 \log (1-2 x)+412 \log (6 x+4)}{5292} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(18375*(1 - 2*x) + 28/(2 + 3*x) - 35937*Log[1 - 2*x] + 412*Log[4 + 6*x])/5292

________________________________________________________________________________________

fricas [A]  time = 0.83, size = 45, normalized size = 1.22 \[ -\frac {110250 \, x^{2} - 412 \, {\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) + 35937 \, {\left (3 \, x + 2\right )} \log \left (2 \, x - 1\right ) + 73500 \, x - 28}{5292 \, {\left (3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5292*(110250*x^2 - 412*(3*x + 2)*log(3*x + 2) + 35937*(3*x + 2)*log(2*x - 1) + 73500*x - 28)/(3*x + 2)

________________________________________________________________________________________

giac [A]  time = 1.04, size = 47, normalized size = 1.27 \[ -\frac {125}{18} \, x + \frac {1}{189 \, {\left (3 \, x + 2\right )}} + \frac {725}{108} \, \log \left (\frac {{\left | 3 \, x + 2 \right |}}{3 \, {\left (3 \, x + 2\right )}^{2}}\right ) - \frac {1331}{196} \, \log \left ({\left | -\frac {7}{3 \, x + 2} + 2 \right |}\right ) - \frac {125}{27} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125/18*x + 1/189/(3*x + 2) + 725/108*log(1/3*abs(3*x + 2)/(3*x + 2)^2) - 1331/196*log(abs(-7/(3*x + 2) + 2))
- 125/27

________________________________________________________________________________________

maple [A]  time = 0.01, size = 30, normalized size = 0.81 \[ -\frac {125 x}{18}-\frac {1331 \ln \left (2 x -1\right )}{196}+\frac {103 \ln \left (3 x +2\right )}{1323}+\frac {1}{567 x +378} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125/18*x+1/189/(3*x+2)+103/1323*ln(3*x+2)-1331/196*ln(2*x-1)

________________________________________________________________________________________

maxima [A]  time = 0.59, size = 29, normalized size = 0.78 \[ -\frac {125}{18} \, x + \frac {1}{189 \, {\left (3 \, x + 2\right )}} + \frac {103}{1323} \, \log \left (3 \, x + 2\right ) - \frac {1331}{196} \, \log \left (2 \, x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125/18*x + 1/189/(3*x + 2) + 103/1323*log(3*x + 2) - 1331/196*log(2*x - 1)

________________________________________________________________________________________

mupad [B]  time = 1.11, size = 23, normalized size = 0.62 \[ \frac {103\,\ln \left (x+\frac {2}{3}\right )}{1323}-\frac {1331\,\ln \left (x-\frac {1}{2}\right )}{196}-\frac {125\,x}{18}+\frac {1}{567\,\left (x+\frac {2}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(103*log(x + 2/3))/1323 - (1331*log(x - 1/2))/196 - (125*x)/18 + 1/(567*(x + 2/3))

________________________________________________________________________________________

sympy [A]  time = 0.15, size = 31, normalized size = 0.84 \[ - \frac {125 x}{18} - \frac {1331 \log {\left (x - \frac {1}{2} \right )}}{196} + \frac {103 \log {\left (x + \frac {2}{3} \right )}}{1323} + \frac {1}{567 x + 378} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125*x/18 - 1331*log(x - 1/2)/196 + 103*log(x + 2/3)/1323 + 1/(567*x + 378)

________________________________________________________________________________________