\(\int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)^2} \, dx\) [1477]

Optimal result

Integrand size = 22, antiderivative size = 37 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)^2} \, dx=-\frac {125 x}{18}+\frac {1}{189 (2+3 x)}-\frac {1331}{196} \log (1-2 x)+\frac {103 \log (2+3 x)}{1323} \]

[Out]

-125/18*x+1/189/(2+3*x)-1331/196*ln(1-2*x)+103/1323*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} \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)^2} \, dx=-\frac {125 x}{18}+\frac {1}{189 (3 x+2)}-\frac {1331}{196} \log (1-2 x)+\frac {103 \log (3 x+2)}{1323} \]

[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 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}{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] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)^2} \, dx=\frac {18375 (1-2 x)+\frac {28}{2+3 x}-35937 \log (1-2 x)+412 \log (4+6 x)}{5292} \]

[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

Maple [A] (verified)

Time = 2.54 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76

method	result	size
risch	\(-\frac {125 x}{18}+\frac {1}{378+567 x}-\frac {1331 \ln \left (-1+2 x \right )}{196}+\frac {103 \ln \left (2+3 x \right )}{1323}\)	\(28\)
default	\(-\frac {125 x}{18}+\frac {1}{378+567 x}-\frac {1331 \ln \left (-1+2 x \right )}{196}+\frac {103 \ln \left (2+3 x \right )}{1323}\)	\(30\)
norman	\(\frac {-\frac {1751}{126} x -\frac {125}{6} x^{2}}{2+3 x}-\frac {1331 \ln \left (-1+2 x \right )}{196}+\frac {103 \ln \left (2+3 x \right )}{1323}\)	\(35\)
parallelrisch	\(\frac {1236 \ln \left (\frac {2}{3}+x \right ) x -107811 \ln \left (x -\frac {1}{2}\right ) x -110250 x^{2}+824 \ln \left (\frac {2}{3}+x \right )-71874 \ln \left (x -\frac {1}{2}\right )-73542 x}{10584+15876 x}\)	\(45\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.22 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)^2} \, dx=-\frac {110250 \, x^{2} - 412 \, {\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) + 35937 \, {\left (3 \, x + 2\right )} \log \left (2 \, x - 1\right ) + 73500 \, x - 28}{5292 \, {\left (3 \, x + 2\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)^2} \, dx=- \frac {125 x}{18} - \frac {1331 \log {\left (x - \frac {1}{2} \right )}}{196} + \frac {103 \log {\left (x + \frac {2}{3} \right )}}{1323} + \frac {1}{567 x + 378} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)^2} \, dx=-\frac {125}{18} \, x + \frac {1}{189 \, {\left (3 \, x + 2\right )}} + \frac {103}{1323} \, \log \left (3 \, x + 2\right ) - \frac {1331}{196} \, \log \left (2 \, x - 1\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.27 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)^2} \, dx=-\frac {125}{18} \, x + \frac {1}{189 \, {\left (3 \, x + 2\right )}} + \frac {725}{108} \, \log \left (\frac {{\left | 3 \, x + 2 \right |}}{3 \, {\left (3 \, x + 2\right )}^{2}}\right ) - \frac {1331}{196} \, \log \left ({\left | -\frac {7}{3 \, x + 2} + 2 \right |}\right ) - \frac {125}{27} \]

[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

Mupad [B] (verification not implemented)

Time = 1.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.62 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)^2} \, dx=\frac {103\,\ln \left (x+\frac {2}{3}\right )}{1323}-\frac {1331\,\ln \left (x-\frac {1}{2}\right )}{196}-\frac {125\,x}{18}+\frac {1}{567\,\left (x+\frac {2}{3}\right )} \]

[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))