\(\int \frac {(2+3 x)^3}{(1-2 x) (3+5 x)} \, dx\) [1490]

Optimal result

Integrand size = 22, antiderivative size = 33 \[ \int \frac {(2+3 x)^3}{(1-2 x) (3+5 x)} \, dx=-\frac {513 x}{100}-\frac {27 x^2}{20}-\frac {343}{88} \log (1-2 x)+\frac {\log (3+5 x)}{1375} \]

[Out]

-513/100*x-27/20*x^2-343/88*ln(1-2*x)+1/1375*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, 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 = {84} \[ \int \frac {(2+3 x)^3}{(1-2 x) (3+5 x)} \, dx=-\frac {27 x^2}{20}-\frac {513 x}{100}-\frac {343}{88} \log (1-2 x)+\frac {\log (5 x+3)}{1375} \]

[In]

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

[Out]

(-513*x)/100 - (27*x^2)/20 - (343*Log[1 - 2*x])/88 + Log[3 + 5*x]/1375

Rule 84

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {513}{100}-\frac {27 x}{10}-\frac {343}{44 (-1+2 x)}+\frac {1}{275 (3+5 x)}\right ) \, dx \\ & = -\frac {513 x}{100}-\frac {27 x^2}{20}-\frac {343}{88} \log (1-2 x)+\frac {\log (3+5 x)}{1375} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {(2+3 x)^3}{(1-2 x) (3+5 x)} \, dx=\frac {-330 \left (94+171 x+45 x^2\right )-42875 \log (3-6 x)+8 \log (-3 (3+5 x))}{11000} \]

[In]

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

[Out]

(-330*(94 + 171*x + 45*x^2) - 42875*Log[3 - 6*x] + 8*Log[-3*(3 + 5*x)])/11000

Maple [A] (verified)

Time = 2.53 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(-\frac {27 x^{2}}{20}-\frac {513 x}{100}+\frac {\ln \left (x +\frac {3}{5}\right )}{1375}-\frac {343 \ln \left (x -\frac {1}{2}\right )}{88}\)	\(22\)
default	\(-\frac {27 x^{2}}{20}-\frac {513 x}{100}+\frac {\ln \left (3+5 x \right )}{1375}-\frac {343 \ln \left (-1+2 x \right )}{88}\)	\(26\)
norman	\(-\frac {27 x^{2}}{20}-\frac {513 x}{100}+\frac {\ln \left (3+5 x \right )}{1375}-\frac {343 \ln \left (-1+2 x \right )}{88}\)	\(26\)
risch	\(-\frac {27 x^{2}}{20}-\frac {513 x}{100}+\frac {\ln \left (3+5 x \right )}{1375}-\frac {343 \ln \left (-1+2 x \right )}{88}\)	\(26\)

[In]

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

[Out]

-27/20*x^2-513/100*x+1/1375*ln(x+3/5)-343/88*ln(x-1/2)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {(2+3 x)^3}{(1-2 x) (3+5 x)} \, dx=-\frac {27}{20} \, x^{2} - \frac {513}{100} \, x + \frac {1}{1375} \, \log \left (5 \, x + 3\right ) - \frac {343}{88} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-27/20*x^2 - 513/100*x + 1/1375*log(5*x + 3) - 343/88*log(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {(2+3 x)^3}{(1-2 x) (3+5 x)} \, dx=- \frac {27 x^{2}}{20} - \frac {513 x}{100} - \frac {343 \log {\left (x - \frac {1}{2} \right )}}{88} + \frac {\log {\left (x + \frac {3}{5} \right )}}{1375} \]

[In]

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

[Out]

-27*x**2/20 - 513*x/100 - 343*log(x - 1/2)/88 + log(x + 3/5)/1375

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {(2+3 x)^3}{(1-2 x) (3+5 x)} \, dx=-\frac {27}{20} \, x^{2} - \frac {513}{100} \, x + \frac {1}{1375} \, \log \left (5 \, x + 3\right ) - \frac {343}{88} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-27/20*x^2 - 513/100*x + 1/1375*log(5*x + 3) - 343/88*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {(2+3 x)^3}{(1-2 x) (3+5 x)} \, dx=-\frac {27}{20} \, x^{2} - \frac {513}{100} \, x + \frac {1}{1375} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {343}{88} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

-27/20*x^2 - 513/100*x + 1/1375*log(abs(5*x + 3)) - 343/88*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 1.38 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.64 \[ \int \frac {(2+3 x)^3}{(1-2 x) (3+5 x)} \, dx=\frac {\ln \left (x+\frac {3}{5}\right )}{1375}-\frac {343\,\ln \left (x-\frac {1}{2}\right )}{88}-\frac {513\,x}{100}-\frac {27\,x^2}{20} \]

[In]

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

[Out]

log(x + 3/5)/1375 - (343*log(x - 1/2))/88 - (513*x)/100 - (27*x^2)/20