\(\int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)} \, dx\) [1395]

Optimal result

Integrand size = 22, antiderivative size = 33 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)} \, dx=\frac {332 x}{225}-\frac {4 x^2}{15}-\frac {343}{27} \log (2+3 x)+\frac {1331}{125} \log (3+5 x) \]

[Out]

332/225*x-4/15*x^2-343/27*ln(2+3*x)+1331/125*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} \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)} \, dx=-\frac {4 x^2}{15}+\frac {332 x}{225}-\frac {343}{27} \log (3 x+2)+\frac {1331}{125} \log (5 x+3) \]

[In]

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

[Out]

(332*x)/225 - (4*x^2)/15 - (343*Log[2 + 3*x])/27 + (1331*Log[3 + 5*x])/125

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 {332}{225}-\frac {8 x}{15}-\frac {343}{9 (2+3 x)}+\frac {1331}{25 (3+5 x)}\right ) \, dx \\ & = \frac {332 x}{225}-\frac {4 x^2}{15}-\frac {343}{27} \log (2+3 x)+\frac {1331}{125} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)} \, dx=\frac {60 \left (62+83 x-15 x^2\right )-42875 \log (2+3 x)+35937 \log (-3 (3+5 x))}{3375} \]

[In]

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

[Out]

(60*(62 + 83*x - 15*x^2) - 42875*Log[2 + 3*x] + 35937*Log[-3*(3 + 5*x)])/3375

Maple [A] (verified)

Time = 2.43 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(-\frac {4 x^{2}}{15}+\frac {332 x}{225}-\frac {343 \ln \left (\frac {2}{3}+x \right )}{27}+\frac {1331 \ln \left (x +\frac {3}{5}\right )}{125}\)	\(22\)
default	\(\frac {332 x}{225}-\frac {4 x^{2}}{15}-\frac {343 \ln \left (2+3 x \right )}{27}+\frac {1331 \ln \left (3+5 x \right )}{125}\)	\(26\)
norman	\(\frac {332 x}{225}-\frac {4 x^{2}}{15}-\frac {343 \ln \left (2+3 x \right )}{27}+\frac {1331 \ln \left (3+5 x \right )}{125}\)	\(26\)
risch	\(\frac {332 x}{225}-\frac {4 x^{2}}{15}-\frac {343 \ln \left (2+3 x \right )}{27}+\frac {1331 \ln \left (3+5 x \right )}{125}\)	\(26\)

[In]

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

[Out]

-4/15*x^2+332/225*x-343/27*ln(2/3+x)+1331/125*ln(x+3/5)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)} \, dx=-\frac {4}{15} \, x^{2} + \frac {332}{225} \, x + \frac {1331}{125} \, \log \left (5 \, x + 3\right ) - \frac {343}{27} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

-4/15*x^2 + 332/225*x + 1331/125*log(5*x + 3) - 343/27*log(3*x + 2)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)} \, dx=- \frac {4 x^{2}}{15} + \frac {332 x}{225} + \frac {1331 \log {\left (x + \frac {3}{5} \right )}}{125} - \frac {343 \log {\left (x + \frac {2}{3} \right )}}{27} \]

[In]

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

[Out]

-4*x**2/15 + 332*x/225 + 1331*log(x + 3/5)/125 - 343*log(x + 2/3)/27

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)} \, dx=-\frac {4}{15} \, x^{2} + \frac {332}{225} \, x + \frac {1331}{125} \, \log \left (5 \, x + 3\right ) - \frac {343}{27} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

-4/15*x^2 + 332/225*x + 1331/125*log(5*x + 3) - 343/27*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)} \, dx=-\frac {4}{15} \, x^{2} + \frac {332}{225} \, x + \frac {1331}{125} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {343}{27} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

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

[Out]

-4/15*x^2 + 332/225*x + 1331/125*log(abs(5*x + 3)) - 343/27*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.64 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)} \, dx=\frac {332\,x}{225}-\frac {343\,\ln \left (x+\frac {2}{3}\right )}{27}+\frac {1331\,\ln \left (x+\frac {3}{5}\right )}{125}-\frac {4\,x^2}{15} \]

[In]

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

[Out]

(332*x)/225 - (343*log(x + 2/3))/27 + (1331*log(x + 3/5))/125 - (4*x^2)/15