\(\int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)^3} \, dx\) [1426]

Optimal result

Integrand size = 22, antiderivative size = 43 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)^3} \, dx=-\frac {1331}{250 (3+5 x)^2}+\frac {4719}{125 (3+5 x)}-\frac {343}{3} \log (2+3 x)+\frac {14289}{125} \log (3+5 x) \]

[Out]

-1331/250/(3+5*x)^2+4719/125/(3+5*x)-343/3*ln(2+3*x)+14289/125*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, 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} \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)^3} \, dx=\frac {4719}{125 (5 x+3)}-\frac {1331}{250 (5 x+3)^2}-\frac {343}{3} \log (3 x+2)+\frac {14289}{125} \log (5 x+3) \]

[In]

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

[Out]

-1331/(250*(3 + 5*x)^2) + 4719/(125*(3 + 5*x)) - (343*Log[2 + 3*x])/3 + (14289*Log[3 + 5*x])/125

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 {343}{2+3 x}+\frac {1331}{25 (3+5 x)^3}-\frac {4719}{25 (3+5 x)^2}+\frac {14289}{25 (3+5 x)}\right ) \, dx \\ & = -\frac {1331}{250 (3+5 x)^2}+\frac {4719}{125 (3+5 x)}-\frac {343}{3} \log (2+3 x)+\frac {14289}{125} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.02 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)^3} \, dx=-\frac {343}{3} \log (2+3 x)+\frac {11 \left (2453+4290 x+2598 (3+5 x)^2 \log (-3 (3+5 x))\right )}{250 (3+5 x)^2} \]

[In]

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

[Out]

(-343*Log[2 + 3*x])/3 + (11*(2453 + 4290*x + 2598*(3 + 5*x)^2*Log[-3*(3 + 5*x)]))/(250*(3 + 5*x)^2)

Maple [A] (verified)

Time = 2.48 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {\frac {4719 x}{25}+\frac {26983}{250}}{\left (3+5 x \right )^{2}}-\frac {343 \ln \left (2+3 x \right )}{3}+\frac {14289 \ln \left (3+5 x \right )}{125}\)	\(32\)
norman	\(\frac {-\frac {12826}{75} x -\frac {26983}{90} x^{2}}{\left (3+5 x \right )^{2}}-\frac {343 \ln \left (2+3 x \right )}{3}+\frac {14289 \ln \left (3+5 x \right )}{125}\)	\(35\)
default	\(-\frac {1331}{250 \left (3+5 x \right )^{2}}+\frac {4719}{125 \left (3+5 x \right )}-\frac {343 \ln \left (2+3 x \right )}{3}+\frac {14289 \ln \left (3+5 x \right )}{125}\)	\(36\)
parallelrisch	\(-\frac {6431250 \ln \left (\frac {2}{3}+x \right ) x^{2}-6430050 \ln \left (x +\frac {3}{5}\right ) x^{2}+7717500 \ln \left (\frac {2}{3}+x \right ) x -7716060 \ln \left (x +\frac {3}{5}\right ) x +674575 x^{2}+2315250 \ln \left (\frac {2}{3}+x \right )-2314818 \ln \left (x +\frac {3}{5}\right )+384780 x}{2250 \left (3+5 x \right )^{2}}\)	\(63\)

[In]

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

[Out]

25*(4719/625*x+26983/6250)/(3+5*x)^2-343/3*ln(2+3*x)+14289/125*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)^3} \, dx=\frac {85734 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (5 \, x + 3\right ) - 85750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (3 \, x + 2\right ) + 141570 \, x + 80949}{750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

[In]

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

[Out]

1/750*(85734*(25*x^2 + 30*x + 9)*log(5*x + 3) - 85750*(25*x^2 + 30*x + 9)*log(3*x + 2) + 141570*x + 80949)/(25
*x^2 + 30*x + 9)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)^3} \, dx=- \frac {- 47190 x - 26983}{6250 x^{2} + 7500 x + 2250} + \frac {14289 \log {\left (x + \frac {3}{5} \right )}}{125} - \frac {343 \log {\left (x + \frac {2}{3} \right )}}{3} \]

[In]

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

[Out]

-(-47190*x - 26983)/(6250*x**2 + 7500*x + 2250) + 14289*log(x + 3/5)/125 - 343*log(x + 2/3)/3

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)^3} \, dx=\frac {121 \, {\left (390 \, x + 223\right )}}{250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac {14289}{125} \, \log \left (5 \, x + 3\right ) - \frac {343}{3} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

121/250*(390*x + 223)/(25*x^2 + 30*x + 9) + 14289/125*log(5*x + 3) - 343/3*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.77 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)^3} \, dx=\frac {121 \, {\left (390 \, x + 223\right )}}{250 \, {\left (5 \, x + 3\right )}^{2}} + \frac {14289}{125} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {343}{3} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

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

[Out]

121/250*(390*x + 223)/(5*x + 3)^2 + 14289/125*log(abs(5*x + 3)) - 343/3*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.67 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)^3} \, dx=\frac {14289\,\ln \left (x+\frac {3}{5}\right )}{125}-\frac {343\,\ln \left (x+\frac {2}{3}\right )}{3}+\frac {\frac {4719\,x}{625}+\frac {26983}{6250}}{x^2+\frac {6\,x}{5}+\frac {9}{25}} \]

[In]

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

[Out]

(14289*log(x + 3/5))/125 - (343*log(x + 2/3))/3 + ((4719*x)/625 + 26983/6250)/((6*x)/5 + x^2 + 9/25)