\(\int \frac {1}{(1-2 x)^2 (2+3 x)^2 (3+5 x)} \, dx\) [1598]

Optimal result

Integrand size = 22, antiderivative size = 53 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^2 (3+5 x)} \, dx=\frac {4}{539 (1-2 x)}+\frac {9}{49 (2+3 x)}-\frac {404 \log (1-2 x)}{41503}-\frac {351}{343} \log (2+3 x)+\frac {125}{121} \log (3+5 x) \]

[Out]

4/539/(1-2*x)+9/49/(2+3*x)-404/41503*ln(1-2*x)-351/343*ln(2+3*x)+125/121*ln(3+5*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 53, 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} \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^2 (3+5 x)} \, dx=\frac {4}{539 (1-2 x)}+\frac {9}{49 (3 x+2)}-\frac {404 \log (1-2 x)}{41503}-\frac {351}{343} \log (3 x+2)+\frac {125}{121} \log (5 x+3) \]

[In]

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

[Out]

4/(539*(1 - 2*x)) + 9/(49*(2 + 3*x)) - (404*Log[1 - 2*x])/41503 - (351*Log[2 + 3*x])/343 + (125*Log[3 + 5*x])/
121

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 {8}{539 (-1+2 x)^2}-\frac {808}{41503 (-1+2 x)}-\frac {27}{49 (2+3 x)^2}-\frac {1053}{343 (2+3 x)}+\frac {625}{121 (3+5 x)}\right ) \, dx \\ & = \frac {4}{539 (1-2 x)}+\frac {9}{49 (2+3 x)}-\frac {404 \log (1-2 x)}{41503}-\frac {351}{343} \log (2+3 x)+\frac {125}{121} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^2 (3+5 x)} \, dx=\frac {-\frac {8239}{-2+x+6 x^2}+\frac {14322 x}{-2+x+6 x^2}-404 \log (5-10 x)-42471 \log (5 (2+3 x))+42875 \log (3+5 x)}{41503} \]

[In]

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

[Out]

(-8239/(-2 + x + 6*x^2) + (14322*x)/(-2 + x + 6*x^2) - 404*Log[5 - 10*x] - 42471*Log[5*(2 + 3*x)] + 42875*Log[
3 + 5*x])/41503

Maple [A] (verified)

Time = 2.67 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {125 \ln \left (3+5 x \right )}{121}-\frac {4}{539 \left (-1+2 x \right )}-\frac {404 \ln \left (-1+2 x \right )}{41503}+\frac {9}{49 \left (2+3 x \right )}-\frac {351 \ln \left (2+3 x \right )}{343}\)	\(44\)
risch	\(\frac {\frac {186 x}{539}-\frac {107}{539}}{\left (-1+2 x \right ) \left (2+3 x \right )}-\frac {404 \ln \left (-1+2 x \right )}{41503}-\frac {351 \ln \left (2+3 x \right )}{343}+\frac {125 \ln \left (3+5 x \right )}{121}\)	\(47\)
norman	\(\frac {-\frac {1116 x^{2}}{539}+\frac {265}{539}}{\left (-1+2 x \right ) \left (2+3 x \right )}-\frac {404 \ln \left (-1+2 x \right )}{41503}-\frac {351 \ln \left (2+3 x \right )}{343}+\frac {125 \ln \left (3+5 x \right )}{121}\)	\(48\)
parallelrisch	\(-\frac {254826 \ln \left (\frac {2}{3}+x \right ) x^{2}-257250 \ln \left (x +\frac {3}{5}\right ) x^{2}+2424 \ln \left (x -\frac {1}{2}\right ) x^{2}-20405+42471 \ln \left (\frac {2}{3}+x \right ) x -42875 \ln \left (x +\frac {3}{5}\right ) x +404 \ln \left (x -\frac {1}{2}\right ) x +85932 x^{2}-84942 \ln \left (\frac {2}{3}+x \right )+85750 \ln \left (x +\frac {3}{5}\right )-808 \ln \left (x -\frac {1}{2}\right )}{41503 \left (-1+2 x \right ) \left (2+3 x \right )}\)	\(90\)

[In]

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

[Out]

125/121*ln(3+5*x)-4/539/(-1+2*x)-404/41503*ln(-1+2*x)+9/49/(2+3*x)-351/343*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.23 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^2 (3+5 x)} \, dx=\frac {42875 \, {\left (6 \, x^{2} + x - 2\right )} \log \left (5 \, x + 3\right ) - 42471 \, {\left (6 \, x^{2} + x - 2\right )} \log \left (3 \, x + 2\right ) - 404 \, {\left (6 \, x^{2} + x - 2\right )} \log \left (2 \, x - 1\right ) + 14322 \, x - 8239}{41503 \, {\left (6 \, x^{2} + x - 2\right )}} \]

[In]

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

[Out]

1/41503*(42875*(6*x^2 + x - 2)*log(5*x + 3) - 42471*(6*x^2 + x - 2)*log(3*x + 2) - 404*(6*x^2 + x - 2)*log(2*x
- 1) + 14322*x - 8239)/(6*x^2 + x - 2)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^2 (3+5 x)} \, dx=\frac {186 x - 107}{3234 x^{2} + 539 x - 1078} - \frac {404 \log {\left (x - \frac {1}{2} \right )}}{41503} + \frac {125 \log {\left (x + \frac {3}{5} \right )}}{121} - \frac {351 \log {\left (x + \frac {2}{3} \right )}}{343} \]

[In]

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

[Out]

(186*x - 107)/(3234*x**2 + 539*x - 1078) - 404*log(x - 1/2)/41503 + 125*log(x + 3/5)/121 - 351*log(x + 2/3)/34
3

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^2 (3+5 x)} \, dx=\frac {186 \, x - 107}{539 \, {\left (6 \, x^{2} + x - 2\right )}} + \frac {125}{121} \, \log \left (5 \, x + 3\right ) - \frac {351}{343} \, \log \left (3 \, x + 2\right ) - \frac {404}{41503} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

1/539*(186*x - 107)/(6*x^2 + x - 2) + 125/121*log(5*x + 3) - 351/343*log(3*x + 2) - 404/41503*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^2 (3+5 x)} \, dx=\frac {9}{49 \, {\left (3 \, x + 2\right )}} + \frac {24}{3773 \, {\left (\frac {7}{3 \, x + 2} - 2\right )}} + \frac {125}{121} \, \log \left ({\left | -\frac {1}{3 \, x + 2} + 5 \right |}\right ) - \frac {404}{41503} \, \log \left ({\left | -\frac {7}{3 \, x + 2} + 2 \right |}\right ) \]

[In]

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

[Out]

9/49/(3*x + 2) + 24/3773/(7/(3*x + 2) - 2) + 125/121*log(abs(-1/(3*x + 2) + 5)) - 404/41503*log(abs(-7/(3*x +
2) + 2))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.66 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^2 (3+5 x)} \, dx=\frac {125\,\ln \left (x+\frac {3}{5}\right )}{121}-\frac {351\,\ln \left (x+\frac {2}{3}\right )}{343}-\frac {404\,\ln \left (x-\frac {1}{2}\right )}{41503}+\frac {\frac {31\,x}{539}-\frac {107}{3234}}{x^2+\frac {x}{6}-\frac {1}{3}} \]

[In]

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

[Out]

(125*log(x + 3/5))/121 - (351*log(x + 2/3))/343 - (404*log(x - 1/2))/41503 + ((31*x)/539 - 107/3234)/(x/6 + x^
2 - 1/3)