3.1511 \(\int \frac {1}{(1-2 x) (2+3 x)^2 (3+5 x)^2} \, dx\)

Optimal. Leaf size=53 \[ -\frac {9}{7 (3 x+2)}-\frac {25}{11 (5 x+3)}-\frac {8 \log (1-2 x)}{5929}+\frac {648}{49} \log (3 x+2)-\frac {1600}{121} \log (5 x+3) \]

[Out]

-9/7/(2+3*x)-25/11/(3+5*x)-8/5929*ln(1-2*x)+648/49*ln(2+3*x)-1600/121*ln(3+5*x)

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Rubi [A]  time = 0.02, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {88} \[ -\frac {9}{7 (3 x+2)}-\frac {25}{11 (5 x+3)}-\frac {8 \log (1-2 x)}{5929}+\frac {648}{49} \log (3 x+2)-\frac {1600}{121} \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-9/(7*(2 + 3*x)) - 25/(11*(3 + 5*x)) - (8*Log[1 - 2*x])/5929 + (648*Log[2 + 3*x])/49 - (1600*Log[3 + 5*x])/121

Rule 88

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*} \int \frac {1}{(1-2 x) (2+3 x)^2 (3+5 x)^2} \, dx &=\int \left (-\frac {16}{5929 (-1+2 x)}+\frac {27}{7 (2+3 x)^2}+\frac {1944}{49 (2+3 x)}+\frac {125}{11 (3+5 x)^2}-\frac {8000}{121 (3+5 x)}\right ) \, dx\\ &=-\frac {9}{7 (2+3 x)}-\frac {25}{11 (3+5 x)}-\frac {8 \log (1-2 x)}{5929}+\frac {648}{49} \log (2+3 x)-\frac {1600}{121} \log (3+5 x)\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 47, normalized size = 0.89 \[ \frac {2 \left (-\frac {7623}{6 x+4}-\frac {13475}{10 x+6}-4 \log (1-2 x)+39204 \log (6 x+4)-39200 \log (10 x+6)\right )}{5929} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-7623/(4 + 6*x) - 13475/(6 + 10*x) - 4*Log[1 - 2*x] + 39204*Log[4 + 6*x] - 39200*Log[6 + 10*x]))/5929

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fricas [A]  time = 0.93, size = 73, normalized size = 1.38 \[ -\frac {78400 \, {\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (5 \, x + 3\right ) - 78408 \, {\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (3 \, x + 2\right ) + 8 \, {\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (2 \, x - 1\right ) + 78540 \, x + 49819}{5929 \, {\left (15 \, x^{2} + 19 \, x + 6\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5929*(78400*(15*x^2 + 19*x + 6)*log(5*x + 3) - 78408*(15*x^2 + 19*x + 6)*log(3*x + 2) + 8*(15*x^2 + 19*x +
6)*log(2*x - 1) + 78540*x + 49819)/(15*x^2 + 19*x + 6)

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giac [A]  time = 0.92, size = 53, normalized size = 1.00 \[ -\frac {25}{11 \, {\left (5 \, x + 3\right )}} + \frac {135}{7 \, {\left (\frac {1}{5 \, x + 3} + 3\right )}} + \frac {648}{49} \, \log \left ({\left | -\frac {1}{5 \, x + 3} - 3 \right |}\right ) - \frac {8}{5929} \, \log \left ({\left | -\frac {11}{5 \, x + 3} + 2 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/11/(5*x + 3) + 135/7/(1/(5*x + 3) + 3) + 648/49*log(abs(-1/(5*x + 3) - 3)) - 8/5929*log(abs(-11/(5*x + 3)
+ 2))

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maple [A]  time = 0.01, size = 44, normalized size = 0.83 \[ -\frac {8 \ln \left (2 x -1\right )}{5929}+\frac {648 \ln \left (3 x +2\right )}{49}-\frac {1600 \ln \left (5 x +3\right )}{121}-\frac {25}{11 \left (5 x +3\right )}-\frac {9}{7 \left (3 x +2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/11/(5*x+3)-1600/121*ln(5*x+3)-9/7/(3*x+2)+648/49*ln(3*x+2)-8/5929*ln(2*x-1)

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maxima [A]  time = 0.55, size = 44, normalized size = 0.83 \[ -\frac {1020 \, x + 647}{77 \, {\left (15 \, x^{2} + 19 \, x + 6\right )}} - \frac {1600}{121} \, \log \left (5 \, x + 3\right ) + \frac {648}{49} \, \log \left (3 \, x + 2\right ) - \frac {8}{5929} \, \log \left (2 \, x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/77*(1020*x + 647)/(15*x^2 + 19*x + 6) - 1600/121*log(5*x + 3) + 648/49*log(3*x + 2) - 8/5929*log(2*x - 1)

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mupad [B]  time = 0.04, size = 36, normalized size = 0.68 \[ \frac {648\,\ln \left (x+\frac {2}{3}\right )}{49}-\frac {8\,\ln \left (x-\frac {1}{2}\right )}{5929}-\frac {1600\,\ln \left (x+\frac {3}{5}\right )}{121}-\frac {\frac {68\,x}{77}+\frac {647}{1155}}{x^2+\frac {19\,x}{15}+\frac {2}{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(648*log(x + 2/3))/49 - (8*log(x - 1/2))/5929 - (1600*log(x + 3/5))/121 - ((68*x)/77 + 647/1155)/((19*x)/15 +
x^2 + 2/5)

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sympy [A]  time = 0.20, size = 44, normalized size = 0.83 \[ - \frac {1020 x + 647}{1155 x^{2} + 1463 x + 462} - \frac {8 \log {\left (x - \frac {1}{2} \right )}}{5929} - \frac {1600 \log {\left (x + \frac {3}{5} \right )}}{121} + \frac {648 \log {\left (x + \frac {2}{3} \right )}}{49} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(1020*x + 647)/(1155*x**2 + 1463*x + 462) - 8*log(x - 1/2)/5929 - 1600*log(x + 3/5)/121 + 648*log(x + 2/3)/49

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