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*Log[1 - 2*x])/5929 + (648*Log[2 + 3*x])/49 - (1600*Log[3 + 5*x])/121

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Rubi [A]  time = 0.0230057, antiderivative size = 53, normalized size of antiderivative = 1., 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*}

Mathematica [A]  time = 0.0248485, 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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Maple [A]  time = 0.009, size = 44, normalized size = 0.8 \begin{align*} -{\frac{8\,\ln \left ( 2\,x-1 \right ) }{5929}}-{\frac{9}{14+21\,x}}+{\frac{648\,\ln \left ( 2+3\,x \right ) }{49}}-{\frac{25}{33+55\,x}}-{\frac{1600\,\ln \left ( 3+5\,x \right ) }{121}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.10766, size = 59, normalized size = 1.11 \begin{align*} -\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 ) \end{align*}

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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Fricas [A]  time = 1.55902, size = 224, normalized size = 4.23 \begin{align*} -\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 )}} \end{align*}

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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Sympy [A]  time = 0.180936, size = 44, normalized size = 0.83 \begin{align*} - \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} \end{align*}

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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Giac [A]  time = 1.46677, size = 72, normalized size = 1.36 \begin{align*} -\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 ) \end{align*}

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))