3.1461 \(\int \frac{(3+5 x)^2}{(1-2 x) (2+3 x)^2} \, dx\)

Optimal. Leaf size=32 \[ -\frac{1}{63 (3 x+2)}-\frac{121}{98} \log (1-2 x)-\frac{68}{441} \log (3 x+2) \]

[Out]

-1/(63*(2 + 3*x)) - (121*Log[1 - 2*x])/98 - (68*Log[2 + 3*x])/441

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Rubi [A]  time = 0.0138162, antiderivative size = 32, 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} \[ -\frac{1}{63 (3 x+2)}-\frac{121}{98} \log (1-2 x)-\frac{68}{441} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(63*(2 + 3*x)) - (121*Log[1 - 2*x])/98 - (68*Log[2 + 3*x])/441

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{(3+5 x)^2}{(1-2 x) (2+3 x)^2} \, dx &=\int \left (-\frac{121}{49 (-1+2 x)}+\frac{1}{21 (2+3 x)^2}-\frac{68}{147 (2+3 x)}\right ) \, dx\\ &=-\frac{1}{63 (2+3 x)}-\frac{121}{98} \log (1-2 x)-\frac{68}{441} \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.019648, size = 30, normalized size = 0.94 \[ \frac{1}{882} \left (-\frac{14}{3 x+2}-1089 \log (1-2 x)-136 \log (6 x+4)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-14/(2 + 3*x) - 1089*Log[1 - 2*x] - 136*Log[4 + 6*x])/882

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Maple [A]  time = 0.007, size = 27, normalized size = 0.8 \begin{align*} -{\frac{121\,\ln \left ( 2\,x-1 \right ) }{98}}-{\frac{1}{126+189\,x}}-{\frac{68\,\ln \left ( 2+3\,x \right ) }{441}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-121/98*ln(2*x-1)-1/63/(2+3*x)-68/441*ln(2+3*x)

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Maxima [A]  time = 1.11057, size = 35, normalized size = 1.09 \begin{align*} -\frac{1}{63 \,{\left (3 \, x + 2\right )}} - \frac{68}{441} \, \log \left (3 \, x + 2\right ) - \frac{121}{98} \, \log \left (2 \, x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/63/(3*x + 2) - 68/441*log(3*x + 2) - 121/98*log(2*x - 1)

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Fricas [A]  time = 1.39137, size = 111, normalized size = 3.47 \begin{align*} -\frac{136 \,{\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) + 1089 \,{\left (3 \, x + 2\right )} \log \left (2 \, x - 1\right ) + 14}{882 \,{\left (3 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/882*(136*(3*x + 2)*log(3*x + 2) + 1089*(3*x + 2)*log(2*x - 1) + 14)/(3*x + 2)

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Sympy [A]  time = 0.13641, size = 27, normalized size = 0.84 \begin{align*} - \frac{121 \log{\left (x - \frac{1}{2} \right )}}{98} - \frac{68 \log{\left (x + \frac{2}{3} \right )}}{441} - \frac{1}{189 x + 126} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-121*log(x - 1/2)/98 - 68*log(x + 2/3)/441 - 1/(189*x + 126)

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Giac [A]  time = 1.90436, size = 58, normalized size = 1.81 \begin{align*} -\frac{1}{63 \,{\left (3 \, x + 2\right )}} + \frac{25}{18} \, \log \left (\frac{{\left | 3 \, x + 2 \right |}}{3 \,{\left (3 \, x + 2\right )}^{2}}\right ) - \frac{121}{98} \, \log \left ({\left | -\frac{7}{3 \, x + 2} + 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/63/(3*x + 2) + 25/18*log(1/3*abs(3*x + 2)/(3*x + 2)^2) - 121/98*log(abs(-7/(3*x + 2) + 2))