3.1508 \(\int \frac{2+3 x}{(1-2 x) (3+5 x)^2} \, dx\)

Optimal. Leaf size=32 \[ -\frac{1}{55 (5 x+3)}-\frac{7}{121} \log (1-2 x)+\frac{7}{121} \log (5 x+3) \]

[Out]

-1/(55*(3 + 5*x)) - (7*Log[1 - 2*x])/121 + (7*Log[3 + 5*x])/121

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Rubi [A]  time = 0.0130007, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {77} \[ -\frac{1}{55 (5 x+3)}-\frac{7}{121} \log (1-2 x)+\frac{7}{121} \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(55*(3 + 5*x)) - (7*Log[1 - 2*x])/121 + (7*Log[3 + 5*x])/121

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{2+3 x}{(1-2 x) (3+5 x)^2} \, dx &=\int \left (-\frac{14}{121 (-1+2 x)}+\frac{1}{11 (3+5 x)^2}+\frac{35}{121 (3+5 x)}\right ) \, dx\\ &=-\frac{1}{55 (3+5 x)}-\frac{7}{121} \log (1-2 x)+\frac{7}{121} \log (3+5 x)\\ \end{align*}

Mathematica [A]  time = 0.0140118, size = 30, normalized size = 0.94 \[ \frac{1}{605} \left (-\frac{11}{5 x+3}-35 \log (5-10 x)+35 \log (5 x+3)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11/(3 + 5*x) - 35*Log[5 - 10*x] + 35*Log[3 + 5*x])/605

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Maple [A]  time = 0.005, size = 27, normalized size = 0.8 \begin{align*} -{\frac{7\,\ln \left ( 2\,x-1 \right ) }{121}}-{\frac{1}{165+275\,x}}+{\frac{7\,\ln \left ( 3+5\,x \right ) }{121}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/121*ln(2*x-1)-1/55/(3+5*x)+7/121*ln(3+5*x)

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Maxima [A]  time = 1.06481, size = 35, normalized size = 1.09 \begin{align*} -\frac{1}{55 \,{\left (5 \, x + 3\right )}} + \frac{7}{121} \, \log \left (5 \, x + 3\right ) - \frac{7}{121} \, \log \left (2 \, x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/55/(5*x + 3) + 7/121*log(5*x + 3) - 7/121*log(2*x - 1)

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Fricas [A]  time = 1.54965, size = 105, normalized size = 3.28 \begin{align*} \frac{35 \,{\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 35 \,{\left (5 \, x + 3\right )} \log \left (2 \, x - 1\right ) - 11}{605 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/605*(35*(5*x + 3)*log(5*x + 3) - 35*(5*x + 3)*log(2*x - 1) - 11)/(5*x + 3)

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Sympy [A]  time = 0.118017, size = 26, normalized size = 0.81 \begin{align*} - \frac{7 \log{\left (x - \frac{1}{2} \right )}}{121} + \frac{7 \log{\left (x + \frac{3}{5} \right )}}{121} - \frac{1}{275 x + 165} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7*log(x - 1/2)/121 + 7*log(x + 3/5)/121 - 1/(275*x + 165)

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Giac [A]  time = 1.77226, size = 34, normalized size = 1.06 \begin{align*} -\frac{1}{55 \,{\left (5 \, x + 3\right )}} - \frac{7}{121} \, \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((2+3*x)/(1-2*x)/(3+5*x)^2,x, algorithm="giac")

[Out]

-1/55/(5*x + 3) - 7/121*log(abs(-11/(5*x + 3) + 2))