3.1201 \(\int \frac{1-2 x}{(2+3 x)^2 (3+5 x)} \, dx\)

Optimal. Leaf size=28 \[ \frac{7}{3 (3 x+2)}-11 \log (3 x+2)+11 \log (5 x+3) \]

[Out]

7/(3*(2 + 3*x)) - 11*Log[2 + 3*x] + 11*Log[3 + 5*x]

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Rubi [A]  time = 0.0135171, antiderivative size = 28, 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} \[ \frac{7}{3 (3 x+2)}-11 \log (3 x+2)+11 \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

7/(3*(2 + 3*x)) - 11*Log[2 + 3*x] + 11*Log[3 + 5*x]

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{1-2 x}{(2+3 x)^2 (3+5 x)} \, dx &=\int \left (-\frac{7}{(2+3 x)^2}-\frac{33}{2+3 x}+\frac{55}{3+5 x}\right ) \, dx\\ &=\frac{7}{3 (2+3 x)}-11 \log (2+3 x)+11 \log (3+5 x)\\ \end{align*}

Mathematica [A]  time = 0.0133397, size = 38, normalized size = 1.36 \[ \frac{-33 (3 x+2) \log (3 x+2)+33 (3 x+2) \log (-3 (5 x+3))+7}{9 x+6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7 - 33*(2 + 3*x)*Log[2 + 3*x] + 33*(2 + 3*x)*Log[-3*(3 + 5*x)])/(6 + 9*x)

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Maple [A]  time = 0.006, size = 27, normalized size = 1. \begin{align*}{\frac{7}{6+9\,x}}-11\,\ln \left ( 2+3\,x \right ) +11\,\ln \left ( 3+5\,x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/3/(2+3*x)-11*ln(2+3*x)+11*ln(3+5*x)

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Maxima [A]  time = 2.40898, size = 35, normalized size = 1.25 \begin{align*} \frac{7}{3 \,{\left (3 \, x + 2\right )}} + 11 \, \log \left (5 \, x + 3\right ) - 11 \, \log \left (3 \, x + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/3/(3*x + 2) + 11*log(5*x + 3) - 11*log(3*x + 2)

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Fricas [A]  time = 1.78787, size = 101, normalized size = 3.61 \begin{align*} \frac{33 \,{\left (3 \, x + 2\right )} \log \left (5 \, x + 3\right ) - 33 \,{\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) + 7}{3 \,{\left (3 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(33*(3*x + 2)*log(5*x + 3) - 33*(3*x + 2)*log(3*x + 2) + 7)/(3*x + 2)

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Sympy [A]  time = 0.109632, size = 22, normalized size = 0.79 \begin{align*} 11 \log{\left (x + \frac{3}{5} \right )} - 11 \log{\left (x + \frac{2}{3} \right )} + \frac{7}{9 x + 6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11*log(x + 3/5) - 11*log(x + 2/3) + 7/(9*x + 6)

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Giac [A]  time = 3.75867, size = 34, normalized size = 1.21 \begin{align*} \frac{7}{3 \,{\left (3 \, x + 2\right )}} + 11 \, \log \left ({\left | -\frac{1}{3 \, x + 2} + 5 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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