3.1215 \(\int \frac{1-2 x}{(2+3 x) (3+5 x)^2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0127324, 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} \[ -\frac{11}{5 (5 x+3)}+7 \log (3 x+2)-7 \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-11/(5*(3 + 5*x)) + 7*Log[2 + 3*x] - 7*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) (3+5 x)^2} \, dx &=\int \left (\frac{21}{2+3 x}+\frac{11}{(3+5 x)^2}-\frac{35}{3+5 x}\right ) \, dx\\ &=-\frac{11}{5 (3+5 x)}+7 \log (2+3 x)-7 \log (3+5 x)\\ \end{align*}

Mathematica [A]  time = 0.0138742, size = 30, normalized size = 1.07 \[ -\frac{11}{5 (5 x+3)}+7 \log (5 (3 x+2))-7 \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 27, normalized size = 1. \begin{align*} -{\frac{11}{15+25\,x}}+7\,\ln \left ( 2+3\,x \right ) -7\,\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)/(3+5*x)^2,x)

[Out]

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

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Maxima [A]  time = 1.02682, size = 35, normalized size = 1.25 \begin{align*} -\frac{11}{5 \,{\left (5 \, x + 3\right )}} - 7 \, \log \left (5 \, x + 3\right ) + 7 \, \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)/(3+5*x)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.52058, size = 104, normalized size = 3.71 \begin{align*} -\frac{35 \,{\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 35 \,{\left (5 \, x + 3\right )} \log \left (3 \, x + 2\right ) + 11}{5 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.109081, size = 22, normalized size = 0.79 \begin{align*} - 7 \log{\left (x + \frac{3}{5} \right )} + 7 \log{\left (x + \frac{2}{3} \right )} - \frac{11}{25 x + 15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7*log(x + 3/5) + 7*log(x + 2/3) - 11/(25*x + 15)

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Giac [A]  time = 1.77144, size = 34, normalized size = 1.21 \begin{align*} -\frac{11}{5 \,{\left (5 \, x + 3\right )}} + 7 \, \log \left ({\left | -\frac{1}{5 \, x + 3} - 3 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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