3.1297 \(\int \frac{(1-2 x)^2}{(2+3 x) (3+5 x)} \, dx\)

Optimal. Leaf size=26 \[ \frac{4 x}{15}-\frac{49}{9} \log (3 x+2)+\frac{121}{25} \log (5 x+3) \]

[Out]

(4*x)/15 - (49*Log[2 + 3*x])/9 + (121*Log[3 + 5*x])/25

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Rubi [A]  time = 0.0121358, antiderivative size = 26, 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 = {72} \[ \frac{4 x}{15}-\frac{49}{9} \log (3 x+2)+\frac{121}{25} \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*x)/15 - (49*Log[2 + 3*x])/9 + (121*Log[3 + 5*x])/25

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

\begin{align*} \int \frac{(1-2 x)^2}{(2+3 x) (3+5 x)} \, dx &=\int \left (\frac{4}{15}-\frac{49}{3 (2+3 x)}+\frac{121}{5 (3+5 x)}\right ) \, dx\\ &=\frac{4 x}{15}-\frac{49}{9} \log (2+3 x)+\frac{121}{25} \log (3+5 x)\\ \end{align*}

Mathematica [A]  time = 0.0090238, size = 27, normalized size = 1.04 \[ \frac{1}{225} (60 x-1225 \log (3 x+2)+1089 \log (-3 (5 x+3))+40) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(40 + 60*x - 1225*Log[2 + 3*x] + 1089*Log[-3*(3 + 5*x)])/225

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Maple [A]  time = 0.006, size = 21, normalized size = 0.8 \begin{align*}{\frac{4\,x}{15}}-{\frac{49\,\ln \left ( 2+3\,x \right ) }{9}}+{\frac{121\,\ln \left ( 3+5\,x \right ) }{25}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*x-49/9*ln(2+3*x)+121/25*ln(3+5*x)

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Maxima [A]  time = 1.01823, size = 27, normalized size = 1.04 \begin{align*} \frac{4}{15} \, x + \frac{121}{25} \, \log \left (5 \, x + 3\right ) - \frac{49}{9} \, \log \left (3 \, x + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*x + 121/25*log(5*x + 3) - 49/9*log(3*x + 2)

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Fricas [A]  time = 1.45382, size = 68, normalized size = 2.62 \begin{align*} \frac{4}{15} \, x + \frac{121}{25} \, \log \left (5 \, x + 3\right ) - \frac{49}{9} \, \log \left (3 \, x + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*x + 121/25*log(5*x + 3) - 49/9*log(3*x + 2)

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Sympy [A]  time = 0.111508, size = 24, normalized size = 0.92 \begin{align*} \frac{4 x}{15} + \frac{121 \log{\left (x + \frac{3}{5} \right )}}{25} - \frac{49 \log{\left (x + \frac{2}{3} \right )}}{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*x/15 + 121*log(x + 3/5)/25 - 49*log(x + 2/3)/9

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Giac [A]  time = 2.39439, size = 30, normalized size = 1.15 \begin{align*} \frac{4}{15} \, x + \frac{121}{25} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac{49}{9} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*x + 121/25*log(abs(5*x + 3)) - 49/9*log(abs(3*x + 2))