3.1441 \(\int \frac{(2+3 x) (3+5 x)}{1-2 x} \, dx\)

Optimal. Leaf size=23 \[ -\frac{15 x^2}{4}-\frac{53 x}{4}-\frac{77}{8} \log (1-2 x) \]

[Out]

(-53*x)/4 - (15*x^2)/4 - (77*Log[1 - 2*x])/8

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Rubi [A]  time = 0.0087319, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used = {77} \[ -\frac{15 x^2}{4}-\frac{53 x}{4}-\frac{77}{8} \log (1-2 x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-53*x)/4 - (15*x^2)/4 - (77*Log[1 - 2*x])/8

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) (3+5 x)}{1-2 x} \, dx &=\int \left (-\frac{53}{4}-\frac{15 x}{2}-\frac{77}{4 (-1+2 x)}\right ) \, dx\\ &=-\frac{53 x}{4}-\frac{15 x^2}{4}-\frac{77}{8} \log (1-2 x)\\ \end{align*}

Mathematica [A]  time = 0.0045626, size = 22, normalized size = 0.96 \[ \frac{1}{16} \left (-60 x^2-212 x-154 \log (1-2 x)+121\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(121 - 212*x - 60*x^2 - 154*Log[1 - 2*x])/16

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Maple [A]  time = 0.001, size = 18, normalized size = 0.8 \begin{align*} -{\frac{15\,{x}^{2}}{4}}-{\frac{53\,x}{4}}-{\frac{77\,\ln \left ( 2\,x-1 \right ) }{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*x^2-53/4*x-77/8*ln(2*x-1)

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Maxima [A]  time = 1.09141, size = 23, normalized size = 1. \begin{align*} -\frac{15}{4} \, x^{2} - \frac{53}{4} \, x - \frac{77}{8} \, \log \left (2 \, x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*x^2 - 53/4*x - 77/8*log(2*x - 1)

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Fricas [A]  time = 1.31319, size = 54, normalized size = 2.35 \begin{align*} -\frac{15}{4} \, x^{2} - \frac{53}{4} \, x - \frac{77}{8} \, \log \left (2 \, x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*x^2 - 53/4*x - 77/8*log(2*x - 1)

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Sympy [A]  time = 0.081297, size = 22, normalized size = 0.96 \begin{align*} - \frac{15 x^{2}}{4} - \frac{53 x}{4} - \frac{77 \log{\left (2 x - 1 \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15*x**2/4 - 53*x/4 - 77*log(2*x - 1)/8

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Giac [A]  time = 2.63064, size = 24, normalized size = 1.04 \begin{align*} -\frac{15}{4} \, x^{2} - \frac{53}{4} \, x - \frac{77}{8} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*x^2 - 53/4*x - 77/8*log(abs(2*x - 1))