3.1545 \(\int \frac{(2+3 x) (3+5 x)}{(1-2 x)^2} \, dx\)

Optimal. Leaf size=27 \[ \frac{15 x}{4}+\frac{77}{8 (1-2 x)}+\frac{17}{2} \log (1-2 x) \]

[Out]

77/(8*(1 - 2*x)) + (15*x)/4 + (17*Log[1 - 2*x])/2

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Rubi [A]  time = 0.0118792, antiderivative size = 27, 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} \[ \frac{15 x}{4}+\frac{77}{8 (1-2 x)}+\frac{17}{2} \log (1-2 x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

77/(8*(1 - 2*x)) + (15*x)/4 + (17*Log[1 - 2*x])/2

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)^2} \, dx &=\int \left (\frac{15}{4}+\frac{77}{4 (-1+2 x)^2}+\frac{17}{-1+2 x}\right ) \, dx\\ &=\frac{77}{8 (1-2 x)}+\frac{15 x}{4}+\frac{17}{2} \log (1-2 x)\\ \end{align*}

Mathematica [A]  time = 0.009845, size = 26, normalized size = 0.96 \[ \frac{1}{8} \left (30 x+\frac{77}{1-2 x}+68 \log (1-2 x)-15\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15 + 77/(1 - 2*x) + 30*x + 68*Log[1 - 2*x])/8

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Maple [A]  time = 0.006, size = 22, normalized size = 0.8 \begin{align*}{\frac{15\,x}{4}}+{\frac{17\,\ln \left ( 2\,x-1 \right ) }{2}}-{\frac{77}{16\,x-8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/4*x+17/2*ln(2*x-1)-77/8/(2*x-1)

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Maxima [A]  time = 1.04119, size = 28, normalized size = 1.04 \begin{align*} \frac{15}{4} \, x - \frac{77}{8 \,{\left (2 \, x - 1\right )}} + \frac{17}{2} \, \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)^2,x, algorithm="maxima")

[Out]

15/4*x - 77/8/(2*x - 1) + 17/2*log(2*x - 1)

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Fricas [A]  time = 1.2741, size = 86, normalized size = 3.19 \begin{align*} \frac{60 \, x^{2} + 68 \,{\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 30 \, x - 77}{8 \,{\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)^2,x, algorithm="fricas")

[Out]

1/8*(60*x^2 + 68*(2*x - 1)*log(2*x - 1) - 30*x - 77)/(2*x - 1)

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Sympy [A]  time = 0.089469, size = 20, normalized size = 0.74 \begin{align*} \frac{15 x}{4} + \frac{17 \log{\left (2 x - 1 \right )}}{2} - \frac{77}{16 x - 8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15*x/4 + 17*log(2*x - 1)/2 - 77/(16*x - 8)

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Giac [A]  time = 1.65171, size = 43, normalized size = 1.59 \begin{align*} \frac{15}{4} \, x - \frac{77}{8 \,{\left (2 \, x - 1\right )}} - \frac{17}{2} \, \log \left (\frac{{\left | 2 \, x - 1 \right |}}{2 \,{\left (2 \, x - 1\right )}^{2}}\right ) - \frac{15}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/4*x - 77/8/(2*x - 1) - 17/2*log(1/2*abs(2*x - 1)/(2*x - 1)^2) - 15/8