3.1637 \(\int \frac{(2+3 x) (3+5 x)}{(1-2 x)^3} \, dx\)

Optimal. Leaf size=33 \[ -\frac{17}{2 (1-2 x)}+\frac{77}{16 (1-2 x)^2}-\frac{15}{8} \log (1-2 x) \]

[Out]

77/(16*(1 - 2*x)^2) - 17/(2*(1 - 2*x)) - (15*Log[1 - 2*x])/8

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Rubi [A]  time = 0.022907, antiderivative size = 33, 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} \[ -\frac{17}{2 (1-2 x)}+\frac{77}{16 (1-2 x)^2}-\frac{15}{8} \log (1-2 x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

77/(16*(1 - 2*x)^2) - 17/(2*(1 - 2*x)) - (15*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)^3} \, dx &=\int \left (-\frac{77}{4 (-1+2 x)^3}-\frac{17}{(-1+2 x)^2}-\frac{15}{4 (-1+2 x)}\right ) \, dx\\ &=\frac{77}{16 (1-2 x)^2}-\frac{17}{2 (1-2 x)}-\frac{15}{8} \log (1-2 x)\\ \end{align*}

Mathematica [A]  time = 0.0064193, size = 33, normalized size = 1. \[ -\frac{17}{2 (1-2 x)}+\frac{77}{16 (1-2 x)^2}-\frac{15}{8} \log (1-2 x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

77/(16*(1 - 2*x)^2) - 17/(2*(1 - 2*x)) - (15*Log[1 - 2*x])/8

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Maple [A]  time = 0.005, size = 28, normalized size = 0.9 \begin{align*} -{\frac{15\,\ln \left ( 2\,x-1 \right ) }{8}}+{\frac{77}{16\, \left ( 2\,x-1 \right ) ^{2}}}+{\frac{17}{4\,x-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/8*ln(2*x-1)+77/16/(2*x-1)^2+17/2/(2*x-1)

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Maxima [A]  time = 1.13722, size = 38, normalized size = 1.15 \begin{align*} \frac{272 \, x - 59}{16 \,{\left (4 \, x^{2} - 4 \, x + 1\right )}} - \frac{15}{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)^3,x, algorithm="maxima")

[Out]

1/16*(272*x - 59)/(4*x^2 - 4*x + 1) - 15/8*log(2*x - 1)

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Fricas [A]  time = 1.27778, size = 100, normalized size = 3.03 \begin{align*} -\frac{30 \,{\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (2 \, x - 1\right ) - 272 \, x + 59}{16 \,{\left (4 \, x^{2} - 4 \, 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)^3,x, algorithm="fricas")

[Out]

-1/16*(30*(4*x^2 - 4*x + 1)*log(2*x - 1) - 272*x + 59)/(4*x^2 - 4*x + 1)

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Sympy [A]  time = 0.109109, size = 24, normalized size = 0.73 \begin{align*} \frac{272 x - 59}{64 x^{2} - 64 x + 16} - \frac{15 \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)**3,x)

[Out]

(272*x - 59)/(64*x**2 - 64*x + 16) - 15*log(2*x - 1)/8

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Giac [A]  time = 3.27684, size = 32, normalized size = 0.97 \begin{align*} \frac{272 \, x - 59}{16 \,{\left (2 \, x - 1\right )}^{2}} - \frac{15}{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)^3,x, algorithm="giac")

[Out]

1/16*(272*x - 59)/(2*x - 1)^2 - 15/8*log(abs(2*x - 1))