3.1457 \(\int \frac{(2+3 x)^2 (3+5 x)^2}{1-2 x} \, dx\)

Optimal. Leaf size=37 \[ -\frac{225 x^4}{8}-\frac{455 x^3}{4}-\frac{3529 x^2}{16}-\frac{5353 x}{16}-\frac{5929}{32} \log (1-2 x) \]

[Out]

(-5353*x)/16 - (3529*x^2)/16 - (455*x^3)/4 - (225*x^4)/8 - (5929*Log[1 - 2*x])/32

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Rubi [A]  time = 0.0164038, antiderivative size = 37, 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 = {88} \[ -\frac{225 x^4}{8}-\frac{455 x^3}{4}-\frac{3529 x^2}{16}-\frac{5353 x}{16}-\frac{5929}{32} \log (1-2 x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5353*x)/16 - (3529*x^2)/16 - (455*x^3)/4 - (225*x^4)/8 - (5929*Log[1 - 2*x])/32

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{(2+3 x)^2 (3+5 x)^2}{1-2 x} \, dx &=\int \left (-\frac{5353}{16}-\frac{3529 x}{8}-\frac{1365 x^2}{4}-\frac{225 x^3}{2}-\frac{5929}{16 (-1+2 x)}\right ) \, dx\\ &=-\frac{5353 x}{16}-\frac{3529 x^2}{16}-\frac{455 x^3}{4}-\frac{225 x^4}{8}-\frac{5929}{32} \log (1-2 x)\\ \end{align*}

Mathematica [A]  time = 0.0092651, size = 32, normalized size = 0.86 \[ \frac{1}{128} \left (-3600 x^4-14560 x^3-28232 x^2-42824 x-23716 \log (1-2 x)+30515\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(30515 - 42824*x - 28232*x^2 - 14560*x^3 - 3600*x^4 - 23716*Log[1 - 2*x])/128

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Maple [A]  time = 0.002, size = 28, normalized size = 0.8 \begin{align*} -{\frac{225\,{x}^{4}}{8}}-{\frac{455\,{x}^{3}}{4}}-{\frac{3529\,{x}^{2}}{16}}-{\frac{5353\,x}{16}}-{\frac{5929\,\ln \left ( 2\,x-1 \right ) }{32}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-225/8*x^4-455/4*x^3-3529/16*x^2-5353/16*x-5929/32*ln(2*x-1)

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Maxima [A]  time = 2.48159, size = 36, normalized size = 0.97 \begin{align*} -\frac{225}{8} \, x^{4} - \frac{455}{4} \, x^{3} - \frac{3529}{16} \, x^{2} - \frac{5353}{16} \, x - \frac{5929}{32} \, \log \left (2 \, x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-225/8*x^4 - 455/4*x^3 - 3529/16*x^2 - 5353/16*x - 5929/32*log(2*x - 1)

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Fricas [A]  time = 1.28844, size = 99, normalized size = 2.68 \begin{align*} -\frac{225}{8} \, x^{4} - \frac{455}{4} \, x^{3} - \frac{3529}{16} \, x^{2} - \frac{5353}{16} \, x - \frac{5929}{32} \, \log \left (2 \, x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-225/8*x^4 - 455/4*x^3 - 3529/16*x^2 - 5353/16*x - 5929/32*log(2*x - 1)

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Sympy [A]  time = 0.090276, size = 36, normalized size = 0.97 \begin{align*} - \frac{225 x^{4}}{8} - \frac{455 x^{3}}{4} - \frac{3529 x^{2}}{16} - \frac{5353 x}{16} - \frac{5929 \log{\left (2 x - 1 \right )}}{32} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-225*x**4/8 - 455*x**3/4 - 3529*x**2/16 - 5353*x/16 - 5929*log(2*x - 1)/32

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Giac [A]  time = 1.33024, size = 38, normalized size = 1.03 \begin{align*} -\frac{225}{8} \, x^{4} - \frac{455}{4} \, x^{3} - \frac{3529}{16} \, x^{2} - \frac{5353}{16} \, x - \frac{5929}{32} \, \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)^2*(3+5*x)^2/(1-2*x),x, algorithm="giac")

[Out]

-225/8*x^4 - 455/4*x^3 - 3529/16*x^2 - 5353/16*x - 5929/32*log(abs(2*x - 1))