3.1560 \(\int \frac{(2+3 x)^2 (3+5 x)^2}{(1-2 x)^2} \, dx\)

Optimal. Leaf size=41 \[ \frac{75 x^3}{4}+\frac{795 x^2}{8}+\frac{5119 x}{16}+\frac{5929}{32 (1-2 x)}+\frac{1309}{4} \log (1-2 x) \]

[Out]

5929/(32*(1 - 2*x)) + (5119*x)/16 + (795*x^2)/8 + (75*x^3)/4 + (1309*Log[1 - 2*x])/4

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Rubi [A]  time = 0.0195414, antiderivative size = 41, 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} \[ \frac{75 x^3}{4}+\frac{795 x^2}{8}+\frac{5119 x}{16}+\frac{5929}{32 (1-2 x)}+\frac{1309}{4} \log (1-2 x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

5929/(32*(1 - 2*x)) + (5119*x)/16 + (795*x^2)/8 + (75*x^3)/4 + (1309*Log[1 - 2*x])/4

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)^2} \, dx &=\int \left (\frac{5119}{16}+\frac{795 x}{4}+\frac{225 x^2}{4}+\frac{5929}{16 (-1+2 x)^2}+\frac{1309}{2 (-1+2 x)}\right ) \, dx\\ &=\frac{5929}{32 (1-2 x)}+\frac{5119 x}{16}+\frac{795 x^2}{8}+\frac{75 x^3}{4}+\frac{1309}{4} \log (1-2 x)\\ \end{align*}

Mathematica [A]  time = 0.0145421, size = 41, normalized size = 1. \[ \frac{300 x^4+1440 x^3+4324 x^2-5554 x+2618 (2 x-1) \log (1-2 x)+15}{16 x-8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15 - 5554*x + 4324*x^2 + 1440*x^3 + 300*x^4 + 2618*(-1 + 2*x)*Log[1 - 2*x])/(-8 + 16*x)

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Maple [A]  time = 0.006, size = 32, normalized size = 0.8 \begin{align*}{\frac{75\,{x}^{3}}{4}}+{\frac{795\,{x}^{2}}{8}}+{\frac{5119\,x}{16}}+{\frac{1309\,\ln \left ( 2\,x-1 \right ) }{4}}-{\frac{5929}{64\,x-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)^2,x)

[Out]

75/4*x^3+795/8*x^2+5119/16*x+1309/4*ln(2*x-1)-5929/32/(2*x-1)

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Maxima [A]  time = 1.0469, size = 42, normalized size = 1.02 \begin{align*} \frac{75}{4} \, x^{3} + \frac{795}{8} \, x^{2} + \frac{5119}{16} \, x - \frac{5929}{32 \,{\left (2 \, x - 1\right )}} + \frac{1309}{4} \, \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)^2,x, algorithm="maxima")

[Out]

75/4*x^3 + 795/8*x^2 + 5119/16*x - 5929/32/(2*x - 1) + 1309/4*log(2*x - 1)

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Fricas [A]  time = 1.26461, size = 132, normalized size = 3.22 \begin{align*} \frac{1200 \, x^{4} + 5760 \, x^{3} + 17296 \, x^{2} + 10472 \,{\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 10238 \, x - 5929}{32 \,{\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)^2,x, algorithm="fricas")

[Out]

1/32*(1200*x^4 + 5760*x^3 + 17296*x^2 + 10472*(2*x - 1)*log(2*x - 1) - 10238*x - 5929)/(2*x - 1)

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Sympy [A]  time = 0.096828, size = 34, normalized size = 0.83 \begin{align*} \frac{75 x^{3}}{4} + \frac{795 x^{2}}{8} + \frac{5119 x}{16} + \frac{1309 \log{\left (2 x - 1 \right )}}{4} - \frac{5929}{64 x - 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)**2,x)

[Out]

75*x**3/4 + 795*x**2/8 + 5119*x/16 + 1309*log(2*x - 1)/4 - 5929/(64*x - 32)

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Giac [A]  time = 3.36188, size = 77, normalized size = 1.88 \begin{align*} \frac{1}{32} \,{\left (2 \, x - 1\right )}^{3}{\left (\frac{1020}{2 \, x - 1} + \frac{6934}{{\left (2 \, x - 1\right )}^{2}} + 75\right )} - \frac{5929}{32 \,{\left (2 \, x - 1\right )}} - \frac{1309}{4} \, \log \left (\frac{{\left | 2 \, x - 1 \right |}}{2 \,{\left (2 \, x - 1\right )}^{2}}\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)^2,x, algorithm="giac")

[Out]

1/32*(2*x - 1)^3*(1020/(2*x - 1) + 6934/(2*x - 1)^2 + 75) - 5929/32/(2*x - 1) - 1309/4*log(1/2*abs(2*x - 1)/(2
*x - 1)^2)