3.1561 \(\int \frac{(2+3 x) (3+5 x)^2}{(1-2 x)^2} \, dx\)

Optimal. Leaf size=34 \[ \frac{75 x^2}{8}+\frac{215 x}{4}+\frac{847}{16 (1-2 x)}+\frac{1133}{16} \log (1-2 x) \]

[Out]

847/(16*(1 - 2*x)) + (215*x)/4 + (75*x^2)/8 + (1133*Log[1 - 2*x])/16

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Rubi [A]  time = 0.0155181, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {77} \[ \frac{75 x^2}{8}+\frac{215 x}{4}+\frac{847}{16 (1-2 x)}+\frac{1133}{16} \log (1-2 x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

847/(16*(1 - 2*x)) + (215*x)/4 + (75*x^2)/8 + (1133*Log[1 - 2*x])/16

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)^2}{(1-2 x)^2} \, dx &=\int \left (\frac{215}{4}+\frac{75 x}{4}+\frac{847}{8 (-1+2 x)^2}+\frac{1133}{8 (-1+2 x)}\right ) \, dx\\ &=\frac{847}{16 (1-2 x)}+\frac{215 x}{4}+\frac{75 x^2}{8}+\frac{1133}{16} \log (1-2 x)\\ \end{align*}

Mathematica [A]  time = 0.010032, size = 36, normalized size = 1.06 \[ \frac{600 x^3+3140 x^2-3590 x+2266 (2 x-1) \log (1-2 x)-759}{64 x-32} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-759 - 3590*x + 3140*x^2 + 600*x^3 + 2266*(-1 + 2*x)*Log[1 - 2*x])/(-32 + 64*x)

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Maple [A]  time = 0.005, size = 27, normalized size = 0.8 \begin{align*}{\frac{75\,{x}^{2}}{8}}+{\frac{215\,x}{4}}+{\frac{1133\,\ln \left ( 2\,x-1 \right ) }{16}}-{\frac{847}{32\,x-16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

75/8*x^2+215/4*x+1133/16*ln(2*x-1)-847/16/(2*x-1)

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Maxima [A]  time = 1.16079, size = 35, normalized size = 1.03 \begin{align*} \frac{75}{8} \, x^{2} + \frac{215}{4} \, x - \frac{847}{16 \,{\left (2 \, x - 1\right )}} + \frac{1133}{16} \, \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)^2/(1-2*x)^2,x, algorithm="maxima")

[Out]

75/8*x^2 + 215/4*x - 847/16/(2*x - 1) + 1133/16*log(2*x - 1)

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Fricas [A]  time = 1.31753, size = 109, normalized size = 3.21 \begin{align*} \frac{300 \, x^{3} + 1570 \, x^{2} + 1133 \,{\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 860 \, x - 847}{16 \,{\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)^2/(1-2*x)^2,x, algorithm="fricas")

[Out]

1/16*(300*x^3 + 1570*x^2 + 1133*(2*x - 1)*log(2*x - 1) - 860*x - 847)/(2*x - 1)

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Sympy [A]  time = 0.097321, size = 27, normalized size = 0.79 \begin{align*} \frac{75 x^{2}}{8} + \frac{215 x}{4} + \frac{1133 \log{\left (2 x - 1 \right )}}{16} - \frac{847}{32 x - 16} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

75*x**2/8 + 215*x/4 + 1133*log(2*x - 1)/16 - 847/(32*x - 16)

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Giac [A]  time = 2.24092, size = 65, normalized size = 1.91 \begin{align*} \frac{5}{32} \,{\left (2 \, x - 1\right )}^{2}{\left (\frac{202}{2 \, x - 1} + 15\right )} - \frac{847}{16 \,{\left (2 \, x - 1\right )}} - \frac{1133}{16} \, \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)*(3+5*x)^2/(1-2*x)^2,x, algorithm="giac")

[Out]

5/32*(2*x - 1)^2*(202/(2*x - 1) + 15) - 847/16/(2*x - 1) - 1133/16*log(1/2*abs(2*x - 1)/(2*x - 1)^2)