3.1507 \(\int \frac{(2+3 x)^2}{(1-2 x) (3+5 x)^2} \, dx\)

Optimal. Leaf size=32 \[ -\frac{1}{275 (5 x+3)}-\frac{49}{242} \log (1-2 x)+\frac{68 \log (5 x+3)}{3025} \]

[Out]

-1/(275*(3 + 5*x)) - (49*Log[1 - 2*x])/242 + (68*Log[3 + 5*x])/3025

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Rubi [A]  time = 0.0140177, antiderivative size = 32, 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} \[ -\frac{1}{275 (5 x+3)}-\frac{49}{242} \log (1-2 x)+\frac{68 \log (5 x+3)}{3025} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(275*(3 + 5*x)) - (49*Log[1 - 2*x])/242 + (68*Log[3 + 5*x])/3025

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}{(1-2 x) (3+5 x)^2} \, dx &=\int \left (-\frac{49}{121 (-1+2 x)}+\frac{1}{55 (3+5 x)^2}+\frac{68}{605 (3+5 x)}\right ) \, dx\\ &=-\frac{1}{275 (3+5 x)}-\frac{49}{242} \log (1-2 x)+\frac{68 \log (3+5 x)}{3025}\\ \end{align*}

Mathematica [A]  time = 0.0165172, size = 30, normalized size = 0.94 \[ \frac{-\frac{22}{5 x+3}-1225 \log (1-2 x)+136 \log (10 x+6)}{6050} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-22/(3 + 5*x) - 1225*Log[1 - 2*x] + 136*Log[6 + 10*x])/6050

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Maple [A]  time = 0.005, size = 27, normalized size = 0.8 \begin{align*} -{\frac{49\,\ln \left ( 2\,x-1 \right ) }{242}}-{\frac{1}{825+1375\,x}}+{\frac{68\,\ln \left ( 3+5\,x \right ) }{3025}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-49/242*ln(2*x-1)-1/275/(3+5*x)+68/3025*ln(3+5*x)

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Maxima [A]  time = 1.06234, size = 35, normalized size = 1.09 \begin{align*} -\frac{1}{275 \,{\left (5 \, x + 3\right )}} + \frac{68}{3025} \, \log \left (5 \, x + 3\right ) - \frac{49}{242} \, \log \left (2 \, x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/275/(5*x + 3) + 68/3025*log(5*x + 3) - 49/242*log(2*x - 1)

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Fricas [A]  time = 1.60118, size = 111, normalized size = 3.47 \begin{align*} \frac{136 \,{\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 1225 \,{\left (5 \, x + 3\right )} \log \left (2 \, x - 1\right ) - 22}{6050 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6050*(136*(5*x + 3)*log(5*x + 3) - 1225*(5*x + 3)*log(2*x - 1) - 22)/(5*x + 3)

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Sympy [A]  time = 0.136497, size = 26, normalized size = 0.81 \begin{align*} - \frac{49 \log{\left (x - \frac{1}{2} \right )}}{242} + \frac{68 \log{\left (x + \frac{3}{5} \right )}}{3025} - \frac{1}{1375 x + 825} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-49*log(x - 1/2)/242 + 68*log(x + 3/5)/3025 - 1/(1375*x + 825)

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Giac [A]  time = 1.21789, size = 58, normalized size = 1.81 \begin{align*} -\frac{1}{275 \,{\left (5 \, x + 3\right )}} + \frac{9}{50} \, \log \left (\frac{{\left | 5 \, x + 3 \right |}}{5 \,{\left (5 \, x + 3\right )}^{2}}\right ) - \frac{49}{242} \, \log \left ({\left | -\frac{11}{5 \, x + 3} + 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/275/(5*x + 3) + 9/50*log(1/5*abs(5*x + 3)/(5*x + 3)^2) - 49/242*log(abs(-11/(5*x + 3) + 2))