3.1596 \(\int \frac{1}{(1-2 x)^2 (3+5 x)} \, dx\)

Optimal. Leaf size=32 \[ \frac{1}{11 (1-2 x)}-\frac{5}{121} \log (1-2 x)+\frac{5}{121} \log (5 x+3) \]

[Out]

1/(11*(1 - 2*x)) - (5*Log[1 - 2*x])/121 + (5*Log[3 + 5*x])/121

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Rubi [A]  time = 0.0123018, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {44} \[ \frac{1}{11 (1-2 x)}-\frac{5}{121} \log (1-2 x)+\frac{5}{121} \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

Int[1/((1 - 2*x)^2*(3 + 5*x)),x]

[Out]

1/(11*(1 - 2*x)) - (5*Log[1 - 2*x])/121 + (5*Log[3 + 5*x])/121

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{1}{(1-2 x)^2 (3+5 x)} \, dx &=\int \left (\frac{2}{11 (-1+2 x)^2}-\frac{10}{121 (-1+2 x)}+\frac{25}{121 (3+5 x)}\right ) \, dx\\ &=\frac{1}{11 (1-2 x)}-\frac{5}{121} \log (1-2 x)+\frac{5}{121} \log (3+5 x)\\ \end{align*}

Mathematica [A]  time = 0.0089032, size = 38, normalized size = 1.19 \[ \frac{(5-10 x) \log (1-2 x)+5 (2 x-1) \log (10 x+6)-11}{121 (2 x-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11 + (5 - 10*x)*Log[1 - 2*x] + 5*(-1 + 2*x)*Log[6 + 10*x])/(121*(-1 + 2*x))

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Maple [A]  time = 0.006, size = 27, normalized size = 0.8 \begin{align*} -{\frac{1}{22\,x-11}}-{\frac{5\,\ln \left ( 2\,x-1 \right ) }{121}}+{\frac{5\,\ln \left ( 3+5\,x \right ) }{121}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/11/(2*x-1)-5/121*ln(2*x-1)+5/121*ln(3+5*x)

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Maxima [A]  time = 1.11908, size = 35, normalized size = 1.09 \begin{align*} -\frac{1}{11 \,{\left (2 \, x - 1\right )}} + \frac{5}{121} \, \log \left (5 \, x + 3\right ) - \frac{5}{121} \, \log \left (2 \, x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/11/(2*x - 1) + 5/121*log(5*x + 3) - 5/121*log(2*x - 1)

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Fricas [A]  time = 1.2648, size = 103, normalized size = 3.22 \begin{align*} \frac{5 \,{\left (2 \, x - 1\right )} \log \left (5 \, x + 3\right ) - 5 \,{\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 11}{121 \,{\left (2 \, x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/121*(5*(2*x - 1)*log(5*x + 3) - 5*(2*x - 1)*log(2*x - 1) - 11)/(2*x - 1)

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Sympy [A]  time = 0.11428, size = 26, normalized size = 0.81 \begin{align*} - \frac{5 \log{\left (x - \frac{1}{2} \right )}}{121} + \frac{5 \log{\left (x + \frac{3}{5} \right )}}{121} - \frac{1}{22 x - 11} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*log(x - 1/2)/121 + 5*log(x + 3/5)/121 - 1/(22*x - 11)

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Giac [A]  time = 2.21576, size = 34, normalized size = 1.06 \begin{align*} -\frac{1}{11 \,{\left (2 \, x - 1\right )}} + \frac{5}{121} \, \log \left ({\left | -\frac{11}{2 \, x - 1} - 5 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/11/(2*x - 1) + 5/121*log(abs(-11/(2*x - 1) - 5))