3.162 \(\int \frac{1}{(5-6 x)^2 x^2} \, dx\)

Optimal. Leaf size=35 \[ \frac{6}{25 (5-6 x)}-\frac{1}{25 x}-\frac{12}{125} \log (5-6 x)+\frac{12 \log (x)}{125} \]

[Out]

6/(25*(5 - 6*x)) - 1/(25*x) - (12*Log[5 - 6*x])/125 + (12*Log[x])/125

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Rubi [A]  time = 0.0142911, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {44} \[ \frac{6}{25 (5-6 x)}-\frac{1}{25 x}-\frac{12}{125} \log (5-6 x)+\frac{12 \log (x)}{125} \]

Antiderivative was successfully verified.

[In]

Int[1/((5 - 6*x)^2*x^2),x]

[Out]

6/(25*(5 - 6*x)) - 1/(25*x) - (12*Log[5 - 6*x])/125 + (12*Log[x])/125

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}{(5-6 x)^2 x^2} \, dx &=\int \left (\frac{1}{25 x^2}+\frac{12}{125 x}+\frac{36}{25 (-5+6 x)^2}-\frac{72}{125 (-5+6 x)}\right ) \, dx\\ &=\frac{6}{25 (5-6 x)}-\frac{1}{25 x}-\frac{12}{125} \log (5-6 x)+\frac{12 \log (x)}{125}\\ \end{align*}

Mathematica [A]  time = 0.0176111, size = 31, normalized size = 0.89 \[ \frac{1}{125} \left (\frac{30}{5-6 x}-\frac{5}{x}-12 \log (5-6 x)+12 \log (x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/((5 - 6*x)^2*x^2),x]

[Out]

(30/(5 - 6*x) - 5/x - 12*Log[5 - 6*x] + 12*Log[x])/125

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Maple [A]  time = 0.008, size = 28, normalized size = 0.8 \begin{align*} -{\frac{6}{-125+150\,x}}-{\frac{12\,\ln \left ( -5+6\,x \right ) }{125}}-{\frac{1}{25\,x}}+{\frac{12\,\ln \left ( x \right ) }{125}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5-6*x)^2/x^2,x)

[Out]

-6/25/(-5+6*x)-12/125*ln(-5+6*x)-1/25/x+12/125*ln(x)

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Maxima [A]  time = 0.928173, size = 42, normalized size = 1.2 \begin{align*} -\frac{12 \, x - 5}{25 \,{\left (6 \, x^{2} - 5 \, x\right )}} - \frac{12}{125} \, \log \left (6 \, x - 5\right ) + \frac{12}{125} \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5-6*x)^2/x^2,x, algorithm="maxima")

[Out]

-1/25*(12*x - 5)/(6*x^2 - 5*x) - 12/125*log(6*x - 5) + 12/125*log(x)

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Fricas [A]  time = 1.93243, size = 124, normalized size = 3.54 \begin{align*} -\frac{12 \,{\left (6 \, x^{2} - 5 \, x\right )} \log \left (6 \, x - 5\right ) - 12 \,{\left (6 \, x^{2} - 5 \, x\right )} \log \left (x\right ) + 60 \, x - 25}{125 \,{\left (6 \, x^{2} - 5 \, x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5-6*x)^2/x^2,x, algorithm="fricas")

[Out]

-1/125*(12*(6*x^2 - 5*x)*log(6*x - 5) - 12*(6*x^2 - 5*x)*log(x) + 60*x - 25)/(6*x^2 - 5*x)

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Sympy [A]  time = 0.118725, size = 29, normalized size = 0.83 \begin{align*} - \frac{12 x - 5}{150 x^{2} - 125 x} + \frac{12 \log{\left (x \right )}}{125} - \frac{12 \log{\left (x - \frac{5}{6} \right )}}{125} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5-6*x)**2/x**2,x)

[Out]

-(12*x - 5)/(150*x**2 - 125*x) + 12*log(x)/125 - 12*log(x - 5/6)/125

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Giac [A]  time = 1.06476, size = 54, normalized size = 1.54 \begin{align*} -\frac{6}{25 \,{\left (6 \, x - 5\right )}} + \frac{6}{125 \,{\left (\frac{5}{6 \, x - 5} + 1\right )}} + \frac{12}{125} \, \log \left ({\left | -\frac{5}{6 \, x - 5} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5-6*x)^2/x^2,x, algorithm="giac")

[Out]

-6/25/(6*x - 5) + 6/125/(5/(6*x - 5) + 1) + 12/125*log(abs(-5/(6*x - 5) - 1))