\(\int \frac {1}{(5-6 x)^2 x^2} \, dx\) [162]

Optimal result

Integrand size = 11, antiderivative size = 35 \[ \int \frac {1}{(5-6 x)^2 x^2} \, dx=\frac {6}{25 (5-6 x)}-\frac {1}{25 x}-\frac {12}{125} \log (5-6 x)+\frac {12 \log (x)}{125} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {46} \[ \int \frac {1}{(5-6 x)^2 x^2} \, dx=\frac {6}{25 (5-6 x)}-\frac {1}{25 x}-\frac {12}{125} \log (5-6 x)+\frac {12 \log (x)}{125} \]

[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 46

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] && Lt
Q[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(5-6 x)^2 x^2} \, dx=\frac {1}{125} \left (\frac {30}{5-6 x}-\frac {5}{x}-12 \log (5-6 x)+12 \log (x)\right ) \]

[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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80

method	result	size
default	\(-\frac {1}{25 x}+\frac {12 \ln \left (x \right )}{125}-\frac {6}{25 \left (6 x -5\right )}-\frac {12 \ln \left (6 x -5\right )}{125}\)	\(28\)
risch	\(\frac {-\frac {12 x}{25}+\frac {1}{5}}{x \left (6 x -5\right )}+\frac {12 \ln \left (x \right )}{125}-\frac {12 \ln \left (6 x -5\right )}{125}\)	\(31\)
norman	\(\frac {\frac {1}{5}-\frac {72 x^{2}}{125}}{x \left (6 x -5\right )}+\frac {12 \ln \left (x \right )}{125}-\frac {12 \ln \left (6 x -5\right )}{125}\)	\(32\)
meijerg	\(-\frac {1}{25 x}+\frac {6}{125}+\frac {12 \ln \left (x \right )}{125}+\frac {12 \ln \left (2\right )}{125}+\frac {12 \ln \left (3\right )}{125}-\frac {12 \ln \left (5\right )}{125}+\frac {12 i \pi }{125}+\frac {108 x}{625 \left (3-\frac {18 x}{5}\right )}-\frac {12 \ln \left (1-\frac {6 x}{5}\right )}{125}\)	\(46\)
parallelrisch	\(\frac {72 x^{2} \ln \left (x \right )-72 \ln \left (x -\frac {5}{6}\right ) x^{2}+25-60 x \ln \left (x \right )+60 \ln \left (x -\frac {5}{6}\right ) x -72 x^{2}}{125 \left (6 x -5\right ) x}\)	\(48\)

[In]

int(1/(-6*x+5)^2/x^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.37 \[ \int \frac {1}{(5-6 x)^2 x^2} \, dx=-\frac {12 \, {\left (6 \, x^{2} - 5 \, x\right )} \log \left (6 \, x - 5\right ) - 12 \, {\left (6 \, x^{2} - 5 \, x\right )} \log \left (x\right ) + 60 \, x - 25}{125 \, {\left (6 \, x^{2} - 5 \, x\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(5-6 x)^2 x^2} \, dx=\frac {5 - 12 x}{150 x^{2} - 125 x} + \frac {12 \log {\left (x \right )}}{125} - \frac {12 \log {\left (x - \frac {5}{6} \right )}}{125} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(5-6 x)^2 x^2} \, dx=-\frac {12 \, x - 5}{25 \, {\left (6 \, x^{2} - 5 \, x\right )}} - \frac {12}{125} \, \log \left (6 \, x - 5\right ) + \frac {12}{125} \, \log \left (x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.14 \[ \int \frac {1}{(5-6 x)^2 x^2} \, dx=-\frac {6}{25 \, {\left (6 \, x - 5\right )}} + \frac {6}{125 \, {\left (\frac {5}{6 \, x - 5} + 1\right )}} + \frac {12}{125} \, \log \left ({\left | -\frac {5}{6 \, x - 5} - 1 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(5-6 x)^2 x^2} \, dx=\frac {1}{5\,x\,\left (6\,x-5\right )}-\frac {12}{25\,\left (6\,x-5\right )}-\frac {12\,\ln \left (\frac {6\,x-5}{x}\right )}{125} \]

[In]

int(1/(x^2*(6*x - 5)^2),x)

[Out]

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