\(\int \frac {-20+(10 x-35 x^2+30 x^3) \log (\frac {2-3 x}{2 x})}{(-2 x+3 x^2) \log (\frac {2-3 x}{2 x})} \, dx\) [6652]

Optimal result

Integrand size = 57, antiderivative size = 22 \[ \int \frac {-20+\left (10 x-35 x^2+30 x^3\right ) \log \left (\frac {2-3 x}{2 x}\right )}{\left (-2 x+3 x^2\right ) \log \left (\frac {2-3 x}{2 x}\right )} \, dx=x \left (-5+5 \left (x-\frac {2 \log \left (\log \left (-\frac {3}{2}+\frac {1}{x}\right )\right )}{x}\right )\right ) \]

[Out]

(5*x-10*ln(ln(1/x-3/2))/x-5)*x

Rubi [F]

\[ \int \frac {-20+\left (10 x-35 x^2+30 x^3\right ) \log \left (\frac {2-3 x}{2 x}\right )}{\left (-2 x+3 x^2\right ) \log \left (\frac {2-3 x}{2 x}\right )} \, dx=\int \frac {-20+\left (10 x-35 x^2+30 x^3\right ) \log \left (\frac {2-3 x}{2 x}\right )}{\left (-2 x+3 x^2\right ) \log \left (\frac {2-3 x}{2 x}\right )} \, dx \]

[In]

Int[(-20 + (10*x - 35*x^2 + 30*x^3)*Log[(2 - 3*x)/(2*x)])/((-2*x + 3*x^2)*Log[(2 - 3*x)/(2*x)]),x]

[Out]

-5*x + 5*x^2 + 10*Defer[Int][1/(x*Log[-3/2 + x^(-1)]), x] - 30*Defer[Int][1/((-2 + 3*x)*Log[-3/2 + x^(-1)]), x
]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-20+\left (10 x-35 x^2+30 x^3\right ) \log \left (\frac {2-3 x}{2 x}\right )}{x (-2+3 x) \log \left (\frac {2-3 x}{2 x}\right )} \, dx \\ & = \int \frac {5 \left (-2+7 x-6 x^2+\frac {4}{x \log \left (-\frac {3}{2}+\frac {1}{x}\right )}\right )}{2-3 x} \, dx \\ & = 5 \int \frac {-2+7 x-6 x^2+\frac {4}{x \log \left (-\frac {3}{2}+\frac {1}{x}\right )}}{2-3 x} \, dx \\ & = 5 \int \left (-1+2 x-\frac {4}{x (-2+3 x) \log \left (-\frac {3}{2}+\frac {1}{x}\right )}\right ) \, dx \\ & = -5 x+5 x^2-20 \int \frac {1}{x (-2+3 x) \log \left (-\frac {3}{2}+\frac {1}{x}\right )} \, dx \\ & = -5 x+5 x^2-20 \int \left (-\frac {1}{2 x \log \left (-\frac {3}{2}+\frac {1}{x}\right )}+\frac {3}{2 (-2+3 x) \log \left (-\frac {3}{2}+\frac {1}{x}\right )}\right ) \, dx \\ & = -5 x+5 x^2+10 \int \frac {1}{x \log \left (-\frac {3}{2}+\frac {1}{x}\right )} \, dx-30 \int \frac {1}{(-2+3 x) \log \left (-\frac {3}{2}+\frac {1}{x}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-20+\left (10 x-35 x^2+30 x^3\right ) \log \left (\frac {2-3 x}{2 x}\right )}{\left (-2 x+3 x^2\right ) \log \left (\frac {2-3 x}{2 x}\right )} \, dx=5 \left (-x+x^2-2 \log \left (\log \left (-\frac {3}{2}+\frac {1}{x}\right )\right )\right ) \]

[In]

Integrate[(-20 + (10*x - 35*x^2 + 30*x^3)*Log[(2 - 3*x)/(2*x)])/((-2*x + 3*x^2)*Log[(2 - 3*x)/(2*x)]),x]

[Out]

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

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(-10 \ln \left (\ln \left (\frac {1}{x}-\frac {3}{2}\right )\right )-5 x +5 x^{2}\)	\(19\)
default	\(-10 \ln \left (\ln \left (\frac {1}{x}-\frac {3}{2}\right )\right )-5 x +5 x^{2}\)	\(19\)
norman	\(-5 x +5 x^{2}-10 \ln \left (\ln \left (\frac {2-3 x}{2 x}\right )\right )\)	\(24\)
risch	\(-5 x +5 x^{2}-10 \ln \left (\ln \left (\frac {2-3 x}{2 x}\right )\right )\)	\(24\)
parts	\(-5 x +5 x^{2}-10 \ln \left (\ln \left (\frac {2-3 x}{2 x}\right )\right )\)	\(24\)
parallelrisch	\(-\frac {80}{9}+5 x^{2}-10 \ln \left (\ln \left (-\frac {-2+3 x}{2 x}\right )\right )-5 x\)	\(25\)

[In]

int(((30*x^3-35*x^2+10*x)*ln(1/2*(2-3*x)/x)-20)/(3*x^2-2*x)/ln(1/2*(2-3*x)/x),x,method=_RETURNVERBOSE)

[Out]

-10*ln(ln(1/x-3/2))-5*x+5*x^2

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {-20+\left (10 x-35 x^2+30 x^3\right ) \log \left (\frac {2-3 x}{2 x}\right )}{\left (-2 x+3 x^2\right ) \log \left (\frac {2-3 x}{2 x}\right )} \, dx=5 \, x^{2} - 5 \, x - 10 \, \log \left (\log \left (-\frac {3 \, x - 2}{2 \, x}\right )\right ) \]

[In]

integrate(((30*x^3-35*x^2+10*x)*log(1/2*(2-3*x)/x)-20)/(3*x^2-2*x)/log(1/2*(2-3*x)/x),x, algorithm="fricas")

[Out]

5*x^2 - 5*x - 10*log(log(-1/2*(3*x - 2)/x))

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-20+\left (10 x-35 x^2+30 x^3\right ) \log \left (\frac {2-3 x}{2 x}\right )}{\left (-2 x+3 x^2\right ) \log \left (\frac {2-3 x}{2 x}\right )} \, dx=5 x^{2} - 5 x - 10 \log {\left (\log {\left (\frac {1 - \frac {3 x}{2}}{x} \right )} \right )} \]

[In]

integrate(((30*x**3-35*x**2+10*x)*ln(1/2*(2-3*x)/x)-20)/(3*x**2-2*x)/ln(1/2*(2-3*x)/x),x)

[Out]

5*x**2 - 5*x - 10*log(log((1 - 3*x/2)/x))

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-20+\left (10 x-35 x^2+30 x^3\right ) \log \left (\frac {2-3 x}{2 x}\right )}{\left (-2 x+3 x^2\right ) \log \left (\frac {2-3 x}{2 x}\right )} \, dx=5 \, x^{2} - 5 \, x - 10 \, \log \left (-\log \left (2\right ) - \log \left (x\right ) + \log \left (-3 \, x + 2\right )\right ) \]

[In]

integrate(((30*x^3-35*x^2+10*x)*log(1/2*(2-3*x)/x)-20)/(3*x^2-2*x)/log(1/2*(2-3*x)/x),x, algorithm="maxima")

[Out]

5*x^2 - 5*x - 10*log(-log(2) - log(x) + log(-3*x + 2))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (18) = 36\).

Time = 0.30 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.41 \[ \int \frac {-20+\left (10 x-35 x^2+30 x^3\right ) \log \left (\frac {2-3 x}{2 x}\right )}{\left (-2 x+3 x^2\right ) \log \left (\frac {2-3 x}{2 x}\right )} \, dx=\frac {10 \, {\left (\frac {3 \, x - 2}{x} - 1\right )}}{\frac {{\left (3 \, x - 2\right )}^{2}}{x^{2}} - \frac {6 \, {\left (3 \, x - 2\right )}}{x} + 9} - 10 \, \log \left (\log \left (-\frac {3 \, x - 2}{2 \, x}\right )\right ) \]

[In]

integrate(((30*x^3-35*x^2+10*x)*log(1/2*(2-3*x)/x)-20)/(3*x^2-2*x)/log(1/2*(2-3*x)/x),x, algorithm="giac")

[Out]

10*((3*x - 2)/x - 1)/((3*x - 2)^2/x^2 - 6*(3*x - 2)/x + 9) - 10*log(log(-1/2*(3*x - 2)/x))

Mupad [B] (verification not implemented)

Time = 12.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {-20+\left (10 x-35 x^2+30 x^3\right ) \log \left (\frac {2-3 x}{2 x}\right )}{\left (-2 x+3 x^2\right ) \log \left (\frac {2-3 x}{2 x}\right )} \, dx=5\,x^2-10\,\ln \left (\ln \left (-\frac {3\,x-2}{2\,x}\right )\right )-5\,x \]

[In]

int(-(log(-((3*x)/2 - 1)/x)*(10*x - 35*x^2 + 30*x^3) - 20)/(log(-((3*x)/2 - 1)/x)*(2*x - 3*x^2)),x)

[Out]

5*x^2 - 10*log(log(-(3*x - 2)/(2*x))) - 5*x