\(\int \frac {-6+8 x+(-16 x-8 x^2) \log (\frac {1}{4} (-3+4 x))+(6-8 x) \log ^2(\frac {1}{4} (-3+4 x))}{-3 x^2+4 x^3+(6 x^2-8 x^3) \log ^2(\frac {1}{4} (-3+4 x))+(-3 x^2+4 x^3) \log ^4(\frac {1}{4} (-3+4 x))} \, dx\) [3234]

Optimal result

Integrand size = 105, antiderivative size = 22 \[ \int \frac {-6+8 x+\left (-16 x-8 x^2\right ) \log \left (\frac {1}{4} (-3+4 x)\right )+(6-8 x) \log ^2\left (\frac {1}{4} (-3+4 x)\right )}{-3 x^2+4 x^3+\left (6 x^2-8 x^3\right ) \log ^2\left (\frac {1}{4} (-3+4 x)\right )+\left (-3 x^2+4 x^3\right ) \log ^4\left (\frac {1}{4} (-3+4 x)\right )} \, dx=3-\frac {2+x}{x-x \log ^2\left (-\frac {3}{4}+x\right )} \]

[Out]

3-1/(x-ln(x-3/4)^2*x)*(2+x)

Rubi [F]

\[ \int \frac {-6+8 x+\left (-16 x-8 x^2\right ) \log \left (\frac {1}{4} (-3+4 x)\right )+(6-8 x) \log ^2\left (\frac {1}{4} (-3+4 x)\right )}{-3 x^2+4 x^3+\left (6 x^2-8 x^3\right ) \log ^2\left (\frac {1}{4} (-3+4 x)\right )+\left (-3 x^2+4 x^3\right ) \log ^4\left (\frac {1}{4} (-3+4 x)\right )} \, dx=\int \frac {-6+8 x+\left (-16 x-8 x^2\right ) \log \left (\frac {1}{4} (-3+4 x)\right )+(6-8 x) \log ^2\left (\frac {1}{4} (-3+4 x)\right )}{-3 x^2+4 x^3+\left (6 x^2-8 x^3\right ) \log ^2\left (\frac {1}{4} (-3+4 x)\right )+\left (-3 x^2+4 x^3\right ) \log ^4\left (\frac {1}{4} (-3+4 x)\right )} \, dx \]

[In]

Int[(-6 + 8*x + (-16*x - 8*x^2)*Log[(-3 + 4*x)/4] + (6 - 8*x)*Log[(-3 + 4*x)/4]^2)/(-3*x^2 + 4*x^3 + (6*x^2 -
8*x^3)*Log[(-3 + 4*x)/4]^2 + (-3*x^2 + 4*x^3)*Log[(-3 + 4*x)/4]^4),x]

[Out]

-11/(6*(1 - Log[-3/4 + x])) - 11/(6*(1 + Log[-3/4 + x])) + (4*Defer[Int][1/(x*(-1 + Log[-3/4 + x])^2), x])/3 -
Defer[Int][1/(x^2*(-1 + Log[-3/4 + x])), x] - (4*Defer[Int][1/(x*(1 + Log[-3/4 + x])^2), x])/3 + Defer[Int][1
/(x^2*(1 + Log[-3/4 + x])), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {2 \left (3-4 x+4 x (2+x) \log \left (-\frac {3}{4}+x\right )+(-3+4 x) \log ^2\left (-\frac {3}{4}+x\right )\right )}{(3-4 x) x^2 \left (1-\log ^2\left (-\frac {3}{4}+x\right )\right )^2} \, dx \\ & = 2 \int \frac {3-4 x+4 x (2+x) \log \left (-\frac {3}{4}+x\right )+(-3+4 x) \log ^2\left (-\frac {3}{4}+x\right )}{(3-4 x) x^2 \left (1-\log ^2\left (-\frac {3}{4}+x\right )\right )^2} \, dx \\ & = 2 \int \left (\frac {-2-x}{x (-3+4 x) \left (-1+\log \left (-\frac {3}{4}+x\right )\right )^2}-\frac {1}{2 x^2 \left (-1+\log \left (-\frac {3}{4}+x\right )\right )}+\frac {2+x}{x (-3+4 x) \left (1+\log \left (-\frac {3}{4}+x\right )\right )^2}+\frac {1}{2 x^2 \left (1+\log \left (-\frac {3}{4}+x\right )\right )}\right ) \, dx \\ & = 2 \int \frac {-2-x}{x (-3+4 x) \left (-1+\log \left (-\frac {3}{4}+x\right )\right )^2} \, dx+2 \int \frac {2+x}{x (-3+4 x) \left (1+\log \left (-\frac {3}{4}+x\right )\right )^2} \, dx-\int \frac {1}{x^2 \left (-1+\log \left (-\frac {3}{4}+x\right )\right )} \, dx+\int \frac {1}{x^2 \left (1+\log \left (-\frac {3}{4}+x\right )\right )} \, dx \\ & = 2 \int \left (\frac {2}{3 x \left (-1+\log \left (-\frac {3}{4}+x\right )\right )^2}-\frac {11}{3 (-3+4 x) \left (-1+\log \left (-\frac {3}{4}+x\right )\right )^2}\right ) \, dx+2 \int \left (-\frac {2}{3 x \left (1+\log \left (-\frac {3}{4}+x\right )\right )^2}+\frac {11}{3 (-3+4 x) \left (1+\log \left (-\frac {3}{4}+x\right )\right )^2}\right ) \, dx-\int \frac {1}{x^2 \left (-1+\log \left (-\frac {3}{4}+x\right )\right )} \, dx+\int \frac {1}{x^2 \left (1+\log \left (-\frac {3}{4}+x\right )\right )} \, dx \\ & = \frac {4}{3} \int \frac {1}{x \left (-1+\log \left (-\frac {3}{4}+x\right )\right )^2} \, dx-\frac {4}{3} \int \frac {1}{x \left (1+\log \left (-\frac {3}{4}+x\right )\right )^2} \, dx-\frac {22}{3} \int \frac {1}{(-3+4 x) \left (-1+\log \left (-\frac {3}{4}+x\right )\right )^2} \, dx+\frac {22}{3} \int \frac {1}{(-3+4 x) \left (1+\log \left (-\frac {3}{4}+x\right )\right )^2} \, dx-\int \frac {1}{x^2 \left (-1+\log \left (-\frac {3}{4}+x\right )\right )} \, dx+\int \frac {1}{x^2 \left (1+\log \left (-\frac {3}{4}+x\right )\right )} \, dx \\ & = \frac {4}{3} \int \frac {1}{x \left (-1+\log \left (-\frac {3}{4}+x\right )\right )^2} \, dx-\frac {4}{3} \int \frac {1}{x \left (1+\log \left (-\frac {3}{4}+x\right )\right )^2} \, dx-\frac {22}{3} \text {Subst}\left (\int \frac {1}{4 x (-1+\log (x))^2} \, dx,x,-\frac {3}{4}+x\right )+\frac {22}{3} \text {Subst}\left (\int \frac {1}{4 x (1+\log (x))^2} \, dx,x,-\frac {3}{4}+x\right )-\int \frac {1}{x^2 \left (-1+\log \left (-\frac {3}{4}+x\right )\right )} \, dx+\int \frac {1}{x^2 \left (1+\log \left (-\frac {3}{4}+x\right )\right )} \, dx \\ & = \frac {4}{3} \int \frac {1}{x \left (-1+\log \left (-\frac {3}{4}+x\right )\right )^2} \, dx-\frac {4}{3} \int \frac {1}{x \left (1+\log \left (-\frac {3}{4}+x\right )\right )^2} \, dx-\frac {11}{6} \text {Subst}\left (\int \frac {1}{x (-1+\log (x))^2} \, dx,x,-\frac {3}{4}+x\right )+\frac {11}{6} \text {Subst}\left (\int \frac {1}{x (1+\log (x))^2} \, dx,x,-\frac {3}{4}+x\right )-\int \frac {1}{x^2 \left (-1+\log \left (-\frac {3}{4}+x\right )\right )} \, dx+\int \frac {1}{x^2 \left (1+\log \left (-\frac {3}{4}+x\right )\right )} \, dx \\ & = \frac {4}{3} \int \frac {1}{x \left (-1+\log \left (-\frac {3}{4}+x\right )\right )^2} \, dx-\frac {4}{3} \int \frac {1}{x \left (1+\log \left (-\frac {3}{4}+x\right )\right )^2} \, dx-\frac {11}{6} \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,-1+\log \left (-\frac {3}{4}+x\right )\right )+\frac {11}{6} \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,1+\log \left (-\frac {3}{4}+x\right )\right )-\int \frac {1}{x^2 \left (-1+\log \left (-\frac {3}{4}+x\right )\right )} \, dx+\int \frac {1}{x^2 \left (1+\log \left (-\frac {3}{4}+x\right )\right )} \, dx \\ & = -\frac {11}{6 \left (1-\log \left (-\frac {3}{4}+x\right )\right )}-\frac {11}{6 \left (1+\log \left (-\frac {3}{4}+x\right )\right )}+\frac {4}{3} \int \frac {1}{x \left (-1+\log \left (-\frac {3}{4}+x\right )\right )^2} \, dx-\frac {4}{3} \int \frac {1}{x \left (1+\log \left (-\frac {3}{4}+x\right )\right )^2} \, dx-\int \frac {1}{x^2 \left (-1+\log \left (-\frac {3}{4}+x\right )\right )} \, dx+\int \frac {1}{x^2 \left (1+\log \left (-\frac {3}{4}+x\right )\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-6+8 x+\left (-16 x-8 x^2\right ) \log \left (\frac {1}{4} (-3+4 x)\right )+(6-8 x) \log ^2\left (\frac {1}{4} (-3+4 x)\right )}{-3 x^2+4 x^3+\left (6 x^2-8 x^3\right ) \log ^2\left (\frac {1}{4} (-3+4 x)\right )+\left (-3 x^2+4 x^3\right ) \log ^4\left (\frac {1}{4} (-3+4 x)\right )} \, dx=-\frac {-2-x}{x \left (-1+\log ^2\left (-\frac {3}{4}+x\right )\right )} \]

[In]

Integrate[(-6 + 8*x + (-16*x - 8*x^2)*Log[(-3 + 4*x)/4] + (6 - 8*x)*Log[(-3 + 4*x)/4]^2)/(-3*x^2 + 4*x^3 + (6*
x^2 - 8*x^3)*Log[(-3 + 4*x)/4]^2 + (-3*x^2 + 4*x^3)*Log[(-3 + 4*x)/4]^4),x]

[Out]

-((-2 - x)/(x*(-1 + Log[-3/4 + x]^2)))

Maple [A] (verified)

Time = 2.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82

method	result	size
norman	\(\frac {2+x}{x \left (\ln \left (x -\frac {3}{4}\right )^{2}-1\right )}\)	\(18\)
risch	\(\frac {2+x}{x \left (\ln \left (x -\frac {3}{4}\right )^{2}-1\right )}\)	\(18\)
derivativedivides	\(-\frac {-2 x -4}{2 x \left (\ln \left (x -\frac {3}{4}\right )^{2}-1\right )}\)	\(21\)
default	\(-\frac {-2 x -4}{2 x \left (\ln \left (x -\frac {3}{4}\right )^{2}-1\right )}\)	\(21\)
parallelrisch	\(\frac {16 x +32}{16 \left (\ln \left (x -\frac {3}{4}\right )^{2}-1\right ) x}\)	\(21\)

[In]

int(((-8*x+6)*ln(x-3/4)^2+(-8*x^2-16*x)*ln(x-3/4)+8*x-6)/((4*x^3-3*x^2)*ln(x-3/4)^4+(-8*x^3+6*x^2)*ln(x-3/4)^2
+4*x^3-3*x^2),x,method=_RETURNVERBOSE)

[Out]

(2+x)/x/(ln(x-3/4)^2-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {-6+8 x+\left (-16 x-8 x^2\right ) \log \left (\frac {1}{4} (-3+4 x)\right )+(6-8 x) \log ^2\left (\frac {1}{4} (-3+4 x)\right )}{-3 x^2+4 x^3+\left (6 x^2-8 x^3\right ) \log ^2\left (\frac {1}{4} (-3+4 x)\right )+\left (-3 x^2+4 x^3\right ) \log ^4\left (\frac {1}{4} (-3+4 x)\right )} \, dx=\frac {x + 2}{x \log \left (x - \frac {3}{4}\right )^{2} - x} \]

[In]

integrate(((-8*x+6)*log(x-3/4)^2+(-8*x^2-16*x)*log(x-3/4)+8*x-6)/((4*x^3-3*x^2)*log(x-3/4)^4+(-8*x^3+6*x^2)*lo
g(x-3/4)^2+4*x^3-3*x^2),x, algorithm="fricas")

[Out]

(x + 2)/(x*log(x - 3/4)^2 - x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {-6+8 x+\left (-16 x-8 x^2\right ) \log \left (\frac {1}{4} (-3+4 x)\right )+(6-8 x) \log ^2\left (\frac {1}{4} (-3+4 x)\right )}{-3 x^2+4 x^3+\left (6 x^2-8 x^3\right ) \log ^2\left (\frac {1}{4} (-3+4 x)\right )+\left (-3 x^2+4 x^3\right ) \log ^4\left (\frac {1}{4} (-3+4 x)\right )} \, dx=\frac {x + 2}{x \log {\left (x - \frac {3}{4} \right )}^{2} - x} \]

[In]

integrate(((-8*x+6)*ln(x-3/4)**2+(-8*x**2-16*x)*ln(x-3/4)+8*x-6)/((4*x**3-3*x**2)*ln(x-3/4)**4+(-8*x**3+6*x**2
)*ln(x-3/4)**2+4*x**3-3*x**2),x)

[Out]

(x + 2)/(x*log(x - 3/4)**2 - x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).

Time = 0.31 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.86 \[ \int \frac {-6+8 x+\left (-16 x-8 x^2\right ) \log \left (\frac {1}{4} (-3+4 x)\right )+(6-8 x) \log ^2\left (\frac {1}{4} (-3+4 x)\right )}{-3 x^2+4 x^3+\left (6 x^2-8 x^3\right ) \log ^2\left (\frac {1}{4} (-3+4 x)\right )+\left (-3 x^2+4 x^3\right ) \log ^4\left (\frac {1}{4} (-3+4 x)\right )} \, dx=-\frac {x + 2}{4 \, x \log \left (2\right ) \log \left (4 \, x - 3\right ) - x \log \left (4 \, x - 3\right )^{2} - {\left (4 \, \log \left (2\right )^{2} - 1\right )} x} \]

[In]

integrate(((-8*x+6)*log(x-3/4)^2+(-8*x^2-16*x)*log(x-3/4)+8*x-6)/((4*x^3-3*x^2)*log(x-3/4)^4+(-8*x^3+6*x^2)*lo
g(x-3/4)^2+4*x^3-3*x^2),x, algorithm="maxima")

[Out]

-(x + 2)/(4*x*log(2)*log(4*x - 3) - x*log(4*x - 3)^2 - (4*log(2)^2 - 1)*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {-6+8 x+\left (-16 x-8 x^2\right ) \log \left (\frac {1}{4} (-3+4 x)\right )+(6-8 x) \log ^2\left (\frac {1}{4} (-3+4 x)\right )}{-3 x^2+4 x^3+\left (6 x^2-8 x^3\right ) \log ^2\left (\frac {1}{4} (-3+4 x)\right )+\left (-3 x^2+4 x^3\right ) \log ^4\left (\frac {1}{4} (-3+4 x)\right )} \, dx=\frac {x + 2}{x \log \left (x - \frac {3}{4}\right )^{2} - x} \]

[In]

integrate(((-8*x+6)*log(x-3/4)^2+(-8*x^2-16*x)*log(x-3/4)+8*x-6)/((4*x^3-3*x^2)*log(x-3/4)^4+(-8*x^3+6*x^2)*lo
g(x-3/4)^2+4*x^3-3*x^2),x, algorithm="giac")

[Out]

(x + 2)/(x*log(x - 3/4)^2 - x)

Mupad [B] (verification not implemented)

Time = 9.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {-6+8 x+\left (-16 x-8 x^2\right ) \log \left (\frac {1}{4} (-3+4 x)\right )+(6-8 x) \log ^2\left (\frac {1}{4} (-3+4 x)\right )}{-3 x^2+4 x^3+\left (6 x^2-8 x^3\right ) \log ^2\left (\frac {1}{4} (-3+4 x)\right )+\left (-3 x^2+4 x^3\right ) \log ^4\left (\frac {1}{4} (-3+4 x)\right )} \, dx=\frac {x+2}{x\,\left ({\ln \left (x-\frac {3}{4}\right )}^2-1\right )} \]

[In]

int((log(x - 3/4)*(16*x + 8*x^2) - 8*x + log(x - 3/4)^2*(8*x - 6) + 6)/(log(x - 3/4)^4*(3*x^2 - 4*x^3) - log(x
- 3/4)^2*(6*x^2 - 8*x^3) + 3*x^2 - 4*x^3),x)

[Out]

(x + 2)/(x*(log(x - 3/4)^2 - 1))