\(\int \frac {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{((-3-x+2 x^2) \log ^2(16)+(-6-2 x+4 x^2) \log (16) \log (1+x)+(-3-x+2 x^2) \log ^2(1+x)) \log ^2(-3+2 x)} \, dx\) [3886]

Optimal result

Integrand size = 92, antiderivative size = 19 \[ \int \frac {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{\left (\left (-3-x+2 x^2\right ) \log ^2(16)+\left (-6-2 x+4 x^2\right ) \log (16) \log (1+x)+\left (-3-x+2 x^2\right ) \log ^2(1+x)\right ) \log ^2(-3+2 x)} \, dx=\frac {3}{(\log (16)+\log (1+x)) \log (-3+2 x)} \]

[Out]

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

Rubi [F]

\[ \int \frac {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{\left (\left (-3-x+2 x^2\right ) \log ^2(16)+\left (-6-2 x+4 x^2\right ) \log (16) \log (1+x)+\left (-3-x+2 x^2\right ) \log ^2(1+x)\right ) \log ^2(-3+2 x)} \, dx=\int \frac {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{\left (\left (-3-x+2 x^2\right ) \log ^2(16)+\left (-6-2 x+4 x^2\right ) \log (16) \log (1+x)+\left (-3-x+2 x^2\right ) \log ^2(1+x)\right ) \log ^2(-3+2 x)} \, dx \]

[In]

Int[((-6 - 6*x)*Log[16] + (-6 - 6*x)*Log[1 + x] + (9 - 6*x)*Log[-3 + 2*x])/(((-3 - x + 2*x^2)*Log[16]^2 + (-6
- 2*x + 4*x^2)*Log[16]*Log[1 + x] + (-3 - x + 2*x^2)*Log[1 + x]^2)*Log[-3 + 2*x]^2),x]

[Out]

3*Log[256]*Defer[Int][1/((3 - 2*x)*Log[-3 + 2*x]^2*Log[16 + 16*x]^2), x] + (3*Log[256]*Defer[Int][1/((-1 - x)*
Log[-3 + 2*x]^2*Log[16 + 16*x]^2), x])/5 + (3*Log[256]*Defer[Int][1/((1 + x)*Log[-3 + 2*x]^2*Log[16 + 16*x]^2)
, x])/5 + 6*Defer[Int][Log[1 + x]/((3 - 2*x)*Log[-3 + 2*x]^2*Log[16 + 16*x]^2), x] + (6*Defer[Int][Log[1 + x]/
((-1 - x)*Log[-3 + 2*x]^2*Log[16 + 16*x]^2), x])/5 + (6*Defer[Int][Log[1 + x]/((1 + x)*Log[-3 + 2*x]^2*Log[16
+ 16*x]^2), x])/5 + (18*Defer[Int][1/((3 - 2*x)*Log[-3 + 2*x]*Log[16 + 16*x]^2), x])/5 + 3*Defer[Int][1/((-1 -
x)*Log[-3 + 2*x]*Log[16 + 16*x]^2), x] + (18*Defer[Int][1/((-3 + 2*x)*Log[-3 + 2*x]*Log[16 + 16*x]^2), x])/5

Rubi steps \begin{align*} \text {integral}& = \int \frac {3 ((1+x) \log (256)+2 (1+x) \log (1+x)+(-3+2 x) \log (-3+2 x))}{\left (3+x-2 x^2\right ) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx \\ & = 3 \int \frac {(1+x) \log (256)+2 (1+x) \log (1+x)+(-3+2 x) \log (-3+2 x)}{\left (3+x-2 x^2\right ) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx \\ & = 3 \int \left (\frac {2 (\log (256)+x \log (256)+2 \log (1+x)+2 x \log (1+x)-3 \log (-3+2 x)+2 x \log (-3+2 x))}{5 (3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {\log (256)+x \log (256)+2 \log (1+x)+2 x \log (1+x)-3 \log (-3+2 x)+2 x \log (-3+2 x)}{5 (1+x) \log ^2(-3+2 x) \log ^2(16+16 x)}\right ) \, dx \\ & = \frac {3}{5} \int \frac {\log (256)+x \log (256)+2 \log (1+x)+2 x \log (1+x)-3 \log (-3+2 x)+2 x \log (-3+2 x)}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {6}{5} \int \frac {\log (256)+x \log (256)+2 \log (1+x)+2 x \log (1+x)-3 \log (-3+2 x)+2 x \log (-3+2 x)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx \\ & = \frac {3}{5} \int \left (\frac {\log (256)}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {x \log (256)}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {2 \log (1+x)}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {2 x \log (1+x)}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {3}{(-1-x) \log (-3+2 x) \log ^2(16+16 x)}+\frac {2 x}{(1+x) \log (-3+2 x) \log ^2(16+16 x)}\right ) \, dx+\frac {6}{5} \int \left (\frac {\log (256)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {x \log (256)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {2 \log (1+x)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {2 x \log (1+x)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {2 x}{(3-2 x) \log (-3+2 x) \log ^2(16+16 x)}+\frac {3}{(-3+2 x) \log (-3+2 x) \log ^2(16+16 x)}\right ) \, dx \\ & = \frac {6}{5} \int \frac {\log (1+x)}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {6}{5} \int \frac {x \log (1+x)}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {6}{5} \int \frac {x}{(1+x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {9}{5} \int \frac {1}{(-1-x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {12}{5} \int \frac {\log (1+x)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {12}{5} \int \frac {x \log (1+x)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {12}{5} \int \frac {x}{(3-2 x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {18}{5} \int \frac {1}{(-3+2 x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (3 \log (256)) \int \frac {1}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (3 \log (256)) \int \frac {x}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (6 \log (256)) \int \frac {1}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (6 \log (256)) \int \frac {x}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx \\ & = \frac {6}{5} \int \left (\frac {\log (1+x)}{\log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {\log (1+x)}{(-1-x) \log ^2(-3+2 x) \log ^2(16+16 x)}\right ) \, dx+\frac {6}{5} \int \left (\frac {1}{\log (-3+2 x) \log ^2(16+16 x)}+\frac {1}{(-1-x) \log (-3+2 x) \log ^2(16+16 x)}\right ) \, dx+\frac {6}{5} \int \frac {\log (1+x)}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {9}{5} \int \frac {1}{(-1-x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {12}{5} \int \left (-\frac {\log (1+x)}{2 \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {3 \log (1+x)}{2 (3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)}\right ) \, dx+\frac {12}{5} \int \left (-\frac {1}{2 \log (-3+2 x) \log ^2(16+16 x)}+\frac {3}{2 (3-2 x) \log (-3+2 x) \log ^2(16+16 x)}\right ) \, dx+\frac {12}{5} \int \frac {\log (1+x)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {18}{5} \int \frac {1}{(-3+2 x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (3 \log (256)) \int \left (\frac {1}{\log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {1}{(-1-x) \log ^2(-3+2 x) \log ^2(16+16 x)}\right ) \, dx+\frac {1}{5} (3 \log (256)) \int \frac {1}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (6 \log (256)) \int \left (-\frac {1}{2 \log ^2(-3+2 x) \log ^2(16+16 x)}+\frac {3}{2 (3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)}\right ) \, dx+\frac {1}{5} (6 \log (256)) \int \frac {1}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx \\ & = \frac {6}{5} \int \frac {\log (1+x)}{(-1-x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {6}{5} \int \frac {\log (1+x)}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {6}{5} \int \frac {1}{(-1-x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {9}{5} \int \frac {1}{(-1-x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {12}{5} \int \frac {\log (1+x)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {18}{5} \int \frac {\log (1+x)}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {18}{5} \int \frac {1}{(3-2 x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {18}{5} \int \frac {1}{(-3+2 x) \log (-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (3 \log (256)) \int \frac {1}{(-1-x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (3 \log (256)) \int \frac {1}{(1+x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (6 \log (256)) \int \frac {1}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx+\frac {1}{5} (9 \log (256)) \int \frac {1}{(3-2 x) \log ^2(-3+2 x) \log ^2(16+16 x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.53 \[ \int \frac {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{\left (\left (-3-x+2 x^2\right ) \log ^2(16)+\left (-6-2 x+4 x^2\right ) \log (16) \log (1+x)+\left (-3-x+2 x^2\right ) \log ^2(1+x)\right ) \log ^2(-3+2 x)} \, dx=\frac {3 (\log (256)+2 \log (1+x))}{2 \log ^2(16 (1+x)) \log (-3+2 x)} \]

[In]

Integrate[((-6 - 6*x)*Log[16] + (-6 - 6*x)*Log[1 + x] + (9 - 6*x)*Log[-3 + 2*x])/(((-3 - x + 2*x^2)*Log[16]^2
+ (-6 - 2*x + 4*x^2)*Log[16]*Log[1 + x] + (-3 - x + 2*x^2)*Log[1 + x]^2)*Log[-3 + 2*x]^2),x]

[Out]

(3*(Log[256] + 2*Log[1 + x]))/(2*Log[16*(1 + x)]^2*Log[-3 + 2*x])

Maple [A] (verified)

Time = 16.94 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16

method	result	size
risch	\(\frac {3}{\left (\ln \left (1+x \right )+4 \ln \left (2\right )\right ) \ln \left (-3+2 x \right )}\)	\(22\)
parallelrisch	\(\frac {3}{\left (\ln \left (1+x \right )+4 \ln \left (2\right )\right ) \ln \left (-3+2 x \right )}\)	\(22\)

[In]

int(((-6*x+9)*ln(-3+2*x)+(-6*x-6)*ln(1+x)+4*(-6*x-6)*ln(2))/((2*x^2-x-3)*ln(1+x)^2+4*(4*x^2-2*x-6)*ln(2)*ln(1+
x)+16*(2*x^2-x-3)*ln(2)^2)/ln(-3+2*x)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{\left (\left (-3-x+2 x^2\right ) \log ^2(16)+\left (-6-2 x+4 x^2\right ) \log (16) \log (1+x)+\left (-3-x+2 x^2\right ) \log ^2(1+x)\right ) \log ^2(-3+2 x)} \, dx=\frac {3}{{\left (4 \, \log \left (2\right ) + \log \left (x + 1\right )\right )} \log \left (2 \, x - 3\right )} \]

[In]

integrate(((-6*x+9)*log(-3+2*x)+(-6*x-6)*log(1+x)+4*(-6*x-6)*log(2))/((2*x^2-x-3)*log(1+x)^2+4*(4*x^2-2*x-6)*l
og(2)*log(1+x)+16*(2*x^2-x-3)*log(2)^2)/log(-3+2*x)^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{\left (\left (-3-x+2 x^2\right ) \log ^2(16)+\left (-6-2 x+4 x^2\right ) \log (16) \log (1+x)+\left (-3-x+2 x^2\right ) \log ^2(1+x)\right ) \log ^2(-3+2 x)} \, dx=\frac {3}{\left (\log {\left (x + 1 \right )} + 4 \log {\left (2 \right )}\right ) \log {\left (2 x - 3 \right )}} \]

[In]

integrate(((-6*x+9)*ln(-3+2*x)+(-6*x-6)*ln(1+x)+4*(-6*x-6)*ln(2))/((2*x**2-x-3)*ln(1+x)**2+4*(4*x**2-2*x-6)*ln
(2)*ln(1+x)+16*(2*x**2-x-3)*ln(2)**2)/ln(-3+2*x)**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{\left (\left (-3-x+2 x^2\right ) \log ^2(16)+\left (-6-2 x+4 x^2\right ) \log (16) \log (1+x)+\left (-3-x+2 x^2\right ) \log ^2(1+x)\right ) \log ^2(-3+2 x)} \, dx=\frac {3}{{\left (4 \, \log \left (2\right ) + \log \left (x + 1\right )\right )} \log \left (2 \, x - 3\right )} \]

[In]

integrate(((-6*x+9)*log(-3+2*x)+(-6*x-6)*log(1+x)+4*(-6*x-6)*log(2))/((2*x^2-x-3)*log(1+x)^2+4*(4*x^2-2*x-6)*l
og(2)*log(1+x)+16*(2*x^2-x-3)*log(2)^2)/log(-3+2*x)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{\left (\left (-3-x+2 x^2\right ) \log ^2(16)+\left (-6-2 x+4 x^2\right ) \log (16) \log (1+x)+\left (-3-x+2 x^2\right ) \log ^2(1+x)\right ) \log ^2(-3+2 x)} \, dx=\frac {3}{4 \, \log \left (2\right ) \log \left (2 \, x - 3\right ) + \log \left (2 \, x - 3\right ) \log \left (x + 1\right )} \]

[In]

integrate(((-6*x+9)*log(-3+2*x)+(-6*x-6)*log(1+x)+4*(-6*x-6)*log(2))/((2*x^2-x-3)*log(1+x)^2+4*(4*x^2-2*x-6)*l
og(2)*log(1+x)+16*(2*x^2-x-3)*log(2)^2)/log(-3+2*x)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.12 (sec) , antiderivative size = 194, normalized size of antiderivative = 10.21 \[ \int \frac {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{\left (\left (-3-x+2 x^2\right ) \log ^2(16)+\left (-6-2 x+4 x^2\right ) \log (16) \log (1+x)+\left (-3-x+2 x^2\right ) \log ^2(1+x)\right ) \log ^2(-3+2 x)} \, dx=\frac {15}{4\,\left (x+1\right )}-\frac {\frac {15\,\left (4\,\ln \left (2\right )+1\right )}{4\,\left (x+1\right )}+\frac {15\,\ln \left (x+1\right )}{4\,\left (x+1\right )}}{\ln \left (x+1\right )+4\,\ln \left (2\right )}+\frac {\frac {3}{\ln \left (x+1\right )+4\,\ln \left (2\right )}+\frac {3\,\ln \left (2\,x-3\right )\,\left (2\,x-3\right )}{2\,\left (8\,\ln \left (x+1\right )\,\ln \left (2\right )+x\,{\ln \left (x+1\right )}^2+16\,x\,{\ln \left (2\right )}^2+{\ln \left (x+1\right )}^2+16\,{\ln \left (2\right )}^2+8\,x\,\ln \left (x+1\right )\,\ln \left (2\right )\right )}}{\ln \left (2\,x-3\right )}+\frac {\frac {3\,\left (10\,\ln \left (2\right )-2\,x+3\right )}{2\,\left (x+1\right )}+\frac {15\,\ln \left (x+1\right )}{4\,\left (x+1\right )}}{{\ln \left (x+1\right )}^2+8\,\ln \left (2\right )\,\ln \left (x+1\right )+16\,{\ln \left (2\right )}^2} \]

[In]

int((4*log(2)*(6*x + 6) + log(2*x - 3)*(6*x - 9) + log(x + 1)*(6*x + 6))/(log(2*x - 3)^2*(log(x + 1)^2*(x - 2*
x^2 + 3) + 16*log(2)^2*(x - 2*x^2 + 3) + 4*log(x + 1)*log(2)*(2*x - 4*x^2 + 6))),x)

[Out]

15/(4*(x + 1)) - ((15*(4*log(2) + 1))/(4*(x + 1)) + (15*log(x + 1))/(4*(x + 1)))/(log(x + 1) + 4*log(2)) + (3/
(log(x + 1) + 4*log(2)) + (3*log(2*x - 3)*(2*x - 3))/(2*(8*log(x + 1)*log(2) + x*log(x + 1)^2 + 16*x*log(2)^2
+ log(x + 1)^2 + 16*log(2)^2 + 8*x*log(x + 1)*log(2))))/log(2*x - 3) + ((3*(10*log(2) - 2*x + 3))/(2*(x + 1))
+ (15*log(x + 1))/(4*(x + 1)))/(8*log(x + 1)*log(2) + log(x + 1)^2 + 16*log(2)^2)