\(\int \frac {-90 x-3 x^2+3 x^3+(-44 x-33 x^2) \log (4+3 x)+(-120-94 x+x^2+3 x^3) \log (4+3 x) \log (\frac {6-x}{(10+2 x) \log (4+3 x)})}{(-120-94 x+x^2+3 x^3) \log (4+3 x) \log ^2(\frac {6-x}{(10+2 x) \log (4+3 x)})} \, dx\) [1830]

Optimal result

Integrand size = 120, antiderivative size = 30 \[ \int \frac {-90 x-3 x^2+3 x^3+\left (-44 x-33 x^2\right ) \log (4+3 x)+\left (-120-94 x+x^2+3 x^3\right ) \log (4+3 x) \log \left (\frac {6-x}{(10+2 x) \log (4+3 x)}\right )}{\left (-120-94 x+x^2+3 x^3\right ) \log (4+3 x) \log ^2\left (\frac {6-x}{(10+2 x) \log (4+3 x)}\right )} \, dx=\frac {x}{\log \left (\frac {\frac {3 (2-x)}{2}+x}{(5+x) \log (4+3 x)}\right )} \] Output:

x/ln((3-1/2*x)/ln(4+3*x)/(5+x))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {-90 x-3 x^2+3 x^3+\left (-44 x-33 x^2\right ) \log (4+3 x)+\left (-120-94 x+x^2+3 x^3\right ) \log (4+3 x) \log \left (\frac {6-x}{(10+2 x) \log (4+3 x)}\right )}{\left (-120-94 x+x^2+3 x^3\right ) \log (4+3 x) \log ^2\left (\frac {6-x}{(10+2 x) \log (4+3 x)}\right )} \, dx=\frac {x}{\log \left (-\frac {-6+x}{2 (5+x) \log (4+3 x)}\right )} \] Input:

Integrate[(-90*x - 3*x^2 + 3*x^3 + (-44*x - 33*x^2)*Log[4 + 3*x] + (-120 - 
 94*x + x^2 + 3*x^3)*Log[4 + 3*x]*Log[(6 - x)/((10 + 2*x)*Log[4 + 3*x])])/ 
((-120 - 94*x + x^2 + 3*x^3)*Log[4 + 3*x]*Log[(6 - x)/((10 + 2*x)*Log[4 + 
3*x])]^2),x]
 

Output:

x/Log[-1/2*(-6 + x)/((5 + x)*Log[4 + 3*x])]
 

Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {3 x^3-3 x^2+\left (-33 x^2-44 x\right ) \log (3 x+4)+\left (3 x^3+x^2-94 x-120\right ) \log (3 x+4) \log \left (\frac {6-x}{(2 x+10) \log (3 x+4)}\right )-90 x}{\left (3 x^3+x^2-94 x-120\right ) \log (3 x+4) \log ^2\left (\frac {6-x}{(2 x+10) \log (3 x+4)}\right )} \, dx\)

\(\Big \downarrow \) 2463

\(\displaystyle \int \left (\frac {3 x^3-3 x^2+\left (-33 x^2-44 x\right ) \log (3 x+4)+\left (3 x^3+x^2-94 x-120\right ) \log (3 x+4) \log \left (\frac {6-x}{(2 x+10) \log (3 x+4)}\right )-90 x}{242 (x-6) \log (3 x+4) \log ^2\left (\frac {6-x}{(2 x+10) \log (3 x+4)}\right )}+\frac {3 x^3-3 x^2+\left (-33 x^2-44 x\right ) \log (3 x+4)+\left (3 x^3+x^2-94 x-120\right ) \log (3 x+4) \log \left (\frac {6-x}{(2 x+10) \log (3 x+4)}\right )-90 x}{121 (x+5) \log (3 x+4) \log ^2\left (\frac {6-x}{(2 x+10) \log (3 x+4)}\right )}-\frac {9 \left (3 x^3-3 x^2+\left (-33 x^2-44 x\right ) \log (3 x+4)+\left (3 x^3+x^2-94 x-120\right ) \log (3 x+4) \log \left (\frac {6-x}{(2 x+10) \log (3 x+4)}\right )-90 x\right )}{242 (3 x+4) \log (3 x+4) \log ^2\left (\frac {6-x}{(2 x+10) \log (3 x+4)}\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -6 \int \frac {1}{(x-6) \log ^2\left (-\frac {x-6}{2 (x+5) \log (3 x+4)}\right )}dx-5 \int \frac {1}{(x+5) \log ^2\left (-\frac {x-6}{2 (x+5) \log (3 x+4)}\right )}dx+\int \frac {1}{\log (3 x+4) \log ^2\left (-\frac {x-6}{2 (x+5) \log (3 x+4)}\right )}dx-4 \int \frac {1}{(3 x+4) \log (3 x+4) \log ^2\left (-\frac {x-6}{2 (x+5) \log (3 x+4)}\right )}dx+\int \frac {1}{\log \left (-\frac {x-6}{2 (x+5) \log (3 x+4)}\right )}dx\)

Input:

Int[(-90*x - 3*x^2 + 3*x^3 + (-44*x - 33*x^2)*Log[4 + 3*x] + (-120 - 94*x 
+ x^2 + 3*x^3)*Log[4 + 3*x]*Log[(6 - x)/((10 + 2*x)*Log[4 + 3*x])])/((-120 
 - 94*x + x^2 + 3*x^3)*Log[4 + 3*x]*Log[(6 - x)/((10 + 2*x)*Log[4 + 3*x])] 
^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 2.82 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87

method	result	size
parallelrisch	\(\frac {x}{\ln \left (-\frac {-6+x}{\left (2 x +10\right ) \ln \left (4+3 x \right )}\right )}\)	\(26\)
risch	\(\frac {2 i x}{2 \pi \operatorname {csgn}\left (\frac {i \left (-6+x \right )}{\left (5+x \right ) \ln \left (4+3 x \right )}\right )^{2}+\pi \,\operatorname {csgn}\left (i \left (-6+x \right )\right ) \operatorname {csgn}\left (\frac {i}{5+x}\right ) \operatorname {csgn}\left (\frac {i \left (-6+x \right )}{5+x}\right )-\pi \,\operatorname {csgn}\left (i \left (-6+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-6+x \right )}{5+x}\right )^{2}-\pi \,\operatorname {csgn}\left (\frac {i}{5+x}\right ) \operatorname {csgn}\left (\frac {i \left (-6+x \right )}{5+x}\right )^{2}+\pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (4+3 x \right )}\right ) \operatorname {csgn}\left (\frac {i \left (-6+x \right )}{5+x}\right ) \operatorname {csgn}\left (\frac {i \left (-6+x \right )}{\left (5+x \right ) \ln \left (4+3 x \right )}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (4+3 x \right )}\right ) \operatorname {csgn}\left (\frac {i \left (-6+x \right )}{\left (5+x \right ) \ln \left (4+3 x \right )}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i \left (-6+x \right )}{5+x}\right )^{3}-\pi \,\operatorname {csgn}\left (\frac {i \left (-6+x \right )}{5+x}\right ) \operatorname {csgn}\left (\frac {i \left (-6+x \right )}{\left (5+x \right ) \ln \left (4+3 x \right )}\right )^{2}-\pi \operatorname {csgn}\left (\frac {i \left (-6+x \right )}{\left (5+x \right ) \ln \left (4+3 x \right )}\right )^{3}-2 \pi +2 i \ln \left (-6+x \right )-2 i \ln \left (5+x \right )-2 i \ln \left (2\right )-2 i \ln \left (\ln \left (4+3 x \right )\right )}\)	\(306\)
default	\(\text {Expression too large to display}\)	\(799\)

Input:

int(((3*x^3+x^2-94*x-120)*ln(4+3*x)*ln((-x+6)/(2*x+10)/ln(4+3*x))+(-33*x^2 
-44*x)*ln(4+3*x)+3*x^3-3*x^2-90*x)/(3*x^3+x^2-94*x-120)/ln(4+3*x)/ln((-x+6 
)/(2*x+10)/ln(4+3*x))^2,x,method=_RETURNVERBOSE)
 

Output:

x/ln(-1/(2*x+10)/ln(4+3*x)*(-6+x))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {-90 x-3 x^2+3 x^3+\left (-44 x-33 x^2\right ) \log (4+3 x)+\left (-120-94 x+x^2+3 x^3\right ) \log (4+3 x) \log \left (\frac {6-x}{(10+2 x) \log (4+3 x)}\right )}{\left (-120-94 x+x^2+3 x^3\right ) \log (4+3 x) \log ^2\left (\frac {6-x}{(10+2 x) \log (4+3 x)}\right )} \, dx=\frac {x}{\log \left (-\frac {x - 6}{2 \, {\left (x + 5\right )} \log \left (3 \, x + 4\right )}\right )} \] Input:

integrate(((3*x^3+x^2-94*x-120)*log(4+3*x)*log((6-x)/(2*x+10)/log(4+3*x))+ 
(-33*x^2-44*x)*log(4+3*x)+3*x^3-3*x^2-90*x)/(3*x^3+x^2-94*x-120)/log(4+3*x 
)/log((6-x)/(2*x+10)/log(4+3*x))^2,x, algorithm="fricas")
 

Output:

x/log(-1/2*(x - 6)/((x + 5)*log(3*x + 4)))
 

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.57 \[ \int \frac {-90 x-3 x^2+3 x^3+\left (-44 x-33 x^2\right ) \log (4+3 x)+\left (-120-94 x+x^2+3 x^3\right ) \log (4+3 x) \log \left (\frac {6-x}{(10+2 x) \log (4+3 x)}\right )}{\left (-120-94 x+x^2+3 x^3\right ) \log (4+3 x) \log ^2\left (\frac {6-x}{(10+2 x) \log (4+3 x)}\right )} \, dx=\frac {x}{\log {\left (\frac {6 - x}{\left (2 x + 10\right ) \log {\left (3 x + 4 \right )}} \right )}} \] Input:

integrate(((3*x**3+x**2-94*x-120)*ln(4+3*x)*ln((6-x)/(2*x+10)/ln(4+3*x))+( 
-33*x**2-44*x)*ln(4+3*x)+3*x**3-3*x**2-90*x)/(3*x**3+x**2-94*x-120)/ln(4+3 
*x)/ln((6-x)/(2*x+10)/ln(4+3*x))**2,x)
 

Output:

x/log((6 - x)/((2*x + 10)*log(3*x + 4)))
 

Maxima [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {-90 x-3 x^2+3 x^3+\left (-44 x-33 x^2\right ) \log (4+3 x)+\left (-120-94 x+x^2+3 x^3\right ) \log (4+3 x) \log \left (\frac {6-x}{(10+2 x) \log (4+3 x)}\right )}{\left (-120-94 x+x^2+3 x^3\right ) \log (4+3 x) \log ^2\left (\frac {6-x}{(10+2 x) \log (4+3 x)}\right )} \, dx=-\frac {x}{\log \left (2\right ) + \log \left (x + 5\right ) - \log \left (-x + 6\right ) + \log \left (\log \left (3 \, x + 4\right )\right )} \] Input:

integrate(((3*x^3+x^2-94*x-120)*log(4+3*x)*log((6-x)/(2*x+10)/log(4+3*x))+ 
(-33*x^2-44*x)*log(4+3*x)+3*x^3-3*x^2-90*x)/(3*x^3+x^2-94*x-120)/log(4+3*x 
)/log((6-x)/(2*x+10)/log(4+3*x))^2,x, algorithm="maxima")
 

Output:

-x/(log(2) + log(x + 5) - log(-x + 6) + log(log(3*x + 4)))
 

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {-90 x-3 x^2+3 x^3+\left (-44 x-33 x^2\right ) \log (4+3 x)+\left (-120-94 x+x^2+3 x^3\right ) \log (4+3 x) \log \left (\frac {6-x}{(10+2 x) \log (4+3 x)}\right )}{\left (-120-94 x+x^2+3 x^3\right ) \log (4+3 x) \log ^2\left (\frac {6-x}{(10+2 x) \log (4+3 x)}\right )} \, dx=-\frac {x}{\log \left (2 \, x \log \left (3 \, x + 4\right ) + 10 \, \log \left (3 \, x + 4\right )\right ) - \log \left (-x + 6\right )} \] Input:

integrate(((3*x^3+x^2-94*x-120)*log(4+3*x)*log((6-x)/(2*x+10)/log(4+3*x))+ 
(-33*x^2-44*x)*log(4+3*x)+3*x^3-3*x^2-90*x)/(3*x^3+x^2-94*x-120)/log(4+3*x 
)/log((6-x)/(2*x+10)/log(4+3*x))^2,x, algorithm="giac")
 

Output:

-x/(log(2*x*log(3*x + 4) + 10*log(3*x + 4)) - log(-x + 6))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {-90 x-3 x^2+3 x^3+\left (-44 x-33 x^2\right ) \log (4+3 x)+\left (-120-94 x+x^2+3 x^3\right ) \log (4+3 x) \log \left (\frac {6-x}{(10+2 x) \log (4+3 x)}\right )}{\left (-120-94 x+x^2+3 x^3\right ) \log (4+3 x) \log ^2\left (\frac {6-x}{(10+2 x) \log (4+3 x)}\right )} \, dx=\int \frac {90\,x+\ln \left (3\,x+4\right )\,\left (33\,x^2+44\,x\right )+3\,x^2-3\,x^3+\ln \left (-\frac {x-6}{\ln \left (3\,x+4\right )\,\left (2\,x+10\right )}\right )\,\ln \left (3\,x+4\right )\,\left (-3\,x^3-x^2+94\,x+120\right )}{{\ln \left (-\frac {x-6}{\ln \left (3\,x+4\right )\,\left (2\,x+10\right )}\right )}^2\,\ln \left (3\,x+4\right )\,\left (-3\,x^3-x^2+94\,x+120\right )} \,d x \] Input:

int((90*x + log(3*x + 4)*(44*x + 33*x^2) + 3*x^2 - 3*x^3 + log(-(x - 6)/(l 
og(3*x + 4)*(2*x + 10)))*log(3*x + 4)*(94*x - x^2 - 3*x^3 + 120))/(log(-(x 
 - 6)/(log(3*x + 4)*(2*x + 10)))^2*log(3*x + 4)*(94*x - x^2 - 3*x^3 + 120) 
),x)
 

Output:

int((90*x + log(3*x + 4)*(44*x + 33*x^2) + 3*x^2 - 3*x^3 + log(-(x - 6)/(l 
og(3*x + 4)*(2*x + 10)))*log(3*x + 4)*(94*x - x^2 - 3*x^3 + 120))/(log(-(x 
 - 6)/(log(3*x + 4)*(2*x + 10)))^2*log(3*x + 4)*(94*x - x^2 - 3*x^3 + 120) 
), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {-90 x-3 x^2+3 x^3+\left (-44 x-33 x^2\right ) \log (4+3 x)+\left (-120-94 x+x^2+3 x^3\right ) \log (4+3 x) \log \left (\frac {6-x}{(10+2 x) \log (4+3 x)}\right )}{\left (-120-94 x+x^2+3 x^3\right ) \log (4+3 x) \log ^2\left (\frac {6-x}{(10+2 x) \log (4+3 x)}\right )} \, dx=\frac {x}{\mathrm {log}\left (\frac {-x +6}{2 \,\mathrm {log}\left (3 x +4\right ) x +10 \,\mathrm {log}\left (3 x +4\right )}\right )} \] Input:

int(((3*x^3+x^2-94*x-120)*log(4+3*x)*log((6-x)/(2*x+10)/log(4+3*x))+(-33*x 
^2-44*x)*log(4+3*x)+3*x^3-3*x^2-90*x)/(3*x^3+x^2-94*x-120)/log(4+3*x)/log( 
(6-x)/(2*x+10)/log(4+3*x))^2,x)
 

Output:

x/log(( - x + 6)/(2*log(3*x + 4)*x + 10*log(3*x + 4)))