3.7.76 \(\int \frac {150-150 x+(-100+100 x) \log (x)+450 x \log ^2(x)+(-50+50 x-300 x \log ^2(x)) \log (x^2)+50 x \log ^2(x) \log ^2(x^2)}{(-9 x+9 x^2) \log ^2(x)+(6 x-6 x^2) \log ^2(x) \log (x^2)+(-x+x^2) \log ^2(x) \log ^2(x^2)} \, dx\) [676]

3.7.76.1 Optimal result

Integrand size = 104, antiderivative size = 32 \[ \int \frac {150-150 x+(-100+100 x) \log (x)+450 x \log ^2(x)+\left (-50+50 x-300 x \log ^2(x)\right ) \log \left (x^2\right )+50 x \log ^2(x) \log ^2\left (x^2\right )}{\left (-9 x+9 x^2\right ) \log ^2(x)+\left (6 x-6 x^2\right ) \log ^2(x) \log \left (x^2\right )+\left (-x+x^2\right ) \log ^2(x) \log ^2\left (x^2\right )} \, dx=10 \left (-3+5 \left (\log \left (\frac {1-x}{3}\right )+\frac {1}{\log (x) \left (3-\log \left (x^2\right )\right )}\right )\right ) \]

output

50*ln(1/3-1/3*x)+50/(3-ln(x^2))/ln(x)-30
 

3.7.76.2 Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {150-150 x+(-100+100 x) \log (x)+450 x \log ^2(x)+\left (-50+50 x-300 x \log ^2(x)\right ) \log \left (x^2\right )+50 x \log ^2(x) \log ^2\left (x^2\right )}{\left (-9 x+9 x^2\right ) \log ^2(x)+\left (6 x-6 x^2\right ) \log ^2(x) \log \left (x^2\right )+\left (-x+x^2\right ) \log ^2(x) \log ^2\left (x^2\right )} \, dx=50 \left (\log (-1+x)+\frac {1}{3 \log (x)-\log (x) \log \left (x^2\right )}\right ) \]

input

Integrate[(150 - 150*x + (-100 + 100*x)*Log[x] + 450*x*Log[x]^2 + (-50 + 5 
0*x - 300*x*Log[x]^2)*Log[x^2] + 50*x*Log[x]^2*Log[x^2]^2)/((-9*x + 9*x^2) 
*Log[x]^2 + (6*x - 6*x^2)*Log[x]^2*Log[x^2] + (-x + x^2)*Log[x]^2*Log[x^2] 
^2),x]
 

output

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

3.7.76.3 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.

\(\displaystyle \int \frac {50 x \log ^2\left (x^2\right ) \log ^2(x)+\left (50 x-300 x \log ^2(x)-50\right ) \log \left (x^2\right )-150 x+450 x \log ^2(x)+(100 x-100) \log (x)+150}{\left (x^2-x\right ) \log ^2\left (x^2\right ) \log ^2(x)+\left (9 x^2-9 x\right ) \log ^2(x)+\left (6 x-6 x^2\right ) \log \left (x^2\right ) \log ^2(x)} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int 50 \left (\frac {1}{x \log ^2(x) \left (\log \left (x^2\right )-3\right )}+\frac {2}{x \log (x) \left (\log \left (x^2\right )-3\right )^2}+\frac {1}{x-1}\right )dx\)

\(\Big \downarrow \) 27

\(\displaystyle 50 \int \left (-\frac {1}{x \log ^2(x) \left (3-\log \left (x^2\right )\right )}+\frac {2}{x \log (x) \left (3-\log \left (x^2\right )\right )^2}+\frac {1}{x-1}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 50 \left (-\int \frac {1}{x \log ^2(x) \left (3-\log \left (x^2\right )\right )}dx+2 \int \frac {1}{x \log (x) \left (3-\log \left (x^2\right )\right )^2}dx+\log (1-x)\right )\)

input

Int[(150 - 150*x + (-100 + 100*x)*Log[x] + 450*x*Log[x]^2 + (-50 + 50*x - 
300*x*Log[x]^2)*Log[x^2] + 50*x*Log[x]^2*Log[x^2]^2)/((-9*x + 9*x^2)*Log[x 
]^2 + (6*x - 6*x^2)*Log[x]^2*Log[x^2] + (-x + x^2)*Log[x]^2*Log[x^2]^2),x]
 

output

$Aborted
 

3.7.76.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7239

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

3.7.76.4 Maple [A] (verified)

Time = 2.86 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16

method	result	size
parallelrisch	\(\frac {-1200+1200 \ln \left (-1+x \right ) \ln \left (x \right ) \ln \left (x^{2}\right )-3600 \ln \left (x \right ) \ln \left (-1+x \right )}{24 \left (\ln \left (x^{2}\right )-3\right ) \ln \left (x \right )}\)	\(37\)
default	\(50 \ln \left (-1+x \right )-\frac {50}{\left (-2 \ln \left (x \right )+\ln \left (x^{2}\right )-3\right ) \ln \left (x \right )}+\frac {100}{\left (-2 \ln \left (x \right )+\ln \left (x^{2}\right )-3\right ) \left (\ln \left (x^{2}\right )-3\right )}\)	\(48\)
parts	\(50 \ln \left (-1+x \right )-\frac {50}{\left (-2 \ln \left (x \right )+\ln \left (x^{2}\right )-3\right ) \ln \left (x \right )}+\frac {100}{\left (-2 \ln \left (x \right )+\ln \left (x^{2}\right )-3\right ) \left (\ln \left (x^{2}\right )-3\right )}\)	\(48\)
risch	\(50 \ln \left (-1+x \right )-\frac {100 i}{\left (\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}+4 i \ln \left (x \right )-6 i\right ) \ln \left (x \right )}\)	\(69\)

input

int((50*x*ln(x)^2*ln(x^2)^2+(-300*x*ln(x)^2+50*x-50)*ln(x^2)+450*x*ln(x)^2 
+(100*x-100)*ln(x)-150*x+150)/((x^2-x)*ln(x)^2*ln(x^2)^2+(-6*x^2+6*x)*ln(x 
)^2*ln(x^2)+(9*x^2-9*x)*ln(x)^2),x,method=_RETURNVERBOSE)
 

output

1/24*(-1200+1200*ln(-1+x)*ln(x)*ln(x^2)-3600*ln(x)*ln(-1+x))/(ln(x^2)-3)/l 
n(x)
 

3.7.76.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09 \[ \int \frac {150-150 x+(-100+100 x) \log (x)+450 x \log ^2(x)+\left (-50+50 x-300 x \log ^2(x)\right ) \log \left (x^2\right )+50 x \log ^2(x) \log ^2\left (x^2\right )}{\left (-9 x+9 x^2\right ) \log ^2(x)+\left (6 x-6 x^2\right ) \log ^2(x) \log \left (x^2\right )+\left (-x+x^2\right ) \log ^2(x) \log ^2\left (x^2\right )} \, dx=\frac {50 \, {\left (2 \, \log \left (x - 1\right ) \log \left (x\right )^{2} - 3 \, \log \left (x - 1\right ) \log \left (x\right ) - 1\right )}}{2 \, \log \left (x\right )^{2} - 3 \, \log \left (x\right )} \]

input

integrate((50*x*log(x)^2*log(x^2)^2+(-300*x*log(x)^2+50*x-50)*log(x^2)+450 
*x*log(x)^2+(100*x-100)*log(x)-150*x+150)/((x^2-x)*log(x)^2*log(x^2)^2+(-6 
*x^2+6*x)*log(x)^2*log(x^2)+(9*x^2-9*x)*log(x)^2),x, algorithm=\
 

output

50*(2*log(x - 1)*log(x)^2 - 3*log(x - 1)*log(x) - 1)/(2*log(x)^2 - 3*log(x 
))
 

3.7.76.6 Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.59 \[ \int \frac {150-150 x+(-100+100 x) \log (x)+450 x \log ^2(x)+\left (-50+50 x-300 x \log ^2(x)\right ) \log \left (x^2\right )+50 x \log ^2(x) \log ^2\left (x^2\right )}{\left (-9 x+9 x^2\right ) \log ^2(x)+\left (6 x-6 x^2\right ) \log ^2(x) \log \left (x^2\right )+\left (-x+x^2\right ) \log ^2(x) \log ^2\left (x^2\right )} \, dx=50 \log {\left (x - 1 \right )} - \frac {50}{2 \log {\left (x \right )}^{2} - 3 \log {\left (x \right )}} \]

input

integrate((50*x*ln(x)**2*ln(x**2)**2+(-300*x*ln(x)**2+50*x-50)*ln(x**2)+45 
0*x*ln(x)**2+(100*x-100)*ln(x)-150*x+150)/((x**2-x)*ln(x)**2*ln(x**2)**2+( 
-6*x**2+6*x)*ln(x)**2*ln(x**2)+(9*x**2-9*x)*ln(x)**2),x)
 

output

50*log(x - 1) - 50/(2*log(x)**2 - 3*log(x))
 

3.7.76.7 Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {150-150 x+(-100+100 x) \log (x)+450 x \log ^2(x)+\left (-50+50 x-300 x \log ^2(x)\right ) \log \left (x^2\right )+50 x \log ^2(x) \log ^2\left (x^2\right )}{\left (-9 x+9 x^2\right ) \log ^2(x)+\left (6 x-6 x^2\right ) \log ^2(x) \log \left (x^2\right )+\left (-x+x^2\right ) \log ^2(x) \log ^2\left (x^2\right )} \, dx=-\frac {50}{2 \, \log \left (x\right )^{2} - 3 \, \log \left (x\right )} + 50 \, \log \left (x - 1\right ) \]

input

integrate((50*x*log(x)^2*log(x^2)^2+(-300*x*log(x)^2+50*x-50)*log(x^2)+450 
*x*log(x)^2+(100*x-100)*log(x)-150*x+150)/((x^2-x)*log(x)^2*log(x^2)^2+(-6 
*x^2+6*x)*log(x)^2*log(x^2)+(9*x^2-9*x)*log(x)^2),x, algorithm=\
 

output

-50/(2*log(x)^2 - 3*log(x)) + 50*log(x - 1)
 

3.7.76.8 Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int \frac {150-150 x+(-100+100 x) \log (x)+450 x \log ^2(x)+\left (-50+50 x-300 x \log ^2(x)\right ) \log \left (x^2\right )+50 x \log ^2(x) \log ^2\left (x^2\right )}{\left (-9 x+9 x^2\right ) \log ^2(x)+\left (6 x-6 x^2\right ) \log ^2(x) \log \left (x^2\right )+\left (-x+x^2\right ) \log ^2(x) \log ^2\left (x^2\right )} \, dx=-\frac {100}{3 \, {\left (2 \, \log \left (x\right ) - 3\right )}} + \frac {50}{3 \, \log \left (x\right )} + 50 \, \log \left (x - 1\right ) \]

input

integrate((50*x*log(x)^2*log(x^2)^2+(-300*x*log(x)^2+50*x-50)*log(x^2)+450 
*x*log(x)^2+(100*x-100)*log(x)-150*x+150)/((x^2-x)*log(x)^2*log(x^2)^2+(-6 
*x^2+6*x)*log(x)^2*log(x^2)+(9*x^2-9*x)*log(x)^2),x, algorithm=\
 

output

-100/3/(2*log(x) - 3) + 50/3/log(x) + 50*log(x - 1)
 

3.7.76.9 Mupad [B] (verification not implemented)

Time = 9.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.66 \[ \int \frac {150-150 x+(-100+100 x) \log (x)+450 x \log ^2(x)+\left (-50+50 x-300 x \log ^2(x)\right ) \log \left (x^2\right )+50 x \log ^2(x) \log ^2\left (x^2\right )}{\left (-9 x+9 x^2\right ) \log ^2(x)+\left (6 x-6 x^2\right ) \log ^2(x) \log \left (x^2\right )+\left (-x+x^2\right ) \log ^2(x) \log ^2\left (x^2\right )} \, dx=50\,\ln \left (x-1\right )-\frac {50}{\ln \left (x\right )\,\left (\ln \left (x^2\right )-3\right )} \]

input

int(-(450*x*log(x)^2 - 150*x + log(x)*(100*x - 100) - log(x^2)*(300*x*log( 
x)^2 - 50*x + 50) + 50*x*log(x^2)^2*log(x)^2 + 150)/(log(x)^2*(9*x - 9*x^2 
) - log(x^2)*log(x)^2*(6*x - 6*x^2) + log(x^2)^2*log(x)^2*(x - x^2)),x)
 

output

50*log(x - 1) - 50/(log(x)*(log(x^2) - 3))