\(\int \frac {1800-600 x-550 x^2+100 x^3+50 x^4+(600-100 x-100 x^2) \log (x)+50 \log ^2(x)+(1500-250 x+50 x^2-50 x^3-50 x^4+(550-50 x) \log (x)+50 \log ^2(x)) \log (x^2)+x^3 \log ^2(x^2)}{x^3 \log ^2(x^2)} \, dx\) [341]

Optimal result

Integrand size = 98, antiderivative size = 26 \[ \int \frac {1800-600 x-550 x^2+100 x^3+50 x^4+\left (600-100 x-100 x^2\right ) \log (x)+50 \log ^2(x)+\left (1500-250 x+50 x^2-50 x^3-50 x^4+(550-50 x) \log (x)+50 \log ^2(x)\right ) \log \left (x^2\right )+x^3 \log ^2\left (x^2\right )}{x^3 \log ^2\left (x^2\right )} \, dx=x-\frac {25 \left (x-\frac {6-x+\log (x)}{x}\right )^2}{\log \left (x^2\right )} \] Output:

x-5*(x-(ln(x)-x+6)/x)*(5*x-5*(ln(x)-x+6)/x)/ln(x^2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.85 \[ \int \frac {1800-600 x-550 x^2+100 x^3+50 x^4+\left (600-100 x-100 x^2\right ) \log (x)+50 \log ^2(x)+\left (1500-250 x+50 x^2-50 x^3-50 x^4+(550-50 x) \log (x)+50 \log ^2(x)\right ) \log \left (x^2\right )+x^3 \log ^2\left (x^2\right )}{x^3 \log ^2\left (x^2\right )} \, dx=\frac {-25 \left (-6+x+x^2\right )^2+50 \left (-6+x+x^2\right ) \log (x)-25 \log ^2(x)+(-25+x) x^2 \log \left (x^2\right )}{x^2 \log \left (x^2\right )} \] Input:

Integrate[(1800 - 600*x - 550*x^2 + 100*x^3 + 50*x^4 + (600 - 100*x - 100* 
x^2)*Log[x] + 50*Log[x]^2 + (1500 - 250*x + 50*x^2 - 50*x^3 - 50*x^4 + (55 
0 - 50*x)*Log[x] + 50*Log[x]^2)*Log[x^2] + x^3*Log[x^2]^2)/(x^3*Log[x^2]^2 
),x]
 

Output:

(-25*(-6 + x + x^2)^2 + 50*(-6 + x + x^2)*Log[x] - 25*Log[x]^2 + (-25 + x) 
*x^2*Log[x^2])/(x^2*Log[x^2])
 

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.

\(\displaystyle \int \frac {50 x^4+100 x^3-550 x^2+\left (-100 x^2-100 x+600\right ) \log (x)+x^3 \log ^2\left (x^2\right )+\left (-50 x^4-50 x^3+50 x^2-250 x+50 \log ^2(x)+(550-50 x) \log (x)+1500\right ) \log \left (x^2\right )-600 x+50 \log ^2(x)+1800}{x^3 \log ^2\left (x^2\right )} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {50 \left (x^2+x-\log (x)-6\right )^2}{x^3 \log ^2\left (x^2\right )}-\frac {50 \left (x^2+\log (x)+5\right ) \left (x^2+x-\log (x)-6\right )}{x^3 \log \left (x^2\right )}+1\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 50 \int \frac {\log ^2(x)}{x^3 \log ^2\left (x^2\right )}dx+50 \int \frac {\log ^2(x)}{x^3 \log \left (x^2\right )}dx-25 \log (x) \operatorname {ExpIntegralEi}\left (-\log \left (x^2\right )\right )+\frac {25}{2} \log \left (x^2\right ) \operatorname {ExpIntegralEi}\left (-\log \left (x^2\right )\right )+\frac {25}{2 x^2}-\frac {25 x^2}{\log \left (x^2\right )}-\frac {50 x}{\log \left (x^2\right )}+\frac {50 \log (x)}{\log \left (x^2\right )}+\frac {275}{\log \left (x^2\right )}+\frac {50 \log (x)}{x \log \left (x^2\right )}+\frac {300}{x \log \left (x^2\right )}-\frac {300 \log (x)}{x^2 \log \left (x^2\right )}-\frac {900}{x^2 \log \left (x^2\right )}+x\)

Input:

Int[(1800 - 600*x - 550*x^2 + 100*x^3 + 50*x^4 + (600 - 100*x - 100*x^2)*L 
og[x] + 50*Log[x]^2 + (1500 - 250*x + 50*x^2 - 50*x^3 - 50*x^4 + (550 - 50 
*x)*Log[x] + 50*Log[x]^2)*Log[x^2] + x^3*Log[x^2]^2)/(x^3*Log[x^2]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 2.36 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.42

method	result	size
parallelrisch	\(\frac {-3600-100 x^{4}+4 x^{3} \ln \left (x^{2}\right )-200 x^{3}+200 x^{2} \ln \left (x \right )+1100 x^{2}+200 x \ln \left (x \right )-100 \ln \left (x \right )^{2}+1200 x -1200 \ln \left (x \right )}{4 x^{2} \ln \left (x^{2}\right )}\)	\(63\)
risch	\(-\frac {25 \ln \left (x \right )}{2 x^{2}}+\frac {-25 i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+50 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-25 i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+8 x^{3}+200 x -1200}{8 x^{2}}-\frac {25 i \left (-\pi ^{2} \operatorname {csgn}\left (i x \right )^{4} \operatorname {csgn}\left (i x^{2}\right )^{2}+4 \pi ^{2} \operatorname {csgn}\left (i x \right )^{3} \operatorname {csgn}\left (i x^{2}\right )^{3}-6 \pi ^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )^{4}+4 \pi ^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{5}-\pi ^{2} \operatorname {csgn}\left (i x^{2}\right )^{6}-8 i \pi \,x^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+48 i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+16 i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-8 i \pi x \operatorname {csgn}\left (i x^{2}\right )^{3}-8 i \pi \,x^{2} \operatorname {csgn}\left (i x^{2}\right )^{3}+48 i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-96 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+16 i \pi \,x^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-8 i \pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+16 x^{4}+32 x^{3}-176 x^{2}-192 x +576\right )}{8 x^{2} \left (\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}+4 i \ln \left (x \right )\right )}\)	\(406\)

Input:

int((x^3*ln(x^2)^2+(50*ln(x)^2+(-50*x+550)*ln(x)-50*x^4-50*x^3+50*x^2-250* 
x+1500)*ln(x^2)+50*ln(x)^2+(-100*x^2-100*x+600)*ln(x)+50*x^4+100*x^3-550*x 
^2-600*x+1800)/x^3/ln(x^2)^2,x,method=_RETURNVERBOSE)
 

Output:

1/4/x^2*(-3600-100*x^4+4*x^3*ln(x^2)-200*x^3+200*x^2*ln(x)+1100*x^2+200*x* 
ln(x)-100*ln(x)^2+1200*x-1200*ln(x))/ln(x^2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.81 \[ \int \frac {1800-600 x-550 x^2+100 x^3+50 x^4+\left (600-100 x-100 x^2\right ) \log (x)+50 \log ^2(x)+\left (1500-250 x+50 x^2-50 x^3-50 x^4+(550-50 x) \log (x)+50 \log ^2(x)\right ) \log \left (x^2\right )+x^3 \log ^2\left (x^2\right )}{x^3 \log ^2\left (x^2\right )} \, dx=-\frac {25 \, x^{4} + 50 \, x^{3} - 275 \, x^{2} - 2 \, {\left (x^{3} + 25 \, x - 150\right )} \log \left (x\right ) + 25 \, \log \left (x\right )^{2} - 300 \, x + 900}{2 \, x^{2} \log \left (x\right )} \] Input:

integrate((x^3*log(x^2)^2+(50*log(x)^2+(-50*x+550)*log(x)-50*x^4-50*x^3+50 
*x^2-250*x+1500)*log(x^2)+50*log(x)^2+(-100*x^2-100*x+600)*log(x)+50*x^4+1 
00*x^3-550*x^2-600*x+1800)/x^3/log(x^2)^2,x, algorithm="fricas")
 

Output:

-1/2*(25*x^4 + 50*x^3 - 275*x^2 - 2*(x^3 + 25*x - 150)*log(x) + 25*log(x)^ 
2 - 300*x + 900)/(x^2*log(x))
 

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.85 \[ \int \frac {1800-600 x-550 x^2+100 x^3+50 x^4+\left (600-100 x-100 x^2\right ) \log (x)+50 \log ^2(x)+\left (1500-250 x+50 x^2-50 x^3-50 x^4+(550-50 x) \log (x)+50 \log ^2(x)\right ) \log \left (x^2\right )+x^3 \log ^2\left (x^2\right )}{x^3 \log ^2\left (x^2\right )} \, dx=x + \frac {25 x - 150}{x^{2}} + \frac {- 25 x^{4} - 50 x^{3} + 275 x^{2} + 300 x - 900}{2 x^{2} \log {\left (x \right )}} - \frac {25 \log {\left (x \right )}}{2 x^{2}} \] Input:

integrate((x**3*ln(x**2)**2+(50*ln(x)**2+(-50*x+550)*ln(x)-50*x**4-50*x**3 
+50*x**2-250*x+1500)*ln(x**2)+50*ln(x)**2+(-100*x**2-100*x+600)*ln(x)+50*x 
**4+100*x**3-550*x**2-600*x+1800)/x**3/ln(x**2)**2,x)
 

Output:

x + (25*x - 150)/x**2 + (-25*x**4 - 50*x**3 + 275*x**2 + 300*x - 900)/(2*x 
**2*log(x)) - 25*log(x)/(2*x**2)
 

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {1800-600 x-550 x^2+100 x^3+50 x^4+\left (600-100 x-100 x^2\right ) \log (x)+50 \log ^2(x)+\left (1500-250 x+50 x^2-50 x^3-50 x^4+(550-50 x) \log (x)+50 \log ^2(x)\right ) \log \left (x^2\right )+x^3 \log ^2\left (x^2\right )}{x^3 \log ^2\left (x^2\right )} \, dx=x - \frac {25 \, {\left (x^{4} + 2 \, x^{3} - 11 \, x^{2} - 2 \, {\left (x - 6\right )} \log \left (x\right ) + \log \left (x\right )^{2} - 12 \, x + 36\right )}}{2 \, x^{2} \log \left (x\right )} \] Input:

integrate((x^3*log(x^2)^2+(50*log(x)^2+(-50*x+550)*log(x)-50*x^4-50*x^3+50 
*x^2-250*x+1500)*log(x^2)+50*log(x)^2+(-100*x^2-100*x+600)*log(x)+50*x^4+1 
00*x^3-550*x^2-600*x+1800)/x^3/log(x^2)^2,x, algorithm="maxima")
 

Output:

x - 25/2*(x^4 + 2*x^3 - 11*x^2 - 2*(x - 6)*log(x) + log(x)^2 - 12*x + 36)/ 
(x^2*log(x))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {1800-600 x-550 x^2+100 x^3+50 x^4+\left (600-100 x-100 x^2\right ) \log (x)+50 \log ^2(x)+\left (1500-250 x+50 x^2-50 x^3-50 x^4+(550-50 x) \log (x)+50 \log ^2(x)\right ) \log \left (x^2\right )+x^3 \log ^2\left (x^2\right )}{x^3 \log ^2\left (x^2\right )} \, dx=x + \frac {25 \, {\left (x - 6\right )}}{x^{2}} - \frac {25 \, \log \left (x\right )}{2 \, x^{2}} - \frac {25 \, {\left (x^{4} + 2 \, x^{3} - 11 \, x^{2} - 12 \, x + 36\right )}}{2 \, x^{2} \log \left (x\right )} \] Input:

integrate((x^3*log(x^2)^2+(50*log(x)^2+(-50*x+550)*log(x)-50*x^4-50*x^3+50 
*x^2-250*x+1500)*log(x^2)+50*log(x)^2+(-100*x^2-100*x+600)*log(x)+50*x^4+1 
00*x^3-550*x^2-600*x+1800)/x^3/log(x^2)^2,x, algorithm="giac")
 

Output:

x + 25*(x - 6)/x^2 - 25/2*log(x)/x^2 - 25/2*(x^4 + 2*x^3 - 11*x^2 - 12*x + 
 36)/(x^2*log(x))
 

Mupad [B] (verification not implemented)

Time = 2.49 (sec) , antiderivative size = 103, normalized size of antiderivative = 3.96 \[ \int \frac {1800-600 x-550 x^2+100 x^3+50 x^4+\left (600-100 x-100 x^2\right ) \log (x)+50 \log ^2(x)+\left (1500-250 x+50 x^2-50 x^3-50 x^4+(550-50 x) \log (x)+50 \log ^2(x)\right ) \log \left (x^2\right )+x^3 \log ^2\left (x^2\right )}{x^3 \log ^2\left (x^2\right )} \, dx=x-\frac {50\,x}{\ln \left (x^2\right )}+\frac {275}{\ln \left (x^2\right )}+\frac {300}{x\,\ln \left (x^2\right )}-\frac {900}{x^2\,\ln \left (x^2\right )}-\frac {25\,x^2}{\ln \left (x^2\right )}+\frac {50\,\ln \left (x\right )}{\ln \left (x^2\right )}-\frac {25\,{\ln \left (x\right )}^2}{x^2\,\ln \left (x^2\right )}+\frac {50\,\ln \left (x\right )}{x\,\ln \left (x^2\right )}-\frac {300\,\ln \left (x\right )}{x^2\,\ln \left (x^2\right )} \] Input:

int((50*log(x)^2 - 600*x - log(x^2)*(250*x - 50*log(x)^2 + log(x)*(50*x - 
550) - 50*x^2 + 50*x^3 + 50*x^4 - 1500) - log(x)*(100*x + 100*x^2 - 600) - 
 550*x^2 + 100*x^3 + 50*x^4 + x^3*log(x^2)^2 + 1800)/(x^3*log(x^2)^2),x)
 

Output:

x - (50*x)/log(x^2) + 275/log(x^2) + 300/(x*log(x^2)) - 900/(x^2*log(x^2)) 
 - (25*x^2)/log(x^2) + (50*log(x))/log(x^2) - (25*log(x)^2)/(x^2*log(x^2)) 
 + (50*log(x))/(x*log(x^2)) - (300*log(x))/(x^2*log(x^2))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.31 \[ \int \frac {1800-600 x-550 x^2+100 x^3+50 x^4+\left (600-100 x-100 x^2\right ) \log (x)+50 \log ^2(x)+\left (1500-250 x+50 x^2-50 x^3-50 x^4+(550-50 x) \log (x)+50 \log ^2(x)\right ) \log \left (x^2\right )+x^3 \log ^2\left (x^2\right )}{x^3 \log ^2\left (x^2\right )} \, dx=\frac {\mathrm {log}\left (x^{2}\right ) x^{3}-25 \mathrm {log}\left (x \right )^{2}+50 \,\mathrm {log}\left (x \right ) x^{2}+50 \,\mathrm {log}\left (x \right ) x -300 \,\mathrm {log}\left (x \right )-25 x^{4}-50 x^{3}+275 x^{2}+300 x -900}{\mathrm {log}\left (x^{2}\right ) x^{2}} \] Input:

int((x^3*log(x^2)^2+(50*log(x)^2+(-50*x+550)*log(x)-50*x^4-50*x^3+50*x^2-2 
50*x+1500)*log(x^2)+50*log(x)^2+(-100*x^2-100*x+600)*log(x)+50*x^4+100*x^3 
-550*x^2-600*x+1800)/x^3/log(x^2)^2,x)
 

Output:

(log(x**2)*x**3 - 25*log(x)**2 + 50*log(x)*x**2 + 50*log(x)*x - 300*log(x) 
 - 25*x**4 - 50*x**3 + 275*x**2 + 300*x - 900)/(log(x**2)*x**2)