\(\int \frac {540 x-315 x^2+50 x^3-25 x^4+e^x (-1125 x^2+1125 x^3-125 x^4+125 x^5)+(-90 x+45 x^2-10 x^3+5 x^4+e^x (450 x^2-450 x^3+50 x^4-50 x^5)) \log (\frac {9+x^2}{x^2})+e^x (-45 x^2+45 x^3-5 x^4+5 x^5) \log ^2(\frac {9+x^2}{x^2})}{27-54 x+30 x^2-6 x^3+3 x^4+e^x (-270 x+270 x^2-30 x^3+30 x^4)+e^{2 x} (675 x^2+75 x^4)+(e^{2 x} (-270 x^2-30 x^4)+e^x (54 x-54 x^2+6 x^3-6 x^4)) \log (\frac {9+x^2}{x^2})+e^{2 x} (27 x^2+3 x^4) \log ^2(\frac {9+x^2}{x^2})} \, dx\) [1374]

Optimal result

Integrand size = 280, antiderivative size = 32 \[ \int \frac {540 x-315 x^2+50 x^3-25 x^4+e^x \left (-1125 x^2+1125 x^3-125 x^4+125 x^5\right )+\left (-90 x+45 x^2-10 x^3+5 x^4+e^x \left (450 x^2-450 x^3+50 x^4-50 x^5\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^x \left (-45 x^2+45 x^3-5 x^4+5 x^5\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )}{27-54 x+30 x^2-6 x^3+3 x^4+e^x \left (-270 x+270 x^2-30 x^3+30 x^4\right )+e^{2 x} \left (675 x^2+75 x^4\right )+\left (e^{2 x} \left (-270 x^2-30 x^4\right )+e^x \left (54 x-54 x^2+6 x^3-6 x^4\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^{2 x} \left (27 x^2+3 x^4\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )} \, dx=\frac {5 x}{3 \left (-e^x+\frac {-1+x}{x \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}\right )} \] Output:

5/3*x/((-1+x)/(ln(9/x^2+1)-5)/x-exp(x))
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int \frac {540 x-315 x^2+50 x^3-25 x^4+e^x \left (-1125 x^2+1125 x^3-125 x^4+125 x^5\right )+\left (-90 x+45 x^2-10 x^3+5 x^4+e^x \left (450 x^2-450 x^3+50 x^4-50 x^5\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^x \left (-45 x^2+45 x^3-5 x^4+5 x^5\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )}{27-54 x+30 x^2-6 x^3+3 x^4+e^x \left (-270 x+270 x^2-30 x^3+30 x^4\right )+e^{2 x} \left (675 x^2+75 x^4\right )+\left (e^{2 x} \left (-270 x^2-30 x^4\right )+e^x \left (54 x-54 x^2+6 x^3-6 x^4\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^{2 x} \left (27 x^2+3 x^4\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )} \, dx=-\frac {5 x^2 \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{3 \left (1-\left (1+5 e^x\right ) x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )} \] Input:

Integrate[(540*x - 315*x^2 + 50*x^3 - 25*x^4 + E^x*(-1125*x^2 + 1125*x^3 - 
 125*x^4 + 125*x^5) + (-90*x + 45*x^2 - 10*x^3 + 5*x^4 + E^x*(450*x^2 - 45 
0*x^3 + 50*x^4 - 50*x^5))*Log[(9 + x^2)/x^2] + E^x*(-45*x^2 + 45*x^3 - 5*x 
^4 + 5*x^5)*Log[(9 + x^2)/x^2]^2)/(27 - 54*x + 30*x^2 - 6*x^3 + 3*x^4 + E^ 
x*(-270*x + 270*x^2 - 30*x^3 + 30*x^4) + E^(2*x)*(675*x^2 + 75*x^4) + (E^( 
2*x)*(-270*x^2 - 30*x^4) + E^x*(54*x - 54*x^2 + 6*x^3 - 6*x^4))*Log[(9 + x 
^2)/x^2] + E^(2*x)*(27*x^2 + 3*x^4)*Log[(9 + x^2)/x^2]^2),x]
 

Output:

(-5*x^2*(-5 + Log[1 + 9/x^2]))/(3*(1 - (1 + 5*E^x)*x + E^x*x*Log[1 + 9/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.

\(\displaystyle \int \frac {-25 x^4+50 x^3-315 x^2+e^x \left (125 x^5-125 x^4+1125 x^3-1125 x^2\right )+e^x \left (5 x^5-5 x^4+45 x^3-45 x^2\right ) \log ^2\left (\frac {x^2+9}{x^2}\right )+\left (5 x^4-10 x^3+45 x^2+e^x \left (-50 x^5+50 x^4-450 x^3+450 x^2\right )-90 x\right ) \log \left (\frac {x^2+9}{x^2}\right )+540 x}{3 x^4-6 x^3+30 x^2+e^{2 x} \left (75 x^4+675 x^2\right )+e^{2 x} \left (3 x^4+27 x^2\right ) \log ^2\left (\frac {x^2+9}{x^2}\right )+e^x \left (30 x^4-30 x^3+270 x^2-270 x\right )+\left (e^{2 x} \left (-30 x^4-270 x^2\right )+e^x \left (-6 x^4+6 x^3-54 x^2+54 x\right )\right ) \log \left (\frac {x^2+9}{x^2}\right )-54 x+27} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {5 x \left (25 e^x x^4-5 \left (5 e^x+1\right ) x^3+5 \left (45 e^x+2\right ) x^2-\left (x^2+9\right ) \left (10 e^x x^2-\left (10 e^x+1\right ) x+2\right ) \log \left (\frac {9}{x^2}+1\right )+e^x \left (x^3-x^2+9 x-9\right ) x \log ^2\left (\frac {9}{x^2}+1\right )-9 \left (25 e^x+7\right ) x+108\right )}{3 \left (x^2+9\right ) \left (e^x x \log \left (\frac {9}{x^2}+1\right )-\left (\left (5 e^x+1\right ) x\right )+1\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {5}{3} \int \frac {x \left (25 e^x x^4-5 \left (1+5 e^x\right ) x^3+5 \left (2+45 e^x\right ) x^2-e^x \left (-x^3+x^2-9 x+9\right ) \log ^2\left (1+\frac {9}{x^2}\right ) x-9 \left (7+25 e^x\right ) x-\left (x^2+9\right ) \left (10 e^x x^2-\left (1+10 e^x\right ) x+2\right ) \log \left (1+\frac {9}{x^2}\right )+108\right )}{\left (x^2+9\right ) \left (-\left (\left (1+5 e^x\right ) x\right )+e^x \log \left (1+\frac {9}{x^2}\right ) x+1\right )^2}dx\)

\(\Big \downarrow \) 7276

\(\displaystyle \frac {5}{3} \int \left (\frac {(x-1) x \left (\log \left (1+\frac {9}{x^2}\right )-5\right )}{-5 e^x x+e^x \log \left (1+\frac {9}{x^2}\right ) x-x+1}+\frac {x \left (\log \left (1+\frac {9}{x^2}\right ) x^4-5 x^4-\log \left (1+\frac {9}{x^2}\right ) x^3+5 x^3+8 \log \left (1+\frac {9}{x^2}\right ) x^2-40 x^2-9 \log \left (1+\frac {9}{x^2}\right ) x+27 x-9 \log \left (1+\frac {9}{x^2}\right )+63\right )}{\left (x^2+9\right ) \left (-5 e^x x+e^x \log \left (1+\frac {9}{x^2}\right ) x-x+1\right )^2}\right )dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \frac {5}{3} \int \left (\frac {(x-1) x \left (\log \left (1+\frac {9}{x^2}\right )-5\right )}{-5 e^x x+e^x \log \left (1+\frac {9}{x^2}\right ) x-x+1}+\frac {x \left (\log \left (1+\frac {9}{x^2}\right ) x^4-5 x^4-\log \left (1+\frac {9}{x^2}\right ) x^3+5 x^3+8 \log \left (1+\frac {9}{x^2}\right ) x^2-40 x^2-9 \log \left (1+\frac {9}{x^2}\right ) x+27 x-9 \log \left (1+\frac {9}{x^2}\right )+63\right )}{\left (x^2+9\right ) \left (-5 e^x x+e^x \log \left (1+\frac {9}{x^2}\right ) x-x+1\right )^2}\right )dx\)

Input:

Int[(540*x - 315*x^2 + 50*x^3 - 25*x^4 + E^x*(-1125*x^2 + 1125*x^3 - 125*x 
^4 + 125*x^5) + (-90*x + 45*x^2 - 10*x^3 + 5*x^4 + E^x*(450*x^2 - 450*x^3 
+ 50*x^4 - 50*x^5))*Log[(9 + x^2)/x^2] + E^x*(-45*x^2 + 45*x^3 - 5*x^4 + 5 
*x^5)*Log[(9 + x^2)/x^2]^2)/(27 - 54*x + 30*x^2 - 6*x^3 + 3*x^4 + E^x*(-27 
0*x + 270*x^2 - 30*x^3 + 30*x^4) + E^(2*x)*(675*x^2 + 75*x^4) + (E^(2*x)*( 
-270*x^2 - 30*x^4) + E^x*(54*x - 54*x^2 + 6*x^3 - 6*x^4))*Log[(9 + x^2)/x^ 
2] + E^(2*x)*(27*x^2 + 3*x^4)*Log[(9 + x^2)/x^2]^2),x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.04 (sec) , antiderivative size = 216, normalized size of antiderivative = 6.75

\[-\frac {5 \,{\mathrm e}^{-x} x}{3}+\frac {10 i x \left (-1+x \right ) {\mathrm e}^{-x}}{3 \left (\pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right ) {\mathrm e}^{x}-2 \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{x}+\pi x \operatorname {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{x}-\pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (i \left (x^{2}+9\right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}+9\right )}{x^{2}}\right ) {\mathrm e}^{x}+\pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+9\right )}{x^{2}}\right )}^{2} {\mathrm e}^{x}+\pi x \,\operatorname {csgn}\left (i \left (x^{2}+9\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+9\right )}{x^{2}}\right )}^{2} {\mathrm e}^{x}-\pi x {\operatorname {csgn}\left (\frac {i \left (x^{2}+9\right )}{x^{2}}\right )}^{3} {\mathrm e}^{x}+4 i x \,{\mathrm e}^{x} \ln \left (x \right )-2 i x \,{\mathrm e}^{x} \ln \left (x^{2}+9\right )+10 i x \,{\mathrm e}^{x}+2 i x -2 i\right )}\]

Input:

int(((5*x^5-5*x^4+45*x^3-45*x^2)*exp(x)*ln((x^2+9)/x^2)^2+((-50*x^5+50*x^4 
-450*x^3+450*x^2)*exp(x)+5*x^4-10*x^3+45*x^2-90*x)*ln((x^2+9)/x^2)+(125*x^ 
5-125*x^4+1125*x^3-1125*x^2)*exp(x)-25*x^4+50*x^3-315*x^2+540*x)/((3*x^4+2 
7*x^2)*exp(x)^2*ln((x^2+9)/x^2)^2+((-30*x^4-270*x^2)*exp(x)^2+(-6*x^4+6*x^ 
3-54*x^2+54*x)*exp(x))*ln((x^2+9)/x^2)+(75*x^4+675*x^2)*exp(x)^2+(30*x^4-3 
0*x^3+270*x^2-270*x)*exp(x)+3*x^4-6*x^3+30*x^2-54*x+27),x)
 

Output:

-5/3*x/exp(x)+10/3*I*x*(-1+x)/(Pi*x*csgn(I*x)^2*csgn(I*x^2)*exp(x)-2*Pi*x* 
csgn(I*x)*csgn(I*x^2)^2*exp(x)+Pi*x*csgn(I*x^2)^3*exp(x)-Pi*x*csgn(I/x^2)* 
csgn(I*(x^2+9))*csgn(I/x^2*(x^2+9))*exp(x)+Pi*x*csgn(I/x^2)*csgn(I/x^2*(x^ 
2+9))^2*exp(x)+Pi*x*csgn(I*(x^2+9))*csgn(I/x^2*(x^2+9))^2*exp(x)-Pi*x*csgn 
(I/x^2*(x^2+9))^3*exp(x)+4*I*x*exp(x)*ln(x)-2*I*x*exp(x)*ln(x^2+9)+10*I*x* 
exp(x)+2*I*x-2*I)/exp(x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50 \[ \int \frac {540 x-315 x^2+50 x^3-25 x^4+e^x \left (-1125 x^2+1125 x^3-125 x^4+125 x^5\right )+\left (-90 x+45 x^2-10 x^3+5 x^4+e^x \left (450 x^2-450 x^3+50 x^4-50 x^5\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^x \left (-45 x^2+45 x^3-5 x^4+5 x^5\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )}{27-54 x+30 x^2-6 x^3+3 x^4+e^x \left (-270 x+270 x^2-30 x^3+30 x^4\right )+e^{2 x} \left (675 x^2+75 x^4\right )+\left (e^{2 x} \left (-270 x^2-30 x^4\right )+e^x \left (54 x-54 x^2+6 x^3-6 x^4\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^{2 x} \left (27 x^2+3 x^4\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )} \, dx=-\frac {5 \, {\left (x^{2} \log \left (\frac {x^{2} + 9}{x^{2}}\right ) - 5 \, x^{2}\right )}}{3 \, {\left (x e^{x} \log \left (\frac {x^{2} + 9}{x^{2}}\right ) - 5 \, x e^{x} - x + 1\right )}} \] Input:

integrate(((5*x^5-5*x^4+45*x^3-45*x^2)*exp(x)*log((x^2+9)/x^2)^2+((-50*x^5 
+50*x^4-450*x^3+450*x^2)*exp(x)+5*x^4-10*x^3+45*x^2-90*x)*log((x^2+9)/x^2) 
+(125*x^5-125*x^4+1125*x^3-1125*x^2)*exp(x)-25*x^4+50*x^3-315*x^2+540*x)/( 
(3*x^4+27*x^2)*exp(x)^2*log((x^2+9)/x^2)^2+((-30*x^4-270*x^2)*exp(x)^2+(-6 
*x^4+6*x^3-54*x^2+54*x)*exp(x))*log((x^2+9)/x^2)+(75*x^4+675*x^2)*exp(x)^2 
+(30*x^4-30*x^3+270*x^2-270*x)*exp(x)+3*x^4-6*x^3+30*x^2-54*x+27),x, algor 
ithm="fricas")
 

Output:

-5/3*(x^2*log((x^2 + 9)/x^2) - 5*x^2)/(x*e^x*log((x^2 + 9)/x^2) - 5*x*e^x 
- x + 1)
 

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int \frac {540 x-315 x^2+50 x^3-25 x^4+e^x \left (-1125 x^2+1125 x^3-125 x^4+125 x^5\right )+\left (-90 x+45 x^2-10 x^3+5 x^4+e^x \left (450 x^2-450 x^3+50 x^4-50 x^5\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^x \left (-45 x^2+45 x^3-5 x^4+5 x^5\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )}{27-54 x+30 x^2-6 x^3+3 x^4+e^x \left (-270 x+270 x^2-30 x^3+30 x^4\right )+e^{2 x} \left (675 x^2+75 x^4\right )+\left (e^{2 x} \left (-270 x^2-30 x^4\right )+e^x \left (54 x-54 x^2+6 x^3-6 x^4\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^{2 x} \left (27 x^2+3 x^4\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )} \, dx=\frac {- 5 x^{2} \log {\left (\frac {x^{2} + 9}{x^{2}} \right )} + 25 x^{2}}{- 3 x + \left (3 x \log {\left (\frac {x^{2} + 9}{x^{2}} \right )} - 15 x\right ) e^{x} + 3} \] Input:

integrate(((5*x**5-5*x**4+45*x**3-45*x**2)*exp(x)*ln((x**2+9)/x**2)**2+((- 
50*x**5+50*x**4-450*x**3+450*x**2)*exp(x)+5*x**4-10*x**3+45*x**2-90*x)*ln( 
(x**2+9)/x**2)+(125*x**5-125*x**4+1125*x**3-1125*x**2)*exp(x)-25*x**4+50*x 
**3-315*x**2+540*x)/((3*x**4+27*x**2)*exp(x)**2*ln((x**2+9)/x**2)**2+((-30 
*x**4-270*x**2)*exp(x)**2+(-6*x**4+6*x**3-54*x**2+54*x)*exp(x))*ln((x**2+9 
)/x**2)+(75*x**4+675*x**2)*exp(x)**2+(30*x**4-30*x**3+270*x**2-270*x)*exp( 
x)+3*x**4-6*x**3+30*x**2-54*x+27),x)
 

Output:

(-5*x**2*log((x**2 + 9)/x**2) + 25*x**2)/(-3*x + (3*x*log((x**2 + 9)/x**2) 
 - 15*x)*exp(x) + 3)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (29) = 58\).

Time = 0.18 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.16 \[ \int \frac {540 x-315 x^2+50 x^3-25 x^4+e^x \left (-1125 x^2+1125 x^3-125 x^4+125 x^5\right )+\left (-90 x+45 x^2-10 x^3+5 x^4+e^x \left (450 x^2-450 x^3+50 x^4-50 x^5\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^x \left (-45 x^2+45 x^3-5 x^4+5 x^5\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )}{27-54 x+30 x^2-6 x^3+3 x^4+e^x \left (-270 x+270 x^2-30 x^3+30 x^4\right )+e^{2 x} \left (675 x^2+75 x^4\right )+\left (e^{2 x} \left (-270 x^2-30 x^4\right )+e^x \left (54 x-54 x^2+6 x^3-6 x^4\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^{2 x} \left (27 x^2+3 x^4\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )} \, dx=-\frac {5 \, {\left (x^{2} e^{x} \log \left (x^{2} + 9\right ) - {\left (2 \, x^{2} \log \left (x\right ) + 5 \, x^{2}\right )} e^{x}\right )}}{3 \, {\left (x e^{\left (2 \, x\right )} \log \left (x^{2} + 9\right ) - {\left (2 \, x \log \left (x\right ) + 5 \, x\right )} e^{\left (2 \, x\right )} - {\left (x - 1\right )} e^{x}\right )}} \] Input:

integrate(((5*x^5-5*x^4+45*x^3-45*x^2)*exp(x)*log((x^2+9)/x^2)^2+((-50*x^5 
+50*x^4-450*x^3+450*x^2)*exp(x)+5*x^4-10*x^3+45*x^2-90*x)*log((x^2+9)/x^2) 
+(125*x^5-125*x^4+1125*x^3-1125*x^2)*exp(x)-25*x^4+50*x^3-315*x^2+540*x)/( 
(3*x^4+27*x^2)*exp(x)^2*log((x^2+9)/x^2)^2+((-30*x^4-270*x^2)*exp(x)^2+(-6 
*x^4+6*x^3-54*x^2+54*x)*exp(x))*log((x^2+9)/x^2)+(75*x^4+675*x^2)*exp(x)^2 
+(30*x^4-30*x^3+270*x^2-270*x)*exp(x)+3*x^4-6*x^3+30*x^2-54*x+27),x, algor 
ithm="maxima")
 

Output:

-5/3*(x^2*e^x*log(x^2 + 9) - (2*x^2*log(x) + 5*x^2)*e^x)/(x*e^(2*x)*log(x^ 
2 + 9) - (2*x*log(x) + 5*x)*e^(2*x) - (x - 1)*e^x)
 

Giac [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50 \[ \int \frac {540 x-315 x^2+50 x^3-25 x^4+e^x \left (-1125 x^2+1125 x^3-125 x^4+125 x^5\right )+\left (-90 x+45 x^2-10 x^3+5 x^4+e^x \left (450 x^2-450 x^3+50 x^4-50 x^5\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^x \left (-45 x^2+45 x^3-5 x^4+5 x^5\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )}{27-54 x+30 x^2-6 x^3+3 x^4+e^x \left (-270 x+270 x^2-30 x^3+30 x^4\right )+e^{2 x} \left (675 x^2+75 x^4\right )+\left (e^{2 x} \left (-270 x^2-30 x^4\right )+e^x \left (54 x-54 x^2+6 x^3-6 x^4\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^{2 x} \left (27 x^2+3 x^4\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )} \, dx=-\frac {5 \, {\left (x^{2} \log \left (\frac {x^{2} + 9}{x^{2}}\right ) - 5 \, x^{2}\right )}}{3 \, {\left (x e^{x} \log \left (\frac {x^{2} + 9}{x^{2}}\right ) - 5 \, x e^{x} - x + 1\right )}} \] Input:

integrate(((5*x^5-5*x^4+45*x^3-45*x^2)*exp(x)*log((x^2+9)/x^2)^2+((-50*x^5 
+50*x^4-450*x^3+450*x^2)*exp(x)+5*x^4-10*x^3+45*x^2-90*x)*log((x^2+9)/x^2) 
+(125*x^5-125*x^4+1125*x^3-1125*x^2)*exp(x)-25*x^4+50*x^3-315*x^2+540*x)/( 
(3*x^4+27*x^2)*exp(x)^2*log((x^2+9)/x^2)^2+((-30*x^4-270*x^2)*exp(x)^2+(-6 
*x^4+6*x^3-54*x^2+54*x)*exp(x))*log((x^2+9)/x^2)+(75*x^4+675*x^2)*exp(x)^2 
+(30*x^4-30*x^3+270*x^2-270*x)*exp(x)+3*x^4-6*x^3+30*x^2-54*x+27),x, algor 
ithm="giac")
 

Output:

-5/3*(x^2*log((x^2 + 9)/x^2) - 5*x^2)/(x*e^x*log((x^2 + 9)/x^2) - 5*x*e^x 
- x + 1)
 

Mupad [B] (verification not implemented)

Time = 3.00 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int \frac {540 x-315 x^2+50 x^3-25 x^4+e^x \left (-1125 x^2+1125 x^3-125 x^4+125 x^5\right )+\left (-90 x+45 x^2-10 x^3+5 x^4+e^x \left (450 x^2-450 x^3+50 x^4-50 x^5\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^x \left (-45 x^2+45 x^3-5 x^4+5 x^5\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )}{27-54 x+30 x^2-6 x^3+3 x^4+e^x \left (-270 x+270 x^2-30 x^3+30 x^4\right )+e^{2 x} \left (675 x^2+75 x^4\right )+\left (e^{2 x} \left (-270 x^2-30 x^4\right )+e^x \left (54 x-54 x^2+6 x^3-6 x^4\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^{2 x} \left (27 x^2+3 x^4\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )} \, dx=\frac {5\,x^2\,\left (\ln \left (\frac {x^2+9}{x^2}\right )-5\right )}{3\,\left (x+5\,x\,{\mathrm {e}}^x-x\,{\mathrm {e}}^x\,\ln \left (\frac {x^2+9}{x^2}\right )-1\right )} \] Input:

int(-(exp(x)*(1125*x^2 - 1125*x^3 + 125*x^4 - 125*x^5) - 540*x - log((x^2 
+ 9)/x^2)*(exp(x)*(450*x^2 - 450*x^3 + 50*x^4 - 50*x^5) - 90*x + 45*x^2 - 
10*x^3 + 5*x^4) + 315*x^2 - 50*x^3 + 25*x^4 + exp(x)*log((x^2 + 9)/x^2)^2* 
(45*x^2 - 45*x^3 + 5*x^4 - 5*x^5))/(log((x^2 + 9)/x^2)*(exp(x)*(54*x - 54* 
x^2 + 6*x^3 - 6*x^4) - exp(2*x)*(270*x^2 + 30*x^4)) - exp(x)*(270*x - 270* 
x^2 + 30*x^3 - 30*x^4) - 54*x + exp(2*x)*(675*x^2 + 75*x^4) + 30*x^2 - 6*x 
^3 + 3*x^4 + exp(2*x)*log((x^2 + 9)/x^2)^2*(27*x^2 + 3*x^4) + 27),x)
 

Output:

(5*x^2*(log((x^2 + 9)/x^2) - 5))/(3*(x + 5*x*exp(x) - x*exp(x)*log((x^2 + 
9)/x^2) - 1))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50 \[ \int \frac {540 x-315 x^2+50 x^3-25 x^4+e^x \left (-1125 x^2+1125 x^3-125 x^4+125 x^5\right )+\left (-90 x+45 x^2-10 x^3+5 x^4+e^x \left (450 x^2-450 x^3+50 x^4-50 x^5\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^x \left (-45 x^2+45 x^3-5 x^4+5 x^5\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )}{27-54 x+30 x^2-6 x^3+3 x^4+e^x \left (-270 x+270 x^2-30 x^3+30 x^4\right )+e^{2 x} \left (675 x^2+75 x^4\right )+\left (e^{2 x} \left (-270 x^2-30 x^4\right )+e^x \left (54 x-54 x^2+6 x^3-6 x^4\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^{2 x} \left (27 x^2+3 x^4\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )} \, dx=\frac {5 x^{2} \left (-\mathrm {log}\left (\frac {x^{2}+9}{x^{2}}\right )+5\right )}{3 e^{x} \mathrm {log}\left (\frac {x^{2}+9}{x^{2}}\right ) x -15 e^{x} x -3 x +3} \] Input:

int(((5*x^5-5*x^4+45*x^3-45*x^2)*exp(x)*log((x^2+9)/x^2)^2+((-50*x^5+50*x^ 
4-450*x^3+450*x^2)*exp(x)+5*x^4-10*x^3+45*x^2-90*x)*log((x^2+9)/x^2)+(125* 
x^5-125*x^4+1125*x^3-1125*x^2)*exp(x)-25*x^4+50*x^3-315*x^2+540*x)/((3*x^4 
+27*x^2)*exp(x)^2*log((x^2+9)/x^2)^2+((-30*x^4-270*x^2)*exp(x)^2+(-6*x^4+6 
*x^3-54*x^2+54*x)*exp(x))*log((x^2+9)/x^2)+(75*x^4+675*x^2)*exp(x)^2+(30*x 
^4-30*x^3+270*x^2-270*x)*exp(x)+3*x^4-6*x^3+30*x^2-54*x+27),x)
 

Output:

(5*x**2*( - log((x**2 + 9)/x**2) + 5))/(3*(e**x*log((x**2 + 9)/x**2)*x - 5 
*e**x*x - x + 1))