\(\int \frac {-32 i \pi +8 e^{3 x} x+e^{3 x} (8 x+24 x^2) \log (x)+(-\pi ^2 (480 x+32 x^2)+i e^{3 x} \pi (-240 x^2-16 x^3)+e^{6 x} (30 x^3+2 x^4)) \log ^2(x)}{(-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3) \log ^2(x)} \, dx\) [560]

Optimal result

Integrand size = 125, antiderivative size = 30 \[ \int \frac {-32 i \pi +8 e^{3 x} x+e^{3 x} \left (8 x+24 x^2\right ) \log (x)+\left (-\pi ^2 \left (480 x+32 x^2\right )+i e^{3 x} \pi \left (-240 x^2-16 x^3\right )+e^{6 x} \left (30 x^3+2 x^4\right )\right ) \log ^2(x)}{\left (-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3\right ) \log ^2(x)} \, dx=(15+x)^2+\frac {2}{\left (i \pi -\frac {1}{4} e^{3 x} x\right ) \log (x)} \] Output:

2/(I*Pi-1/4*x*exp(3*x))/ln(x)+(x+15)^2
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {-32 i \pi +8 e^{3 x} x+e^{3 x} \left (8 x+24 x^2\right ) \log (x)+\left (-\pi ^2 \left (480 x+32 x^2\right )+i e^{3 x} \pi \left (-240 x^2-16 x^3\right )+e^{6 x} \left (30 x^3+2 x^4\right )\right ) \log ^2(x)}{\left (-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3\right ) \log ^2(x)} \, dx=2 \left (15 x+\frac {x^2}{2}-\frac {4 i}{\left (4 \pi +i e^{3 x} x\right ) \log (x)}\right ) \] Input:

Integrate[((-32*I)*Pi + 8*E^(3*x)*x + E^(3*x)*(8*x + 24*x^2)*Log[x] + (-(P 
i^2*(480*x + 32*x^2)) + I*E^(3*x)*Pi*(-240*x^2 - 16*x^3) + E^(6*x)*(30*x^3 
 + 2*x^4))*Log[x]^2)/((-16*Pi^2*x - (8*I)*E^(3*x)*Pi*x^2 + E^(6*x)*x^3)*Lo 
g[x]^2),x]
 

Output:

2*(15*x + x^2/2 - (4*I)/((4*Pi + I*E^(3*x)*x)*Log[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.

\(\displaystyle \int \frac {e^{3 x} \left (24 x^2+8 x\right ) \log (x)+\left (-\pi ^2 \left (32 x^2+480 x\right )+e^{6 x} \left (2 x^4+30 x^3\right )+i \pi e^{3 x} \left (-16 x^3-240 x^2\right )\right ) \log ^2(x)+8 e^{3 x} x-32 i \pi }{\left (e^{6 x} x^3-8 i \pi e^{3 x} x^2-16 \pi ^2 x\right ) \log ^2(x)} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int 2 \left (x+\frac {4}{\left (e^{3 x} x-4 i \pi \right ) x \log ^2(x)}+\frac {4 e^{3 x} (3 x+1)}{\left (e^{3 x} x-4 i \pi \right )^2 \log (x)}+15\right )dx\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \int \left (x+\frac {4 e^{3 x} (3 x+1)}{\left (4 i \pi -e^{3 x} x\right )^2 \log (x)}+15-\frac {4}{\left (4 i \pi -e^{3 x} x\right ) \log ^2(x) x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (-4 \int \frac {1}{x \left (4 i \pi -e^{3 x} x\right ) \log ^2(x)}dx-4 \int \frac {e^{3 x}}{\left (i e^{3 x} x+4 \pi \right )^2 \log (x)}dx+12 \int \frac {e^{3 x} x}{\left (e^{3 x} x-4 i \pi \right )^2 \log (x)}dx+\frac {x^2}{2}+15 x\right )\)

Input:

Int[((-32*I)*Pi + 8*E^(3*x)*x + E^(3*x)*(8*x + 24*x^2)*Log[x] + (-(Pi^2*(4 
80*x + 32*x^2)) + I*E^(3*x)*Pi*(-240*x^2 - 16*x^3) + E^(6*x)*(30*x^3 + 2*x 
^4))*Log[x]^2)/((-16*Pi^2*x - (8*I)*E^(3*x)*Pi*x^2 + E^(6*x)*x^3)*Log[x]^2 
),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 38.92 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97

method	result	size
risch	\(x^{2}+30 x -\frac {8 i}{\left (i x \,{\mathrm e}^{3 x}+4 \pi \right ) \ln \left (x \right )}\)	\(29\)
parallelrisch	\(\frac {i \ln \left (x \right ) {\mathrm e}^{3 x} x^{3}+30 i \ln \left (x \right ) {\mathrm e}^{3 x} x^{2}+4 \ln \left (x \right ) \pi \,x^{2}+120 \ln \left (x \right ) \pi x -8 i}{\left (i x \,{\mathrm e}^{3 x}+4 \pi \right ) \ln \left (x \right )}\)	\(61\)

Input:

int((((2*x^4+30*x^3)*exp(3*x)^2+I*(-16*x^3-240*x^2)*Pi*exp(3*x)-(32*x^2+48 
0*x)*Pi^2)*ln(x)^2+(24*x^2+8*x)*exp(3*x)*ln(x)+8*x*exp(3*x)-32*I*Pi)/(x^3* 
exp(3*x)^2-8*I*x^2*Pi*exp(3*x)-16*x*Pi^2)/ln(x)^2,x,method=_RETURNVERBOSE)
 

Output:

x^2+30*x-8*I/(I*x*exp(3*x)+4*Pi)/ln(x)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.57 \[ \int \frac {-32 i \pi +8 e^{3 x} x+e^{3 x} \left (8 x+24 x^2\right ) \log (x)+\left (-\pi ^2 \left (480 x+32 x^2\right )+i e^{3 x} \pi \left (-240 x^2-16 x^3\right )+e^{6 x} \left (30 x^3+2 x^4\right )\right ) \log ^2(x)}{\left (-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3\right ) \log ^2(x)} \, dx=\frac {{\left (-4 i \, \pi x^{2} - 120 i \, \pi x + {\left (x^{3} + 30 \, x^{2}\right )} e^{\left (3 \, x\right )}\right )} \log \left (x\right ) - 8}{{\left (-4 i \, \pi + x e^{\left (3 \, x\right )}\right )} \log \left (x\right )} \] Input:

integrate((((2*x^4+30*x^3)*exp(3*x)^2+I*(-16*x^3-240*x^2)*pi*exp(3*x)-(32* 
x^2+480*x)*pi^2)*log(x)^2+(24*x^2+8*x)*exp(3*x)*log(x)+8*x*exp(3*x)-32*I*p 
i)/(x^3*exp(3*x)^2-8*I*x^2*pi*exp(3*x)-16*x*pi^2)/log(x)^2,x, algorithm="f 
ricas")
 

Output:

((-4*I*pi*x^2 - 120*I*pi*x + (x^3 + 30*x^2)*e^(3*x))*log(x) - 8)/((-4*I*pi 
 + x*e^(3*x))*log(x))
 

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {-32 i \pi +8 e^{3 x} x+e^{3 x} \left (8 x+24 x^2\right ) \log (x)+\left (-\pi ^2 \left (480 x+32 x^2\right )+i e^{3 x} \pi \left (-240 x^2-16 x^3\right )+e^{6 x} \left (30 x^3+2 x^4\right )\right ) \log ^2(x)}{\left (-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3\right ) \log ^2(x)} \, dx=x^{2} + 30 x - \frac {8}{x e^{3 x} \log {\left (x \right )} - 4 i \pi \log {\left (x \right )}} \] Input:

integrate((((2*x**4+30*x**3)*exp(3*x)**2+I*(-16*x**3-240*x**2)*pi*exp(3*x) 
-(32*x**2+480*x)*pi**2)*ln(x)**2+(24*x**2+8*x)*exp(3*x)*ln(x)+8*x*exp(3*x) 
-32*I*pi)/(x**3*exp(3*x)**2-8*I*x**2*pi*exp(3*x)-16*x*pi**2)/ln(x)**2,x)
 

Output:

x**2 + 30*x - 8/(x*exp(3*x)*log(x) - 4*I*pi*log(x))
 

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.67 \[ \int \frac {-32 i \pi +8 e^{3 x} x+e^{3 x} \left (8 x+24 x^2\right ) \log (x)+\left (-\pi ^2 \left (480 x+32 x^2\right )+i e^{3 x} \pi \left (-240 x^2-16 x^3\right )+e^{6 x} \left (30 x^3+2 x^4\right )\right ) \log ^2(x)}{\left (-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3\right ) \log ^2(x)} \, dx=\frac {{\left (x^{3} + 30 \, x^{2}\right )} e^{\left (3 \, x\right )} \log \left (x\right ) - 4 \, {\left (i \, \pi x^{2} + 30 i \, \pi x\right )} \log \left (x\right ) - 8}{x e^{\left (3 \, x\right )} \log \left (x\right ) - 4 i \, \pi \log \left (x\right )} \] Input:

integrate((((2*x^4+30*x^3)*exp(3*x)^2+I*(-16*x^3-240*x^2)*pi*exp(3*x)-(32* 
x^2+480*x)*pi^2)*log(x)^2+(24*x^2+8*x)*exp(3*x)*log(x)+8*x*exp(3*x)-32*I*p 
i)/(x^3*exp(3*x)^2-8*I*x^2*pi*exp(3*x)-16*x*pi^2)/log(x)^2,x, algorithm="m 
axima")
 

Output:

((x^3 + 30*x^2)*e^(3*x)*log(x) - 4*(I*pi*x^2 + 30*I*pi*x)*log(x) - 8)/(x*e 
^(3*x)*log(x) - 4*I*pi*log(x))
 

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (25) = 50\).

Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.90 \[ \int \frac {-32 i \pi +8 e^{3 x} x+e^{3 x} \left (8 x+24 x^2\right ) \log (x)+\left (-\pi ^2 \left (480 x+32 x^2\right )+i e^{3 x} \pi \left (-240 x^2-16 x^3\right )+e^{6 x} \left (30 x^3+2 x^4\right )\right ) \log ^2(x)}{\left (-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3\right ) \log ^2(x)} \, dx=-\frac {2 \, {\left (-i \, x^{3} e^{\left (3 \, x\right )} \log \left (x\right ) - 4 \, \pi x^{2} \log \left (x\right ) - 30 i \, x^{2} e^{\left (3 \, x\right )} \log \left (x\right ) - 120 \, \pi x \log \left (x\right ) + 8 i\right )}}{2 i \, x e^{\left (3 \, x\right )} \log \left (x\right ) + 8 \, \pi \log \left (x\right )} \] Input:

integrate((((2*x^4+30*x^3)*exp(3*x)^2+I*(-16*x^3-240*x^2)*pi*exp(3*x)-(32* 
x^2+480*x)*pi^2)*log(x)^2+(24*x^2+8*x)*exp(3*x)*log(x)+8*x*exp(3*x)-32*I*p 
i)/(x^3*exp(3*x)^2-8*I*x^2*pi*exp(3*x)-16*x*pi^2)/log(x)^2,x, algorithm="g 
iac")
 

Output:

-2*(-I*x^3*e^(3*x)*log(x) - 4*pi*x^2*log(x) - 30*I*x^2*e^(3*x)*log(x) - 12 
0*pi*x*log(x) + 8*I)/(2*I*x*e^(3*x)*log(x) + 8*pi*log(x))
 

Mupad [B] (verification not implemented)

Time = 1.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {-32 i \pi +8 e^{3 x} x+e^{3 x} \left (8 x+24 x^2\right ) \log (x)+\left (-\pi ^2 \left (480 x+32 x^2\right )+i e^{3 x} \pi \left (-240 x^2-16 x^3\right )+e^{6 x} \left (30 x^3+2 x^4\right )\right ) \log ^2(x)}{\left (-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3\right ) \log ^2(x)} \, dx=30\,x+\frac {8}{\ln \left (x\right )\,\left (-x\,{\mathrm {e}}^{3\,x}+\Pi \,4{}\mathrm {i}\right )}+x^2 \] Input:

int((Pi*32i - 8*x*exp(3*x) + log(x)^2*(Pi^2*(480*x + 32*x^2) - exp(6*x)*(3 
0*x^3 + 2*x^4) + Pi*exp(3*x)*(240*x^2 + 16*x^3)*1i) - exp(3*x)*log(x)*(8*x 
 + 24*x^2))/(log(x)^2*(16*Pi^2*x - x^3*exp(6*x) + Pi*x^2*exp(3*x)*8i)),x)
 

Output:

30*x + 8/(log(x)*(Pi*4i - x*exp(3*x))) + x^2
 

Reduce [F]

\[ \int \frac {-32 i \pi +8 e^{3 x} x+e^{3 x} \left (8 x+24 x^2\right ) \log (x)+\left (-\pi ^2 \left (480 x+32 x^2\right )+i e^{3 x} \pi \left (-240 x^2-16 x^3\right )+e^{6 x} \left (30 x^3+2 x^4\right )\right ) \log ^2(x)}{\left (-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3\right ) \log ^2(x)} \, dx=8 \left (\int \frac {e^{3 x}}{e^{6 x} \mathrm {log}\left (x \right )^{2} x^{2}-8 e^{3 x} \mathrm {log}\left (x \right )^{2} i \pi x -16 \mathrm {log}\left (x \right )^{2} \pi ^{2}}d x \right )+8 \left (\int \frac {e^{3 x}}{e^{6 x} \mathrm {log}\left (x \right ) x^{2}-8 e^{3 x} \mathrm {log}\left (x \right ) i \pi x -16 \,\mathrm {log}\left (x \right ) \pi ^{2}}d x \right )+2 \left (\int \frac {e^{6 x} x^{3}}{e^{6 x} x^{2}-8 e^{3 x} i \pi x -16 \pi ^{2}}d x \right )+30 \left (\int \frac {e^{6 x} x^{2}}{e^{6 x} x^{2}-8 e^{3 x} i \pi x -16 \pi ^{2}}d x \right )-16 \left (\int \frac {e^{3 x} x^{2}}{e^{6 x} x^{2}-8 e^{3 x} i \pi x -16 \pi ^{2}}d x \right ) i \pi -240 \left (\int \frac {e^{3 x} x}{e^{6 x} x^{2}-8 e^{3 x} i \pi x -16 \pi ^{2}}d x \right ) i \pi +24 \left (\int \frac {e^{3 x} x}{e^{6 x} \mathrm {log}\left (x \right ) x^{2}-8 e^{3 x} \mathrm {log}\left (x \right ) i \pi x -16 \,\mathrm {log}\left (x \right ) \pi ^{2}}d x \right )-32 \left (\int \frac {x}{e^{6 x} x^{2}-8 e^{3 x} i \pi x -16 \pi ^{2}}d x \right ) \pi ^{2}-32 \left (\int \frac {1}{e^{6 x} \mathrm {log}\left (x \right )^{2} x^{3}-8 e^{3 x} \mathrm {log}\left (x \right )^{2} i \pi \,x^{2}-16 \mathrm {log}\left (x \right )^{2} \pi ^{2} x}d x \right ) i \pi -480 \left (\int \frac {1}{e^{6 x} x^{2}-8 e^{3 x} i \pi x -16 \pi ^{2}}d x \right ) \pi ^{2} \] Input:

int((((2*x^4+30*x^3)*exp(3*x)^2+I*(-16*x^3-240*x^2)*Pi*exp(3*x)-(32*x^2+48 
0*x)*Pi^2)*log(x)^2+(24*x^2+8*x)*exp(3*x)*log(x)+8*x*exp(3*x)-32*I*Pi)/(x^ 
3*exp(3*x)^2-8*I*x^2*Pi*exp(3*x)-16*x*Pi^2)/log(x)^2,x)
 

Output:

2*(4*int(e**(3*x)/(e**(6*x)*log(x)**2*x**2 - 8*e**(3*x)*log(x)**2*i*pi*x - 
 16*log(x)**2*pi**2),x) + 4*int(e**(3*x)/(e**(6*x)*log(x)*x**2 - 8*e**(3*x 
)*log(x)*i*pi*x - 16*log(x)*pi**2),x) + int((e**(6*x)*x**3)/(e**(6*x)*x**2 
 - 8*e**(3*x)*i*pi*x - 16*pi**2),x) + 15*int((e**(6*x)*x**2)/(e**(6*x)*x** 
2 - 8*e**(3*x)*i*pi*x - 16*pi**2),x) - 8*int((e**(3*x)*x**2)/(e**(6*x)*x** 
2 - 8*e**(3*x)*i*pi*x - 16*pi**2),x)*i*pi - 120*int((e**(3*x)*x)/(e**(6*x) 
*x**2 - 8*e**(3*x)*i*pi*x - 16*pi**2),x)*i*pi + 12*int((e**(3*x)*x)/(e**(6 
*x)*log(x)*x**2 - 8*e**(3*x)*log(x)*i*pi*x - 16*log(x)*pi**2),x) - 16*int( 
x/(e**(6*x)*x**2 - 8*e**(3*x)*i*pi*x - 16*pi**2),x)*pi**2 - 16*int(1/(e**( 
6*x)*log(x)**2*x**3 - 8*e**(3*x)*log(x)**2*i*pi*x**2 - 16*log(x)**2*pi**2* 
x),x)*i*pi - 240*int(1/(e**(6*x)*x**2 - 8*e**(3*x)*i*pi*x - 16*pi**2),x)*p 
i**2)