\(\int \frac {e^{\frac {3}{2 \log (x)}} (-3 x^2-3 (i \pi -\log (\frac {5}{2}))-4 x^2 \log ^2(x))}{(2 x^5+4 x^3 (i \pi -\log (\frac {5}{2}))+2 x (i \pi -\log (\frac {5}{2}))^2) \log ^2(x)} \, dx\) [103]

Optimal result

Integrand size = 86, antiderivative size = 28 \[ \int \frac {e^{\frac {3}{2 \log (x)}} \left (-3 x^2-3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )-4 x^2 \log ^2(x)\right )}{\left (2 x^5+4 x^3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )+2 x \left (i \pi -\log \left (\frac {5}{2}\right )\right )^2\right ) \log ^2(x)} \, dx=\frac {e^{\frac {3}{2 \log (x)}}}{i \pi +x^2-\log \left (\frac {5}{2}\right )} \] Output:

exp(3/2/ln(x))/(ln(2/5)+I*Pi+x^2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {3}{2 \log (x)}} \left (-3 x^2-3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )-4 x^2 \log ^2(x)\right )}{\left (2 x^5+4 x^3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )+2 x \left (i \pi -\log \left (\frac {5}{2}\right )\right )^2\right ) \log ^2(x)} \, dx=\frac {e^{\frac {3}{2 \log (x)}}}{i \pi +x^2-\log \left (\frac {5}{2}\right )} \] Input:

Integrate[(E^(3/(2*Log[x]))*(-3*x^2 - 3*(I*Pi - Log[5/2]) - 4*x^2*Log[x]^2 
))/((2*x^5 + 4*x^3*(I*Pi - Log[5/2]) + 2*x*(I*Pi - Log[5/2])^2)*Log[x]^2), 
x]
 

Output:

E^(3/(2*Log[x]))/(I*Pi + x^2 - Log[5/2])
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {2026, 1380, 27, 2726}

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.

\(\displaystyle \int \frac {e^{\frac {3}{2 \log (x)}} \left (-3 x^2-4 x^2 \log ^2(x)-3 \left (-\log \left (\frac {5}{2}\right )+i \pi \right )\right )}{\left (2 x^5+4 x^3 \left (-\log \left (\frac {5}{2}\right )+i \pi \right )+2 x \left (-\log \left (\frac {5}{2}\right )+i \pi \right )^2\right ) \log ^2(x)} \, dx\)

\(\Big \downarrow \) 2026

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

\(\Big \downarrow \) 1380

\(\displaystyle 2 \int -\frac {e^{\frac {3}{2 \log (x)}} \left (4 \log ^2(x) x^2+3 x^2-\log \left (\frac {125}{8}\right )+3 i \pi \right )}{4 x \left (x^2-\log \left (\frac {5}{2}\right )+i \pi \right )^2 \log ^2(x)}dx\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{2} \int \frac {e^{\frac {3}{2 \log (x)}} \left (4 \log ^2(x) x^2+3 x^2-\log \left (\frac {125}{8}\right )+3 i \pi \right )}{x \left (x^2-\log \left (\frac {5}{2}\right )+i \pi \right )^2 \log ^2(x)}dx\)

\(\Big \downarrow \) 2726

\(\displaystyle \frac {e^{\frac {3}{2 \log (x)}} \left (3 x^2+3 i \pi -\log \left (\frac {125}{8}\right )\right )}{3 \left (x^2+i \pi -\log \left (\frac {5}{2}\right )\right )^2}\)

Input:

Int[(E^(3/(2*Log[x]))*(-3*x^2 - 3*(I*Pi - Log[5/2]) - 4*x^2*Log[x]^2))/((2 
*x^5 + 4*x^3*(I*Pi - Log[5/2]) + 2*x*(I*Pi - Log[5/2])^2)*Log[x]^2),x]
 

Output:

(E^(3/(2*Log[x]))*((3*I)*Pi + 3*x^2 - Log[125/8]))/(3*(I*Pi + x^2 - Log[5/ 
2])^2)
 

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 1380

Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S 
imp[1/c^p   Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] 
&& EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 

rule 2026

Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p 
*r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ 
erQ[p] &&  !MonomialQ[Px, x] && (ILtQ[p, 0] ||  !PolyQ[u, x])
 

rule 2726

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, 
 x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
 

Maple [A] (verified)

Time = 8.38 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75

method	result	size
parallelrisch	\(\frac {{\mathrm e}^{\frac {3}{2 \ln \left (x \right )}}}{\ln \left (\frac {2}{5}\right )+i \pi +x^{2}}\)	\(21\)
risch	\(\frac {{\mathrm e}^{\frac {3}{2 \ln \left (x \right )}}}{i \pi +x^{2}+\ln \left (2\right )-\ln \left (5\right )}\)	\(25\)

Input:

int((-4*x^2*ln(x)^2-3*ln(2/5)-3*I*Pi-3*x^2)*exp(3/2/ln(x))/(2*x*(ln(2/5)+I 
*Pi)^2+4*x^3*(ln(2/5)+I*Pi)+2*x^5)/ln(x)^2,x,method=_RETURNVERBOSE)
 

Output:

exp(3/2/ln(x))/(ln(2/5)+I*Pi+x^2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\frac {3}{2 \log (x)}} \left (-3 x^2-3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )-4 x^2 \log ^2(x)\right )}{\left (2 x^5+4 x^3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )+2 x \left (i \pi -\log \left (\frac {5}{2}\right )\right )^2\right ) \log ^2(x)} \, dx=\frac {e^{\left (\frac {3}{2 \, \log \left (x\right )}\right )}}{i \, \pi + x^{2} + \log \left (\frac {2}{5}\right )} \] Input:

integrate((-4*x^2*log(x)^2-3*log(2/5)-3*I*pi-3*x^2)*exp(3/2/log(x))/(2*x*( 
log(2/5)+I*pi)^2+4*x^3*(log(2/5)+I*pi)+2*x^5)/log(x)^2,x, algorithm="frica 
s")
 

Output:

e^(3/2/log(x))/(I*pi + x^2 + log(2/5))
 

Sympy [A] (verification not implemented)

Time = 0.65 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {3}{2 \log (x)}} \left (-3 x^2-3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )-4 x^2 \log ^2(x)\right )}{\left (2 x^5+4 x^3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )+2 x \left (i \pi -\log \left (\frac {5}{2}\right )\right )^2\right ) \log ^2(x)} \, dx=\frac {e^{\frac {3}{2 \log {\left (x \right )}}}}{x^{2} - \log {\left (5 \right )} + \log {\left (2 \right )} + i \pi } \] Input:

integrate((-4*x**2*ln(x)**2-3*ln(2/5)-3*I*pi-3*x**2)*exp(3/2/ln(x))/(2*x*( 
ln(2/5)+I*pi)**2+4*x**3*(ln(2/5)+I*pi)+2*x**5)/ln(x)**2,x)
 

Output:

exp(3/(2*log(x)))/(x**2 - log(5) + log(2) + I*pi)
 

Maxima [F]

\[ \int \frac {e^{\frac {3}{2 \log (x)}} \left (-3 x^2-3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )-4 x^2 \log ^2(x)\right )}{\left (2 x^5+4 x^3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )+2 x \left (i \pi -\log \left (\frac {5}{2}\right )\right )^2\right ) \log ^2(x)} \, dx=\int { \frac {{\left (-3 i \, \pi - 4 \, x^{2} \log \left (x\right )^{2} - 3 \, x^{2} - 3 \, \log \left (\frac {2}{5}\right )\right )} e^{\left (\frac {3}{2 \, \log \left (x\right )}\right )}}{2 \, {\left (x^{5} + 2 \, {\left (i \, \pi + \log \left (\frac {2}{5}\right )\right )} x^{3} + {\left (i \, \pi + \log \left (\frac {2}{5}\right )\right )}^{2} x\right )} \log \left (x\right )^{2}} \,d x } \] Input:

integrate((-4*x^2*log(x)^2-3*log(2/5)-3*I*pi-3*x^2)*exp(3/2/log(x))/(2*x*( 
log(2/5)+I*pi)^2+4*x^3*(log(2/5)+I*pi)+2*x^5)/log(x)^2,x, algorithm="maxim 
a")
 

Output:

x^2*e^(3/2/log(x))/(x^4 - 2*(-I*pi + log(5) - log(2))*x^2 - pi^2 - 2*pi*(I 
*log(5) - I*log(2)) + log(5)^2 - 2*log(5)*log(2) + log(2)^2) + I*pi*e^(3/2 
/log(x))/(x^4 - 2*(-I*pi + log(5) - log(2))*x^2 - pi^2 - 2*pi*(I*log(5) - 
I*log(2)) + log(5)^2 - 2*log(5)*log(2) + log(2)^2) + e^(3/2/log(x))*log(2/ 
5)/(x^4 - 2*(-I*pi + log(5) - log(2))*x^2 - pi^2 - 2*pi*(I*log(5) - I*log( 
2)) + log(5)^2 - 2*log(5)*log(2) + log(2)^2) - 2*integrate(x*e^(3/2/log(x) 
)/(x^4 - 2*(-I*pi + log(5) - log(2))*x^2 - pi^2 - 2*pi*(I*log(5) - I*log(2 
)) + log(5)^2 - 2*log(5)*log(2) + log(2)^2), x)
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {3}{2 \log (x)}} \left (-3 x^2-3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )-4 x^2 \log ^2(x)\right )}{\left (2 x^5+4 x^3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )+2 x \left (i \pi -\log \left (\frac {5}{2}\right )\right )^2\right ) \log ^2(x)} \, dx=-\frac {i \, e^{\left (\frac {3}{2 \, \log \left (x\right )}\right )}}{\pi - i \, x^{2} + i \, \log \left (5\right ) - i \, \log \left (2\right )} \] Input:

integrate((-4*x^2*log(x)^2-3*log(2/5)-3*I*pi-3*x^2)*exp(3/2/log(x))/(2*x*( 
log(2/5)+I*pi)^2+4*x^3*(log(2/5)+I*pi)+2*x^5)/log(x)^2,x, algorithm="giac" 
)
 

Output:

-I*e^(3/2/log(x))/(pi - I*x^2 + I*log(5) - I*log(2))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {3}{2 \log (x)}} \left (-3 x^2-3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )-4 x^2 \log ^2(x)\right )}{\left (2 x^5+4 x^3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )+2 x \left (i \pi -\log \left (\frac {5}{2}\right )\right )^2\right ) \log ^2(x)} \, dx=\int -\frac {{\mathrm {e}}^{\frac {3}{2\,\ln \left (x\right )}}\,\left (4\,x^2\,{\ln \left (x\right )}^2+3\,x^2+\Pi \,3{}\mathrm {i}+3\,\ln \left (\frac {2}{5}\right )\right )}{{\ln \left (x\right )}^2\,\left (2\,x\,{\left (\ln \left (\frac {2}{5}\right )+\Pi \,1{}\mathrm {i}\right )}^2+4\,x^3\,\left (\ln \left (\frac {2}{5}\right )+\Pi \,1{}\mathrm {i}\right )+2\,x^5\right )} \,d x \] Input:

int(-(exp(3/(2*log(x)))*(Pi*3i + 3*log(2/5) + 4*x^2*log(x)^2 + 3*x^2))/(lo 
g(x)^2*(2*x*(Pi*1i + log(2/5))^2 + 4*x^3*(Pi*1i + log(2/5)) + 2*x^5)),x)
 

Output:

int(-(exp(3/(2*log(x)))*(Pi*3i + 3*log(2/5) + 4*x^2*log(x)^2 + 3*x^2))/(lo 
g(x)^2*(2*x*(Pi*1i + log(2/5))^2 + 4*x^3*(Pi*1i + log(2/5)) + 2*x^5)), x)
 

Reduce [F]

\[ \int \frac {e^{\frac {3}{2 \log (x)}} \left (-3 x^2-3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )-4 x^2 \log ^2(x)\right )}{\left (2 x^5+4 x^3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )+2 x \left (i \pi -\log \left (\frac {5}{2}\right )\right )^2\right ) \log ^2(x)} \, dx=\text {too large to display} \] Input:

int((-4*x^2*log(x)^2-3*log(2/5)-3*I*Pi-3*x^2)*exp(3/2/log(x))/(2*x*(log(2/ 
5)+I*Pi)^2+4*x^3*(log(2/5)+I*Pi)+2*x^5)/log(x)^2,x)
 

Output:

( - 3*int(e**(3/(2*log(x)))/(log(2/5)**2*log(x)**2*x + 2*log(2/5)*log(x)** 
2*i*pi*x + 2*log(2/5)*log(x)**2*x**3 + 2*log(x)**2*i*pi*x**3 - log(x)**2*p 
i**2*x + log(x)**2*x**5),x)*log(2/5)**8 - 24*int(e**(3/(2*log(x)))/(log(2/ 
5)**2*log(x)**2*x + 2*log(2/5)*log(x)**2*i*pi*x + 2*log(2/5)*log(x)**2*x** 
3 + 2*log(x)**2*i*pi*x**3 - log(x)**2*pi**2*x + log(x)**2*x**5),x)*log(2/5 
)**7*i*pi + 84*int(e**(3/(2*log(x)))/(log(2/5)**2*log(x)**2*x + 2*log(2/5) 
*log(x)**2*i*pi*x + 2*log(2/5)*log(x)**2*x**3 + 2*log(x)**2*i*pi*x**3 - lo 
g(x)**2*pi**2*x + log(x)**2*x**5),x)*log(2/5)**6*pi**2 + 168*int(e**(3/(2* 
log(x)))/(log(2/5)**2*log(x)**2*x + 2*log(2/5)*log(x)**2*i*pi*x + 2*log(2/ 
5)*log(x)**2*x**3 + 2*log(x)**2*i*pi*x**3 - log(x)**2*pi**2*x + log(x)**2* 
x**5),x)*log(2/5)**5*i*pi**3 - 210*int(e**(3/(2*log(x)))/(log(2/5)**2*log( 
x)**2*x + 2*log(2/5)*log(x)**2*i*pi*x + 2*log(2/5)*log(x)**2*x**3 + 2*log( 
x)**2*i*pi*x**3 - log(x)**2*pi**2*x + log(x)**2*x**5),x)*log(2/5)**4*pi**4 
 - 168*int(e**(3/(2*log(x)))/(log(2/5)**2*log(x)**2*x + 2*log(2/5)*log(x)* 
*2*i*pi*x + 2*log(2/5)*log(x)**2*x**3 + 2*log(x)**2*i*pi*x**3 - log(x)**2* 
pi**2*x + log(x)**2*x**5),x)*log(2/5)**3*i*pi**5 + 84*int(e**(3/(2*log(x)) 
)/(log(2/5)**2*log(x)**2*x + 2*log(2/5)*log(x)**2*i*pi*x + 2*log(2/5)*log( 
x)**2*x**3 + 2*log(x)**2*i*pi*x**3 - log(x)**2*pi**2*x + log(x)**2*x**5),x 
)*log(2/5)**2*pi**6 + 24*int(e**(3/(2*log(x)))/(log(2/5)**2*log(x)**2*x + 
2*log(2/5)*log(x)**2*i*pi*x + 2*log(2/5)*log(x)**2*x**3 + 2*log(x)**2*i...