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

Optimal. Leaf size=28 \[ \frac {e^{\frac {3}{2 \log (x)}}}{i \pi +x^2-\log \left (\frac {5}{2}\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 0.46, antiderivative size = 48, normalized size of antiderivative = 1.71, number of steps used = 3, number of rules used = 3, integrand size = 86, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {1594, 28, 2288} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

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

Rule 28

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

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2288

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\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 )}{x \left (2 x^4+4 x^2 \left (i \pi -\log \left (\frac {5}{2}\right )\right )+2 \left (i \pi -\log \left (\frac {5}{2}\right )\right )^2\right ) \log ^2(x)} \, dx\\ &=2 \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 )}{x \left (2 x^2+2 \left (i \pi -\log \left (\frac {5}{2}\right )\right )\right )^2 \log ^2(x)} \, dx\\ &=\frac {e^{\frac {3}{2 \log (x)}} \left (3 i \pi +3 x^2-\log \left (\frac {125}{8}\right )\right )}{3 \left (i \pi +x^2-\log \left (\frac {5}{2}\right )\right )^2}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.15, size = 28, normalized size = 1.00 \begin {gather*} \frac {e^{\frac {3}{2 \log (x)}}}{i \pi +x^2-\log \left (\frac {5}{2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.97, size = 19, normalized size = 0.68 \begin {gather*} \frac {e^{\left (\frac {3}{2 \, \log \relax (x)}\right )}}{i \, \pi + x^{2} + \log \left (\frac {2}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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*p
i)+2*x^5)/log(x)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.42, size = 26, normalized size = 0.93 \begin {gather*} -\frac {i \, e^{\left (\frac {3}{2 \, \log \relax (x)}\right )}}{\pi - i \, x^{2} + i \, \log \relax (5) - i \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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*p
i)+2*x^5)/log(x)^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.11, size = 28, normalized size = 1.00

method	result	size
risch	\(-\frac {{\mathrm e}^{\frac {3}{2 \ln \relax (x )}}}{-i \pi -x^{2}+\ln \relax (5)-\ln \relax (2)}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)/l
n(x)^2,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {x^{2} e^{\left (\frac {3}{2 \, \log \relax (x)}\right )}}{x^{4} - 2 \, {\left (-i \, \pi + \log \relax (5) - \log \relax (2)\right )} x^{2} - \pi ^{2} - 2 \, \pi {\left (i \, \log \relax (5) - i \, \log \relax (2)\right )} + \log \relax (5)^{2} - 2 \, \log \relax (5) \log \relax (2) + \log \relax (2)^{2}} + \frac {i \, \pi e^{\left (\frac {3}{2 \, \log \relax (x)}\right )}}{x^{4} - 2 \, {\left (-i \, \pi + \log \relax (5) - \log \relax (2)\right )} x^{2} - \pi ^{2} - 2 \, \pi {\left (i \, \log \relax (5) - i \, \log \relax (2)\right )} + \log \relax (5)^{2} - 2 \, \log \relax (5) \log \relax (2) + \log \relax (2)^{2}} + \frac {e^{\left (\frac {3}{2 \, \log \relax (x)}\right )} \log \left (\frac {2}{5}\right )}{x^{4} - 2 \, {\left (-i \, \pi + \log \relax (5) - \log \relax (2)\right )} x^{2} - \pi ^{2} - 2 \, \pi {\left (i \, \log \relax (5) - i \, \log \relax (2)\right )} + \log \relax (5)^{2} - 2 \, \log \relax (5) \log \relax (2) + \log \relax (2)^{2}} - 2 \, \int \frac {x e^{\left (\frac {3}{2 \, \log \relax (x)}\right )}}{x^{4} - 2 \, {\left (-i \, \pi + \log \relax (5) - \log \relax (2)\right )} x^{2} - \pi ^{2} - 2 \, \pi {\left (i \, \log \relax (5) - i \, \log \relax (2)\right )} + \log \relax (5)^{2} - 2 \, \log \relax (5) \log \relax (2) + \log \relax (2)^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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*p
i)+2*x^5)/log(x)^2,x, algorithm="maxima")

[Out]

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*l
og(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/lo
g(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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^{\frac {3}{2\,\ln \relax (x)}}\,\left (4\,x^2\,{\ln \relax (x)}^2+3\,x^2+\Pi \,3{}\mathrm {i}+3\,\ln \left (\frac {2}{5}\right )\right )}{{\ln \relax (x)}^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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 1.83, size = 22, normalized size = 0.79 \begin {gather*} - \frac {e^{\frac {3}{2 \log {\relax (x )}}}}{- x^{2} - \log {\relax (2 )} + \log {\relax (5 )} - i \pi } \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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*p
i)+2*x**5)/ln(x)**2,x)

[Out]

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

________________________________________________________________________________________