3.35.3 \(\int \frac {e^{\frac {\log ^2(2 x)}{25 x^4-10 x^6+x^8}} (-375 x^4-250 x^5+225 x^6+150 x^7-45 x^8-30 x^9+3 x^{10}+2 x^{11}+(-30-10 x+6 x^2+2 x^3) \log (2 x)+(60+20 x-24 x^2-8 x^3) \log ^2(2 x))}{-125 x^4+75 x^6-15 x^8+x^{10}} \, dx\)

Optimal. Leaf size=26 \[ e^{\frac {\log ^2(2 x)}{x^4 \left (5-x^2\right )^2}} x (3+x) \]

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Rubi [B]  time = 0.94, antiderivative size = 160, normalized size of antiderivative = 6.15, number of steps used = 1, number of rules used = 1, integrand size = 130, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.008, Rules used = {2288} \begin {gather*} \frac {e^{\frac {\log ^2(2 x)}{x^8-10 x^6+25 x^4}} \left (\left (-x^3-3 x^2+5 x+15\right ) \log (2 x)-2 \left (-2 x^3-6 x^2+5 x+15\right ) \log ^2(2 x)\right )}{\left (-x^{10}+15 x^8-75 x^6+125 x^4\right ) \left (\frac {\log (2 x)}{x \left (x^8-10 x^6+25 x^4\right )}-\frac {2 \left (2 x^7-15 x^5+25 x^3\right ) \log ^2(2 x)}{\left (x^8-10 x^6+25 x^4\right )^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(Log[2*x]^2/(25*x^4 - 10*x^6 + x^8))*(-375*x^4 - 250*x^5 + 225*x^6 + 150*x^7 - 45*x^8 - 30*x^9 + 3*x^10
+ 2*x^11 + (-30 - 10*x + 6*x^2 + 2*x^3)*Log[2*x] + (60 + 20*x - 24*x^2 - 8*x^3)*Log[2*x]^2))/(-125*x^4 + 75*x
^6 - 15*x^8 + x^10),x]

[Out]

(E^(Log[2*x]^2/(25*x^4 - 10*x^6 + x^8))*((15 + 5*x - 3*x^2 - x^3)*Log[2*x] - 2*(15 + 5*x - 6*x^2 - 2*x^3)*Log[
2*x]^2))/((125*x^4 - 75*x^6 + 15*x^8 - x^10)*(Log[2*x]/(x*(25*x^4 - 10*x^6 + x^8)) - (2*(25*x^3 - 15*x^5 + 2*x
^7)*Log[2*x]^2)/(25*x^4 - 10*x^6 + x^8)^2))

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} &=\frac {e^{\frac {\log ^2(2 x)}{25 x^4-10 x^6+x^8}} \left (\left (15+5 x-3 x^2-x^3\right ) \log (2 x)-2 \left (15+5 x-6 x^2-2 x^3\right ) \log ^2(2 x)\right )}{\left (125 x^4-75 x^6+15 x^8-x^{10}\right ) \left (\frac {\log (2 x)}{x \left (25 x^4-10 x^6+x^8\right )}-\frac {2 \left (25 x^3-15 x^5+2 x^7\right ) \log ^2(2 x)}{\left (25 x^4-10 x^6+x^8\right )^2}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 24, normalized size = 0.92 \begin {gather*} e^{\frac {\log ^2(2 x)}{x^4 \left (-5+x^2\right )^2}} x (3+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(Log[2*x]^2/(25*x^4 - 10*x^6 + x^8))*(-375*x^4 - 250*x^5 + 225*x^6 + 150*x^7 - 45*x^8 - 30*x^9 +
3*x^10 + 2*x^11 + (-30 - 10*x + 6*x^2 + 2*x^3)*Log[2*x] + (60 + 20*x - 24*x^2 - 8*x^3)*Log[2*x]^2))/(-125*x^4
+ 75*x^6 - 15*x^8 + x^10),x]

[Out]

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

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fricas [A]  time = 0.56, size = 32, normalized size = 1.23 \begin {gather*} {\left (x^{2} + 3 \, x\right )} e^{\left (\frac {\log \left (2 \, x\right )^{2}}{x^{8} - 10 \, x^{6} + 25 \, x^{4}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^3-24*x^2+20*x+60)*log(2*x)^2+(2*x^3+6*x^2-10*x-30)*log(2*x)+2*x^11+3*x^10-30*x^9-45*x^8+150*x
^7+225*x^6-250*x^5-375*x^4)*exp(log(2*x)^2/(x^8-10*x^6+25*x^4))/(x^10-15*x^8+75*x^6-125*x^4),x, algorithm="fri
cas")

[Out]

(x^2 + 3*x)*e^(log(2*x)^2/(x^8 - 10*x^6 + 25*x^4))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, x^{11} + 3 \, x^{10} - 30 \, x^{9} - 45 \, x^{8} + 150 \, x^{7} + 225 \, x^{6} - 250 \, x^{5} - 375 \, x^{4} - 4 \, {\left (2 \, x^{3} + 6 \, x^{2} - 5 \, x - 15\right )} \log \left (2 \, x\right )^{2} + 2 \, {\left (x^{3} + 3 \, x^{2} - 5 \, x - 15\right )} \log \left (2 \, x\right )\right )} e^{\left (\frac {\log \left (2 \, x\right )^{2}}{x^{8} - 10 \, x^{6} + 25 \, x^{4}}\right )}}{x^{10} - 15 \, x^{8} + 75 \, x^{6} - 125 \, x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^3-24*x^2+20*x+60)*log(2*x)^2+(2*x^3+6*x^2-10*x-30)*log(2*x)+2*x^11+3*x^10-30*x^9-45*x^8+150*x
^7+225*x^6-250*x^5-375*x^4)*exp(log(2*x)^2/(x^8-10*x^6+25*x^4))/(x^10-15*x^8+75*x^6-125*x^4),x, algorithm="gia
c")

[Out]

integrate((2*x^11 + 3*x^10 - 30*x^9 - 45*x^8 + 150*x^7 + 225*x^6 - 250*x^5 - 375*x^4 - 4*(2*x^3 + 6*x^2 - 5*x
- 15)*log(2*x)^2 + 2*(x^3 + 3*x^2 - 5*x - 15)*log(2*x))*e^(log(2*x)^2/(x^8 - 10*x^6 + 25*x^4))/(x^10 - 15*x^8
+ 75*x^6 - 125*x^4), x)

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maple [A]  time = 0.03, size = 24, normalized size = 0.92

method	result	size
risch	\(\left (3+x \right ) x \,{\mathrm e}^{\frac {\ln \left (2 x \right )^{2}}{x^{4} \left (x^{2}-5\right )^{2}}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*x^3-24*x^2+20*x+60)*ln(2*x)^2+(2*x^3+6*x^2-10*x-30)*ln(2*x)+2*x^11+3*x^10-30*x^9-45*x^8+150*x^7+225*x
^6-250*x^5-375*x^4)*exp(ln(2*x)^2/(x^8-10*x^6+25*x^4))/(x^10-15*x^8+75*x^6-125*x^4),x,method=_RETURNVERBOSE)

[Out]

(3+x)*x*exp(ln(2*x)^2/x^4/(x^2-5)^2)

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maxima [B]  time = 1.01, size = 157, normalized size = 6.04 \begin {gather*} {\left (x^{2} + 3 \, x\right )} e^{\left (\frac {\log \relax (2)^{2}}{25 \, {\left (x^{4} - 10 \, x^{2} + 25\right )}} - \frac {2 \, \log \relax (2)^{2}}{125 \, {\left (x^{2} - 5\right )}} + \frac {2 \, \log \relax (2) \log \relax (x)}{25 \, {\left (x^{4} - 10 \, x^{2} + 25\right )}} - \frac {4 \, \log \relax (2) \log \relax (x)}{125 \, {\left (x^{2} - 5\right )}} + \frac {\log \relax (x)^{2}}{25 \, {\left (x^{4} - 10 \, x^{2} + 25\right )}} - \frac {2 \, \log \relax (x)^{2}}{125 \, {\left (x^{2} - 5\right )}} + \frac {2 \, \log \relax (2)^{2}}{125 \, x^{2}} + \frac {4 \, \log \relax (2) \log \relax (x)}{125 \, x^{2}} + \frac {2 \, \log \relax (x)^{2}}{125 \, x^{2}} + \frac {\log \relax (2)^{2}}{25 \, x^{4}} + \frac {2 \, \log \relax (2) \log \relax (x)}{25 \, x^{4}} + \frac {\log \relax (x)^{2}}{25 \, x^{4}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^3-24*x^2+20*x+60)*log(2*x)^2+(2*x^3+6*x^2-10*x-30)*log(2*x)+2*x^11+3*x^10-30*x^9-45*x^8+150*x
^7+225*x^6-250*x^5-375*x^4)*exp(log(2*x)^2/(x^8-10*x^6+25*x^4))/(x^10-15*x^8+75*x^6-125*x^4),x, algorithm="max
ima")

[Out]

(x^2 + 3*x)*e^(1/25*log(2)^2/(x^4 - 10*x^2 + 25) - 2/125*log(2)^2/(x^2 - 5) + 2/25*log(2)*log(x)/(x^4 - 10*x^2
+ 25) - 4/125*log(2)*log(x)/(x^2 - 5) + 1/25*log(x)^2/(x^4 - 10*x^2 + 25) - 2/125*log(x)^2/(x^2 - 5) + 2/125*
log(2)^2/x^2 + 4/125*log(2)*log(x)/x^2 + 2/125*log(x)^2/x^2 + 1/25*log(2)^2/x^4 + 2/25*log(2)*log(x)/x^4 + 1/2
5*log(x)^2/x^4)

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mupad [B]  time = 2.31, size = 71, normalized size = 2.73 \begin {gather*} x\,x^{\frac {2\,\ln \relax (2)}{x^8-10\,x^6+25\,x^4}}\,{\mathrm {e}}^{\frac {{\ln \relax (x)}^2}{x^8-10\,x^6+25\,x^4}}\,{\mathrm {e}}^{\frac {{\ln \relax (2)}^2}{x^8-10\,x^6+25\,x^4}}\,\left (x+3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(2*x)^2/(25*x^4 - 10*x^6 + x^8))*(log(2*x)*(10*x - 6*x^2 - 2*x^3 + 30) - log(2*x)^2*(20*x - 24*x^2
- 8*x^3 + 60) + 375*x^4 + 250*x^5 - 225*x^6 - 150*x^7 + 45*x^8 + 30*x^9 - 3*x^10 - 2*x^11))/(125*x^4 - 75*x^6
+ 15*x^8 - x^10),x)

[Out]

x*x^((2*log(2))/(25*x^4 - 10*x^6 + x^8))*exp(log(x)^2/(25*x^4 - 10*x^6 + x^8))*exp(log(2)^2/(25*x^4 - 10*x^6 +
x^8))*(x + 3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x**3-24*x**2+20*x+60)*ln(2*x)**2+(2*x**3+6*x**2-10*x-30)*ln(2*x)+2*x**11+3*x**10-30*x**9-45*x**
8+150*x**7+225*x**6-250*x**5-375*x**4)*exp(ln(2*x)**2/(x**8-10*x**6+25*x**4))/(x**10-15*x**8+75*x**6-125*x**4)
,x)

[Out]

Timed out

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