3.14.27 \(\int (e^{2 e^{2 x} \log ^2(\frac {2}{x})} (-10 e^{10}+40 e^{10+2 x} \log (\frac {2}{x})-40 e^{10+2 x} x \log ^2(\frac {2}{x}))+e^{4 e^{2 x} \log ^2(\frac {2}{x})} (2 e^{20} x-8 e^{20+2 x} x \log (\frac {2}{x})+8 e^{20+2 x} x^2 \log ^2(\frac {2}{x}))) \, dx\)

Optimal. Leaf size=26 \[ \left (5-e^{10+2 e^{2 x} \log ^2\left (\frac {2}{x}\right )} x\right )^2 \]

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Rubi [B]  time = 0.75, antiderivative size = 173, normalized size of antiderivative = 6.65, number of steps used = 3, number of rules used = 1, integrand size = 119, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.008, Rules used = {2288} \begin {gather*} \frac {e^{4 e^{2 x} \log ^2\left (\frac {2}{x}\right )} \left (e^{2 x+20} x \log \left (\frac {2}{x}\right )-e^{2 x+20} x^2 \log ^2\left (\frac {2}{x}\right )\right )}{\frac {e^{2 x} \log \left (\frac {2}{x}\right )}{x}-e^{2 x} \log ^2\left (\frac {2}{x}\right )}-\frac {10 e^{2 e^{2 x} \log ^2\left (\frac {2}{x}\right )} \left (e^{2 x+10} \log \left (\frac {2}{x}\right )-e^{2 x+10} x \log ^2\left (\frac {2}{x}\right )\right )}{\frac {e^{2 x} \log \left (\frac {2}{x}\right )}{x}-e^{2 x} \log ^2\left (\frac {2}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(2*E^(2*x)*Log[2/x]^2)*(-10*E^10 + 40*E^(10 + 2*x)*Log[2/x] - 40*E^(10 + 2*x)*x*Log[2/x]^2) + E^(4*E^(2*
x)*Log[2/x]^2)*(2*E^20*x - 8*E^(20 + 2*x)*x*Log[2/x] + 8*E^(20 + 2*x)*x^2*Log[2/x]^2),x]

[Out]

(-10*E^(2*E^(2*x)*Log[2/x]^2)*(E^(10 + 2*x)*Log[2/x] - E^(10 + 2*x)*x*Log[2/x]^2))/((E^(2*x)*Log[2/x])/x - E^(
2*x)*Log[2/x]^2) + (E^(4*E^(2*x)*Log[2/x]^2)*(E^(20 + 2*x)*x*Log[2/x] - E^(20 + 2*x)*x^2*Log[2/x]^2))/((E^(2*x
)*Log[2/x])/x - E^(2*x)*Log[2/x]^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} &=\int e^{2 e^{2 x} \log ^2\left (\frac {2}{x}\right )} \left (-10 e^{10}+40 e^{10+2 x} \log \left (\frac {2}{x}\right )-40 e^{10+2 x} x \log ^2\left (\frac {2}{x}\right )\right ) \, dx+\int e^{4 e^{2 x} \log ^2\left (\frac {2}{x}\right )} \left (2 e^{20} x-8 e^{20+2 x} x \log \left (\frac {2}{x}\right )+8 e^{20+2 x} x^2 \log ^2\left (\frac {2}{x}\right )\right ) \, dx\\ &=-\frac {10 e^{2 e^{2 x} \log ^2\left (\frac {2}{x}\right )} \left (e^{10+2 x} \log \left (\frac {2}{x}\right )-e^{10+2 x} x \log ^2\left (\frac {2}{x}\right )\right )}{\frac {e^{2 x} \log \left (\frac {2}{x}\right )}{x}-e^{2 x} \log ^2\left (\frac {2}{x}\right )}+\frac {e^{4 e^{2 x} \log ^2\left (\frac {2}{x}\right )} \left (e^{20+2 x} x \log \left (\frac {2}{x}\right )-e^{20+2 x} x^2 \log ^2\left (\frac {2}{x}\right )\right )}{\frac {e^{2 x} \log \left (\frac {2}{x}\right )}{x}-e^{2 x} \log ^2\left (\frac {2}{x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 2.18, size = 46, normalized size = 1.77 \begin {gather*} e^{2 \left (5+e^{2 x} \log ^2\left (\frac {2}{x}\right )\right )} x \left (-10+e^{2 \left (5+e^{2 x} \log ^2\left (\frac {2}{x}\right )\right )} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(2*E^(2*x)*Log[2/x]^2)*(-10*E^10 + 40*E^(10 + 2*x)*Log[2/x] - 40*E^(10 + 2*x)*x*Log[2/x]^2) + E^(4
*E^(2*x)*Log[2/x]^2)*(2*E^20*x - 8*E^(20 + 2*x)*x*Log[2/x] + 8*E^(20 + 2*x)*x^2*Log[2/x]^2),x]

[Out]

E^(2*(5 + E^(2*x)*Log[2/x]^2))*x*(-10 + E^(2*(5 + E^(2*x)*Log[2/x]^2))*x)

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fricas [A]  time = 0.62, size = 42, normalized size = 1.62 \begin {gather*} x^{2} e^{\left (4 \, e^{\left (2 \, x\right )} \log \left (\frac {2}{x}\right )^{2} + 20\right )} - 10 \, x e^{\left (2 \, e^{\left (2 \, x\right )} \log \left (\frac {2}{x}\right )^{2} + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^2*exp(5)^4*exp(x)^2*log(2/x)^2-8*x*exp(5)^4*exp(x)^2*log(2/x)+2*x*exp(5)^4)*exp(2*exp(x)^2*log(
2/x)^2)^2+(-40*x*exp(5)^2*exp(x)^2*log(2/x)^2+40*exp(5)^2*exp(x)^2*log(2/x)-10*exp(5)^2)*exp(2*exp(x)^2*log(2/
x)^2),x, algorithm="fricas")

[Out]

x^2*e^(4*e^(2*x)*log(2/x)^2 + 20) - 10*x*e^(2*e^(2*x)*log(2/x)^2 + 10)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int 2 \, {\left (4 \, x^{2} e^{\left (2 \, x + 20\right )} \log \left (\frac {2}{x}\right )^{2} - 4 \, x e^{\left (2 \, x + 20\right )} \log \left (\frac {2}{x}\right ) + x e^{20}\right )} e^{\left (4 \, e^{\left (2 \, x\right )} \log \left (\frac {2}{x}\right )^{2}\right )} - 10 \, {\left (4 \, x e^{\left (2 \, x + 10\right )} \log \left (\frac {2}{x}\right )^{2} - 4 \, e^{\left (2 \, x + 10\right )} \log \left (\frac {2}{x}\right ) + e^{10}\right )} e^{\left (2 \, e^{\left (2 \, x\right )} \log \left (\frac {2}{x}\right )^{2}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^2*exp(5)^4*exp(x)^2*log(2/x)^2-8*x*exp(5)^4*exp(x)^2*log(2/x)+2*x*exp(5)^4)*exp(2*exp(x)^2*log(
2/x)^2)^2+(-40*x*exp(5)^2*exp(x)^2*log(2/x)^2+40*exp(5)^2*exp(x)^2*log(2/x)-10*exp(5)^2)*exp(2*exp(x)^2*log(2/
x)^2),x, algorithm="giac")

[Out]

integrate(2*(4*x^2*e^(2*x + 20)*log(2/x)^2 - 4*x*e^(2*x + 20)*log(2/x) + x*e^20)*e^(4*e^(2*x)*log(2/x)^2) - 10
*(4*x*e^(2*x + 10)*log(2/x)^2 - 4*e^(2*x + 10)*log(2/x) + e^10)*e^(2*e^(2*x)*log(2/x)^2), x)

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maple [F]  time = 180.00, size = 0, normalized size = 0.00 \[\int \left (8 x^{2} {\mathrm e}^{20} {\mathrm e}^{2 x} \ln \left (\frac {2}{x}\right )^{2}-8 x \,{\mathrm e}^{20} {\mathrm e}^{2 x} \ln \left (\frac {2}{x}\right )+2 x \,{\mathrm e}^{20}\right ) {\mathrm e}^{4 \,{\mathrm e}^{2 x} \ln \left (\frac {2}{x}\right )^{2}}+\left (-40 x \,{\mathrm e}^{10} {\mathrm e}^{2 x} \ln \left (\frac {2}{x}\right )^{2}+40 \,{\mathrm e}^{10} {\mathrm e}^{2 x} \ln \left (\frac {2}{x}\right )-10 \,{\mathrm e}^{10}\right ) {\mathrm e}^{2 \,{\mathrm e}^{2 x} \ln \left (\frac {2}{x}\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^2*exp(5)^4*exp(x)^2*ln(2/x)^2-8*x*exp(5)^4*exp(x)^2*ln(2/x)+2*x*exp(5)^4)*exp(2*exp(x)^2*ln(2/x)^2)^2
+(-40*x*exp(5)^2*exp(x)^2*ln(2/x)^2+40*exp(5)^2*exp(x)^2*ln(2/x)-10*exp(5)^2)*exp(2*exp(x)^2*ln(2/x)^2),x)

[Out]

int((8*x^2*exp(5)^4*exp(x)^2*ln(2/x)^2-8*x*exp(5)^4*exp(x)^2*ln(2/x)+2*x*exp(5)^4)*exp(2*exp(x)^2*ln(2/x)^2)^2
+(-40*x*exp(5)^2*exp(x)^2*ln(2/x)^2+40*exp(5)^2*exp(x)^2*ln(2/x)-10*exp(5)^2)*exp(2*exp(x)^2*ln(2/x)^2),x)

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maxima [B]  time = 0.72, size = 74, normalized size = 2.85 \begin {gather*} x^{2} e^{\left (4 \, e^{\left (2 \, x\right )} \log \relax (2)^{2} - 8 \, e^{\left (2 \, x\right )} \log \relax (2) \log \relax (x) + 4 \, e^{\left (2 \, x\right )} \log \relax (x)^{2} + 20\right )} - 10 \, x e^{\left (2 \, e^{\left (2 \, x\right )} \log \relax (2)^{2} - 4 \, e^{\left (2 \, x\right )} \log \relax (2) \log \relax (x) + 2 \, e^{\left (2 \, x\right )} \log \relax (x)^{2} + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^2*exp(5)^4*exp(x)^2*log(2/x)^2-8*x*exp(5)^4*exp(x)^2*log(2/x)+2*x*exp(5)^4)*exp(2*exp(x)^2*log(
2/x)^2)^2+(-40*x*exp(5)^2*exp(x)^2*log(2/x)^2+40*exp(5)^2*exp(x)^2*log(2/x)-10*exp(5)^2)*exp(2*exp(x)^2*log(2/
x)^2),x, algorithm="maxima")

[Out]

x^2*e^(4*e^(2*x)*log(2)^2 - 8*e^(2*x)*log(2)*log(x) + 4*e^(2*x)*log(x)^2 + 20) - 10*x*e^(2*e^(2*x)*log(2)^2 -
4*e^(2*x)*log(2)*log(x) + 2*e^(2*x)*log(x)^2 + 10)

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mupad [B]  time = 1.16, size = 87, normalized size = 3.35 \begin {gather*} -x\,{\mathrm {e}}^{10}\,{\mathrm {e}}^{2\,{\ln \left (\frac {1}{x}\right )}^2\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}\,{\ln \relax (2)}^2}\,\left (\frac {10}{{\left (\frac {1}{x}\right )}^{4\,{\mathrm {e}}^{2\,x}\,\ln \relax (2)}}-x\,{\mathrm {e}}^{10}\,{\mathrm {e}}^{2\,{\ln \left (\frac {1}{x}\right )}^2\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}\,{\ln \relax (2)}^2}\right )\,{\left (\frac {1}{x}\right )}^{8\,{\mathrm {e}}^{2\,x}\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(4*exp(2*x)*log(2/x)^2)*(2*x*exp(20) - 8*x*exp(2*x)*exp(20)*log(2/x) + 8*x^2*exp(2*x)*exp(20)*log(2/x)^
2) - exp(2*exp(2*x)*log(2/x)^2)*(10*exp(10) - 40*exp(2*x)*exp(10)*log(2/x) + 40*x*exp(2*x)*exp(10)*log(2/x)^2)
,x)

[Out]

-x*exp(10)*exp(2*log(1/x)^2*exp(2*x))*exp(2*exp(2*x)*log(2)^2)*(10/(1/x)^(4*exp(2*x)*log(2)) - x*exp(10)*exp(2
*log(1/x)^2*exp(2*x))*exp(2*exp(2*x)*log(2)^2))*(1/x)^(8*exp(2*x)*log(2))

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sympy [A]  time = 20.19, size = 42, normalized size = 1.62 \begin {gather*} x^{2} e^{20} e^{4 e^{2 x} \log {\left (\frac {2}{x} \right )}^{2}} - 10 x e^{10} e^{2 e^{2 x} \log {\left (\frac {2}{x} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x**2*exp(5)**4*exp(x)**2*ln(2/x)**2-8*x*exp(5)**4*exp(x)**2*ln(2/x)+2*x*exp(5)**4)*exp(2*exp(x)**
2*ln(2/x)**2)**2+(-40*x*exp(5)**2*exp(x)**2*ln(2/x)**2+40*exp(5)**2*exp(x)**2*ln(2/x)-10*exp(5)**2)*exp(2*exp(
x)**2*ln(2/x)**2),x)

[Out]

x**2*exp(20)*exp(4*exp(2*x)*log(2/x)**2) - 10*x*exp(10)*exp(2*exp(2*x)*log(2/x)**2)

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