3.79.83 \(\int \frac {-4 x^3-8 e^{4 e^{2 x}+2 x} x^3-2 x^4+e^{2 e^{2 x}} (-2 x^3+e^{2 x} (-16 x^3-8 x^4))+(-32-x^3) \log ^2(2)}{x^3 \log ^2(2)} \, dx\)

Optimal. Leaf size=29 \[ \frac {16}{x^2}-x-\frac {\left (2+e^{2 e^{2 x}}+x\right )^2}{\log ^2(2)} \]

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Rubi [F]  time = 0.32, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-4 x^3-8 e^{4 e^{2 x}+2 x} x^3-2 x^4+e^{2 e^{2 x}} \left (-2 x^3+e^{2 x} \left (-16 x^3-8 x^4\right )\right )+\left (-32-x^3\right ) \log ^2(2)}{x^3 \log ^2(2)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4*x^3 - 8*E^(4*E^(2*x) + 2*x)*x^3 - 2*x^4 + E^(2*E^(2*x))*(-2*x^3 + E^(2*x)*(-16*x^3 - 8*x^4)) + (-32 -
x^3)*Log[2]^2)/(x^3*Log[2]^2),x]

[Out]

16/x^2 - x*(1 + 4/Log[2]^2) - x^2/Log[2]^2 - ExpIntegralEi[2*E^(2*x)]/Log[2]^2 - (16*Defer[Int][E^(2*(E^(2*x)
+ x)), x])/Log[2]^2 - (8*Defer[Int][E^(2*(2*E^(2*x) + x)), x])/Log[2]^2 - (8*Defer[Int][E^(2*(E^(2*x) + x))*x,
x])/Log[2]^2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-4 x^3-8 e^{4 e^{2 x}+2 x} x^3-2 x^4+e^{2 e^{2 x}} \left (-2 x^3+e^{2 x} \left (-16 x^3-8 x^4\right )\right )+\left (-32-x^3\right ) \log ^2(2)}{x^3} \, dx}{\log ^2(2)}\\ &=\frac {\int \left (8 e^{2 \left (e^{2 x}+x\right )} \left (-2-e^{2 e^{2 x}}-x\right )+\frac {-2 e^{2 e^{2 x}} x^3-2 x^4-32 \log ^2(2)-4 x^3 \left (1+\frac {\log ^2(2)}{4}\right )}{x^3}\right ) \, dx}{\log ^2(2)}\\ &=\frac {\int \frac {-2 e^{2 e^{2 x}} x^3-2 x^4-32 \log ^2(2)-4 x^3 \left (1+\frac {\log ^2(2)}{4}\right )}{x^3} \, dx}{\log ^2(2)}+\frac {8 \int e^{2 \left (e^{2 x}+x\right )} \left (-2-e^{2 e^{2 x}}-x\right ) \, dx}{\log ^2(2)}\\ &=\frac {\int \left (-2 e^{2 e^{2 x}}+\frac {-2 x^4-32 \log ^2(2)-x^3 \left (4+\log ^2(2)\right )}{x^3}\right ) \, dx}{\log ^2(2)}+\frac {8 \int \left (-2 e^{2 \left (e^{2 x}+x\right )}-e^{2 e^{2 x}+2 \left (e^{2 x}+x\right )}-e^{2 \left (e^{2 x}+x\right )} x\right ) \, dx}{\log ^2(2)}\\ &=\frac {\int \frac {-2 x^4-32 \log ^2(2)-x^3 \left (4+\log ^2(2)\right )}{x^3} \, dx}{\log ^2(2)}-\frac {2 \int e^{2 e^{2 x}} \, dx}{\log ^2(2)}-\frac {8 \int e^{2 e^{2 x}+2 \left (e^{2 x}+x\right )} \, dx}{\log ^2(2)}-\frac {8 \int e^{2 \left (e^{2 x}+x\right )} x \, dx}{\log ^2(2)}-\frac {16 \int e^{2 \left (e^{2 x}+x\right )} \, dx}{\log ^2(2)}\\ &=\frac {\int \left (-2 x-\frac {32 \log ^2(2)}{x^3}-4 \left (1+\frac {\log ^2(2)}{4}\right )\right ) \, dx}{\log ^2(2)}-\frac {\operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,e^{2 x}\right )}{\log ^2(2)}-\frac {8 \int e^{2 \left (2 e^{2 x}+x\right )} \, dx}{\log ^2(2)}-\frac {8 \int e^{2 \left (e^{2 x}+x\right )} x \, dx}{\log ^2(2)}-\frac {16 \int e^{2 \left (e^{2 x}+x\right )} \, dx}{\log ^2(2)}\\ &=\frac {16}{x^2}-x \left (1+\frac {4}{\log ^2(2)}\right )-\frac {x^2}{\log ^2(2)}-\frac {\text {Ei}\left (2 e^{2 x}\right )}{\log ^2(2)}-\frac {8 \int e^{2 \left (2 e^{2 x}+x\right )} \, dx}{\log ^2(2)}-\frac {8 \int e^{2 \left (e^{2 x}+x\right )} x \, dx}{\log ^2(2)}-\frac {16 \int e^{2 \left (e^{2 x}+x\right )} \, dx}{\log ^2(2)}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.13, size = 59, normalized size = 2.03 \begin {gather*} -\frac {e^{4 e^{2 x}} x^2+x^4+2 e^{2 e^{2 x}} x^2 (2+x)-16 \log ^2(2)+x^3 \left (4+\log ^2(2)\right )}{x^2 \log ^2(2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*x^3 - 8*E^(4*E^(2*x) + 2*x)*x^3 - 2*x^4 + E^(2*E^(2*x))*(-2*x^3 + E^(2*x)*(-16*x^3 - 8*x^4)) + (
-32 - x^3)*Log[2]^2)/(x^3*Log[2]^2),x]

[Out]

-((E^(4*E^(2*x))*x^2 + x^4 + 2*E^(2*E^(2*x))*x^2*(2 + x) - 16*Log[2]^2 + x^3*(4 + Log[2]^2))/(x^2*Log[2]^2))

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fricas [B]  time = 0.97, size = 57, normalized size = 1.97 \begin {gather*} -\frac {x^{4} + 4 \, x^{3} + x^{2} e^{\left (4 \, e^{\left (2 \, x\right )}\right )} + {\left (x^{3} - 16\right )} \log \relax (2)^{2} + 2 \, {\left (x^{3} + 2 \, x^{2}\right )} e^{\left (2 \, e^{\left (2 \, x\right )}\right )}}{x^{2} \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x^3*exp(x)^2*exp(exp(x)^2)^4+((-8*x^4-16*x^3)*exp(x)^2-2*x^3)*exp(exp(x)^2)^2+(-x^3-32)*log(2)^2
-2*x^4-4*x^3)/x^3/log(2)^2,x, algorithm="fricas")

[Out]

-(x^4 + 4*x^3 + x^2*e^(4*e^(2*x)) + (x^3 - 16)*log(2)^2 + 2*(x^3 + 2*x^2)*e^(2*e^(2*x)))/(x^2*log(2)^2)

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giac [B]  time = 0.21, size = 100, normalized size = 3.45 \begin {gather*} -\frac {{\left (x^{3} e^{\left (2 \, x\right )} \log \relax (2)^{2} + x^{4} e^{\left (2 \, x\right )} + 4 \, x^{3} e^{\left (2 \, x\right )} + 2 \, x^{3} e^{\left (2 \, x + 2 \, e^{\left (2 \, x\right )}\right )} + x^{2} e^{\left (2 \, x + 4 \, e^{\left (2 \, x\right )}\right )} + 4 \, x^{2} e^{\left (2 \, x + 2 \, e^{\left (2 \, x\right )}\right )} - 16 \, e^{\left (2 \, x\right )} \log \relax (2)^{2}\right )} e^{\left (-2 \, x\right )}}{x^{2} \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x^3*exp(x)^2*exp(exp(x)^2)^4+((-8*x^4-16*x^3)*exp(x)^2-2*x^3)*exp(exp(x)^2)^2+(-x^3-32)*log(2)^2
-2*x^4-4*x^3)/x^3/log(2)^2,x, algorithm="giac")

[Out]

-(x^3*e^(2*x)*log(2)^2 + x^4*e^(2*x) + 4*x^3*e^(2*x) + 2*x^3*e^(2*x + 2*e^(2*x)) + x^2*e^(2*x + 4*e^(2*x)) + 4
*x^2*e^(2*x + 2*e^(2*x)) - 16*e^(2*x)*log(2)^2)*e^(-2*x)/(x^2*log(2)^2)

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maple [B]  time = 0.05, size = 56, normalized size = 1.93

method	result	size
risch	\(-x -\frac {x^{2}}{\ln \relax (2)^{2}}-\frac {4 x}{\ln \relax (2)^{2}}+\frac {16}{x^{2}}-\frac {{\mathrm e}^{4 \,{\mathrm e}^{2 x}}}{\ln \relax (2)^{2}}+\frac {\left (-2 x -4\right ) {\mathrm e}^{2 \,{\mathrm e}^{2 x}}}{\ln \relax (2)^{2}}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*x^3*exp(x)^2*exp(exp(x)^2)^4+((-8*x^4-16*x^3)*exp(x)^2-2*x^3)*exp(exp(x)^2)^2+(-x^3-32)*ln(2)^2-2*x^4-
4*x^3)/x^3/ln(2)^2,x,method=_RETURNVERBOSE)

[Out]

-x-x^2/ln(2)^2-4*x/ln(2)^2+16/x^2-1/ln(2)^2*exp(4*exp(2*x))+1/ln(2)^2*(-2*x-4)*exp(2*exp(2*x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {x \log \relax (2)^{2} + x^{2} + 2 \, x e^{\left (2 \, e^{\left (2 \, x\right )}\right )} + 4 \, x - \frac {16 \, \log \relax (2)^{2}}{x^{2}} + e^{\left (4 \, e^{\left (2 \, x\right )}\right )} + 4 \, e^{\left (2 \, e^{\left (2 \, x\right )}\right )}}{\log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x^3*exp(x)^2*exp(exp(x)^2)^4+((-8*x^4-16*x^3)*exp(x)^2-2*x^3)*exp(exp(x)^2)^2+(-x^3-32)*log(2)^2
-2*x^4-4*x^3)/x^3/log(2)^2,x, algorithm="maxima")

[Out]

-(x*log(2)^2 + x^2 + 2*x*e^(2*e^(2*x)) + 4*x - 16*log(2)^2/x^2 + Ei(2*e^(2*x)) + e^(4*e^(2*x)) + 4*e^(2*e^(2*x
)) - 2*integrate(e^(2*e^(2*x)), x))/log(2)^2

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mupad [B]  time = 0.28, size = 64, normalized size = 2.21 \begin {gather*} \frac {16}{x^2}-\frac {{\mathrm {e}}^{4\,{\mathrm {e}}^{2\,x}}}{{\ln \relax (2)}^2}-{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}}\,\left (\frac {2\,x}{{\ln \relax (2)}^2}+\frac {4}{{\ln \relax (2)}^2}\right )-\frac {x^2}{{\ln \relax (2)}^2}-\frac {x\,\left ({\ln \relax (2)}^2+4\right )}{{\ln \relax (2)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*exp(2*x))*(exp(2*x)*(16*x^3 + 8*x^4) + 2*x^3) + log(2)^2*(x^3 + 32) + 4*x^3 + 2*x^4 + 8*x^3*exp(4*
exp(2*x))*exp(2*x))/(x^3*log(2)^2),x)

[Out]

16/x^2 - exp(4*exp(2*x))/log(2)^2 - exp(2*exp(2*x))*((2*x)/log(2)^2 + 4/log(2)^2) - x^2/log(2)^2 - (x*(log(2)^
2 + 4))/log(2)^2

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sympy [B]  time = 0.32, size = 70, normalized size = 2.41 \begin {gather*} \frac {\left (- 2 x \log {\relax (2 )}^{2} - 4 \log {\relax (2 )}^{2}\right ) e^{2 e^{2 x}} - e^{4 e^{2 x}} \log {\relax (2 )}^{2}}{\log {\relax (2 )}^{4}} + \frac {- x^{2} - x \left (\log {\relax (2 )}^{2} + 4\right ) + \frac {16 \log {\relax (2 )}^{2}}{x^{2}}}{\log {\relax (2 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x**3*exp(x)**2*exp(exp(x)**2)**4+((-8*x**4-16*x**3)*exp(x)**2-2*x**3)*exp(exp(x)**2)**2+(-x**3-3
2)*ln(2)**2-2*x**4-4*x**3)/x**3/ln(2)**2,x)

[Out]

((-2*x*log(2)**2 - 4*log(2)**2)*exp(2*exp(2*x)) - exp(4*exp(2*x))*log(2)**2)/log(2)**4 + (-x**2 - x*(log(2)**2
+ 4) + 16*log(2)**2/x**2)/log(2)**2

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