3.50.56 \(\int \frac {x^2+e^{\frac {2}{x^2}} (4+x^2)+(x^2+e^{\frac {2}{x^2}} x^2) \log (\frac {1+e^{\frac {2}{x^2}}}{5 x})+\log ^2(\frac {1+e^{\frac {2}{x^2}}}{5 x}) (x^2+e^{\frac {2}{x^2}} x^2+e^x (x^2+x^3+e^{\frac {2}{x^2}} (x^2+x^3))+(x^2+e^{\frac {2}{x^2}} x^2) \log (x))}{(x^2+e^{\frac {2}{x^2}} x^2) \log ^2(\frac {1+e^{\frac {2}{x^2}}}{5 x})} \, dx\)

Optimal. Leaf size=27 \[ x \left (e^x+\frac {1}{\log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )}+\log (x)\right ) \]

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Rubi [F]  time = 1.97, 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 {x^2+e^{\frac {2}{x^2}} \left (4+x^2\right )+\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )+\log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right ) \left (x^2+e^{\frac {2}{x^2}} x^2+e^x \left (x^2+x^3+e^{\frac {2}{x^2}} \left (x^2+x^3\right )\right )+\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log (x)\right )}{\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x^2 + E^(2/x^2)*(4 + x^2) + (x^2 + E^(2/x^2)*x^2)*Log[(1 + E^(2/x^2))/(5*x)] + Log[(1 + E^(2/x^2))/(5*x)]
^2*(x^2 + E^(2/x^2)*x^2 + E^x*(x^2 + x^3 + E^(2/x^2)*(x^2 + x^3)) + (x^2 + E^(2/x^2)*x^2)*Log[x]))/((x^2 + E^(
2/x^2)*x^2)*Log[(1 + E^(2/x^2))/(5*x)]^2),x]

[Out]

E^x*x + x*Log[x] + Defer[Int][Log[(1 + E^(2/x^2))/(5*x)]^(-2), x] + Defer[Int][Log[(1 + E^(2/x^2))/(5*x)]^(-1)
, x] - 4*Defer[Subst][Defer[Int][Log[((1 + E^(2*x^2))*x)/5]^(-2), x], x, x^(-1)] + 4*Defer[Subst][Defer[Int][1
/((1 + E^(2*x^2))*Log[((1 + E^(2*x^2))*x)/5]^2), x], x, x^(-1)]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1+e^x+e^x x+\frac {x^2+e^{\frac {2}{x^2}} \left (4+x^2\right )}{\left (1+e^{\frac {2}{x^2}}\right ) x^2 \log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )}+\frac {1}{\log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )}+\log (x)\right ) \, dx\\ &=x+\int e^x \, dx+\int e^x x \, dx+\int \frac {x^2+e^{\frac {2}{x^2}} \left (4+x^2\right )}{\left (1+e^{\frac {2}{x^2}}\right ) x^2 \log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx+\int \frac {1}{\log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx+\int \log (x) \, dx\\ &=e^x+e^x x+x \log (x)-\int e^x \, dx+\int \left (-\frac {4}{\left (1+e^{\frac {2}{x^2}}\right ) x^2 \log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )}+\frac {4+x^2}{x^2 \log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )}\right ) \, dx+\int \frac {1}{\log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx\\ &=e^x x+x \log (x)-4 \int \frac {1}{\left (1+e^{\frac {2}{x^2}}\right ) x^2 \log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx+\int \frac {4+x^2}{x^2 \log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx+\int \frac {1}{\log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx\\ &=e^x x+x \log (x)+4 \operatorname {Subst}\left (\int \frac {1}{\left (1+e^{2 x^2}\right ) \log ^2\left (\frac {1}{5} \left (1+e^{2 x^2}\right ) x\right )} \, dx,x,\frac {1}{x}\right )+\int \left (\frac {1}{\log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )}+\frac {4}{x^2 \log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )}\right ) \, dx+\int \frac {1}{\log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx\\ &=e^x x+x \log (x)+4 \int \frac {1}{x^2 \log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx+4 \operatorname {Subst}\left (\int \frac {1}{\left (1+e^{2 x^2}\right ) \log ^2\left (\frac {1}{5} \left (1+e^{2 x^2}\right ) x\right )} \, dx,x,\frac {1}{x}\right )+\int \frac {1}{\log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx+\int \frac {1}{\log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx\\ &=e^x x+x \log (x)-4 \operatorname {Subst}\left (\int \frac {1}{\log ^2\left (\frac {1}{5} \left (1+e^{2 x^2}\right ) x\right )} \, dx,x,\frac {1}{x}\right )+4 \operatorname {Subst}\left (\int \frac {1}{\left (1+e^{2 x^2}\right ) \log ^2\left (\frac {1}{5} \left (1+e^{2 x^2}\right ) x\right )} \, dx,x,\frac {1}{x}\right )+\int \frac {1}{\log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx+\int \frac {1}{\log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.39, size = 27, normalized size = 1.00 \begin {gather*} x \left (e^x+\frac {1}{\log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )}+\log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + E^(2/x^2)*(4 + x^2) + (x^2 + E^(2/x^2)*x^2)*Log[(1 + E^(2/x^2))/(5*x)] + Log[(1 + E^(2/x^2))/
(5*x)]^2*(x^2 + E^(2/x^2)*x^2 + E^x*(x^2 + x^3 + E^(2/x^2)*(x^2 + x^3)) + (x^2 + E^(2/x^2)*x^2)*Log[x]))/((x^2
+ E^(2/x^2)*x^2)*Log[(1 + E^(2/x^2))/(5*x)]^2),x]

[Out]

x*(E^x + Log[(1 + E^(2/x^2))/(5*x)]^(-1) + Log[x])

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fricas [A]  time = 1.12, size = 43, normalized size = 1.59 \begin {gather*} \frac {{\left (x e^{x} + x \log \relax (x)\right )} \log \left (\frac {e^{\left (\frac {2}{x^{2}}\right )} + 1}{5 \, x}\right ) + x}{\log \left (\frac {e^{\left (\frac {2}{x^{2}}\right )} + 1}{5 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2*exp(2/x^2)+x^2)*log(x)+((x^3+x^2)*exp(2/x^2)+x^3+x^2)*exp(x)+x^2*exp(2/x^2)+x^2)*log(1/5*(exp
(2/x^2)+1)/x)^2+(x^2*exp(2/x^2)+x^2)*log(1/5*(exp(2/x^2)+1)/x)+(x^2+4)*exp(2/x^2)+x^2)/(x^2*exp(2/x^2)+x^2)/lo
g(1/5*(exp(2/x^2)+1)/x)^2,x, algorithm="fricas")

[Out]

((x*e^x + x*log(x))*log(1/5*(e^(2/x^2) + 1)/x) + x)/log(1/5*(e^(2/x^2) + 1)/x)

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giac [B]  time = 0.66, size = 75, normalized size = 2.78 \begin {gather*} \frac {x e^{x} \log \relax (5) + x e^{x} \log \relax (x) + x \log \relax (5) \log \relax (x) + x \log \relax (x)^{2} - x e^{x} \log \left (e^{\left (\frac {2}{x^{2}}\right )} + 1\right ) - x \log \relax (x) \log \left (e^{\left (\frac {2}{x^{2}}\right )} + 1\right ) - x}{\log \relax (5) + \log \relax (x) - \log \left (e^{\left (\frac {2}{x^{2}}\right )} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2*exp(2/x^2)+x^2)*log(x)+((x^3+x^2)*exp(2/x^2)+x^3+x^2)*exp(x)+x^2*exp(2/x^2)+x^2)*log(1/5*(exp
(2/x^2)+1)/x)^2+(x^2*exp(2/x^2)+x^2)*log(1/5*(exp(2/x^2)+1)/x)+(x^2+4)*exp(2/x^2)+x^2)/(x^2*exp(2/x^2)+x^2)/lo
g(1/5*(exp(2/x^2)+1)/x)^2,x, algorithm="giac")

[Out]

(x*e^x*log(5) + x*e^x*log(x) + x*log(5)*log(x) + x*log(x)^2 - x*e^x*log(e^(2/x^2) + 1) - x*log(x)*log(e^(2/x^2
) + 1) - x)/(log(5) + log(x) - log(e^(2/x^2) + 1))

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maple [C]  time = 0.41, size = 153, normalized size = 5.67

method	result	size
risch	\({\mathrm e}^{x} x +x \ln \relax (x )+\frac {2 i x}{\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{\frac {2}{x^{2}}}+1\right )\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{\frac {2}{x^{2}}}+1\right )}{x}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{\frac {2}{x^{2}}}+1\right )}{x}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{\frac {2}{x^{2}}}+1\right )\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{\frac {2}{x^{2}}}+1\right )}{x}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{\frac {2}{x^{2}}}+1\right )}{x}\right )^{3}-2 i \ln \relax (5)-2 i \ln \relax (x )+2 i \ln \left ({\mathrm e}^{\frac {2}{x^{2}}}+1\right )}\)	\(153\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^2*exp(2/x^2)+x^2)*ln(x)+((x^3+x^2)*exp(2/x^2)+x^3+x^2)*exp(x)+x^2*exp(2/x^2)+x^2)*ln(1/5*(exp(2/x^2)+
1)/x)^2+(x^2*exp(2/x^2)+x^2)*ln(1/5*(exp(2/x^2)+1)/x)+(x^2+4)*exp(2/x^2)+x^2)/(x^2*exp(2/x^2)+x^2)/ln(1/5*(exp
(2/x^2)+1)/x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.50, size = 67, normalized size = 2.48 \begin {gather*} \frac {x \log \relax (5) \log \relax (x) + x \log \relax (x)^{2} + {\left (x \log \relax (5) + x \log \relax (x)\right )} e^{x} - {\left (x e^{x} + x \log \relax (x)\right )} \log \left (e^{\left (\frac {2}{x^{2}}\right )} + 1\right ) - x}{\log \relax (5) + \log \relax (x) - \log \left (e^{\left (\frac {2}{x^{2}}\right )} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2*exp(2/x^2)+x^2)*log(x)+((x^3+x^2)*exp(2/x^2)+x^3+x^2)*exp(x)+x^2*exp(2/x^2)+x^2)*log(1/5*(exp
(2/x^2)+1)/x)^2+(x^2*exp(2/x^2)+x^2)*log(1/5*(exp(2/x^2)+1)/x)+(x^2+4)*exp(2/x^2)+x^2)/(x^2*exp(2/x^2)+x^2)/lo
g(1/5*(exp(2/x^2)+1)/x)^2,x, algorithm="maxima")

[Out]

(x*log(5)*log(x) + x*log(x)^2 + (x*log(5) + x*log(x))*e^x - (x*e^x + x*log(x))*log(e^(2/x^2) + 1) - x)/(log(5)
+ log(x) - log(e^(2/x^2) + 1))

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mupad [B]  time = 3.63, size = 144, normalized size = 5.33 \begin {gather*} \frac {x+\frac {x^3\,\ln \left (\frac {\frac {{\mathrm {e}}^{\frac {2}{x^2}}}{5}+\frac {1}{5}}{x}\right )\,\left ({\mathrm {e}}^{\frac {2}{x^2}}+1\right )}{4\,{\mathrm {e}}^{\frac {2}{x^2}}+x^2\,{\mathrm {e}}^{\frac {2}{x^2}}+x^2}}{\ln \left (\frac {\frac {{\mathrm {e}}^{\frac {2}{x^2}}}{5}+\frac {1}{5}}{x}\right )}-x+\frac {4\,x}{x^2+4}+x\,{\mathrm {e}}^x+x\,\ln \relax (x)-\frac {4\,\left (3\,x^{10}+4\,x^8\right )}{\left (x^2+4\right )\,\left ({\mathrm {e}}^{\frac {2}{x^2}}\,\left (x^2+4\right )+x^2\right )\,\left (3\,x^7+4\,x^5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x + (x^3*log((exp(2/x^2)/5 + 1/5)/x)*(exp(2/x^2) + 1))/(4*exp(2/x^2) + x^2*exp(2/x^2) + x^2))/log((exp(2/x^2)
/5 + 1/5)/x) - x + (4*x)/(x^2 + 4) + x*exp(x) + x*log(x) - (4*(4*x^8 + 3*x^10))/((x^2 + 4)*(exp(2/x^2)*(x^2 +
4) + x^2)*(4*x^5 + 3*x^7))

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sympy [A]  time = 0.74, size = 26, normalized size = 0.96 \begin {gather*} x e^{x} + x \log {\relax (x )} + \frac {x}{\log {\left (\frac {\frac {e^{\frac {2}{x^{2}}}}{5} + \frac {1}{5}}{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**2*exp(2/x**2)+x**2)*ln(x)+((x**3+x**2)*exp(2/x**2)+x**3+x**2)*exp(x)+x**2*exp(2/x**2)+x**2)*ln
(1/5*(exp(2/x**2)+1)/x)**2+(x**2*exp(2/x**2)+x**2)*ln(1/5*(exp(2/x**2)+1)/x)+(x**2+4)*exp(2/x**2)+x**2)/(x**2*
exp(2/x**2)+x**2)/ln(1/5*(exp(2/x**2)+1)/x)**2,x)

[Out]

x*exp(x) + x*log(x) + x/log((exp(2/x**2)/5 + 1/5)/x)

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