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

Optimal. Leaf size=24 \[ e^x \left (-\log (x)+e^3 \left (4 x-\log \left (\log \left (x^2\right )\right )\right )\right ) \]

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Rubi [F]  time = 0.51, 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 {-2 e^{3+x}+\left (e^x \left (-1+e^3 \left (4 x+4 x^2\right )\right )-e^x x \log (x)\right ) \log \left (x^2\right )-e^{3+x} x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 24, normalized size = 1.00 \begin {gather*} e^x \left (-\log (x)+e^3 \left (4 x-\log \left (\log \left (x^2\right )\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.68, size = 30, normalized size = 1.25 \begin {gather*} {\left (4 \, x e^{\left (x + 6\right )} - e^{\left (x + 3\right )} \log \relax (x) - e^{\left (x + 6\right )} \log \left (2 \, \log \relax (x)\right )\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*x*e^(x + 6) - e^(x + 3)*log(x) - e^(x + 6)*log(2*log(x)))*e^(-3)

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giac [A]  time = 0.22, size = 25, normalized size = 1.04 \begin {gather*} 4 \, x e^{\left (x + 3\right )} - e^{x} \log \relax (x) - e^{\left (x + 3\right )} \log \left (\log \left (x^{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.26, size = 55, normalized size = 2.29

method	result	size
risch	\(4 \,{\mathrm e}^{3+x} x -\ln \left (2 \ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}\right ) {\mathrm e}^{3+x}-{\mathrm e}^{x} \ln \relax (x )\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.50, size = 43, normalized size = 1.79 \begin {gather*} 4 \, {\left (x e^{3} - e^{3}\right )} e^{x} - e^{\left (x + 3\right )} \log \relax (2) - e^{x} \log \relax (x) - e^{\left (x + 3\right )} \log \left (\log \relax (x)\right ) + 4 \, e^{\left (x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.78, size = 25, normalized size = 1.04 \begin {gather*} 4\,x\,{\mathrm {e}}^3\,{\mathrm {e}}^x-{\mathrm {e}}^x\,\ln \relax (x)-{\mathrm {e}}^3\,{\mathrm {e}}^x\,\ln \left (\ln \left (x^2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*x*exp(3)*exp(x) - exp(x)*log(x) - exp(3)*exp(x)*log(log(x^2))

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sympy [A]  time = 16.23, size = 22, normalized size = 0.92 \begin {gather*} \left (4 x e^{3} - \log {\relax (x )} - e^{3} \log {\left (2 \log {\relax (x )} \right )}\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*x*exp(3) - log(x) - exp(3)*log(2*log(x)))*exp(x)

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