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

Optimal. Leaf size=15 \[ x+\frac {1}{\log \left (\log \left (e^x x+x^3\right )\right )} \]

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Rubi [F]  time = 2.54, 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 {e^x (-1-x)-3 x^2+\left (e^x x+x^3\right ) \log \left (e^x x+x^3\right ) \log ^2\left (\log \left (e^x x+x^3\right )\right )}{\left (e^x x+x^3\right ) \log \left (e^x x+x^3\right ) \log ^2\left (\log \left (e^x x+x^3\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 15, normalized size = 1.00 \begin {gather*} x+\frac {1}{\log \left (\log \left (x \left (e^x+x^2\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x + Log[Log[x*(E^x + x^2)]]^(-1)

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fricas [A]  time = 0.60, size = 27, normalized size = 1.80 \begin {gather*} \frac {x \log \left (\log \left (x^{3} + x e^{x}\right )\right ) + 1}{\log \left (\log \left (x^{3} + x e^{x}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*log(log(x^3 + x*e^x)) + 1)/log(log(x^3 + x*e^x))

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giac [A]  time = 0.64, size = 27, normalized size = 1.80 \begin {gather*} \frac {x \log \left (\log \left (x^{3} + x e^{x}\right )\right ) + 1}{\log \left (\log \left (x^{3} + x e^{x}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*log(log(x^3 + x*e^x)) + 1)/log(log(x^3 + x*e^x))

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maple [C]  time = 0.13, size = 74, normalized size = 4.93

method	result	size
risch	\(x +\frac {1}{\ln \left (\ln \relax (x )+\ln \left (x^{2}+{\mathrm e}^{x}\right )-\frac {i \pi \,\mathrm {csgn}\left (i x \left (x^{2}+{\mathrm e}^{x}\right )\right ) \left (-\mathrm {csgn}\left (i x \left (x^{2}+{\mathrm e}^{x}\right )\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (i x \left (x^{2}+{\mathrm e}^{x}\right )\right )+\mathrm {csgn}\left (i \left (x^{2}+{\mathrm e}^{x}\right )\right )\right )}{2}\right )}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.45, size = 29, normalized size = 1.93 \begin {gather*} \frac {x \log \left (\log \left (x^{2} + e^{x}\right ) + \log \relax (x)\right ) + 1}{\log \left (\log \left (x^{2} + e^{x}\right ) + \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.33, size = 14, normalized size = 0.93 \begin {gather*} x+\frac {1}{\ln \left (\ln \left (x\,{\mathrm {e}}^x+x^3\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 1/log(log(x*exp(x) + x^3))

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sympy [A]  time = 1.28, size = 14, normalized size = 0.93 \begin {gather*} x + \frac {1}{\log {\left (\log {\left (x^{3} + x e^{x} \right )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 1/log(log(x**3 + x*exp(x)))

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