3.97.97 \(\int -\frac {e^{5-\frac {e^5 (3 x-\frac {(-15+3 x) \log (3+\log (x))}{e^5})}{\log (3+\log (x))}} (-3+(9+3 \log (x)) \log (3+\log (x))-\frac {(9+3 \log (x)) \log ^2(3+\log (x))}{e^5})}{(3+\log (x)) \log ^2(3+\log (x))} \, dx\)

Optimal. Leaf size=20 \[ e^{3 \left (-5+x-\frac {e^5 x}{\log (3+\log (x))}\right )} \]

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Rubi [F]  time = 1.49, 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 {\exp \left (5-\frac {e^5 \left (3 x-\frac {(-15+3 x) \log (3+\log (x))}{e^5}\right )}{\log (3+\log (x))}\right ) \left (-3+(9+3 \log (x)) \log (3+\log (x))-\frac {(9+3 \log (x)) \log ^2(3+\log (x))}{e^5}\right )}{(3+\log (x)) \log ^2(3+\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[-((E^(5 - (E^5*(3*x - ((-15 + 3*x)*Log[3 + Log[x]])/E^5))/Log[3 + Log[x]])*(-3 + (9 + 3*Log[x])*Log[3 + Lo
g[x]] - ((9 + 3*Log[x])*Log[3 + Log[x]]^2)/E^5))/((3 + Log[x])*Log[3 + Log[x]]^2)),x]

[Out]

3*Defer[Int][E^(3*(-5 + x - (E^5*x)/Log[3 + Log[x]])), x] + 3*Defer[Int][E^(-10 + 3*x - (3*E^5*x)/Log[3 + Log[
x]])/((3 + Log[x])*Log[3 + Log[x]]^2), x] - 3*Defer[Int][E^(-10 + 3*x - (3*E^5*x)/Log[3 + Log[x]])/Log[3 + Log
[x]], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\int \frac {\exp \left (\frac {-3 e^5 x-10 \log (3+\log (x))+3 x \log (3+\log (x))}{\log (3+\log (x))}\right ) \left (-3+(9+3 \log (x)) \log (3+\log (x))-\frac {(9+3 \log (x)) \log ^2(3+\log (x))}{e^5}\right )}{(3+\log (x)) \log ^2(3+\log (x))} \, dx\\ &=-\int \left (-3 \exp \left (-5+\frac {-3 e^5 x-10 \log (3+\log (x))+3 x \log (3+\log (x))}{\log (3+\log (x))}\right )-\frac {3 \exp \left (\frac {-3 e^5 x-10 \log (3+\log (x))+3 x \log (3+\log (x))}{\log (3+\log (x))}\right )}{(3+\log (x)) \log ^2(3+\log (x))}+\frac {3 \exp \left (\frac {-3 e^5 x-10 \log (3+\log (x))+3 x \log (3+\log (x))}{\log (3+\log (x))}\right )}{\log (3+\log (x))}\right ) \, dx\\ &=3 \int \exp \left (-5+\frac {-3 e^5 x-10 \log (3+\log (x))+3 x \log (3+\log (x))}{\log (3+\log (x))}\right ) \, dx+3 \int \frac {\exp \left (\frac {-3 e^5 x-10 \log (3+\log (x))+3 x \log (3+\log (x))}{\log (3+\log (x))}\right )}{(3+\log (x)) \log ^2(3+\log (x))} \, dx-3 \int \frac {\exp \left (\frac {-3 e^5 x-10 \log (3+\log (x))+3 x \log (3+\log (x))}{\log (3+\log (x))}\right )}{\log (3+\log (x))} \, dx\\ &=3 \int e^{3 \left (-5+x-\frac {e^5 x}{\log (3+\log (x))}\right )} \, dx+3 \int \frac {e^{-10+3 x-\frac {3 e^5 x}{\log (3+\log (x))}}}{(3+\log (x)) \log ^2(3+\log (x))} \, dx-3 \int \frac {e^{-10+3 x-\frac {3 e^5 x}{\log (3+\log (x))}}}{\log (3+\log (x))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 20, normalized size = 1.00 \begin {gather*} e^{-15+3 x-\frac {3 e^5 x}{\log (3+\log (x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-((E^(5 - (E^5*(3*x - ((-15 + 3*x)*Log[3 + Log[x]])/E^5))/Log[3 + Log[x]])*(-3 + (9 + 3*Log[x])*Log[
3 + Log[x]] - ((9 + 3*Log[x])*Log[3 + Log[x]]^2)/E^5))/((3 + Log[x])*Log[3 + Log[x]]^2)),x]

[Out]

E^(-15 + 3*x - (3*E^5*x)/Log[3 + Log[x]])

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fricas [B]  time = 0.73, size = 51, normalized size = 2.55 \begin {gather*} -e^{\left (-\frac {3 \, x e^{5} - {\left (3 \, x - 10\right )} \log \left (\log \relax (x) + 3\right ) + \log \left (-\log \left (\log \relax (x) + 3\right )\right ) \log \left (\log \relax (x) + 3\right )}{\log \left (\log \relax (x) + 3\right )} - 5\right )} \log \left (\log \relax (x) + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*log(x)+9)*log(3+log(x))*exp(log(-log(3+log(x)))-5)+(3*log(x)+9)*log(3+log(x))-3)*exp(((3*x-15)*e
xp(log(-log(3+log(x)))-5)+3*x)/exp(log(-log(3+log(x)))-5))/(3+log(x))/log(3+log(x))/exp(log(-log(3+log(x)))-5)
,x, algorithm="fricas")

[Out]

-e^(-(3*x*e^5 - (3*x - 10)*log(log(x) + 3) + log(-log(log(x) + 3))*log(log(x) + 3))/log(log(x) + 3) - 5)*log(l
og(x) + 3)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*log(x)+9)*log(3+log(x))*exp(log(-log(3+log(x)))-5)+(3*log(x)+9)*log(3+log(x))-3)*exp(((3*x-15)*e
xp(log(-log(3+log(x)))-5)+3*x)/exp(log(-log(3+log(x)))-5))/(3+log(x))/log(3+log(x))/exp(log(-log(3+log(x)))-5)
,x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Evaluation time:
0.59Unable to divide, perhaps due to rounding error%%%{27,[0,1,0,3]%%%}+%%%{81,[0,0,0,3]%%%} / %%%{27,[0,2,0,3
]%%%}+%%%{162,

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maple [A]  time = 0.06, size = 35, normalized size = 1.75

method	result	size
risch	\({\mathrm e}^{\frac {3 \left (\ln \left (3+\ln \relax (x )\right ) {\mathrm e}^{-5} x -5 \ln \left (3+\ln \relax (x )\right ) {\mathrm e}^{-5}-x \right ) {\mathrm e}^{5}}{\ln \left (3+\ln \relax (x )\right )}}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*ln(x)+9)*ln(3+ln(x))*exp(ln(-ln(3+ln(x)))-5)+(3*ln(x)+9)*ln(3+ln(x))-3)*exp(((3*x-15)*exp(ln(-ln(3+ln(
x)))-5)+3*x)/exp(ln(-ln(3+ln(x)))-5))/(3+ln(x))/ln(3+ln(x))/exp(ln(-ln(3+ln(x)))-5),x,method=_RETURNVERBOSE)

[Out]

exp(3*(ln(3+ln(x))*exp(-5)*x-5*ln(3+ln(x))*exp(-5)-x)*exp(5)/ln(3+ln(x)))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*log(x)+9)*log(3+log(x))*exp(log(-log(3+log(x)))-5)+(3*log(x)+9)*log(3+log(x))-3)*exp(((3*x-15)*e
xp(log(-log(3+log(x)))-5)+3*x)/exp(log(-log(3+log(x)))-5))/(3+log(x))/log(3+log(x))/exp(log(-log(3+log(x)))-5)
,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 6.23, size = 20, normalized size = 1.00 \begin {gather*} {\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{-15}\,{\mathrm {e}}^{-\frac {3\,x\,{\mathrm {e}}^5}{\ln \left (\ln \relax (x)+3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(5 - log(-log(log(x) + 3)))*(3*x + exp(log(-log(log(x) + 3)) - 5)*(3*x - 15)))*exp(5 - log(-log(lo
g(x) + 3)))*(log(log(x) + 3)*(3*log(x) + 9) + exp(log(-log(log(x) + 3)) - 5)*log(log(x) + 3)*(3*log(x) + 9) -
3))/(log(log(x) + 3)*(log(x) + 3)),x)

[Out]

exp(3*x)*exp(-15)*exp(-(3*x*exp(5))/log(log(x) + 3))

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sympy [A]  time = 0.72, size = 31, normalized size = 1.55 \begin {gather*} e^{- \frac {\left (3 x - \frac {\left (3 x - 15\right ) \log {\left (\log {\relax (x )} + 3 \right )}}{e^{5}}\right ) e^{5}}{\log {\left (\log {\relax (x )} + 3 \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*ln(x)+9)*ln(3+ln(x))*exp(ln(-ln(3+ln(x)))-5)+(3*ln(x)+9)*ln(3+ln(x))-3)*exp(((3*x-15)*exp(ln(-ln
(3+ln(x)))-5)+3*x)/exp(ln(-ln(3+ln(x)))-5))/(3+ln(x))/ln(3+ln(x))/exp(ln(-ln(3+ln(x)))-5),x)

[Out]

exp(-(3*x - (3*x - 15)*exp(-5)*log(log(x) + 3))*exp(5)/log(log(x) + 3))

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