3.5.78 \(\int \frac {e^{\frac {5 x-5 e^{2+x+25 x^2+50 x^3+25 x^4} x-5 x \log (\log (3 x))}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}} (-5+(-5 e^{4+2 x+50 x^2+100 x^3+50 x^4}+e^{2+x+25 x^2+50 x^3+25 x^4} (5-5 x-250 x^2-750 x^3-500 x^4)) \log (3 x)+(5-10 e^{2+x+25 x^2+50 x^3+25 x^4}) \log (3 x) \log (\log (3 x))-5 \log (3 x) \log ^2(\log (3 x)))}{e^{4+2 x+50 x^2+100 x^3+50 x^4} \log (3 x)+2 e^{2+x+25 x^2+50 x^3+25 x^4} \log (3 x) \log (\log (3 x))+\log (3 x) \log ^2(\log (3 x))} \, dx\)

Optimal. Leaf size=32 \[ e^{5 \left (-x+\frac {x}{e^{2+x+25 \left (x+x^2\right )^2}+\log (\log (3 x))}\right )} \]

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Rubi [F]  time = 152.41, 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 (\frac {5 x-5 e^{2+x+25 x^2+50 x^3+25 x^4} x-5 x \log (\log (3 x))}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \left (-5+\left (-5 e^{4+2 x+50 x^2+100 x^3+50 x^4}+e^{2+x+25 x^2+50 x^3+25 x^4} \left (5-5 x-250 x^2-750 x^3-500 x^4\right )\right ) \log (3 x)+\left (5-10 e^{2+x+25 x^2+50 x^3+25 x^4}\right ) \log (3 x) \log (\log (3 x))-5 \log (3 x) \log ^2(\log (3 x))\right )}{e^{4+2 x+50 x^2+100 x^3+50 x^4} \log (3 x)+2 e^{2+x+25 x^2+50 x^3+25 x^4} \log (3 x) \log (\log (3 x))+\log (3 x) \log ^2(\log (3 x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((5*x - 5*E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4)*x - 5*x*Log[Log[3*x]])/(E^(2 + x + 25*x^2 + 50*x^3 + 25*
x^4) + Log[Log[3*x]]))*(-5 + (-5*E^(4 + 2*x + 50*x^2 + 100*x^3 + 50*x^4) + E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4
)*(5 - 5*x - 250*x^2 - 750*x^3 - 500*x^4))*Log[3*x] + (5 - 10*E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4))*Log[3*x]*L
og[Log[3*x]] - 5*Log[3*x]*Log[Log[3*x]]^2))/(E^(4 + 2*x + 50*x^2 + 100*x^3 + 50*x^4)*Log[3*x] + 2*E^(2 + x + 2
5*x^2 + 50*x^3 + 25*x^4)*Log[3*x]*Log[Log[3*x]] + Log[3*x]*Log[Log[3*x]]^2),x]

[Out]

-5*Defer[Int][E^((-5*x*(-1 + E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4) + Log[Log[3*x]]))/(E^(2 + x + 25*x^2 + 50*x^
3 + 25*x^4) + Log[Log[3*x]])), x] - 5*Defer[Int][1/(E^((5*x*(-1 + E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4) + Log[L
og[3*x]]))/(E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4) + Log[Log[3*x]]))*Log[3*x]*(E^(2 + x + 25*x^2 + 50*x^3 + 25*x
^4) + Log[Log[3*x]])^2), x] + 5*Defer[Int][(x*Log[Log[3*x]])/(E^((5*x*(-1 + E^(2 + x + 25*x^2 + 50*x^3 + 25*x^
4) + Log[Log[3*x]]))/(E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4) + Log[Log[3*x]]))*(E^(2 + x + 25*x^2 + 50*x^3 + 25*
x^4) + Log[Log[3*x]])^2), x] + 250*Defer[Int][(x^2*Log[Log[3*x]])/(E^((5*x*(-1 + E^(2 + x + 25*x^2 + 50*x^3 +
25*x^4) + Log[Log[3*x]]))/(E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4) + Log[Log[3*x]]))*(E^(2 + x + 25*x^2 + 50*x^3
+ 25*x^4) + Log[Log[3*x]])^2), x] + 750*Defer[Int][(x^3*Log[Log[3*x]])/(E^((5*x*(-1 + E^(2 + x + 25*x^2 + 50*x
^3 + 25*x^4) + Log[Log[3*x]]))/(E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4) + Log[Log[3*x]]))*(E^(2 + x + 25*x^2 + 50
*x^3 + 25*x^4) + Log[Log[3*x]])^2), x] + 500*Defer[Int][(x^4*Log[Log[3*x]])/(E^((5*x*(-1 + E^(2 + x + 25*x^2 +
50*x^3 + 25*x^4) + Log[Log[3*x]]))/(E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4) + Log[Log[3*x]]))*(E^(2 + x + 25*x^2
+ 50*x^3 + 25*x^4) + Log[Log[3*x]])^2), x] + 5*Defer[Int][1/(E^((5*x*(-1 + E^(2 + x + 25*x^2 + 50*x^3 + 25*x^
4) + Log[Log[3*x]]))/(E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4) + Log[Log[3*x]]))*(E^(2 + x + 25*x^2 + 50*x^3 + 25*
x^4) + Log[Log[3*x]])), x] - 5*Defer[Int][x/(E^((5*x*(-1 + E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4) + Log[Log[3*x]
]))/(E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4) + Log[Log[3*x]]))*(E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4) + Log[Log[3*
x]])), x] - 250*Defer[Int][x^2/(E^((5*x*(-1 + E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4) + Log[Log[3*x]]))/(E^(2 + x
+ 25*x^2 + 50*x^3 + 25*x^4) + Log[Log[3*x]]))*(E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4) + Log[Log[3*x]])), x] - 7
50*Defer[Int][x^3/(E^((5*x*(-1 + E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4) + Log[Log[3*x]]))/(E^(2 + x + 25*x^2 + 5
0*x^3 + 25*x^4) + Log[Log[3*x]]))*(E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4) + Log[Log[3*x]])), x] - 500*Defer[Int]
[x^4/(E^((5*x*(-1 + E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4) + Log[Log[3*x]]))/(E^(2 + x + 25*x^2 + 50*x^3 + 25*x^
4) + Log[Log[3*x]]))*(E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4) + Log[Log[3*x]])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \left (-5+\left (-5 e^{4+2 x+50 x^2+100 x^3+50 x^4}+e^{2+x+25 x^2+50 x^3+25 x^4} \left (5-5 x-250 x^2-750 x^3-500 x^4\right )\right ) \log (3 x)+\left (5-10 e^{2+x+25 x^2+50 x^3+25 x^4}\right ) \log (3 x) \log (\log (3 x))-5 \log (3 x) \log ^2(\log (3 x))\right )}{\log (3 x) \left (e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )^2} \, dx\\ &=\int \left (-5 \exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right )-\frac {5 \exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \left (-1+x+50 x^2+150 x^3+100 x^4\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}+\frac {5 \exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \left (-1+x \log (3 x) \log (\log (3 x))+50 x^2 \log (3 x) \log (\log (3 x))+150 x^3 \log (3 x) \log (\log (3 x))+100 x^4 \log (3 x) \log (\log (3 x))\right )}{\log (3 x) \left (e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )^2}\right ) \, dx\\ &=-\left (5 \int \exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \, dx\right )-5 \int \frac {\exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \left (-1+x+50 x^2+150 x^3+100 x^4\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))} \, dx+5 \int \frac {\exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \left (-1+x \log (3 x) \log (\log (3 x))+50 x^2 \log (3 x) \log (\log (3 x))+150 x^3 \log (3 x) \log (\log (3 x))+100 x^4 \log (3 x) \log (\log (3 x))\right )}{\log (3 x) \left (e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )^2} \, dx\\ &=-\left (5 \int \exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \, dx\right )+5 \int \frac {\exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \left (-1+x \left (1+50 x+150 x^2+100 x^3\right ) \log (3 x) \log (\log (3 x))\right )}{\log (3 x) \left (e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )^2} \, dx-5 \int \left (-\frac {\exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}+\frac {\exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) x}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}+\frac {50 \exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) x^2}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}+\frac {150 \exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) x^3}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}+\frac {100 \exp \left (-\frac {5 x \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))\right )}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) x^4}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}\right ) \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [B]  time = 1.70, size = 107, normalized size = 3.34 \begin {gather*} e^{-5 x-\frac {5 \left (-1+e^{2+x+25 x^2+50 x^3+25 x^4}\right ) x}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}} \log ^{\frac {5 x}{\log (\log (3 x))}-\frac {5 x}{e^{2+x+25 x^2+50 x^3+25 x^4}+\log (\log (3 x))}}(3 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((5*x - 5*E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4)*x - 5*x*Log[Log[3*x]])/(E^(2 + x + 25*x^2 + 50*x^3
+ 25*x^4) + Log[Log[3*x]]))*(-5 + (-5*E^(4 + 2*x + 50*x^2 + 100*x^3 + 50*x^4) + E^(2 + x + 25*x^2 + 50*x^3 +
25*x^4)*(5 - 5*x - 250*x^2 - 750*x^3 - 500*x^4))*Log[3*x] + (5 - 10*E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4))*Log[
3*x]*Log[Log[3*x]] - 5*Log[3*x]*Log[Log[3*x]]^2))/(E^(4 + 2*x + 50*x^2 + 100*x^3 + 50*x^4)*Log[3*x] + 2*E^(2 +
x + 25*x^2 + 50*x^3 + 25*x^4)*Log[3*x]*Log[Log[3*x]] + Log[3*x]*Log[Log[3*x]]^2),x]

[Out]

E^(-5*x - (5*(-1 + E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4))*x)/(E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4) + Log[Log[3*
x]]))*Log[3*x]^((5*x)/Log[Log[3*x]] - (5*x)/(E^(2 + x + 25*x^2 + 50*x^3 + 25*x^4) + Log[Log[3*x]]))

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fricas [B]  time = 0.84, size = 62, normalized size = 1.94 \begin {gather*} e^{\left (-\frac {5 \, {\left (x e^{\left (25 \, x^{4} + 50 \, x^{3} + 25 \, x^{2} + x + 2\right )} + x \log \left (\log \left (3 \, x\right )\right ) - x\right )}}{e^{\left (25 \, x^{4} + 50 \, x^{3} + 25 \, x^{2} + x + 2\right )} + \log \left (\log \left (3 \, x\right )\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(3*x)*log(log(3*x))^2+(-10*exp(25*x^4+50*x^3+25*x^2+x+2)+5)*log(3*x)*log(log(3*x))+(-5*exp(25
*x^4+50*x^3+25*x^2+x+2)^2+(-500*x^4-750*x^3-250*x^2-5*x+5)*exp(25*x^4+50*x^3+25*x^2+x+2))*log(3*x)-5)*exp((-5*
x*log(log(3*x))-5*x*exp(25*x^4+50*x^3+25*x^2+x+2)+5*x)/(log(log(3*x))+exp(25*x^4+50*x^3+25*x^2+x+2)))/(log(3*x
)*log(log(3*x))^2+2*exp(25*x^4+50*x^3+25*x^2+x+2)*log(3*x)*log(log(3*x))+exp(25*x^4+50*x^3+25*x^2+x+2)^2*log(3
*x)),x, algorithm="fricas")

[Out]

e^(-5*(x*e^(25*x^4 + 50*x^3 + 25*x^2 + x + 2) + x*log(log(3*x)) - x)/(e^(25*x^4 + 50*x^3 + 25*x^2 + x + 2) + l
og(log(3*x))))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(3*x)*log(log(3*x))^2+(-10*exp(25*x^4+50*x^3+25*x^2+x+2)+5)*log(3*x)*log(log(3*x))+(-5*exp(25
*x^4+50*x^3+25*x^2+x+2)^2+(-500*x^4-750*x^3-250*x^2-5*x+5)*exp(25*x^4+50*x^3+25*x^2+x+2))*log(3*x)-5)*exp((-5*
x*log(log(3*x))-5*x*exp(25*x^4+50*x^3+25*x^2+x+2)+5*x)/(log(log(3*x))+exp(25*x^4+50*x^3+25*x^2+x+2)))/(log(3*x
)*log(log(3*x))^2+2*exp(25*x^4+50*x^3+25*x^2+x+2)*log(3*x)*log(log(3*x))+exp(25*x^4+50*x^3+25*x^2+x+2)^2*log(3
*x)),x, algorithm="giac")

[Out]

undef

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maple [A]  time = 0.08, size = 58, normalized size = 1.81

method	result	size
risch	\({\mathrm e}^{-\frac {5 x \left (\ln \left (\ln \left (3 x \right )\right )+{\mathrm e}^{25 x^{4}+50 x^{3}+25 x^{2}+x +2}-1\right )}{\ln \left (\ln \left (3 x \right )\right )+{\mathrm e}^{25 x^{4}+50 x^{3}+25 x^{2}+x +2}}}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5*ln(3*x)*ln(ln(3*x))^2+(-10*exp(25*x^4+50*x^3+25*x^2+x+2)+5)*ln(3*x)*ln(ln(3*x))+(-5*exp(25*x^4+50*x^3+
25*x^2+x+2)^2+(-500*x^4-750*x^3-250*x^2-5*x+5)*exp(25*x^4+50*x^3+25*x^2+x+2))*ln(3*x)-5)*exp((-5*x*ln(ln(3*x))
-5*x*exp(25*x^4+50*x^3+25*x^2+x+2)+5*x)/(ln(ln(3*x))+exp(25*x^4+50*x^3+25*x^2+x+2)))/(ln(3*x)*ln(ln(3*x))^2+2*
exp(25*x^4+50*x^3+25*x^2+x+2)*ln(3*x)*ln(ln(3*x))+exp(25*x^4+50*x^3+25*x^2+x+2)^2*ln(3*x)),x,method=_RETURNVER
BOSE)

[Out]

exp(-5*x*(ln(ln(3*x))+exp(25*x^4+50*x^3+25*x^2+x+2)-1)/(ln(ln(3*x))+exp(25*x^4+50*x^3+25*x^2+x+2)))

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maxima [B]  time = 0.97, size = 120, normalized size = 3.75 \begin {gather*} e^{\left (-\frac {5 \, x e^{\left (25 \, x^{4} + 50 \, x^{3} + 25 \, x^{2} + x + 2\right )}}{e^{\left (25 \, x^{4} + 50 \, x^{3} + 25 \, x^{2} + x + 2\right )} + \log \left (\log \relax (3) + \log \relax (x)\right )} - \frac {5 \, x \log \left (\log \relax (3) + \log \relax (x)\right )}{e^{\left (25 \, x^{4} + 50 \, x^{3} + 25 \, x^{2} + x + 2\right )} + \log \left (\log \relax (3) + \log \relax (x)\right )} + \frac {5 \, x}{e^{\left (25 \, x^{4} + 50 \, x^{3} + 25 \, x^{2} + x + 2\right )} + \log \left (\log \relax (3) + \log \relax (x)\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(3*x)*log(log(3*x))^2+(-10*exp(25*x^4+50*x^3+25*x^2+x+2)+5)*log(3*x)*log(log(3*x))+(-5*exp(25
*x^4+50*x^3+25*x^2+x+2)^2+(-500*x^4-750*x^3-250*x^2-5*x+5)*exp(25*x^4+50*x^3+25*x^2+x+2))*log(3*x)-5)*exp((-5*
x*log(log(3*x))-5*x*exp(25*x^4+50*x^3+25*x^2+x+2)+5*x)/(log(log(3*x))+exp(25*x^4+50*x^3+25*x^2+x+2)))/(log(3*x
)*log(log(3*x))^2+2*exp(25*x^4+50*x^3+25*x^2+x+2)*log(3*x)*log(log(3*x))+exp(25*x^4+50*x^3+25*x^2+x+2)^2*log(3
*x)),x, algorithm="maxima")

[Out]

e^(-5*x*e^(25*x^4 + 50*x^3 + 25*x^2 + x + 2)/(e^(25*x^4 + 50*x^3 + 25*x^2 + x + 2) + log(log(3) + log(x))) - 5
*x*log(log(3) + log(x))/(e^(25*x^4 + 50*x^3 + 25*x^2 + x + 2) + log(log(3) + log(x))) + 5*x/(e^(25*x^4 + 50*x^
3 + 25*x^2 + x + 2) + log(log(3) + log(x))))

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mupad [B]  time = 1.07, size = 138, normalized size = 4.31 \begin {gather*} \frac {{\mathrm {e}}^{\frac {5\,x}{\ln \left (\ln \relax (3)+\ln \relax (x)\right )+{\mathrm {e}}^2\,{\mathrm {e}}^{25\,x^2}\,{\mathrm {e}}^{25\,x^4}\,{\mathrm {e}}^{50\,x^3}\,{\mathrm {e}}^x}}\,{\mathrm {e}}^{-\frac {5\,x\,{\mathrm {e}}^2\,{\mathrm {e}}^{25\,x^2}\,{\mathrm {e}}^{25\,x^4}\,{\mathrm {e}}^{50\,x^3}\,{\mathrm {e}}^x}{\ln \left (\ln \relax (3)+\ln \relax (x)\right )+{\mathrm {e}}^2\,{\mathrm {e}}^{25\,x^2}\,{\mathrm {e}}^{25\,x^4}\,{\mathrm {e}}^{50\,x^3}\,{\mathrm {e}}^x}}}{{\left (\ln \relax (3)+\ln \relax (x)\right )}^{\frac {5\,x}{\ln \left (\ln \relax (3)+\ln \relax (x)\right )+{\mathrm {e}}^2\,{\mathrm {e}}^{25\,x^2}\,{\mathrm {e}}^{25\,x^4}\,{\mathrm {e}}^{50\,x^3}\,{\mathrm {e}}^x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(5*x*log(log(3*x)) - 5*x + 5*x*exp(x + 25*x^2 + 50*x^3 + 25*x^4 + 2))/(log(log(3*x)) + exp(x + 25*x
^2 + 50*x^3 + 25*x^4 + 2)))*(5*log(3*x)*log(log(3*x))^2 + log(3*x)*(5*exp(2*x + 50*x^2 + 100*x^3 + 50*x^4 + 4)
+ exp(x + 25*x^2 + 50*x^3 + 25*x^4 + 2)*(5*x + 250*x^2 + 750*x^3 + 500*x^4 - 5)) + log(3*x)*log(log(3*x))*(10
*exp(x + 25*x^2 + 50*x^3 + 25*x^4 + 2) - 5) + 5))/(log(3*x)*log(log(3*x))^2 + log(3*x)*exp(2*x + 50*x^2 + 100*
x^3 + 50*x^4 + 4) + 2*log(3*x)*exp(x + 25*x^2 + 50*x^3 + 25*x^4 + 2)*log(log(3*x))),x)

[Out]

(exp((5*x)/(log(log(3) + log(x)) + exp(2)*exp(25*x^2)*exp(25*x^4)*exp(50*x^3)*exp(x)))*exp(-(5*x*exp(2)*exp(25
*x^2)*exp(25*x^4)*exp(50*x^3)*exp(x))/(log(log(3) + log(x)) + exp(2)*exp(25*x^2)*exp(25*x^4)*exp(50*x^3)*exp(x
))))/(log(3) + log(x))^((5*x)/(log(log(3) + log(x)) + exp(2)*exp(25*x^2)*exp(25*x^4)*exp(50*x^3)*exp(x)))

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sympy [B]  time = 35.11, size = 65, normalized size = 2.03 \begin {gather*} e^{\frac {- 5 x e^{25 x^{4} + 50 x^{3} + 25 x^{2} + x + 2} - 5 x \log {\left (\log {\left (3 x \right )} \right )} + 5 x}{e^{25 x^{4} + 50 x^{3} + 25 x^{2} + x + 2} + \log {\left (\log {\left (3 x \right )} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*ln(3*x)*ln(ln(3*x))**2+(-10*exp(25*x**4+50*x**3+25*x**2+x+2)+5)*ln(3*x)*ln(ln(3*x))+(-5*exp(25*x
**4+50*x**3+25*x**2+x+2)**2+(-500*x**4-750*x**3-250*x**2-5*x+5)*exp(25*x**4+50*x**3+25*x**2+x+2))*ln(3*x)-5)*e
xp((-5*x*ln(ln(3*x))-5*x*exp(25*x**4+50*x**3+25*x**2+x+2)+5*x)/(ln(ln(3*x))+exp(25*x**4+50*x**3+25*x**2+x+2)))
/(ln(3*x)*ln(ln(3*x))**2+2*exp(25*x**4+50*x**3+25*x**2+x+2)*ln(3*x)*ln(ln(3*x))+exp(25*x**4+50*x**3+25*x**2+x+
2)**2*ln(3*x)),x)

[Out]

exp((-5*x*exp(25*x**4 + 50*x**3 + 25*x**2 + x + 2) - 5*x*log(log(3*x)) + 5*x)/(exp(25*x**4 + 50*x**3 + 25*x**2
+ x + 2) + log(log(3*x))))

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