3.49.19 \(\int \frac {-18 e^6 x^2+(6 e^3 x^2+e^6 (54 x+18 x^2) \log (3+x)) \log (\log (3+x))+e^3 (-36 x-12 x^2) \log (3+x) \log ^2(\log (3+x))+(27+e^{16-x} (-27-9 x)+15 x+2 x^2) \log (3+x) \log ^3(\log (3+x))}{(27+9 x) \log (3+x) \log ^3(\log (3+x))} \, dx\)

Optimal. Leaf size=30 \[ e^{16-x}+x+\left (\frac {x}{3}-\frac {e^3 x}{\log (\log (3+x))}\right )^2 \]

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Rubi [F]  time = 2.43, 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 {-18 e^6 x^2+\left (6 e^3 x^2+e^6 \left (54 x+18 x^2\right ) \log (3+x)\right ) \log (\log (3+x))+e^3 \left (-36 x-12 x^2\right ) \log (3+x) \log ^2(\log (3+x))+\left (27+e^{16-x} (-27-9 x)+15 x+2 x^2\right ) \log (3+x) \log ^3(\log (3+x))}{(27+9 x) \log (3+x) \log ^3(\log (3+x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-18*E^6*x^2 + (6*E^3*x^2 + E^6*(54*x + 18*x^2)*Log[3 + x])*Log[Log[3 + x]] + E^3*(-36*x - 12*x^2)*Log[3 +
x]*Log[Log[3 + x]]^2 + (27 + E^(16 - x)*(-27 - 9*x) + 15*x + 2*x^2)*Log[3 + x]*Log[Log[3 + x]]^3)/((27 + 9*x)
*Log[3 + x]*Log[Log[3 + x]]^3),x]

[Out]

E^(16 - x) + x + x^2/9 + (9*E^6)/Log[Log[3 + x]]^2 - (6*E^3)/Log[Log[3 + x]] - 2*E^6*Defer[Int][x/(Log[3 + x]*
Log[Log[3 + x]]^3), x] + 2*E^6*Defer[Int][x/Log[Log[3 + x]]^2, x] + (2*E^3*Defer[Int][x/(Log[3 + x]*Log[Log[3
+ x]]^2), x])/3 - (4*E^3*Defer[Int][x/Log[Log[3 + x]], x])/3 + 6*E^6*Defer[Subst][Defer[Int][1/(Log[x]*Log[Log
[x]]^3), x], x, 3 + x] - 2*E^3*Defer[Subst][Defer[Int][1/(Log[x]*Log[Log[x]]^2), x], x, 3 + x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1}{9} \left (9-9 e^{16-x}+2 x+\frac {18 e^6 x}{\log ^2(\log (3+x))}-\frac {12 e^3 x}{\log (\log (3+x))}+\frac {6 x^2 \left (-3 e^6+e^3 \log (\log (3+x))\right )}{(3+x) \log (3+x) \log ^3(\log (3+x))}\right ) \, dx\\ &=\frac {1}{9} \int \left (9-9 e^{16-x}+2 x+\frac {18 e^6 x}{\log ^2(\log (3+x))}-\frac {12 e^3 x}{\log (\log (3+x))}+\frac {6 x^2 \left (-3 e^6+e^3 \log (\log (3+x))\right )}{(3+x) \log (3+x) \log ^3(\log (3+x))}\right ) \, dx\\ &=x+\frac {x^2}{9}+\frac {2}{3} \int \frac {x^2 \left (-3 e^6+e^3 \log (\log (3+x))\right )}{(3+x) \log (3+x) \log ^3(\log (3+x))} \, dx-\frac {1}{3} \left (4 e^3\right ) \int \frac {x}{\log (\log (3+x))} \, dx+\left (2 e^6\right ) \int \frac {x}{\log ^2(\log (3+x))} \, dx-\int e^{16-x} \, dx\\ &=e^{16-x}+x+\frac {x^2}{9}+\frac {2}{3} \int \frac {e^3 x^2 \left (-3 e^3+\log (\log (3+x))\right )}{(3+x) \log (3+x) \log ^3(\log (3+x))} \, dx-\frac {1}{3} \left (4 e^3\right ) \int \frac {x}{\log (\log (3+x))} \, dx+\left (2 e^6\right ) \int \frac {x}{\log ^2(\log (3+x))} \, dx\\ &=e^{16-x}+x+\frac {x^2}{9}+\frac {1}{3} \left (2 e^3\right ) \int \frac {x^2 \left (-3 e^3+\log (\log (3+x))\right )}{(3+x) \log (3+x) \log ^3(\log (3+x))} \, dx-\frac {1}{3} \left (4 e^3\right ) \int \frac {x}{\log (\log (3+x))} \, dx+\left (2 e^6\right ) \int \frac {x}{\log ^2(\log (3+x))} \, dx\\ &=e^{16-x}+x+\frac {x^2}{9}+\frac {1}{3} \left (2 e^3\right ) \int \left (-\frac {3 e^3 x^2}{(3+x) \log (3+x) \log ^3(\log (3+x))}+\frac {x^2}{(3+x) \log (3+x) \log ^2(\log (3+x))}\right ) \, dx-\frac {1}{3} \left (4 e^3\right ) \int \frac {x}{\log (\log (3+x))} \, dx+\left (2 e^6\right ) \int \frac {x}{\log ^2(\log (3+x))} \, dx\\ &=e^{16-x}+x+\frac {x^2}{9}+\frac {1}{3} \left (2 e^3\right ) \int \frac {x^2}{(3+x) \log (3+x) \log ^2(\log (3+x))} \, dx-\frac {1}{3} \left (4 e^3\right ) \int \frac {x}{\log (\log (3+x))} \, dx-\left (2 e^6\right ) \int \frac {x^2}{(3+x) \log (3+x) \log ^3(\log (3+x))} \, dx+\left (2 e^6\right ) \int \frac {x}{\log ^2(\log (3+x))} \, dx\\ &=e^{16-x}+x+\frac {x^2}{9}+\frac {1}{3} \left (2 e^3\right ) \int \left (-\frac {3}{\log (3+x) \log ^2(\log (3+x))}+\frac {x}{\log (3+x) \log ^2(\log (3+x))}+\frac {9}{(3+x) \log (3+x) \log ^2(\log (3+x))}\right ) \, dx-\frac {1}{3} \left (4 e^3\right ) \int \frac {x}{\log (\log (3+x))} \, dx-\left (2 e^6\right ) \int \left (-\frac {3}{\log (3+x) \log ^3(\log (3+x))}+\frac {x}{\log (3+x) \log ^3(\log (3+x))}+\frac {9}{(3+x) \log (3+x) \log ^3(\log (3+x))}\right ) \, dx+\left (2 e^6\right ) \int \frac {x}{\log ^2(\log (3+x))} \, dx\\ &=e^{16-x}+x+\frac {x^2}{9}+\frac {1}{3} \left (2 e^3\right ) \int \frac {x}{\log (3+x) \log ^2(\log (3+x))} \, dx-\frac {1}{3} \left (4 e^3\right ) \int \frac {x}{\log (\log (3+x))} \, dx-\left (2 e^3\right ) \int \frac {1}{\log (3+x) \log ^2(\log (3+x))} \, dx+\left (6 e^3\right ) \int \frac {1}{(3+x) \log (3+x) \log ^2(\log (3+x))} \, dx-\left (2 e^6\right ) \int \frac {x}{\log (3+x) \log ^3(\log (3+x))} \, dx+\left (2 e^6\right ) \int \frac {x}{\log ^2(\log (3+x))} \, dx+\left (6 e^6\right ) \int \frac {1}{\log (3+x) \log ^3(\log (3+x))} \, dx-\left (18 e^6\right ) \int \frac {1}{(3+x) \log (3+x) \log ^3(\log (3+x))} \, dx\\ &=e^{16-x}+x+\frac {x^2}{9}+\frac {9 e^6}{\log ^2(\log (3+x))}-\frac {6 e^3}{\log (\log (3+x))}+\frac {1}{3} \left (2 e^3\right ) \int \frac {x}{\log (3+x) \log ^2(\log (3+x))} \, dx-\frac {1}{3} \left (4 e^3\right ) \int \frac {x}{\log (\log (3+x))} \, dx-\left (2 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{\log (x) \log ^2(\log (x))} \, dx,x,3+x\right )-\left (2 e^6\right ) \int \frac {x}{\log (3+x) \log ^3(\log (3+x))} \, dx+\left (2 e^6\right ) \int \frac {x}{\log ^2(\log (3+x))} \, dx+\left (6 e^6\right ) \operatorname {Subst}\left (\int \frac {1}{\log (x) \log ^3(\log (x))} \, dx,x,3+x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 50, normalized size = 1.67 \begin {gather*} \frac {1}{9} \left (9 e^{16-x}+9 x+x^2+\frac {9 e^6 x^2}{\log ^2(\log (3+x))}-\frac {6 e^3 x^2}{\log (\log (3+x))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-18*E^6*x^2 + (6*E^3*x^2 + E^6*(54*x + 18*x^2)*Log[3 + x])*Log[Log[3 + x]] + E^3*(-36*x - 12*x^2)*L
og[3 + x]*Log[Log[3 + x]]^2 + (27 + E^(16 - x)*(-27 - 9*x) + 15*x + 2*x^2)*Log[3 + x]*Log[Log[3 + x]]^3)/((27
+ 9*x)*Log[3 + x]*Log[Log[3 + x]]^3),x]

[Out]

(9*E^(16 - x) + 9*x + x^2 + (9*E^6*x^2)/Log[Log[3 + x]]^2 - (6*E^3*x^2)/Log[Log[3 + x]])/9

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fricas [B]  time = 0.75, size = 53, normalized size = 1.77 \begin {gather*} -\frac {6 \, x^{2} e^{3} \log \left (\log \left (x + 3\right )\right ) - 9 \, x^{2} e^{6} - {\left (x^{2} + 9 \, x + 9 \, e^{\left (-x + 16\right )}\right )} \log \left (\log \left (x + 3\right )\right )^{2}}{9 \, \log \left (\log \left (x + 3\right )\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-9*x-27)*exp(16-x)+2*x^2+15*x+27)*log(3+x)*log(log(3+x))^3+(-12*x^2-36*x)*exp(3)*log(3+x)*log(log
(3+x))^2+((18*x^2+54*x)*exp(3)^2*log(3+x)+6*x^2*exp(3))*log(log(3+x))-18*x^2*exp(3)^2)/(9*x+27)/log(3+x)/log(l
og(3+x))^3,x, algorithm="fricas")

[Out]

-1/9*(6*x^2*e^3*log(log(x + 3)) - 9*x^2*e^6 - (x^2 + 9*x + 9*e^(-x + 16))*log(log(x + 3))^2)/log(log(x + 3))^2

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giac [B]  time = 0.27, size = 66, normalized size = 2.20 \begin {gather*} -\frac {6 \, x^{2} e^{3} \log \left (\log \left (x + 3\right )\right ) - x^{2} \log \left (\log \left (x + 3\right )\right )^{2} - 9 \, x^{2} e^{6} - 9 \, x \log \left (\log \left (x + 3\right )\right )^{2} - 9 \, e^{\left (-x + 16\right )} \log \left (\log \left (x + 3\right )\right )^{2}}{9 \, \log \left (\log \left (x + 3\right )\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-9*x-27)*exp(16-x)+2*x^2+15*x+27)*log(3+x)*log(log(3+x))^3+(-12*x^2-36*x)*exp(3)*log(3+x)*log(log
(3+x))^2+((18*x^2+54*x)*exp(3)^2*log(3+x)+6*x^2*exp(3))*log(log(3+x))-18*x^2*exp(3)^2)/(9*x+27)/log(3+x)/log(l
og(3+x))^3,x, algorithm="giac")

[Out]

-1/9*(6*x^2*e^3*log(log(x + 3)) - x^2*log(log(x + 3))^2 - 9*x^2*e^6 - 9*x*log(log(x + 3))^2 - 9*e^(-x + 16)*lo
g(log(x + 3))^2)/log(log(x + 3))^2

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maple [A]  time = 0.11, size = 40, normalized size = 1.33

method	result	size
risch	\(\frac {x^{2}}{9}+x +{\mathrm e}^{16-x}+\frac {x^{2} {\mathrm e}^{3} \left (3 \,{\mathrm e}^{3}-2 \ln \left (\ln \left (3+x \right )\right )\right )}{3 \ln \left (\ln \left (3+x \right )\right )^{2}}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-9*x-27)*exp(16-x)+2*x^2+15*x+27)*ln(3+x)*ln(ln(3+x))^3+(-12*x^2-36*x)*exp(3)*ln(3+x)*ln(ln(3+x))^2+((1
8*x^2+54*x)*exp(3)^2*ln(3+x)+6*x^2*exp(3))*ln(ln(3+x))-18*x^2*exp(3)^2)/(9*x+27)/ln(3+x)/ln(ln(3+x))^3,x,metho
d=_RETURNVERBOSE)

[Out]

1/9*x^2+x+exp(16-x)+1/3*x^2*exp(3)*(3*exp(3)-2*ln(ln(3+x)))/ln(ln(3+x))^2

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maxima [B]  time = 0.41, size = 61, normalized size = 2.03 \begin {gather*} -\frac {{\left (6 \, x^{2} e^{\left (x + 3\right )} \log \left (\log \left (x + 3\right )\right ) - 9 \, x^{2} e^{\left (x + 6\right )} - {\left ({\left (x^{2} + 9 \, x\right )} e^{x} + 9 \, e^{16}\right )} \log \left (\log \left (x + 3\right )\right )^{2}\right )} e^{\left (-x\right )}}{9 \, \log \left (\log \left (x + 3\right )\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-9*x-27)*exp(16-x)+2*x^2+15*x+27)*log(3+x)*log(log(3+x))^3+(-12*x^2-36*x)*exp(3)*log(3+x)*log(log
(3+x))^2+((18*x^2+54*x)*exp(3)^2*log(3+x)+6*x^2*exp(3))*log(log(3+x))-18*x^2*exp(3)^2)/(9*x+27)/log(3+x)/log(l
og(3+x))^3,x, algorithm="maxima")

[Out]

-1/9*(6*x^2*e^(x + 3)*log(log(x + 3)) - 9*x^2*e^(x + 6) - ((x^2 + 9*x)*e^x + 9*e^16)*log(log(x + 3))^2)*e^(-x)
/log(log(x + 3))^2

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mupad [B]  time = 4.30, size = 281, normalized size = 9.37 \begin {gather*} x+{\mathrm {e}}^{16-x}+\frac {x^2}{9}+6\,{\ln \left (x+3\right )}^2\,{\mathrm {e}}^3\,\ln \left (\ln \left (x+3\right )\right )-\frac {2\,x^2\,{\mathrm {e}}^3}{3\,\ln \left (\ln \left (x+3\right )\right )}+\frac {x^2\,{\mathrm {e}}^6}{{\ln \left (\ln \left (x+3\right )\right )}^2}+2\,x\,\ln \left (x+3\right )\,{\mathrm {e}}^3\,\ln \left (\ln \left (x+3\right )\right )+6\,x\,{\ln \left (x+3\right )}^2\,{\mathrm {e}}^3\,\ln \left (\ln \left (x+3\right )\right )+\frac {2\,x^2\,\ln \left (x+3\right )\,{\mathrm {e}}^3\,\ln \left (\ln \left (x+3\right )\right )}{3}-\frac {54\,{\ln \left (x+3\right )}^2\,{\mathrm {e}}^3\,\ln \left (\ln \left (x+3\right )\right )}{3\,x+9}+\frac {4\,x^2\,{\ln \left (x+3\right )}^2\,{\mathrm {e}}^3\,\ln \left (\ln \left (x+3\right )\right )}{3}-\frac {72\,x\,{\ln \left (x+3\right )}^2\,{\mathrm {e}}^3\,\ln \left (\ln \left (x+3\right )\right )}{3\,x+9}-\frac {12\,x^2\,\ln \left (x+3\right )\,{\mathrm {e}}^3\,\ln \left (\ln \left (x+3\right )\right )}{3\,x+9}-\frac {2\,x^3\,\ln \left (x+3\right )\,{\mathrm {e}}^3\,\ln \left (\ln \left (x+3\right )\right )}{3\,x+9}-\frac {30\,x^2\,{\ln \left (x+3\right )}^2\,{\mathrm {e}}^3\,\ln \left (\ln \left (x+3\right )\right )}{3\,x+9}-\frac {4\,x^3\,{\ln \left (x+3\right )}^2\,{\mathrm {e}}^3\,\ln \left (\ln \left (x+3\right )\right )}{3\,x+9}-\frac {18\,x\,\ln \left (x+3\right )\,{\mathrm {e}}^3\,\ln \left (\ln \left (x+3\right )\right )}{3\,x+9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(x + 3))*(6*x^2*exp(3) + log(x + 3)*exp(6)*(54*x + 18*x^2)) - 18*x^2*exp(6) + log(x + 3)*log(log(x
+ 3))^3*(15*x - exp(16 - x)*(9*x + 27) + 2*x^2 + 27) - log(x + 3)*exp(3)*log(log(x + 3))^2*(36*x + 12*x^2))/(
log(x + 3)*log(log(x + 3))^3*(9*x + 27)),x)

[Out]

x + exp(16 - x) + x^2/9 + 6*log(x + 3)^2*exp(3)*log(log(x + 3)) - (2*x^2*exp(3))/(3*log(log(x + 3))) + (x^2*ex
p(6))/log(log(x + 3))^2 + 2*x*log(x + 3)*exp(3)*log(log(x + 3)) + 6*x*log(x + 3)^2*exp(3)*log(log(x + 3)) + (2
*x^2*log(x + 3)*exp(3)*log(log(x + 3)))/3 - (54*log(x + 3)^2*exp(3)*log(log(x + 3)))/(3*x + 9) + (4*x^2*log(x
+ 3)^2*exp(3)*log(log(x + 3)))/3 - (72*x*log(x + 3)^2*exp(3)*log(log(x + 3)))/(3*x + 9) - (12*x^2*log(x + 3)*e
xp(3)*log(log(x + 3)))/(3*x + 9) - (2*x^3*log(x + 3)*exp(3)*log(log(x + 3)))/(3*x + 9) - (30*x^2*log(x + 3)^2*
exp(3)*log(log(x + 3)))/(3*x + 9) - (4*x^3*log(x + 3)^2*exp(3)*log(log(x + 3)))/(3*x + 9) - (18*x*log(x + 3)*e
xp(3)*log(log(x + 3)))/(3*x + 9)

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sympy [A]  time = 0.54, size = 44, normalized size = 1.47 \begin {gather*} \frac {x^{2}}{9} + x + \frac {- 2 x^{2} e^{3} \log {\left (\log {\left (x + 3 \right )} \right )} + 3 x^{2} e^{6}}{3 \log {\left (\log {\left (x + 3 \right )} \right )}^{2}} + e^{16 - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-9*x-27)*exp(16-x)+2*x**2+15*x+27)*ln(3+x)*ln(ln(3+x))**3+(-12*x**2-36*x)*exp(3)*ln(3+x)*ln(ln(3+
x))**2+((18*x**2+54*x)*exp(3)**2*ln(3+x)+6*x**2*exp(3))*ln(ln(3+x))-18*x**2*exp(3)**2)/(9*x+27)/ln(3+x)/ln(ln(
3+x))**3,x)

[Out]

x**2/9 + x + (-2*x**2*exp(3)*log(log(x + 3)) + 3*x**2*exp(6))/(3*log(log(x + 3))**2) + exp(16 - x)

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