3.64.6 \(\int \frac {192 e^{6+4 e^3}+192 e^{6+2 x}+e^{3+x} (-48-4 x^2) \log (3)+3 \log ^2(3)+e^{2 e^3} (-384 e^{6+x}+48 e^3 \log (3))}{64 e^{6+4 e^3} x^2+64 e^{6+2 x} x^2-16 e^{3+x} x^2 \log (3)+x^2 \log ^2(3)+e^{2 e^3} (-128 e^{6+x} x^2+16 e^3 x^2 \log (3))} \, dx\)

Optimal. Leaf size=36 \[ \frac {-3+\frac {x}{2-\frac {\log (3)}{4 e^3 \left (-e^{2 e^3}+e^x\right )}}}{x} \]

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Rubi [B]  time = 1.93, antiderivative size = 164, normalized size of antiderivative = 4.56, number of steps used = 11, number of rules used = 8, integrand size = 144, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {6, 6741, 6742, 2282, 44, 36, 29, 31} \begin {gather*} -\frac {3}{x}+\frac {x \log (3)}{16 e^{3+2 e^3}+\log (9)}-\frac {x \log (3)}{2 \left (8 e^{3+2 e^3}+\log (3)\right )}+\frac {\log (3) \log \left (-8 e^{x+3}+8 e^{3+2 e^3}+\log (3)\right )}{2 \left (8 e^{3+2 e^3}+\log (3)\right )}-\frac {\log (3) \log \left (-16 e^{x+3}+16 e^{3+2 e^3}+\log (9)\right )}{16 e^{3+2 e^3}+\log (9)}-\frac {\log (3)}{2 \left (-8 e^{x+3}+8 e^{3+2 e^3}+\log (3)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(192*E^(6 + 4*E^3) + 192*E^(6 + 2*x) + E^(3 + x)*(-48 - 4*x^2)*Log[3] + 3*Log[3]^2 + E^(2*E^3)*(-384*E^(6
+ x) + 48*E^3*Log[3]))/(64*E^(6 + 4*E^3)*x^2 + 64*E^(6 + 2*x)*x^2 - 16*E^(3 + x)*x^2*Log[3] + x^2*Log[3]^2 + E
^(2*E^3)*(-128*E^(6 + x)*x^2 + 16*E^3*x^2*Log[3])),x]

[Out]

-3/x - (x*Log[3])/(2*(8*E^(3 + 2*E^3) + Log[3])) - Log[3]/(2*(8*E^(3 + 2*E^3) - 8*E^(3 + x) + Log[3])) + (x*Lo
g[3])/(16*E^(3 + 2*E^3) + Log[9]) + (Log[3]*Log[8*E^(3 + 2*E^3) - 8*E^(3 + x) + Log[3]])/(2*(8*E^(3 + 2*E^3) +
Log[3])) - (Log[3]*Log[16*E^(3 + 2*E^3) - 16*E^(3 + x) + Log[9]])/(16*E^(3 + 2*E^3) + Log[9])

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {192 e^{6+4 e^3}+192 e^{6+2 x}+e^{3+x} \left (-48-4 x^2\right ) \log (3)+3 \log ^2(3)+e^{2 e^3} \left (-384 e^{6+x}+48 e^3 \log (3)\right )}{64 e^{6+2 x} x^2-16 e^{3+x} x^2 \log (3)+e^{2 e^3} \left (-128 e^{6+x} x^2+16 e^3 x^2 \log (3)\right )+x^2 \left (64 e^{6+4 e^3}+\log ^2(3)\right )} \, dx\\ &=\int \frac {192 e^{6+2 x}+e^{3+x} \left (-48-4 x^2\right ) \log (3)+e^{2 e^3} \left (-384 e^{6+x}+48 e^3 \log (3)\right )+192 e^{6+4 e^3} \left (1+\frac {1}{64} e^{-6-4 e^3} \log ^2(3)\right )}{x^2 \left (8 e^{3+x}-8 e^{3+2 e^3} \left (1+\frac {1}{8} e^{-3-2 e^3} \log (3)\right )\right )^2} \, dx\\ &=\int \left (\frac {3}{x^2}+\frac {\left (-8 e^{3+2 e^3}-\log (3)\right ) \log (3)}{2 \left (8 e^{3+x}-8 e^{3+2 e^3} \left (1+\frac {1}{8} e^{-3-2 e^3} \log (3)\right )\right )^2}+\frac {\log (3)}{-16 e^{3+x}+16 e^{3+2 e^3} \left (1+\frac {1}{8} e^{-3-2 e^3} \log (3)\right )}\right ) \, dx\\ &=-\frac {3}{x}+\log (3) \int \frac {1}{-16 e^{3+x}+16 e^{3+2 e^3} \left (1+\frac {1}{8} e^{-3-2 e^3} \log (3)\right )} \, dx-\frac {1}{2} \left (\log (3) \left (8 e^{3+2 e^3}+\log (3)\right )\right ) \int \frac {1}{\left (8 e^{3+x}-8 e^{3+2 e^3} \left (1+\frac {1}{8} e^{-3-2 e^3} \log (3)\right )\right )^2} \, dx\\ &=-\frac {3}{x}+\log (3) \operatorname {Subst}\left (\int \frac {1}{x \left (16 e^{3+2 e^3}-16 x+\log (9)\right )} \, dx,x,e^{3+x}\right )-\frac {1}{2} \left (\log (3) \left (8 e^{3+2 e^3}+\log (3)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (8 e^{3+2 e^3}-8 x+\log (3)\right )^2} \, dx,x,e^{3+x}\right )\\ &=-\frac {3}{x}-\frac {1}{2} \left (\log (3) \left (8 e^{3+2 e^3}+\log (3)\right )\right ) \operatorname {Subst}\left (\int \left (\frac {1}{x \left (8 e^{3+2 e^3}+\log (3)\right )^2}+\frac {8}{\left (8 e^{3+2 e^3}+\log (3)\right ) \left (8 e^{3+2 e^3}-8 x+\log (3)\right )^2}+\frac {8}{\left (8 e^{3+2 e^3}+\log (3)\right )^2 \left (8 e^{3+2 e^3}-8 x+\log (3)\right )}\right ) \, dx,x,e^{3+x}\right )+\frac {\log (3) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{3+x}\right )}{16 e^{3+2 e^3}+\log (9)}+\frac {(16 \log (3)) \operatorname {Subst}\left (\int \frac {1}{16 e^{3+2 e^3}-16 x+\log (9)} \, dx,x,e^{3+x}\right )}{16 e^{3+2 e^3}+\log (9)}\\ &=-\frac {3}{x}-\frac {x \log (3)}{2 \left (8 e^{3+2 e^3}+\log (3)\right )}-\frac {\log (3)}{2 \left (8 e^{3+2 e^3}-8 e^{3+x}+\log (3)\right )}+\frac {x \log (3)}{16 e^{3+2 e^3}+\log (9)}+\frac {\log (3) \log \left (8 e^{3+2 e^3}-8 e^{3+x}+\log (3)\right )}{2 \left (8 e^{3+2 e^3}+\log (3)\right )}-\frac {\log (3) \log \left (16 e^{3+2 e^3}-16 e^{3+x}+\log (9)\right )}{16 e^{3+2 e^3}+\log (9)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 33, normalized size = 0.92 \begin {gather*} -\frac {3}{x}-\frac {\log (3)}{16 e^{3+2 e^3}-16 e^{3+x}+\log (9)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(192*E^(6 + 4*E^3) + 192*E^(6 + 2*x) + E^(3 + x)*(-48 - 4*x^2)*Log[3] + 3*Log[3]^2 + E^(2*E^3)*(-384
*E^(6 + x) + 48*E^3*Log[3]))/(64*E^(6 + 4*E^3)*x^2 + 64*E^(6 + 2*x)*x^2 - 16*E^(3 + x)*x^2*Log[3] + x^2*Log[3]
^2 + E^(2*E^3)*(-128*E^(6 + x)*x^2 + 16*E^3*x^2*Log[3])),x]

[Out]

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

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fricas [A]  time = 0.56, size = 52, normalized size = 1.44 \begin {gather*} -\frac {{\left (x + 6\right )} e^{3} \log \relax (3) - 48 \, e^{\left (x + 6\right )} + 48 \, e^{\left (2 \, e^{3} + 6\right )}}{2 \, {\left (x e^{3} \log \relax (3) - 8 \, x e^{\left (x + 6\right )} + 8 \, x e^{\left (2 \, e^{3} + 6\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((192*exp(3)^2*exp(exp(3))^4+(-384*exp(3)^2*exp(x)+48*exp(3)*log(3))*exp(exp(3))^2+192*exp(3)^2*exp(x
)^2+(-4*x^2-48)*exp(3)*log(3)*exp(x)+3*log(3)^2)/(64*x^2*exp(3)^2*exp(exp(3))^4+(-128*x^2*exp(3)^2*exp(x)+16*x
^2*exp(3)*log(3))*exp(exp(3))^2+64*x^2*exp(3)^2*exp(x)^2-16*x^2*exp(3)*log(3)*exp(x)+x^2*log(3)^2),x, algorith
m="fricas")

[Out]

-1/2*((x + 6)*e^3*log(3) - 48*e^(x + 6) + 48*e^(2*e^3 + 6))/(x*e^3*log(3) - 8*x*e^(x + 6) + 8*x*e^(2*e^3 + 6))

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giac [A]  time = 0.14, size = 51, normalized size = 1.42 \begin {gather*} \frac {x \log \relax (3) - 48 \, e^{\left (x + 3\right )} + 48 \, e^{\left (2 \, e^{3} + 3\right )} + 6 \, \log \relax (3)}{2 \, {\left (8 \, x e^{\left (x + 3\right )} - 8 \, x e^{\left (2 \, e^{3} + 3\right )} - x \log \relax (3)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((192*exp(3)^2*exp(exp(3))^4+(-384*exp(3)^2*exp(x)+48*exp(3)*log(3))*exp(exp(3))^2+192*exp(3)^2*exp(x
)^2+(-4*x^2-48)*exp(3)*log(3)*exp(x)+3*log(3)^2)/(64*x^2*exp(3)^2*exp(exp(3))^4+(-128*x^2*exp(3)^2*exp(x)+16*x
^2*exp(3)*log(3))*exp(exp(3))^2+64*x^2*exp(3)^2*exp(x)^2-16*x^2*exp(3)*log(3)*exp(x)+x^2*log(3)^2),x, algorith
m="giac")

[Out]

1/2*(x*log(3) - 48*e^(x + 3) + 48*e^(2*e^3 + 3) + 6*log(3))/(8*x*e^(x + 3) - 8*x*e^(2*e^3 + 3) - x*log(3))

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maple [A]  time = 0.38, size = 31, normalized size = 0.86

method	result	size
risch	\(-\frac {3}{x}-\frac {\ln \relax (3)}{2 \left (8 \,{\mathrm e}^{3+2 \,{\mathrm e}^{3}}-8 \,{\mathrm e}^{3+x}+\ln \relax (3)\right )}\)	\(31\)
norman	\(\frac {-\frac {x \ln \relax (3)}{2}+24 \,{\mathrm e}^{x} {\mathrm e}^{3}-24 \,{\mathrm e}^{3} {\mathrm e}^{2 \,{\mathrm e}^{3}}-3 \ln \relax (3)}{x \left (8 \,{\mathrm e}^{3} {\mathrm e}^{2 \,{\mathrm e}^{3}}-8 \,{\mathrm e}^{x} {\mathrm e}^{3}+\ln \relax (3)\right )}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((192*exp(3)^2*exp(exp(3))^4+(-384*exp(3)^2*exp(x)+48*exp(3)*ln(3))*exp(exp(3))^2+192*exp(3)^2*exp(x)^2+(-4
*x^2-48)*exp(3)*ln(3)*exp(x)+3*ln(3)^2)/(64*x^2*exp(3)^2*exp(exp(3))^4+(-128*x^2*exp(3)^2*exp(x)+16*x^2*exp(3)
*ln(3))*exp(exp(3))^2+64*x^2*exp(3)^2*exp(x)^2-16*x^2*exp(3)*ln(3)*exp(x)+x^2*ln(3)^2),x,method=_RETURNVERBOSE
)

[Out]

-3/x-1/2*ln(3)/(8*exp(3+2*exp(3))-8*exp(3+x)+ln(3))

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maxima [A]  time = 0.51, size = 50, normalized size = 1.39 \begin {gather*} -\frac {x \log \relax (3) - 48 \, e^{\left (x + 3\right )} + 48 \, e^{\left (2 \, e^{3} + 3\right )} + 6 \, \log \relax (3)}{2 \, {\left (x {\left (8 \, e^{\left (2 \, e^{3} + 3\right )} + \log \relax (3)\right )} - 8 \, x e^{\left (x + 3\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((192*exp(3)^2*exp(exp(3))^4+(-384*exp(3)^2*exp(x)+48*exp(3)*log(3))*exp(exp(3))^2+192*exp(3)^2*exp(x
)^2+(-4*x^2-48)*exp(3)*log(3)*exp(x)+3*log(3)^2)/(64*x^2*exp(3)^2*exp(exp(3))^4+(-128*x^2*exp(3)^2*exp(x)+16*x
^2*exp(3)*log(3))*exp(exp(3))^2+64*x^2*exp(3)^2*exp(x)^2-16*x^2*exp(3)*log(3)*exp(x)+x^2*log(3)^2),x, algorith
m="maxima")

[Out]

-1/2*(x*log(3) - 48*e^(x + 3) + 48*e^(2*e^3 + 3) + 6*log(3))/(x*(8*e^(2*e^3 + 3) + log(3)) - 8*x*e^(x + 3))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {192\,{\mathrm {e}}^{4\,{\mathrm {e}}^3+6}+192\,{\mathrm {e}}^{2\,x+6}-{\mathrm {e}}^{2\,{\mathrm {e}}^3}\,\left (384\,{\mathrm {e}}^{x+6}-48\,{\mathrm {e}}^3\,\ln \relax (3)\right )+3\,{\ln \relax (3)}^2-{\mathrm {e}}^{x+3}\,\ln \relax (3)\,\left (4\,x^2+48\right )}{x^2\,{\ln \relax (3)}^2-{\mathrm {e}}^{2\,{\mathrm {e}}^3}\,\left (128\,x^2\,{\mathrm {e}}^{x+6}-16\,x^2\,{\mathrm {e}}^3\,\ln \relax (3)\right )+64\,x^2\,{\mathrm {e}}^{4\,{\mathrm {e}}^3+6}+64\,x^2\,{\mathrm {e}}^{2\,x+6}-16\,x^2\,{\mathrm {e}}^{x+3}\,\ln \relax (3)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((192*exp(2*x)*exp(6) + exp(2*exp(3))*(48*exp(3)*log(3) - 384*exp(6)*exp(x)) + 3*log(3)^2 + 192*exp(4*exp(3
))*exp(6) - exp(3)*exp(x)*log(3)*(4*x^2 + 48))/(x^2*log(3)^2 + exp(2*exp(3))*(16*x^2*exp(3)*log(3) - 128*x^2*e
xp(6)*exp(x)) + 64*x^2*exp(4*exp(3))*exp(6) + 64*x^2*exp(2*x)*exp(6) - 16*x^2*exp(3)*exp(x)*log(3)),x)

[Out]

int((192*exp(4*exp(3) + 6) + 192*exp(2*x + 6) - exp(2*exp(3))*(384*exp(x + 6) - 48*exp(3)*log(3)) + 3*log(3)^2
- exp(x + 3)*log(3)*(4*x^2 + 48))/(x^2*log(3)^2 - exp(2*exp(3))*(128*x^2*exp(x + 6) - 16*x^2*exp(3)*log(3)) +
64*x^2*exp(4*exp(3) + 6) + 64*x^2*exp(2*x + 6) - 16*x^2*exp(x + 3)*log(3)), x)

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sympy [A]  time = 0.18, size = 31, normalized size = 0.86 \begin {gather*} \frac {\log {\relax (3 )}}{16 e^{3} e^{x} - 16 e^{3} e^{2 e^{3}} - 2 \log {\relax (3 )}} - \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((192*exp(3)**2*exp(exp(3))**4+(-384*exp(3)**2*exp(x)+48*exp(3)*ln(3))*exp(exp(3))**2+192*exp(3)**2*e
xp(x)**2+(-4*x**2-48)*exp(3)*ln(3)*exp(x)+3*ln(3)**2)/(64*x**2*exp(3)**2*exp(exp(3))**4+(-128*x**2*exp(3)**2*e
xp(x)+16*x**2*exp(3)*ln(3))*exp(exp(3))**2+64*x**2*exp(3)**2*exp(x)**2-16*x**2*exp(3)*ln(3)*exp(x)+x**2*ln(3)*
*2),x)

[Out]

log(3)/(16*exp(3)*exp(x) - 16*exp(3)*exp(2*exp(3)) - 2*log(3)) - 3/x

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