3.102.56 \(\int \frac {e^2 (-4 e^{4+\frac {e^2}{8 x^2}}+288 e^{-2+2 x} x^3+e^{\frac {e^2}{16 x^2}} (-12 e^{2+x}+96 e^x x^3))}{144 x^3} \, dx\)

Optimal. Leaf size=24 \[ \left (\frac {1}{3} e^{2+\frac {e^2}{16 x^2}}+e^x\right )^2 \]

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Rubi [A]  time = 0.23, antiderivative size = 43, normalized size of antiderivative = 1.79, number of steps used = 6, number of rules used = 5, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {12, 14, 2194, 2209, 6706} \begin {gather*} \frac {1}{9} e^{\frac {e^2}{8 x^2}+4}+\frac {2}{3} e^{\frac {e^2}{16 x^2}+x+2}+e^{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^2*(-4*E^(4 + E^2/(8*x^2)) + 288*E^(-2 + 2*x)*x^3 + E^(E^2/(16*x^2))*(-12*E^(2 + x) + 96*E^x*x^3)))/(144
*x^3),x]

[Out]

E^(4 + E^2/(8*x^2))/9 + E^(2*x) + (2*E^(2 + E^2/(16*x^2) + x))/3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{144} e^2 \int \frac {-4 e^{4+\frac {e^2}{8 x^2}}+288 e^{-2+2 x} x^3+e^{\frac {e^2}{16 x^2}} \left (-12 e^{2+x}+96 e^x x^3\right )}{x^3} \, dx\\ &=\frac {1}{144} e^2 \int \left (288 e^{-2+2 x}-\frac {4 e^{4+\frac {e^2}{8 x^2}}}{x^3}+\frac {12 e^{\frac {e^2}{16 x^2}+x} \left (-e^2+8 x^3\right )}{x^3}\right ) \, dx\\ &=-\left (\frac {1}{36} e^2 \int \frac {e^{4+\frac {e^2}{8 x^2}}}{x^3} \, dx\right )+\frac {1}{12} e^2 \int \frac {e^{\frac {e^2}{16 x^2}+x} \left (-e^2+8 x^3\right )}{x^3} \, dx+\left (2 e^2\right ) \int e^{-2+2 x} \, dx\\ &=\frac {1}{9} e^{4+\frac {e^2}{8 x^2}}+e^{2 x}+\frac {2}{3} e^{2+\frac {e^2}{16 x^2}+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 26, normalized size = 1.08 \begin {gather*} \frac {1}{9} \left (e^{2+\frac {e^2}{16 x^2}}+3 e^x\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(-4*E^(4 + E^2/(8*x^2)) + 288*E^(-2 + 2*x)*x^3 + E^(E^2/(16*x^2))*(-12*E^(2 + x) + 96*E^x*x^3))
)/(144*x^3),x]

[Out]

(E^(2 + E^2/(16*x^2)) + 3*E^x)^2/9

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fricas [B]  time = 0.56, size = 58, normalized size = 2.42 \begin {gather*} \frac {256 \, x^{4} e^{\left (\frac {e^{2}}{8 \, x^{2}} + 4\right )} + 96 \, x^{2} e^{\left (x + \frac {e^{2}}{16 \, x^{2}} + \log \left (16 \, x^{2}\right ) + 2\right )} + 9 \, e^{\left (2 \, x + 2 \, \log \left (16 \, x^{2}\right )\right )}}{2304 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-4*exp(2)^2*exp(1/exp(log(16*x^2)-2))^2+(6*x*exp(2)*exp(x)*exp(log(16*x^2)-2)-12*exp(2)*exp(x))
*exp(1/exp(log(16*x^2)-2))+18*x*exp(x)^2*exp(log(16*x^2)-2))/x/exp(log(16*x^2)-2),x, algorithm="fricas")

[Out]

1/2304*(256*x^4*e^(1/8*e^2/x^2 + 4) + 96*x^2*e^(x + 1/16*e^2/x^2 + log(16*x^2) + 2) + 9*e^(2*x + 2*log(16*x^2)
))/x^4

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giac [A]  time = 0.16, size = 38, normalized size = 1.58 \begin {gather*} e^{\left (2 \, x\right )} + \frac {2}{3} \, e^{\left (\frac {16 \, x^{3} + 32 \, x^{2} + e^{2}}{16 \, x^{2}}\right )} + \frac {1}{9} \, e^{\left (\frac {e^{2}}{8 \, x^{2}} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-4*exp(2)^2*exp(1/exp(log(16*x^2)-2))^2+(6*x*exp(2)*exp(x)*exp(log(16*x^2)-2)-12*exp(2)*exp(x))
*exp(1/exp(log(16*x^2)-2))+18*x*exp(x)^2*exp(log(16*x^2)-2))/x/exp(log(16*x^2)-2),x, algorithm="giac")

[Out]

e^(2*x) + 2/3*e^(1/16*(16*x^3 + 32*x^2 + e^2)/x^2) + 1/9*e^(1/8*e^2/x^2 + 4)

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maple [B]  time = 0.45, size = 52, normalized size = 2.17

method	result	size
default	\(\frac {2 \,{\mathrm e}^{2} {\mathrm e}^{x} {\mathrm e}^{\frac {{\mathrm e}^{2}}{16 x^{2}}}}{3}+{\mathrm e}^{2 x}+\frac {{\mathrm e}^{4} {\mathrm e}^{\frac {{\mathrm e}^{2}}{8 x^{2}}}}{9}\)	\(52\)
risch	\(\frac {{\mathrm e}^{-\frac {{\mathrm e}^{2} {\mathrm e}^{\frac {i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}} {\mathrm e}^{\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}}{2}}-32 x^{2}}{8 x^{2}}}}{9}+\frac {2 \,{\mathrm e}^{-\frac {{\mathrm e}^{2} {\mathrm e}^{\frac {i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}} {\mathrm e}^{\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}}{2}}-16 x^{3}-32 x^{2}}{16 x^{2}}}}{3}+{\mathrm e}^{2 x}\)	\(111\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*(-4*exp(2)^2*exp(1/exp(ln(16*x^2)-2))^2+(6*x*exp(2)*exp(x)*exp(ln(16*x^2)-2)-12*exp(2)*exp(x))*exp(1/e
xp(ln(16*x^2)-2))+18*x*exp(x)^2*exp(ln(16*x^2)-2))/x/exp(ln(16*x^2)-2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.42, size = 39, normalized size = 1.62 \begin {gather*} \frac {1}{9} \, {\left (3 \, {\left (3 \, e^{\left (2 \, x\right )} + 2 \, e^{\left (x + \frac {e^{2}}{16 \, x^{2}} + 2\right )}\right )} e^{\left (-2\right )} + e^{\left (\frac {e^{2}}{8 \, x^{2}} + 2\right )}\right )} e^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-4*exp(2)^2*exp(1/exp(log(16*x^2)-2))^2+(6*x*exp(2)*exp(x)*exp(log(16*x^2)-2)-12*exp(2)*exp(x))
*exp(1/exp(log(16*x^2)-2))+18*x*exp(x)^2*exp(log(16*x^2)-2))/x/exp(log(16*x^2)-2),x, algorithm="maxima")

[Out]

1/9*(3*(3*e^(2*x) + 2*e^(x + 1/16*e^2/x^2 + 2))*e^(-2) + e^(1/8*e^2/x^2 + 2))*e^2

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mupad [B]  time = 7.93, size = 23, normalized size = 0.96 \begin {gather*} \frac {{\mathrm {e}}^{-4}\,{\left (3\,{\mathrm {e}}^{x+2}+{\mathrm {e}}^{\frac {{\mathrm {e}}^2}{16\,x^2}+4}\right )}^2}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2 - log(16*x^2))*((exp(exp(2 - log(16*x^2)))*(12*exp(2)*exp(x) - 6*x*exp(2)*exp(log(16*x^2) - 2)*exp
(x)))/9 + (4*exp(4)*exp(2*exp(2 - log(16*x^2))))/9 - 2*x*exp(2*x)*exp(log(16*x^2) - 2)))/x,x)

[Out]

(exp(-4)*(3*exp(x + 2) + exp(exp(2)/(16*x^2) + 4))^2)/9

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sympy [B]  time = 2.54, size = 39, normalized size = 1.62 \begin {gather*} e^{2 x} + \frac {2 e^{2} e^{x} e^{\frac {e^{2}}{16 x^{2}}}}{3} + \frac {e^{4} e^{\frac {e^{2}}{8 x^{2}}}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-4*exp(2)**2*exp(1/exp(ln(16*x**2)-2))**2+(6*x*exp(2)*exp(x)*exp(ln(16*x**2)-2)-12*exp(2)*exp(x
))*exp(1/exp(ln(16*x**2)-2))+18*x*exp(x)**2*exp(ln(16*x**2)-2))/x/exp(ln(16*x**2)-2),x)

[Out]

exp(2*x) + 2*exp(2)*exp(x)*exp(exp(2)/(16*x**2))/3 + exp(4)*exp(exp(2)/(8*x**2))/9

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