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

Optimal. Leaf size=29 \[ -8-e^{3 (-5+x)}+\frac {1}{3} e^{\frac {-e+x+x^2}{x}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 25, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 14, 2194, 6706} \begin {gather*} \frac {1}{3} e^{x-\frac {e}{x}+1}-e^{3 x-15} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(1 - E/x + x)/3 - E^(-15 + 3*x)

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 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}{3} \int \frac {-9 e^{-15+3 x} x^2+e^{\frac {-e+x+x^2}{x}} \left (e+x^2\right )}{x^2} \, dx\\ &=\frac {1}{3} \int \left (-9 e^{-15+3 x}+\frac {e^{1-\frac {e}{x}+x} \left (e+x^2\right )}{x^2}\right ) \, dx\\ &=\frac {1}{3} \int \frac {e^{1-\frac {e}{x}+x} \left (e+x^2\right )}{x^2} \, dx-3 \int e^{-15+3 x} \, dx\\ &=\frac {1}{3} e^{1-\frac {e}{x}+x}-e^{-15+3 x}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 25, normalized size = 0.86 \begin {gather*} \frac {1}{3} e^{1-\frac {e}{x}+x}-e^{-15+3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(1 - E/x + x)/3 - E^(-15 + 3*x)

________________________________________________________________________________________

fricas [A]  time = 1.03, size = 25, normalized size = 0.86 \begin {gather*} -e^{\left (3 \, x - 15\right )} + \frac {1}{3} \, e^{\left (\frac {x^{2} + x - e}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-9*x^2*exp(3*x-15)+(exp(1)+x^2)*exp((-exp(1)+x^2+x)/x))/x^2,x, algorithm="fricas")

[Out]

-e^(3*x - 15) + 1/3*e^((x^2 + x - e)/x)

________________________________________________________________________________________

giac [A]  time = 0.35, size = 29, normalized size = 1.00 \begin {gather*} -\frac {1}{3} \, {\left (3 \, e^{\left (3 \, x\right )} - e^{\left (\frac {x^{2} + x - e}{x} + 15\right )}\right )} e^{\left (-15\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-9*x^2*exp(3*x-15)+(exp(1)+x^2)*exp((-exp(1)+x^2+x)/x))/x^2,x, algorithm="giac")

[Out]

-1/3*(3*e^(3*x) - e^((x^2 + x - e)/x + 15))*e^(-15)

________________________________________________________________________________________

maple [A]  time = 0.07, size = 29, normalized size = 1.00

method	result	size
risch	\(-{\mathrm e}^{3 x -15}+\frac {{\mathrm e}^{-\frac {-x^{2}+{\mathrm e}-x}{x}}}{3}\)	\(29\)
norman	\(\frac {\frac {x \,{\mathrm e}^{\frac {-{\mathrm e}+x^{2}+x}{x}}}{3}-x \,{\mathrm e}^{3 x -15}}{x}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(-9*x^2*exp(3*x-15)+(exp(1)+x^2)*exp((-exp(1)+x^2+x)/x))/x^2,x,method=_RETURNVERBOSE)

[Out]

-exp(3*x-15)+1/3*exp(-(-x^2+exp(1)-x)/x)

________________________________________________________________________________________

maxima [A]  time = 0.63, size = 24, normalized size = 0.83 \begin {gather*} -\frac {1}{3} \, {\left (3 \, e^{\left (3 \, x\right )} - e^{\left (x - \frac {e}{x} + 16\right )}\right )} e^{\left (-15\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-9*x^2*exp(3*x-15)+(exp(1)+x^2)*exp((-exp(1)+x^2+x)/x))/x^2,x, algorithm="maxima")

[Out]

-1/3*(3*e^(3*x) - e^(x - e/x + 16))*e^(-15)

________________________________________________________________________________________

mupad [B]  time = 0.71, size = 23, normalized size = 0.79 \begin {gather*} \frac {{\mathrm {e}}^{-\frac {\mathrm {e}}{x}}\,\mathrm {e}\,{\mathrm {e}}^x}{3}-{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{-15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp((x - exp(1) + x^2)/x)*(exp(1) + x^2))/3 - 3*x^2*exp(3*x - 15))/x^2,x)

[Out]

(exp(-exp(1)/x)*exp(1)*exp(x))/3 - exp(3*x)*exp(-15)

________________________________________________________________________________________

sympy [A]  time = 0.37, size = 19, normalized size = 0.66 \begin {gather*} \frac {e^{\frac {x^{2} + x - e}{x}}}{3} - e^{3 x - 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-9*x**2*exp(3*x-15)+(exp(1)+x**2)*exp((-exp(1)+x**2+x)/x))/x**2,x)

[Out]

exp((x**2 + x - E)/x)/3 - exp(3*x - 15)

________________________________________________________________________________________