\(\int \frac {\frac {e^5}{x}+16 e^{32 e^6+16 x} x}{x} \, dx\) [5488]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 26, antiderivative size = 20 \[ \int \frac {\frac {e^5}{x}+16 e^{32 e^6+16 x} x}{x} \, dx=e^{32 e^6+16 x}-\frac {e^5}{x} \]

[Out]

exp(16*exp(3)^2+8*x)^2-exp(5-ln(x))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {14, 2225} \[ \int \frac {\frac {e^5}{x}+16 e^{32 e^6+16 x} x}{x} \, dx=e^{16 x+32 e^6}-\frac {e^5}{x} \]

[In]

Int[(E^5/x + 16*E^(32*E^6 + 16*x)*x)/x,x]

[Out]

E^(32*E^6 + 16*x) - E^5/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 2225

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]

Rubi steps \begin{align*} \text {integral}& = \int \left (16 e^{32 e^6+16 x}+\frac {e^5}{x^2}\right ) \, dx \\ & = -\frac {e^5}{x}+16 \int e^{32 e^6+16 x} \, dx \\ & = e^{32 e^6+16 x}-\frac {e^5}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\frac {e^5}{x}+16 e^{32 e^6+16 x} x}{x} \, dx=e^{32 e^6+16 x}-\frac {e^5}{x} \]

[In]

Integrate[(E^5/x + 16*E^(32*E^6 + 16*x)*x)/x,x]

[Out]

E^(32*E^6 + 16*x) - E^5/x

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90

method result size
risch \(-\frac {{\mathrm e}^{5}}{x}+{\mathrm e}^{32 \,{\mathrm e}^{6}+16 x}\) \(18\)
default \({\mathrm e}^{32 \,{\mathrm e}^{6}+16 x}-{\mathrm e}^{5-\ln \left (x \right )}\) \(24\)
parallelrisch \({\mathrm e}^{32 \,{\mathrm e}^{6}+16 x}-{\mathrm e}^{5-\ln \left (x \right )}\) \(24\)
parts \({\mathrm e}^{32 \,{\mathrm e}^{6}+16 x}-{\mathrm e}^{5-\ln \left (x \right )}\) \(24\)
norman \(\frac {{\mathrm e}^{32 \,{\mathrm e}^{6}+16 x} x -{\mathrm e}^{5}}{x}\) \(25\)

[In]

int((exp(5-ln(x))+16*x*exp(16*exp(3)^2+8*x)^2)/x,x,method=_RETURNVERBOSE)

[Out]

-exp(5)/x+exp(32*exp(6)+16*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\frac {e^5}{x}+16 e^{32 e^6+16 x} x}{x} \, dx=\frac {x e^{\left (16 \, x + 32 \, e^{6}\right )} - e^{5}}{x} \]

[In]

integrate((exp(5-log(x))+16*x*exp(16*exp(3)^2+8*x)^2)/x,x, algorithm="fricas")

[Out]

(x*e^(16*x + 32*e^6) - e^5)/x

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {\frac {e^5}{x}+16 e^{32 e^6+16 x} x}{x} \, dx=e^{16 x + 32 e^{6}} - \frac {e^{5}}{x} \]

[In]

integrate((exp(5-ln(x))+16*x*exp(16*exp(3)**2+8*x)**2)/x,x)

[Out]

exp(16*x + 32*exp(6)) - exp(5)/x

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {\frac {e^5}{x}+16 e^{32 e^6+16 x} x}{x} \, dx=-\frac {e^{5}}{x} + e^{\left (16 \, x + 32 \, e^{6}\right )} \]

[In]

integrate((exp(5-log(x))+16*x*exp(16*exp(3)^2+8*x)^2)/x,x, algorithm="maxima")

[Out]

-e^5/x + e^(16*x + 32*e^6)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\frac {e^5}{x}+16 e^{32 e^6+16 x} x}{x} \, dx=\frac {x e^{\left (16 \, x + 32 \, e^{6}\right )} - e^{5}}{x} \]

[In]

integrate((exp(5-log(x))+16*x*exp(16*exp(3)^2+8*x)^2)/x,x, algorithm="giac")

[Out]

(x*e^(16*x + 32*e^6) - e^5)/x

Mupad [B] (verification not implemented)

Time = 12.52 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {\frac {e^5}{x}+16 e^{32 e^6+16 x} x}{x} \, dx={\mathrm {e}}^{32\,{\mathrm {e}}^6}\,{\mathrm {e}}^{16\,x}-\frac {{\mathrm {e}}^5}{x} \]

[In]

int((exp(5 - log(x)) + 16*x*exp(16*x + 32*exp(6)))/x,x)

[Out]

exp(32*exp(6))*exp(16*x) - exp(5)/x