\(\int \frac {-8+4 e^{6+2 x} x^2}{3 x^2} \, dx\) [9660]

   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 = 21, antiderivative size = 22 \[ \int \frac {-8+4 e^{6+2 x} x^2}{3 x^2} \, dx=-5+\frac {2 \left (4+\left (6+e^{6+2 x}\right ) x\right )}{3 x} \]

[Out]

2/3/x*(4+(exp(3)^2*exp(x)^2+6)*x)-5

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 14, 2225} \[ \int \frac {-8+4 e^{6+2 x} x^2}{3 x^2} \, dx=\frac {2}{3} e^{2 x+6}+\frac {8}{3 x} \]

[In]

Int[(-8 + 4*E^(6 + 2*x)*x^2)/(3*x^2),x]

[Out]

(2*E^(6 + 2*x))/3 + 8/(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 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}& = \frac {1}{3} \int \frac {-8+4 e^{6+2 x} x^2}{x^2} \, dx \\ & = \frac {1}{3} \int \left (4 e^{6+2 x}-\frac {8}{x^2}\right ) \, dx \\ & = \frac {8}{3 x}+\frac {4}{3} \int e^{6+2 x} \, dx \\ & = \frac {2}{3} e^{6+2 x}+\frac {8}{3 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-8+4 e^{6+2 x} x^2}{3 x^2} \, dx=\frac {4}{3} \left (\frac {1}{2} e^{6+2 x}+\frac {2}{x}\right ) \]

[In]

Integrate[(-8 + 4*E^(6 + 2*x)*x^2)/(3*x^2),x]

[Out]

(4*(E^(6 + 2*x)/2 + 2/x))/3

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68

method result size
risch \(\frac {2 \,{\mathrm e}^{2 x +6}}{3}+\frac {8}{3 x}\) \(15\)
default \(\frac {2 \,{\mathrm e}^{6} {\mathrm e}^{2 x}}{3}+\frac {8}{3 x}\) \(17\)
parts \(\frac {2 \,{\mathrm e}^{6} {\mathrm e}^{2 x}}{3}+\frac {8}{3 x}\) \(17\)
norman \(\frac {\frac {8}{3}+\frac {2 x \,{\mathrm e}^{6} {\mathrm e}^{2 x}}{3}}{x}\) \(18\)
parallelrisch \(\frac {8+2 x \,{\mathrm e}^{6} {\mathrm e}^{2 x}}{3 x}\) \(19\)

[In]

int(1/3*(4*x^2*exp(3)^2*exp(x)^2-8)/x^2,x,method=_RETURNVERBOSE)

[Out]

2/3*exp(2*x+6)+8/3/x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {-8+4 e^{6+2 x} x^2}{3 x^2} \, dx=\frac {2 \, {\left (x e^{\left (2 \, x + 6\right )} + 4\right )}}{3 \, x} \]

[In]

integrate(1/3*(4*x^2*exp(3)^2*exp(x)^2-8)/x^2,x, algorithm="fricas")

[Out]

2/3*(x*e^(2*x + 6) + 4)/x

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {-8+4 e^{6+2 x} x^2}{3 x^2} \, dx=\frac {2 e^{6} e^{2 x}}{3} + \frac {8}{3 x} \]

[In]

integrate(1/3*(4*x**2*exp(3)**2*exp(x)**2-8)/x**2,x)

[Out]

2*exp(6)*exp(2*x)/3 + 8/(3*x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {-8+4 e^{6+2 x} x^2}{3 x^2} \, dx=\frac {8}{3 \, x} + \frac {2}{3} \, e^{\left (2 \, x + 6\right )} \]

[In]

integrate(1/3*(4*x^2*exp(3)^2*exp(x)^2-8)/x^2,x, algorithm="maxima")

[Out]

8/3/x + 2/3*e^(2*x + 6)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {-8+4 e^{6+2 x} x^2}{3 x^2} \, dx=\frac {2 \, {\left (x e^{\left (2 \, x + 6\right )} + 4\right )}}{3 \, x} \]

[In]

integrate(1/3*(4*x^2*exp(3)^2*exp(x)^2-8)/x^2,x, algorithm="giac")

[Out]

2/3*(x*e^(2*x + 6) + 4)/x

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {-8+4 e^{6+2 x} x^2}{3 x^2} \, dx=\frac {2\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^6}{3}+\frac {8}{3\,x} \]

[In]

int(((4*x^2*exp(2*x)*exp(6))/3 - 8/3)/x^2,x)

[Out]

(2*exp(2*x)*exp(6))/3 + 8/(3*x)