\(\int \frac {-e-48 x^3+6 e^2 x^3+18 e^x x^3}{6 x^3} \, dx\) [2889]

Optimal result

Integrand size = 32, antiderivative size = 28 \[ \int \frac {-e-48 x^3+6 e^2 x^3+18 e^x x^3}{6 x^3} \, dx=-3 \left (2-e^x\right )+\frac {e}{12 x^2}-\left (8-e^2\right ) x \]

[Out]

1/12*exp(1)/x^2+3*exp(x)-6-(8-exp(2))*x

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6, 12, 14, 2225} \[ \int \frac {-e-48 x^3+6 e^2 x^3+18 e^x x^3}{6 x^3} \, dx=\frac {e}{12 x^2}-\left (\left (8-e^2\right ) x\right )+3 e^x \]

[In]

Int[(-E - 48*x^3 + 6*E^2*x^3 + 18*E^x*x^3)/(6*x^3),x]

[Out]

3*E^x + E/(12*x^2) - (8 - E^2)*x

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 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}& = \int \frac {-e+18 e^x x^3+\left (-48+6 e^2\right ) x^3}{6 x^3} \, dx \\ & = \frac {1}{6} \int \frac {-e+18 e^x x^3+\left (-48+6 e^2\right ) x^3}{x^3} \, dx \\ & = \frac {1}{6} \int \left (18 e^x+\frac {-e-6 \left (8-e^2\right ) x^3}{x^3}\right ) \, dx \\ & = \frac {1}{6} \int \frac {-e-6 \left (8-e^2\right ) x^3}{x^3} \, dx+3 \int e^x \, dx \\ & = 3 e^x+\frac {1}{6} \int \left (6 \left (-8+e^2\right )-\frac {e}{x^3}\right ) \, dx \\ & = 3 e^x+\frac {e}{12 x^2}-\left (8-e^2\right ) x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-e-48 x^3+6 e^2 x^3+18 e^x x^3}{6 x^3} \, dx=3 e^x+\frac {e}{12 x^2}-8 x+e^2 x \]

[In]

Integrate[(-E - 48*x^3 + 6*E^2*x^3 + 18*E^x*x^3)/(6*x^3),x]

[Out]

3*E^x + E/(12*x^2) - 8*x + E^2*x

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71

method	result	size
default	\(-8 x +\frac {{\mathrm e}}{12 x^{2}}+{\mathrm e}^{2} x +3 \,{\mathrm e}^{x}\)	\(20\)
risch	\(-8 x +\frac {{\mathrm e}}{12 x^{2}}+{\mathrm e}^{2} x +3 \,{\mathrm e}^{x}\)	\(20\)
parts	\(-8 x +\frac {{\mathrm e}}{12 x^{2}}+{\mathrm e}^{2} x +3 \,{\mathrm e}^{x}\)	\(20\)
norman	\(\frac {\left (-8+{\mathrm e}^{2}\right ) x^{3}+3 \,{\mathrm e}^{x} x^{2}+\frac {{\mathrm e}}{12}}{x^{2}}\)	\(25\)
parallelrisch	\(\frac {12 x^{3} {\mathrm e}^{2}+36 \,{\mathrm e}^{x} x^{2}-96 x^{3}+{\mathrm e}}{12 x^{2}}\)	\(28\)

[In]

int(1/6*(18*exp(x)*x^3+6*x^3*exp(2)-exp(1)-48*x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

-8*x+1/12*exp(1)/x^2+exp(2)*x+3*exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {-e-48 x^3+6 e^2 x^3+18 e^x x^3}{6 x^3} \, dx=\frac {12 \, x^{3} e^{2} - 96 \, x^{3} + 36 \, x^{2} e^{x} + e}{12 \, x^{2}} \]

[In]

integrate(1/6*(18*exp(x)*x^3+6*x^3*exp(2)-exp(1)-48*x^3)/x^3,x, algorithm="fricas")

[Out]

1/12*(12*x^3*e^2 - 96*x^3 + 36*x^2*e^x + e)/x^2

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-e-48 x^3+6 e^2 x^3+18 e^x x^3}{6 x^3} \, dx=- \frac {x \left (48 - 6 e^{2}\right )}{6} + 3 e^{x} + \frac {e}{12 x^{2}} \]

[In]

integrate(1/6*(18*exp(x)*x**3+6*x**3*exp(2)-exp(1)-48*x**3)/x**3,x)

[Out]

-x*(48 - 6*exp(2))/6 + 3*exp(x) + E/(12*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {-e-48 x^3+6 e^2 x^3+18 e^x x^3}{6 x^3} \, dx=x e^{2} - 8 \, x + \frac {e}{12 \, x^{2}} + 3 \, e^{x} \]

[In]

integrate(1/6*(18*exp(x)*x^3+6*x^3*exp(2)-exp(1)-48*x^3)/x^3,x, algorithm="maxima")

[Out]

x*e^2 - 8*x + 1/12*e/x^2 + 3*e^x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {-e-48 x^3+6 e^2 x^3+18 e^x x^3}{6 x^3} \, dx=\frac {12 \, x^{3} e^{2} - 96 \, x^{3} + 36 \, x^{2} e^{x} + e}{12 \, x^{2}} \]

[In]

integrate(1/6*(18*exp(x)*x^3+6*x^3*exp(2)-exp(1)-48*x^3)/x^3,x, algorithm="giac")

[Out]

1/12*(12*x^3*e^2 - 96*x^3 + 36*x^2*e^x + e)/x^2

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {-e-48 x^3+6 e^2 x^3+18 e^x x^3}{6 x^3} \, dx=3\,{\mathrm {e}}^x+x\,\left ({\mathrm {e}}^2-8\right )+\frac {\mathrm {e}}{12\,x^2} \]

[In]

int(-(exp(1)/6 - 3*x^3*exp(x) - x^3*exp(2) + 8*x^3)/x^3,x)

[Out]

3*exp(x) + x*(exp(2) - 8) + exp(1)/(12*x^2)