\(\int \frac {e^3+e^{-2 x-15 x^2-2 x^3+x^4} (-4 x^2-60 x^3-12 x^4+8 x^5)}{2 x^2} \, dx\) [7702]

Optimal result

Integrand size = 52, antiderivative size = 28 \[ \int \frac {e^3+e^{-2 x-15 x^2-2 x^3+x^4} \left (-4 x^2-60 x^3-12 x^4+8 x^5\right )}{2 x^2} \, dx=e^{-x (2+(5-x) x (3+x))}-\frac {e^3}{2 x} \]

[Out]

exp(-(2+x*(3+x)*(5-x))*x)-1/2*exp(3)/x

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {12, 14, 6838} \[ \int \frac {e^3+e^{-2 x-15 x^2-2 x^3+x^4} \left (-4 x^2-60 x^3-12 x^4+8 x^5\right )}{2 x^2} \, dx=e^{x^4-2 x^3-15 x^2-2 x}-\frac {e^3}{2 x} \]

[In]

Int[(E^3 + E^(-2*x - 15*x^2 - 2*x^3 + x^4)*(-4*x^2 - 60*x^3 - 12*x^4 + 8*x^5))/(2*x^2),x]

[Out]

E^(-2*x - 15*x^2 - 2*x^3 + x^4) - E^3/(2*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 6838

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{align*} \text {integral}& = \frac {1}{2} \int \frac {e^3+e^{-2 x-15 x^2-2 x^3+x^4} \left (-4 x^2-60 x^3-12 x^4+8 x^5\right )}{x^2} \, dx \\ & = \frac {1}{2} \int \left (\frac {e^3}{x^2}+4 e^{-2 x-15 x^2-2 x^3+x^4} \left (-1-15 x-3 x^2+2 x^3\right )\right ) \, dx \\ & = -\frac {e^3}{2 x}+2 \int e^{-2 x-15 x^2-2 x^3+x^4} \left (-1-15 x-3 x^2+2 x^3\right ) \, dx \\ & = e^{-2 x-15 x^2-2 x^3+x^4}-\frac {e^3}{2 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^3+e^{-2 x-15 x^2-2 x^3+x^4} \left (-4 x^2-60 x^3-12 x^4+8 x^5\right )}{2 x^2} \, dx=e^{x \left (-2-15 x-2 x^2+x^3\right )}-\frac {e^3}{2 x} \]

[In]

Integrate[(E^3 + E^(-2*x - 15*x^2 - 2*x^3 + x^4)*(-4*x^2 - 60*x^3 - 12*x^4 + 8*x^5))/(2*x^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89

method	result	size
risch	\(-\frac {{\mathrm e}^{3}}{2 x}+{\mathrm e}^{x \left (x^{3}-2 x^{2}-15 x -2\right )}\)	\(25\)
parts	\(-\frac {{\mathrm e}^{3}}{2 x}+{\mathrm e}^{x^{4}-2 x^{3}-15 x^{2}-2 x}\)	\(27\)
norman	\(\frac {{\mathrm e}^{x^{4}-2 x^{3}-15 x^{2}-2 x} x -\frac {{\mathrm e}^{3}}{2}}{x}\)	\(30\)
parallelrisch	\(-\frac {-2 \,{\mathrm e}^{x^{4}-2 x^{3}-15 x^{2}-2 x} x +{\mathrm e}^{3}}{2 x}\)	\(30\)

[In]

int(1/2*((8*x^5-12*x^4-60*x^3-4*x^2)*exp(x^4-2*x^3-15*x^2-2*x)+exp(3))/x^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {e^3+e^{-2 x-15 x^2-2 x^3+x^4} \left (-4 x^2-60 x^3-12 x^4+8 x^5\right )}{2 x^2} \, dx=\frac {2 \, x e^{\left (x^{4} - 2 \, x^{3} - 15 \, x^{2} - 2 \, x\right )} - e^{3}}{2 \, x} \]

[In]

integrate(1/2*((8*x^5-12*x^4-60*x^3-4*x^2)*exp(x^4-2*x^3-15*x^2-2*x)+exp(3))/x^2,x, algorithm="fricas")

[Out]

1/2*(2*x*e^(x^4 - 2*x^3 - 15*x^2 - 2*x) - e^3)/x

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {e^3+e^{-2 x-15 x^2-2 x^3+x^4} \left (-4 x^2-60 x^3-12 x^4+8 x^5\right )}{2 x^2} \, dx=e^{x^{4} - 2 x^{3} - 15 x^{2} - 2 x} - \frac {e^{3}}{2 x} \]

[In]

integrate(1/2*((8*x**5-12*x**4-60*x**3-4*x**2)*exp(x**4-2*x**3-15*x**2-2*x)+exp(3))/x**2,x)

[Out]

exp(x**4 - 2*x**3 - 15*x**2 - 2*x) - exp(3)/(2*x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^3+e^{-2 x-15 x^2-2 x^3+x^4} \left (-4 x^2-60 x^3-12 x^4+8 x^5\right )}{2 x^2} \, dx=-\frac {e^{3}}{2 \, x} + e^{\left (x^{4} - 2 \, x^{3} - 15 \, x^{2} - 2 \, x\right )} \]

[In]

integrate(1/2*((8*x^5-12*x^4-60*x^3-4*x^2)*exp(x^4-2*x^3-15*x^2-2*x)+exp(3))/x^2,x, algorithm="maxima")

[Out]

-1/2*e^3/x + e^(x^4 - 2*x^3 - 15*x^2 - 2*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {e^3+e^{-2 x-15 x^2-2 x^3+x^4} \left (-4 x^2-60 x^3-12 x^4+8 x^5\right )}{2 x^2} \, dx=\frac {2 \, x e^{\left (x^{4} - 2 \, x^{3} - 15 \, x^{2} - 2 \, x\right )} - e^{3}}{2 \, x} \]

[In]

integrate(1/2*((8*x^5-12*x^4-60*x^3-4*x^2)*exp(x^4-2*x^3-15*x^2-2*x)+exp(3))/x^2,x, algorithm="giac")

[Out]

1/2*(2*x*e^(x^4 - 2*x^3 - 15*x^2 - 2*x) - e^3)/x

Mupad [B] (verification not implemented)

Time = 13.93 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^3+e^{-2 x-15 x^2-2 x^3+x^4} \left (-4 x^2-60 x^3-12 x^4+8 x^5\right )}{2 x^2} \, dx={\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{x^4}\,{\mathrm {e}}^{-2\,x^3}\,{\mathrm {e}}^{-15\,x^2}-\frac {{\mathrm {e}}^3}{2\,x} \]

[In]

int((exp(3)/2 - (exp(x^4 - 15*x^2 - 2*x^3 - 2*x)*(4*x^2 + 60*x^3 + 12*x^4 - 8*x^5))/2)/x^2,x)

[Out]

exp(-2*x)*exp(x^4)*exp(-2*x^3)*exp(-15*x^2) - exp(3)/(2*x)