3.87.85 \(\int \frac {e^{-14-2 x-x^3} (e^6+4 e^{3+x} x+4 e^{2 x} x^2) (-6 e^x x^4+e^3 (-2-2 x-3 x^3))}{x^2 (e^3 x+2 e^x x^2)} \, dx\)

Optimal. Leaf size=25 \[ e^{-14-x^3} \left (2+\frac {e^{3-x}}{x}\right )^2 \]

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Rubi [B]  time = 1.43, antiderivative size = 79, normalized size of antiderivative = 3.16, number of steps used = 6, number of rules used = 4, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {6688, 6742, 2209, 2288} \begin {gather*} 4 e^{-x^3-14}+\frac {4 e^{-x^3-x-11} \left (3 x^3+x\right )}{x^2 \left (3 x^2+1\right )}+\frac {e^{-x^3-2 x-8} \left (3 x^3+2 x\right )}{x^3 \left (3 x^2+2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-14 - 2*x - x^3)*(E^6 + 4*E^(3 + x)*x + 4*E^(2*x)*x^2)*(-6*E^x*x^4 + E^3*(-2 - 2*x - 3*x^3)))/(x^2*(E^
3*x + 2*E^x*x^2)),x]

[Out]

4*E^(-14 - x^3) + (4*E^(-11 - x - x^3)*(x + 3*x^3))/(x^2*(1 + 3*x^2)) + (E^(-8 - 2*x - x^3)*(2*x + 3*x^3))/(x^
3*(2 + 3*x^2))

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-14-2 x-x^3} \left (e^3+2 e^x x\right ) \left (-6 e^x x^4-e^3 \left (2+2 x+3 x^3\right )\right )}{x^3} \, dx\\ &=\int \left (-12 e^{-14-x^3} x^2-\frac {4 e^{-11-x-x^3} \left (1+x+3 x^3\right )}{x^2}-\frac {e^{-8-2 x-x^3} \left (2+2 x+3 x^3\right )}{x^3}\right ) \, dx\\ &=-\left (4 \int \frac {e^{-11-x-x^3} \left (1+x+3 x^3\right )}{x^2} \, dx\right )-12 \int e^{-14-x^3} x^2 \, dx-\int \frac {e^{-8-2 x-x^3} \left (2+2 x+3 x^3\right )}{x^3} \, dx\\ &=4 e^{-14-x^3}+\frac {4 e^{-11-x-x^3} \left (x+3 x^3\right )}{x^2 \left (1+3 x^2\right )}+\frac {e^{-8-2 x-x^3} \left (2 x+3 x^3\right )}{x^3 \left (2+3 x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 28, normalized size = 1.12 \begin {gather*} \frac {e^{-14-2 x-x^3} \left (e^3+2 e^x x\right )^2}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-14 - 2*x - x^3)*(E^6 + 4*E^(3 + x)*x + 4*E^(2*x)*x^2)*(-6*E^x*x^4 + E^3*(-2 - 2*x - 3*x^3)))/(x
^2*(E^3*x + 2*E^x*x^2)),x]

[Out]

(E^(-14 - 2*x - x^3)*(E^3 + 2*E^x*x)^2)/x^2

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fricas [A]  time = 0.50, size = 40, normalized size = 1.60 \begin {gather*} e^{\left (-x^{3} + \log \left (\frac {{\left (4 \, x^{2} e^{\left (2 \, x + 6\right )} + 4 \, x e^{\left (x + 9\right )} + e^{12}\right )} e^{\left (-2 \, x - 6\right )}}{x^{2}}\right ) - 14\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*exp(x)*x^4+(-3*x^3-2*x-2)*exp(3))*exp(log((4*exp(x)^2*x^2+4*x*exp(3)*exp(x)+exp(3)^2)/exp(x)^2/x
^2)-x^3-14)/(2*exp(x)*x^2+x*exp(3)),x, algorithm="fricas")

[Out]

e^(-x^3 + log((4*x^2*e^(2*x + 6) + 4*x*e^(x + 9) + e^12)*e^(-2*x - 6)/x^2) - 14)

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giac [B]  time = 0.24, size = 51, normalized size = 2.04 \begin {gather*} \frac {{\left (4 \, x^{2} e^{\left (2 \, x^{3} + 2 \, x\right )} + 4 \, x e^{\left (2 \, x^{3} + x + 3\right )} + e^{\left (2 \, x^{3} + 6\right )}\right )} e^{\left (-3 \, x^{3} - 2 \, x - 14\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*exp(x)*x^4+(-3*x^3-2*x-2)*exp(3))*exp(log((4*exp(x)^2*x^2+4*x*exp(3)*exp(x)+exp(3)^2)/exp(x)^2/x
^2)-x^3-14)/(2*exp(x)*x^2+x*exp(3)),x, algorithm="giac")

[Out]

(4*x^2*e^(2*x^3 + 2*x) + 4*x*e^(2*x^3 + x + 3) + e^(2*x^3 + 6))*e^(-3*x^3 - 2*x - 14)/x^2

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maple [C]  time = 0.30, size = 516, normalized size = 20.64

method	result	size
risch	\(\frac {\left (4 \,{\mathrm e}^{2 x} x^{2}+4 \,{\mathrm e}^{3+x} x +{\mathrm e}^{6}\right ) {\mathrm e}^{-2 x -14+i \pi \mathrm {csgn}\left (i \left (2 \,{\mathrm e}^{x} x +{\mathrm e}^{3}\right )^{2}\right )^{2} \mathrm {csgn}\left (i \left (2 \,{\mathrm e}^{x} x +{\mathrm e}^{3}\right )\right )+\frac {i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}+\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}}{2}+\frac {i \pi \,\mathrm {csgn}\left (i \left (2 \,{\mathrm e}^{x} x +{\mathrm e}^{3}\right )^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-2 x} \left (2 \,{\mathrm e}^{x} x +{\mathrm e}^{3}\right )^{2}\right )^{2}}{2}-\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-2 x} \left (2 \,{\mathrm e}^{x} x +{\mathrm e}^{3}\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \left (2 \,{\mathrm e}^{x} x +{\mathrm e}^{3}\right )^{2} {\mathrm e}^{-2 x}}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i}{x^{2}}\right )}{2}+\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2}}{2}-\frac {i \pi \mathrm {csgn}\left (\frac {i \left (2 \,{\mathrm e}^{x} x +{\mathrm e}^{3}\right )^{2} {\mathrm e}^{-2 x}}{x^{2}}\right )^{3}}{2}-i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )-\frac {i \pi \,\mathrm {csgn}\left (i \left (2 \,{\mathrm e}^{x} x +{\mathrm e}^{3}\right )^{2}\right ) \mathrm {csgn}\left (i \left (2 \,{\mathrm e}^{x} x +{\mathrm e}^{3}\right )\right )^{2}}{2}+\frac {i \pi \mathrm {csgn}\left (\frac {i \left (2 \,{\mathrm e}^{x} x +{\mathrm e}^{3}\right )^{2} {\mathrm e}^{-2 x}}{x^{2}}\right )^{2} \mathrm {csgn}\left (\frac {i}{x^{2}}\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (i \left (2 \,{\mathrm e}^{x} x +{\mathrm e}^{3}\right )^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-2 x} \left (2 \,{\mathrm e}^{x} x +{\mathrm e}^{3}\right )^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-2 x}\right )}{2}+\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-2 x} \left (2 \,{\mathrm e}^{x} x +{\mathrm e}^{3}\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \left (2 \,{\mathrm e}^{x} x +{\mathrm e}^{3}\right )^{2} {\mathrm e}^{-2 x}}{x^{2}}\right )^{2}}{2}-\frac {i \pi \mathrm {csgn}\left (i {\mathrm e}^{-2 x} \left (2 \,{\mathrm e}^{x} x +{\mathrm e}^{3}\right )^{2}\right )^{3}}{2}-i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )-\frac {i \pi \mathrm {csgn}\left (i \left (2 \,{\mathrm e}^{x} x +{\mathrm e}^{3}\right )^{2}\right )^{3}}{2}+\frac {i \pi \mathrm {csgn}\left (i {\mathrm e}^{-2 x} \left (2 \,{\mathrm e}^{x} x +{\mathrm e}^{3}\right )^{2}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-2 x}\right )}{2}+\frac {i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}}{2}-x^{3}}}{x^{2}}\)	\(516\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-6*exp(x)*x^4+(-3*x^3-2*x-2)*exp(3))*exp(ln((4*exp(x)^2*x^2+4*x*exp(3)*exp(x)+exp(3)^2)/exp(x)^2/x^2)-x^3
-14)/(2*exp(x)*x^2+x*exp(3)),x,method=_RETURNVERBOSE)

[Out]

(4*exp(2*x)*x^2+4*exp(3+x)*x+exp(6))/x^2*exp(-2*x-14+I*Pi*csgn(I*(2*exp(x)*x+exp(3))^2)^2*csgn(I*(2*exp(x)*x+e
xp(3)))+1/2*I*Pi*csgn(I*x^2)^3+1/2*I*Pi*csgn(I*x^2)*csgn(I*x)^2+1/2*I*Pi*csgn(I*(2*exp(x)*x+exp(3))^2)*csgn(I*
exp(-2*x)*(2*exp(x)*x+exp(3))^2)^2-1/2*I*Pi*csgn(I*exp(-2*x)*(2*exp(x)*x+exp(3))^2)*csgn(I/x^2*(2*exp(x)*x+exp
(3))^2*exp(-2*x))*csgn(I/x^2)+1/2*I*Pi*csgn(I*exp(2*x))*csgn(I*exp(x))^2-1/2*I*Pi*csgn(I/x^2*(2*exp(x)*x+exp(3
))^2*exp(-2*x))^3-I*Pi*csgn(I*exp(2*x))^2*csgn(I*exp(x))-1/2*I*Pi*csgn(I*(2*exp(x)*x+exp(3))^2)*csgn(I*(2*exp(
x)*x+exp(3)))^2+1/2*I*Pi*csgn(I/x^2*(2*exp(x)*x+exp(3))^2*exp(-2*x))^2*csgn(I/x^2)-1/2*I*Pi*csgn(I*(2*exp(x)*x
+exp(3))^2)*csgn(I*exp(-2*x)*(2*exp(x)*x+exp(3))^2)*csgn(I*exp(-2*x))+1/2*I*Pi*csgn(I*exp(-2*x)*(2*exp(x)*x+ex
p(3))^2)*csgn(I/x^2*(2*exp(x)*x+exp(3))^2*exp(-2*x))^2-1/2*I*Pi*csgn(I*exp(-2*x)*(2*exp(x)*x+exp(3))^2)^3-I*Pi
*csgn(I*x^2)^2*csgn(I*x)-1/2*I*Pi*csgn(I*(2*exp(x)*x+exp(3))^2)^3+1/2*I*Pi*csgn(I*exp(-2*x)*(2*exp(x)*x+exp(3)
)^2)^2*csgn(I*exp(-2*x))+1/2*I*Pi*csgn(I*exp(2*x))^3-x^3)

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maxima [A]  time = 0.45, size = 34, normalized size = 1.36 \begin {gather*} \frac {{\left (4 \, x^{2} e^{\left (2 \, x\right )} + 4 \, x e^{\left (x + 3\right )} + e^{6}\right )} e^{\left (-x^{3} - 2 \, x - 14\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*exp(x)*x^4+(-3*x^3-2*x-2)*exp(3))*exp(log((4*exp(x)^2*x^2+4*x*exp(3)*exp(x)+exp(3)^2)/exp(x)^2/x
^2)-x^3-14)/(2*exp(x)*x^2+x*exp(3)),x, algorithm="maxima")

[Out]

(4*x^2*e^(2*x) + 4*x*e^(x + 3) + e^6)*e^(-x^3 - 2*x - 14)/x^2

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mupad [B]  time = 5.65, size = 44, normalized size = 1.76 \begin {gather*} 4\,{\mathrm {e}}^{-14}\,{\mathrm {e}}^{-x^3}+\frac {{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{-8}\,{\mathrm {e}}^{-x^3}}{x^2}+\frac {4\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-11}\,{\mathrm {e}}^{-x^3}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log((exp(-2*x)*(exp(6) + 4*x^2*exp(2*x) + 4*x*exp(3)*exp(x)))/x^2) - x^3 - 14)*(6*x^4*exp(x) + exp(3
)*(2*x + 3*x^3 + 2)))/(2*x^2*exp(x) + x*exp(3)),x)

[Out]

4*exp(-14)*exp(-x^3) + (exp(-2*x)*exp(-8)*exp(-x^3))/x^2 + (4*exp(-x)*exp(-11)*exp(-x^3))/x

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sympy [A]  time = 0.26, size = 34, normalized size = 1.36 \begin {gather*} \frac {\left (4 x^{2} + 4 x e^{3} e^{- x} + e^{6} e^{- 2 x}\right ) e^{- x^{3} - 14}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*exp(x)*x**4+(-3*x**3-2*x-2)*exp(3))*exp(ln((4*exp(x)**2*x**2+4*x*exp(3)*exp(x)+exp(3)**2)/exp(x)
**2/x**2)-x**3-14)/(2*exp(x)*x**2+x*exp(3)),x)

[Out]

(4*x**2 + 4*x*exp(3)*exp(-x) + exp(6)*exp(-2*x))*exp(-x**3 - 14)/x**2

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