3.14.30 \(\int \frac {1}{16} e^{-16-8 x-x^2} (15 e^{16+8 x+x^2}+40 x-160 x^2-40 x^3) \, dx\)

Optimal. Leaf size=22 \[ \frac {15 x}{16}+\frac {5}{4} e^{-(4+x)^2} x^2 \]

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Rubi [B]  time = 0.30, antiderivative size = 49, normalized size of antiderivative = 2.23, number of steps used = 11, number of rules used = 6, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 6688, 2226, 2205, 2209, 2212} \begin {gather*} \frac {5}{4} e^{-(x+4)^2} (x+4)^2-10 e^{-(x+4)^2} (x+4)+20 e^{-(x+4)^2}+\frac {15 x}{16} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-16 - 8*x - x^2)*(15*E^(16 + 8*x + x^2) + 40*x - 160*x^2 - 40*x^3))/16,x]

[Out]

20/E^(4 + x)^2 + (15*x)/16 - (10*(4 + x))/E^(4 + x)^2 + (5*(4 + x)^2)/(4*E^(4 + x)^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2205

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erf[(c + d*x)*Rt[-(b*Log[F]),
2]])/(2*d*Rt[-(b*Log[F]), 2]), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

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 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2226

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{16} \int e^{-16-8 x-x^2} \left (15 e^{16+8 x+x^2}+40 x-160 x^2-40 x^3\right ) \, dx\\ &=\frac {1}{16} \int \left (15-40 e^{-(4+x)^2} x \left (-1+4 x+x^2\right )\right ) \, dx\\ &=\frac {15 x}{16}-\frac {5}{2} \int e^{-(4+x)^2} x \left (-1+4 x+x^2\right ) \, dx\\ &=\frac {15 x}{16}-\frac {5}{2} \int \left (4 e^{-(4+x)^2}+15 e^{-(4+x)^2} (4+x)-8 e^{-(4+x)^2} (4+x)^2+e^{-(4+x)^2} (4+x)^3\right ) \, dx\\ &=\frac {15 x}{16}-\frac {5}{2} \int e^{-(4+x)^2} (4+x)^3 \, dx-10 \int e^{-(4+x)^2} \, dx+20 \int e^{-(4+x)^2} (4+x)^2 \, dx-\frac {75}{2} \int e^{-(4+x)^2} (4+x) \, dx\\ &=\frac {75}{4} e^{-(4+x)^2}+\frac {15 x}{16}-10 e^{-(4+x)^2} (4+x)+\frac {5}{4} e^{-(4+x)^2} (4+x)^2-5 \sqrt {\pi } \text {erf}(4+x)-\frac {5}{2} \int e^{-(4+x)^2} (4+x) \, dx+10 \int e^{-(4+x)^2} \, dx\\ &=20 e^{-(4+x)^2}+\frac {15 x}{16}-10 e^{-(4+x)^2} (4+x)+\frac {5}{4} e^{-(4+x)^2} (4+x)^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 22, normalized size = 1.00 \begin {gather*} \frac {5}{16} \left (3 x+4 e^{-(4+x)^2} x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-16 - 8*x - x^2)*(15*E^(16 + 8*x + x^2) + 40*x - 160*x^2 - 40*x^3))/16,x]

[Out]

(5*(3*x + (4*x^2)/E^(4 + x)^2))/16

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fricas [A]  time = 0.60, size = 31, normalized size = 1.41 \begin {gather*} \frac {5}{16} \, {\left (4 \, x^{2} + 3 \, x e^{\left (x^{2} + 8 \, x + 16\right )}\right )} e^{\left (-x^{2} - 8 \, x - 16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(15*exp(x^2+8*x+16)-40*x^3-160*x^2+40*x)/exp(x^2+8*x+16),x, algorithm="fricas")

[Out]

5/16*(4*x^2 + 3*x*e^(x^2 + 8*x + 16))*e^(-x^2 - 8*x - 16)

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giac [A]  time = 0.29, size = 27, normalized size = 1.23 \begin {gather*} \frac {5}{4} \, {\left ({\left (x + 4\right )}^{2} - 8 \, x - 16\right )} e^{\left (-x^{2} - 8 \, x - 16\right )} + \frac {15}{16} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(15*exp(x^2+8*x+16)-40*x^3-160*x^2+40*x)/exp(x^2+8*x+16),x, algorithm="giac")

[Out]

5/4*((x + 4)^2 - 8*x - 16)*e^(-x^2 - 8*x - 16) + 15/16*x

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maple [A]  time = 0.03, size = 18, normalized size = 0.82

method	result	size
risch	\(\frac {15 x}{16}+\frac {5 x^{2} {\mathrm e}^{-\left (4+x \right )^{2}}}{4}\)	\(18\)
default	\(\frac {15 x}{16}+\frac {5 x^{2} {\mathrm e}^{-x^{2}-8 x -16}}{4}\)	\(21\)
norman	\(\left (\frac {5 x^{2}}{4}+\frac {15 x \,{\mathrm e}^{x^{2}+8 x +16}}{16}\right ) {\mathrm e}^{-x^{2}-8 x -16}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*(15*exp(x^2+8*x+16)-40*x^3-160*x^2+40*x)/exp(x^2+8*x+16),x,method=_RETURNVERBOSE)

[Out]

15/16*x+5/4*x^2*exp(-(4+x)^2)

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maxima [C]  time = 0.54, size = 69, normalized size = 3.14 \begin {gather*} -\frac {10 \, {\left (x + 4\right )}^{3} \Gamma \left (\frac {3}{2}, {\left (x + 4\right )}^{2}\right )}{{\left ({\left (x + 4\right )}^{2}\right )}^{\frac {3}{2}}} - \frac {5 \, \sqrt {\pi } {\left (x + 4\right )} {\left (\operatorname {erf}\left (\sqrt {{\left (x + 4\right )}^{2}}\right ) - 1\right )}}{\sqrt {{\left (x + 4\right )}^{2}}} + \frac {15}{16} \, x + \frac {75}{4} \, e^{\left (-{\left (x + 4\right )}^{2}\right )} + \frac {5}{4} \, \Gamma \left (2, {\left (x + 4\right )}^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(15*exp(x^2+8*x+16)-40*x^3-160*x^2+40*x)/exp(x^2+8*x+16),x, algorithm="maxima")

[Out]

-10*(x + 4)^3*gamma(3/2, (x + 4)^2)/((x + 4)^2)^(3/2) - 5*sqrt(pi)*(x + 4)*(erf(sqrt((x + 4)^2)) - 1)/sqrt((x
+ 4)^2) + 15/16*x + 75/4*e^(-(x + 4)^2) + 5/4*gamma(2, (x + 4)^2)

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mupad [B]  time = 0.95, size = 21, normalized size = 0.95 \begin {gather*} \frac {15\,x}{16}+\frac {5\,x^2\,{\mathrm {e}}^{-8\,x}\,{\mathrm {e}}^{-16}\,{\mathrm {e}}^{-x^2}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(- 8*x - x^2 - 16)*((5*x)/2 + (15*exp(8*x + x^2 + 16))/16 - 10*x^2 - (5*x^3)/2),x)

[Out]

(15*x)/16 + (5*x^2*exp(-8*x)*exp(-16)*exp(-x^2))/4

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sympy [A]  time = 0.12, size = 22, normalized size = 1.00 \begin {gather*} \frac {5 x^{2} e^{- x^{2} - 8 x - 16}}{4} + \frac {15 x}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(15*exp(x**2+8*x+16)-40*x**3-160*x**2+40*x)/exp(x**2+8*x+16),x)

[Out]

5*x**2*exp(-x**2 - 8*x - 16)/4 + 15*x/16

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