3.83.12 \(\int e^{-225 x^2} (-1+e^{2+x+225 x^2}-1800 x+450 x^2) \, dx\)

Optimal. Leaf size=20 \[ 3+e^{2+x}+e^{-225 x^2} (4-x) \]

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Rubi [A]  time = 0.15, antiderivative size = 25, normalized size of antiderivative = 1.25, number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {6742, 2205, 2194, 2209, 2212} \begin {gather*} -e^{-225 x^2} x+4 e^{-225 x^2}+e^{x+2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + E^(2 + x + 225*x^2) - 1800*x + 450*x^2)/E^(225*x^2),x]

[Out]

4/E^(225*x^2) + E^(2 + x) - x/E^(225*x^2)

Rule 2194

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]

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-e^{-225 x^2}+e^{2+x}-1800 e^{-225 x^2} x+450 e^{-225 x^2} x^2\right ) \, dx\\ &=450 \int e^{-225 x^2} x^2 \, dx-1800 \int e^{-225 x^2} x \, dx-\int e^{-225 x^2} \, dx+\int e^{2+x} \, dx\\ &=4 e^{-225 x^2}+e^{2+x}-e^{-225 x^2} x-\frac {1}{30} \sqrt {\pi } \text {erf}(15 x)+\int e^{-225 x^2} \, dx\\ &=4 e^{-225 x^2}+e^{2+x}-e^{-225 x^2} x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 25, normalized size = 1.25 \begin {gather*} 4 e^{-225 x^2}+e^{2+x}-e^{-225 x^2} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + E^(2 + x + 225*x^2) - 1800*x + 450*x^2)/E^(225*x^2),x]

[Out]

4/E^(225*x^2) + E^(2 + x) - x/E^(225*x^2)

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fricas [A]  time = 0.59, size = 16, normalized size = 0.80 \begin {gather*} -{\left (x - 4\right )} e^{\left (-225 \, x^{2}\right )} + e^{\left (x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2+x)*exp(225*x^2)+450*x^2-1800*x-1)/exp(225*x^2),x, algorithm="fricas")

[Out]

-(x - 4)*e^(-225*x^2) + e^(x + 2)

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giac [A]  time = 0.18, size = 16, normalized size = 0.80 \begin {gather*} -{\left (x - 4\right )} e^{\left (-225 \, x^{2}\right )} + e^{\left (x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2+x)*exp(225*x^2)+450*x^2-1800*x-1)/exp(225*x^2),x, algorithm="giac")

[Out]

-(x - 4)*e^(-225*x^2) + e^(x + 2)

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maple [A]  time = 0.03, size = 18, normalized size = 0.90

method	result	size
risch	\({\mathrm e}^{2+x}+\left (-x +4\right ) {\mathrm e}^{-225 x^{2}}\)	\(18\)
default	\({\mathrm e}^{2} {\mathrm e}^{x}+4 \,{\mathrm e}^{-225 x^{2}}-x \,{\mathrm e}^{-225 x^{2}}\)	\(24\)
norman	\(\left (4+{\mathrm e}^{2+x} {\mathrm e}^{225 x^{2}}-x \right ) {\mathrm e}^{-225 x^{2}}\)	\(26\)
meijerg	\(-{\mathrm e}^{2} \left (1-{\mathrm e}^{x}\right )-x \,{\mathrm e}^{-225 x^{2}}-4+4 \,{\mathrm e}^{-225 x^{2}}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2+x)*exp(225*x^2)+450*x^2-1800*x-1)/exp(225*x^2),x,method=_RETURNVERBOSE)

[Out]

exp(2+x)+(-x+4)*exp(-225*x^2)

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maxima [A]  time = 0.45, size = 22, normalized size = 1.10 \begin {gather*} -x e^{\left (-225 \, x^{2}\right )} + 4 \, e^{\left (-225 \, x^{2}\right )} + e^{\left (x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2+x)*exp(225*x^2)+450*x^2-1800*x-1)/exp(225*x^2),x, algorithm="maxima")

[Out]

-x*e^(-225*x^2) + 4*e^(-225*x^2) + e^(x + 2)

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mupad [B]  time = 5.30, size = 22, normalized size = 1.10 \begin {gather*} {\mathrm {e}}^{x+2}+4\,{\mathrm {e}}^{-225\,x^2}-x\,{\mathrm {e}}^{-225\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-225*x^2)*(1800*x - exp(x + 2)*exp(225*x^2) - 450*x^2 + 1),x)

[Out]

exp(x + 2) + 4*exp(-225*x^2) - x*exp(-225*x^2)

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sympy [A]  time = 0.28, size = 14, normalized size = 0.70 \begin {gather*} \left (4 - x\right ) e^{- 225 x^{2}} + e^{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2+x)*exp(225*x**2)+450*x**2-1800*x-1)/exp(225*x**2),x)

[Out]

(4 - x)*exp(-225*x**2) + exp(x + 2)

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