3.81.63 \(\int \frac {1}{20} e^{4-x^2} (1+20 e^{-4+x^2}+50 x-2 x^2+10 x \log (4)) \, dx\)

Optimal. Leaf size=24 \[ x-\frac {1}{4} e^{4-x^2} \left (5-\frac {x}{5}+\log (4)\right ) \]

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Rubi [A]  time = 0.24, antiderivative size = 33, normalized size of antiderivative = 1.38, number of steps used = 8, number of rules used = 6, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {6, 12, 6742, 2205, 2212, 2209} \begin {gather*} \frac {1}{20} e^{4-x^2} x-\frac {1}{4} e^{4-x^2} (5+\log (4))+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(4 - x^2)*(1 + 20*E^(-4 + x^2) + 50*x - 2*x^2 + 10*x*Log[4]))/20,x]

[Out]

x + (E^(4 - x^2)*x)/20 - (E^(4 - x^2)*(5 + Log[4]))/4

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 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 \frac {1}{20} e^{4-x^2} \left (1+20 e^{-4+x^2}-2 x^2+x (50+10 \log (4))\right ) \, dx\\ &=\frac {1}{20} \int e^{4-x^2} \left (1+20 e^{-4+x^2}-2 x^2+x (50+10 \log (4))\right ) \, dx\\ &=\frac {1}{20} \int \left (20+e^{4-x^2}-2 e^{4-x^2} x^2+10 e^{4-x^2} x (5+\log (4))\right ) \, dx\\ &=x+\frac {1}{20} \int e^{4-x^2} \, dx-\frac {1}{10} \int e^{4-x^2} x^2 \, dx+\frac {1}{2} (5+\log (4)) \int e^{4-x^2} x \, dx\\ &=x+\frac {1}{20} e^{4-x^2} x+\frac {1}{40} e^4 \sqrt {\pi } \text {erf}(x)-\frac {1}{4} e^{4-x^2} (5+\log (4))-\frac {1}{20} \int e^{4-x^2} \, dx\\ &=x+\frac {1}{20} e^{4-x^2} x-\frac {1}{4} e^{4-x^2} (5+\log (4))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 33, normalized size = 1.38 \begin {gather*} x+\frac {1}{20} e^{4-x^2} x-\frac {1}{4} e^{4-x^2} (5+\log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(4 - x^2)*(1 + 20*E^(-4 + x^2) + 50*x - 2*x^2 + 10*x*Log[4]))/20,x]

[Out]

x + (E^(4 - x^2)*x)/20 - (E^(4 - x^2)*(5 + Log[4]))/4

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fricas [A]  time = 0.56, size = 34, normalized size = 1.42 \begin {gather*} \frac {1}{5} \, {\left (5 \, x e^{\left (x^{2} + 2 \, \log \relax (2) - 4\right )} + x - 10 \, \log \relax (2) - 25\right )} e^{\left (-x^{2} - 2 \, \log \relax (2) + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*exp(2*log(2)+x^2-4)+20*x*log(2)-2*x^2+50*x+1)/exp(2*log(2)+x^2-4),x, algorithm="fricas")

[Out]

1/5*(5*x*e^(x^2 + 2*log(2) - 4) + x - 10*log(2) - 25)*e^(-x^2 - 2*log(2) + 4)

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giac [A]  time = 0.17, size = 19, normalized size = 0.79 \begin {gather*} \frac {1}{20} \, {\left (x - 10 \, \log \relax (2) - 25\right )} e^{\left (-x^{2} + 4\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*exp(2*log(2)+x^2-4)+20*x*log(2)-2*x^2+50*x+1)/exp(2*log(2)+x^2-4),x, algorithm="giac")

[Out]

1/20*(x - 10*log(2) - 25)*e^(-x^2 + 4) + x

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maple [A]  time = 0.03, size = 21, normalized size = 0.88

method	result	size
risch	\(x +\frac {\left (-10 \ln \relax (2)-25+x \right ) {\mathrm e}^{-\left (x -2\right ) \left (2+x \right )}}{20}\)	\(21\)
norman	\(\frac {\left (x \,{\mathrm e}^{2 \ln \relax (2)+x^{2}-4}+\frac {x}{5}-2 \ln \relax (2)-5\right ) {\mathrm e}^{-x^{2}+4}}{4}\)	\(35\)
default	\(x +\frac {{\mathrm e}^{4} \sqrt {\pi }\, \erf \relax (x )}{40}-\frac {5 \,{\mathrm e}^{-x^{2}} {\mathrm e}^{4}}{4}-\frac {{\mathrm e}^{4} \left (-\frac {x \,{\mathrm e}^{-x^{2}}}{2}+\frac {\sqrt {\pi }\, \erf \relax (x )}{4}\right )}{10}-\frac {{\mathrm e}^{4} \ln \relax (2) {\mathrm e}^{-x^{2}}}{2}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(5*exp(2*ln(2)+x^2-4)+20*x*ln(2)-2*x^2+50*x+1)/exp(2*ln(2)+x^2-4),x,method=_RETURNVERBOSE)

[Out]

x+1/20*(-10*ln(2)-25+x)*exp(-(x-2)*(2+x))

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maxima [A]  time = 0.36, size = 35, normalized size = 1.46 \begin {gather*} \frac {1}{20} \, x e^{\left (-x^{2} + 4\right )} - \frac {1}{2} \, e^{\left (-x^{2} + 4\right )} \log \relax (2) + x - \frac {5}{4} \, e^{\left (-x^{2} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*exp(2*log(2)+x^2-4)+20*x*log(2)-2*x^2+50*x+1)/exp(2*log(2)+x^2-4),x, algorithm="maxima")

[Out]

1/20*x*e^(-x^2 + 4) - 1/2*e^(-x^2 + 4)*log(2) + x - 5/4*e^(-x^2 + 4)

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mupad [B]  time = 0.15, size = 35, normalized size = 1.46 \begin {gather*} x-\frac {5\,{\mathrm {e}}^{4-x^2}}{4}-\frac {{\mathrm {e}}^{4-x^2}\,\ln \relax (2)}{2}+\frac {x\,{\mathrm {e}}^{4-x^2}}{20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(4 - x^2 - 2*log(2))*(10*x + exp(2*log(2) + x^2 - 4) + 4*x*log(2) - (2*x^2)/5 + 1/5),x)

[Out]

x - (5*exp(4 - x^2))/4 - (exp(4 - x^2)*log(2))/2 + (x*exp(4 - x^2))/20

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sympy [A]  time = 0.14, size = 17, normalized size = 0.71 \begin {gather*} x + \frac {\left (x - 25 - 10 \log {\relax (2 )}\right ) e^{4 - x^{2}}}{20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*exp(2*ln(2)+x**2-4)+20*x*ln(2)-2*x**2+50*x+1)/exp(2*ln(2)+x**2-4),x)

[Out]

x + (x - 25 - 10*log(2))*exp(4 - x**2)/20

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