3.52.98 \(\int \frac {e^{-x^2} (e^{x^2} (-e^{3+x} \log (5)+(1-4 x) \log (5))+(2 x-2 x^3) \log (\log (4)))}{\log (5)} \, dx\) [5198]

Optimal. Leaf size=33 \[ -e^{3+x}+x-x \left (2 x-\frac {e^{-x^2} x \log (\log (4))}{\log (5)}\right ) \]

[Out]

x-exp(3+x)-x*(2*x-ln(2*ln(2))/ln(5)/exp(x^2)*x)

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Rubi [A]
time = 0.24, antiderivative size = 32, normalized size of antiderivative = 0.97, number of steps used = 10, number of rules used = 6, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {12, 6874, 2225, 2258, 2240, 2243} \begin {gather*} -2 x^2+\frac {e^{-x^2} x^2 \log (\log (4))}{\log (5)}+x-e^{x+3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x^2*(-(E^(3 + x)*Log[5]) + (1 - 4*x)*Log[5]) + (2*x - 2*x^3)*Log[Log[4]])/(E^x^2*Log[5]),x]

[Out]

-E^(3 + x) + x - 2*x^2 + (x^2*Log[Log[4]])/(E^x^2*Log[5])

Rule 12

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

Rule 2225

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 2240

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 2243

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 2258

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 6874

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int e^{-x^2} \left (e^{x^2} \left (-e^{3+x} \log (5)+(1-4 x) \log (5)\right )+\left (2 x-2 x^3\right ) \log (\log (4))\right ) \, dx}{\log (5)}\\ &=\frac {\int \left (-\left (\left (-1+e^{3+x}+4 x\right ) \log (5)\right )-2 e^{-x^2} x \left (-1+x^2\right ) \log (\log (4))\right ) \, dx}{\log (5)}\\ &=-\frac {(2 \log (\log (4))) \int e^{-x^2} x \left (-1+x^2\right ) \, dx}{\log (5)}-\int \left (-1+e^{3+x}+4 x\right ) \, dx\\ &=x-2 x^2-\frac {(2 \log (\log (4))) \int \left (-e^{-x^2} x+e^{-x^2} x^3\right ) \, dx}{\log (5)}-\int e^{3+x} \, dx\\ &=-e^{3+x}+x-2 x^2+\frac {(2 \log (\log (4))) \int e^{-x^2} x \, dx}{\log (5)}-\frac {(2 \log (\log (4))) \int e^{-x^2} x^3 \, dx}{\log (5)}\\ &=-e^{3+x}+x-2 x^2-\frac {e^{-x^2} \log (\log (4))}{\log (5)}+\frac {e^{-x^2} x^2 \log (\log (4))}{\log (5)}-\frac {(2 \log (\log (4))) \int e^{-x^2} x \, dx}{\log (5)}\\ &=-e^{3+x}+x-2 x^2+\frac {e^{-x^2} x^2 \log (\log (4))}{\log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.09, size = 32, normalized size = 0.97 \begin {gather*} -e^{3+x}+x-2 x^2+\frac {e^{-x^2} x^2 \log (\log (4))}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x^2*(-(E^(3 + x)*Log[5]) + (1 - 4*x)*Log[5]) + (2*x - 2*x^3)*Log[Log[4]])/(E^x^2*Log[5]),x]

[Out]

-E^(3 + x) + x - 2*x^2 + (x^2*Log[Log[4]])/(E^x^2*Log[5])

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Maple [A]
time = 0.22, size = 51, normalized size = 1.55

method	result	size
risch	\(-2 x^{2}+x +\frac {\left (\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right ) x^{2} {\mathrm e}^{-x^{2}}}{\ln \left (5\right )}-{\mathrm e}^{3+x}\)	\(34\)
norman	\(\left ({\mathrm e}^{x^{2}} x +\frac {\left (\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right ) x^{2}}{\ln \left (5\right )}-2 x^{2} {\mathrm e}^{x^{2}}-{\mathrm e}^{x^{2}} {\mathrm e}^{3+x}\right ) {\mathrm e}^{-x^{2}}\)	\(48\)
default	\(\frac {-2 x^{2} \ln \left (5\right )-{\mathrm e}^{x} \ln \left (5\right ) {\mathrm e}^{3}+\ln \left (2\right ) {\mathrm e}^{-x^{2}} x^{2}+\ln \left (\ln \left (2\right )\right ) {\mathrm e}^{-x^{2}} x^{2}+x \ln \left (5\right )}{\ln \left (5\right )}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^3+2*x)*ln(2*ln(2))+(-ln(5)*exp(3+x)+(-4*x+1)*ln(5))*exp(x^2))/ln(5)/exp(x^2),x,method=_RETURNVERBOS
E)

[Out]

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

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Maxima [A]
time = 0.29, size = 56, normalized size = 1.70 \begin {gather*} -\frac {2 \, x^{2} \log \left (5\right ) - {\left (x^{2} + 1\right )} e^{\left (-x^{2}\right )} \log \left (2 \, \log \left (2\right )\right ) - x \log \left (5\right ) + e^{\left (x + 3\right )} \log \left (5\right ) + e^{\left (-x^{2}\right )} \log \left (2 \, \log \left (2\right )\right )}{\log \left (5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+2*x)*log(2*log(2))+(-log(5)*exp(3+x)+(-4*x+1)*log(5))*exp(x^2))/log(5)/exp(x^2),x, algorith
m="maxima")

[Out]

-(2*x^2*log(5) - (x^2 + 1)*e^(-x^2)*log(2*log(2)) - x*log(5) + e^(x + 3)*log(5) + e^(-x^2)*log(2*log(2)))/log(
5)

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Fricas [A]
time = 0.37, size = 47, normalized size = 1.42 \begin {gather*} \frac {{\left (x^{2} \log \left (2 \, \log \left (2\right )\right ) - {\left ({\left (2 \, x^{2} - x\right )} \log \left (5\right ) + e^{\left (x + 3\right )} \log \left (5\right )\right )} e^{\left (x^{2}\right )}\right )} e^{\left (-x^{2}\right )}}{\log \left (5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+2*x)*log(2*log(2))+(-log(5)*exp(3+x)+(-4*x+1)*log(5))*exp(x^2))/log(5)/exp(x^2),x, algorith
m="fricas")

[Out]

(x^2*log(2*log(2)) - ((2*x^2 - x)*log(5) + e^(x + 3)*log(5))*e^(x^2))*e^(-x^2)/log(5)

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Sympy [A]
time = 0.14, size = 34, normalized size = 1.03 \begin {gather*} - 2 x^{2} + x + \frac {\left (x^{2} \log {\left (\log {\left (2 \right )} \right )} + x^{2} \log {\left (2 \right )}\right ) e^{- x^{2}}}{\log {\left (5 \right )}} - e^{x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**3+2*x)*ln(2*ln(2))+(-ln(5)*exp(3+x)+(-4*x+1)*ln(5))*exp(x**2))/ln(5)/exp(x**2),x)

[Out]

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

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Giac [A]
time = 0.39, size = 48, normalized size = 1.45 \begin {gather*} -\frac {2 \, x^{2} \log \left (5\right ) - {\left (x^{2} \log \left (2\right ) + x^{2} \log \left (\log \left (2\right )\right )\right )} e^{\left (-x^{2}\right )} - x \log \left (5\right ) + e^{\left (x + 3\right )} \log \left (5\right )}{\log \left (5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+2*x)*log(2*log(2))+(-log(5)*exp(3+x)+(-4*x+1)*log(5))*exp(x^2))/log(5)/exp(x^2),x, algorith
m="giac")

[Out]

-(2*x^2*log(5) - (x^2*log(2) + x^2*log(log(2)))*e^(-x^2) - x*log(5) + e^(x + 3)*log(5))/log(5)

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Mupad [B]
time = 0.14, size = 46, normalized size = 1.39 \begin {gather*} x-{\mathrm {e}}^{x+3}-2\,x^2+\frac {x^2\,{\mathrm {e}}^{-x^2}\,\ln \left (2\right )}{\ln \left (5\right )}+\frac {x^2\,{\mathrm {e}}^{-x^2}\,\ln \left (\ln \left (2\right )\right )}{\ln \left (5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x^2)*(log(2*log(2))*(2*x - 2*x^3) - exp(x^2)*(log(5)*(4*x - 1) + exp(x + 3)*log(5))))/log(5),x)

[Out]

x - exp(x + 3) - 2*x^2 + (x^2*exp(-x^2)*log(2))/log(5) + (x^2*exp(-x^2)*log(log(2)))/log(5)

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