3.67.13 \(\int \frac {e^{-\frac {x^{10}+(-8 x^6+2 x^7) \log ^2(2)+(16 x^2-8 x^3+x^4) \log ^4(2)}{\log ^4(2)}} (-30 x^9+(144 x^5-42 x^6) \log ^2(2)+2 e^{x^2+\frac {x^{10}+(-8 x^6+2 x^7) \log ^2(2)+(16 x^2-8 x^3+x^4) \log ^4(2)}{\log ^4(2)}} x \log ^4(2)+(-96 x+72 x^2-12 x^3) \log ^4(2))}{\log ^4(2)} \, dx\)

Optimal. Leaf size=31 \[ e^{x^2}+3 e^{-x^2 \left (4-x-\frac {x^4}{\log ^2(2)}\right )^2} \]

________________________________________________________________________________________

Rubi [A]  time = 17.53, antiderivative size = 37, normalized size of antiderivative = 1.19, number of steps used = 8, number of rules used = 6, integrand size = 150, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {12, 6741, 6742, 2225, 2209, 6706} \begin {gather*} 3 \exp \left (-\frac {x^2 \left (x^4+x \log ^2(2)-4 \log ^2(2)\right )^2}{\log ^4(2)}\right )+e^{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-30*x^9 + (144*x^5 - 42*x^6)*Log[2]^2 + 2*E^(x^2 + (x^10 + (-8*x^6 + 2*x^7)*Log[2]^2 + (16*x^2 - 8*x^3 +
x^4)*Log[2]^4)/Log[2]^4)*x*Log[2]^4 + (-96*x + 72*x^2 - 12*x^3)*Log[2]^4)/(E^((x^10 + (-8*x^6 + 2*x^7)*Log[2]^
2 + (16*x^2 - 8*x^3 + x^4)*Log[2]^4)/Log[2]^4)*Log[2]^4),x]

[Out]

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

Rule 12

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

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 2225

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && BinomialQ[v, x] &&  !(LinearMatchQ[u, x] && BinomialMatchQ[v, x])

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \exp \left (-\frac {x^{10}+\left (-8 x^6+2 x^7\right ) \log ^2(2)+\left (16 x^2-8 x^3+x^4\right ) \log ^4(2)}{\log ^4(2)}\right ) \left (-30 x^9+\left (144 x^5-42 x^6\right ) \log ^2(2)+2 \exp \left (x^2+\frac {x^{10}+\left (-8 x^6+2 x^7\right ) \log ^2(2)+\left (16 x^2-8 x^3+x^4\right ) \log ^4(2)}{\log ^4(2)}\right ) x \log ^4(2)+\left (-96 x+72 x^2-12 x^3\right ) \log ^4(2)\right ) \, dx}{\log ^4(2)}\\ &=\frac {\int 2 \exp \left (-\frac {x^2 \left (x^4-4 \log ^2(2)+x \log ^2(2)\right )^2}{\log ^4(2)}\right ) x \left (-15 x^8+72 x^4 \log ^2(2)-21 x^5 \log ^2(2)-48 \log ^4(2)+\exp \left (x^2 \left (17-8 x+x^2+\frac {x^8}{\log ^4(2)}-\frac {8 x^4}{\log ^2(2)}+\frac {2 x^5}{\log ^2(2)}\right )\right ) \log ^4(2)+36 x \log ^4(2)-6 x^2 \log ^4(2)\right ) \, dx}{\log ^4(2)}\\ &=\frac {2 \int \exp \left (-\frac {x^2 \left (x^4-4 \log ^2(2)+x \log ^2(2)\right )^2}{\log ^4(2)}\right ) x \left (-15 x^8+72 x^4 \log ^2(2)-21 x^5 \log ^2(2)-48 \log ^4(2)+\exp \left (x^2 \left (17-8 x+x^2+\frac {x^8}{\log ^4(2)}-\frac {8 x^4}{\log ^2(2)}+\frac {2 x^5}{\log ^2(2)}\right )\right ) \log ^4(2)+36 x \log ^4(2)-6 x^2 \log ^4(2)\right ) \, dx}{\log ^4(2)}\\ &=\frac {2 \int \left (\exp \left (17 x^2-8 x^3+x^4+\frac {x^{10}}{\log ^4(2)}-\frac {8 x^6}{\log ^2(2)}+\frac {2 x^7}{\log ^2(2)}-\frac {x^2 \left (x^4-4 \log ^2(2)+x \log ^2(2)\right )^2}{\log ^4(2)}\right ) x \log ^4(2)-3 \exp \left (-\frac {x^2 \left (x^4-4 \log ^2(2)+x \log ^2(2)\right )^2}{\log ^4(2)}\right ) x \left (5 x^8-24 x^4 \log ^2(2)+7 x^5 \log ^2(2)+16 \log ^4(2)-12 x \log ^4(2)+2 x^2 \log ^4(2)\right )\right ) \, dx}{\log ^4(2)}\\ &=2 \int \exp \left (17 x^2-8 x^3+x^4+\frac {x^{10}}{\log ^4(2)}-\frac {8 x^6}{\log ^2(2)}+\frac {2 x^7}{\log ^2(2)}-\frac {x^2 \left (x^4-4 \log ^2(2)+x \log ^2(2)\right )^2}{\log ^4(2)}\right ) x \, dx-\frac {6 \int \exp \left (-\frac {x^2 \left (x^4-4 \log ^2(2)+x \log ^2(2)\right )^2}{\log ^4(2)}\right ) x \left (5 x^8-24 x^4 \log ^2(2)+7 x^5 \log ^2(2)+16 \log ^4(2)-12 x \log ^4(2)+2 x^2 \log ^4(2)\right ) \, dx}{\log ^4(2)}\\ &=3 \exp \left (-\frac {x^2 \left (x^4-4 \log ^2(2)+x \log ^2(2)\right )^2}{\log ^4(2)}\right )+2 \int e^{x^2} x \, dx\\ &=e^{x^2}+3 \exp \left (-\frac {x^2 \left (x^4-4 \log ^2(2)+x \log ^2(2)\right )^2}{\log ^4(2)}\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.25, size = 56, normalized size = 1.81 \begin {gather*} e^{-16 x^2} \left (e^{17 x^2}+3 e^{x^3 \left (8-x-\frac {x^7}{\log ^4(2)}+\frac {8 x^3}{\log ^2(2)}-\frac {2 x^4}{\log ^2(2)}\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-30*x^9 + (144*x^5 - 42*x^6)*Log[2]^2 + 2*E^(x^2 + (x^10 + (-8*x^6 + 2*x^7)*Log[2]^2 + (16*x^2 - 8*
x^3 + x^4)*Log[2]^4)/Log[2]^4)*x*Log[2]^4 + (-96*x + 72*x^2 - 12*x^3)*Log[2]^4)/(E^((x^10 + (-8*x^6 + 2*x^7)*L
og[2]^2 + (16*x^2 - 8*x^3 + x^4)*Log[2]^4)/Log[2]^4)*Log[2]^4),x]

[Out]

(E^(17*x^2) + 3*E^(x^3*(8 - x - x^7/Log[2]^4 + (8*x^3)/Log[2]^2 - (2*x^4)/Log[2]^2)))/E^(16*x^2)

________________________________________________________________________________________

fricas [B]  time = 0.57, size = 137, normalized size = 4.42 \begin {gather*} e^{\left (\frac {x^{10} + {\left (x^{4} - 8 \, x^{3} + 17 \, x^{2}\right )} \log \relax (2)^{4} + 2 \, {\left (x^{7} - 4 \, x^{6}\right )} \log \relax (2)^{2}}{\log \relax (2)^{4}} - \frac {x^{10} + {\left (x^{4} - 8 \, x^{3} + 16 \, x^{2}\right )} \log \relax (2)^{4} + 2 \, {\left (x^{7} - 4 \, x^{6}\right )} \log \relax (2)^{2}}{\log \relax (2)^{4}}\right )} + 3 \, e^{\left (-\frac {x^{10} + {\left (x^{4} - 8 \, x^{3} + 16 \, x^{2}\right )} \log \relax (2)^{4} + 2 \, {\left (x^{7} - 4 \, x^{6}\right )} \log \relax (2)^{2}}{\log \relax (2)^{4}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(2)^4*exp(x^2)*exp(((x^4-8*x^3+16*x^2)*log(2)^4+(2*x^7-8*x^6)*log(2)^2+x^10)/log(2)^4)+(-12*
x^3+72*x^2-96*x)*log(2)^4+(-42*x^6+144*x^5)*log(2)^2-30*x^9)/log(2)^4/exp(((x^4-8*x^3+16*x^2)*log(2)^4+(2*x^7-
8*x^6)*log(2)^2+x^10)/log(2)^4),x, algorithm="fricas")

[Out]

e^((x^10 + (x^4 - 8*x^3 + 17*x^2)*log(2)^4 + 2*(x^7 - 4*x^6)*log(2)^2)/log(2)^4 - (x^10 + (x^4 - 8*x^3 + 16*x^
2)*log(2)^4 + 2*(x^7 - 4*x^6)*log(2)^2)/log(2)^4) + 3*e^(-(x^10 + (x^4 - 8*x^3 + 16*x^2)*log(2)^4 + 2*(x^7 - 4
*x^6)*log(2)^2)/log(2)^4)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left (15 \, x^{9} - x e^{\left (x^{2} + \frac {x^{10} + {\left (x^{4} - 8 \, x^{3} + 16 \, x^{2}\right )} \log \relax (2)^{4} + 2 \, {\left (x^{7} - 4 \, x^{6}\right )} \log \relax (2)^{2}}{\log \relax (2)^{4}}\right )} \log \relax (2)^{4} + 6 \, {\left (x^{3} - 6 \, x^{2} + 8 \, x\right )} \log \relax (2)^{4} + 3 \, {\left (7 \, x^{6} - 24 \, x^{5}\right )} \log \relax (2)^{2}\right )} e^{\left (-\frac {x^{10} + {\left (x^{4} - 8 \, x^{3} + 16 \, x^{2}\right )} \log \relax (2)^{4} + 2 \, {\left (x^{7} - 4 \, x^{6}\right )} \log \relax (2)^{2}}{\log \relax (2)^{4}}\right )}}{\log \relax (2)^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(2)^4*exp(x^2)*exp(((x^4-8*x^3+16*x^2)*log(2)^4+(2*x^7-8*x^6)*log(2)^2+x^10)/log(2)^4)+(-12*
x^3+72*x^2-96*x)*log(2)^4+(-42*x^6+144*x^5)*log(2)^2-30*x^9)/log(2)^4/exp(((x^4-8*x^3+16*x^2)*log(2)^4+(2*x^7-
8*x^6)*log(2)^2+x^10)/log(2)^4),x, algorithm="giac")

[Out]

integrate(-2*(15*x^9 - x*e^(x^2 + (x^10 + (x^4 - 8*x^3 + 16*x^2)*log(2)^4 + 2*(x^7 - 4*x^6)*log(2)^2)/log(2)^4
)*log(2)^4 + 6*(x^3 - 6*x^2 + 8*x)*log(2)^4 + 3*(7*x^6 - 24*x^5)*log(2)^2)*e^(-(x^10 + (x^4 - 8*x^3 + 16*x^2)*
log(2)^4 + 2*(x^7 - 4*x^6)*log(2)^2)/log(2)^4)/log(2)^4, x)

________________________________________________________________________________________

maple [A]  time = 0.10, size = 36, normalized size = 1.16

method	result	size
risch	\({\mathrm e}^{x^{2}}+3 \,{\mathrm e}^{-\frac {x^{2} \left (x^{4}+x \ln \relax (2)^{2}-4 \ln \relax (2)^{2}\right )^{2}}{\ln \relax (2)^{4}}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*ln(2)^4*exp(x^2)*exp(((x^4-8*x^3+16*x^2)*ln(2)^4+(2*x^7-8*x^6)*ln(2)^2+x^10)/ln(2)^4)+(-12*x^3+72*x^2
-96*x)*ln(2)^4+(-42*x^6+144*x^5)*ln(2)^2-30*x^9)/ln(2)^4/exp(((x^4-8*x^3+16*x^2)*ln(2)^4+(2*x^7-8*x^6)*ln(2)^2
+x^10)/ln(2)^4),x,method=_RETURNVERBOSE)

[Out]

exp(x^2)+3*exp(-x^2*(x^4+x*ln(2)^2-4*ln(2)^2)^2/ln(2)^4)

________________________________________________________________________________________

maxima [B]  time = 0.66, size = 65, normalized size = 2.10 \begin {gather*} \frac {3 \, e^{\left (-\frac {x^{10}}{\log \relax (2)^{4}} - \frac {2 \, x^{7}}{\log \relax (2)^{2}} - x^{4} + \frac {8 \, x^{6}}{\log \relax (2)^{2}} + 8 \, x^{3} - 16 \, x^{2}\right )} \log \relax (2)^{4} + e^{\left (x^{2}\right )} \log \relax (2)^{4}}{\log \relax (2)^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(2)^4*exp(x^2)*exp(((x^4-8*x^3+16*x^2)*log(2)^4+(2*x^7-8*x^6)*log(2)^2+x^10)/log(2)^4)+(-12*
x^3+72*x^2-96*x)*log(2)^4+(-42*x^6+144*x^5)*log(2)^2-30*x^9)/log(2)^4/exp(((x^4-8*x^3+16*x^2)*log(2)^4+(2*x^7-
8*x^6)*log(2)^2+x^10)/log(2)^4),x, algorithm="maxima")

[Out]

(3*e^(-x^10/log(2)^4 - 2*x^7/log(2)^2 - x^4 + 8*x^6/log(2)^2 + 8*x^3 - 16*x^2)*log(2)^4 + e^(x^2)*log(2)^4)/lo
g(2)^4

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\mathrm {e}}^{-\frac {{\ln \relax (2)}^4\,\left (x^4-8\,x^3+16\,x^2\right )+x^{10}-{\ln \relax (2)}^2\,\left (8\,x^6-2\,x^7\right )}{{\ln \relax (2)}^4}}\,\left ({\ln \relax (2)}^4\,\left (12\,x^3-72\,x^2+96\,x\right )+30\,x^9-{\ln \relax (2)}^2\,\left (144\,x^5-42\,x^6\right )-2\,x\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{\frac {{\ln \relax (2)}^4\,\left (x^4-8\,x^3+16\,x^2\right )+x^{10}-{\ln \relax (2)}^2\,\left (8\,x^6-2\,x^7\right )}{{\ln \relax (2)}^4}}\,{\ln \relax (2)}^4\right )}{{\ln \relax (2)}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(log(2)^4*(16*x^2 - 8*x^3 + x^4) + x^10 - log(2)^2*(8*x^6 - 2*x^7))/log(2)^4)*(log(2)^4*(96*x - 72*
x^2 + 12*x^3) + 30*x^9 - log(2)^2*(144*x^5 - 42*x^6) - 2*x*exp(x^2)*exp((log(2)^4*(16*x^2 - 8*x^3 + x^4) + x^1
0 - log(2)^2*(8*x^6 - 2*x^7))/log(2)^4)*log(2)^4))/log(2)^4,x)

[Out]

int(-(exp(-(log(2)^4*(16*x^2 - 8*x^3 + x^4) + x^10 - log(2)^2*(8*x^6 - 2*x^7))/log(2)^4)*(log(2)^4*(96*x - 72*
x^2 + 12*x^3) + 30*x^9 - log(2)^2*(144*x^5 - 42*x^6) - 2*x*exp(x^2)*exp((log(2)^4*(16*x^2 - 8*x^3 + x^4) + x^1
0 - log(2)^2*(8*x^6 - 2*x^7))/log(2)^4)*log(2)^4))/log(2)^4, x)

________________________________________________________________________________________

sympy [B]  time = 0.44, size = 49, normalized size = 1.58 \begin {gather*} e^{x^{2}} + 3 e^{- \frac {x^{10} + \left (2 x^{7} - 8 x^{6}\right ) \log {\relax (2 )}^{2} + \left (x^{4} - 8 x^{3} + 16 x^{2}\right ) \log {\relax (2 )}^{4}}{\log {\relax (2 )}^{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*ln(2)**4*exp(x**2)*exp(((x**4-8*x**3+16*x**2)*ln(2)**4+(2*x**7-8*x**6)*ln(2)**2+x**10)/ln(2)**4
)+(-12*x**3+72*x**2-96*x)*ln(2)**4+(-42*x**6+144*x**5)*ln(2)**2-30*x**9)/ln(2)**4/exp(((x**4-8*x**3+16*x**2)*l
n(2)**4+(2*x**7-8*x**6)*ln(2)**2+x**10)/ln(2)**4),x)

[Out]

exp(x**2) + 3*exp(-(x**10 + (2*x**7 - 8*x**6)*log(2)**2 + (x**4 - 8*x**3 + 16*x**2)*log(2)**4)/log(2)**4)

________________________________________________________________________________________