3.77.10 \(\int e^{-1+36 x^2+4 e^{\frac {2 x^2}{e}} x^2-12 x^3+x^4+(12 x^3-2 x^4) \log ^2(25)+x^4 \log ^4(25)+e^{\frac {x^2}{e}} (24 x^2-4 x^3+4 x^3 \log ^2(25))} (e (72 x-36 x^2+4 x^3)+e^{\frac {2 x^2}{e}} (8 e x+16 x^3)+e (36 x^2-8 x^3) \log ^2(25)+4 e x^3 \log ^4(25)+e^{\frac {x^2}{e}} (48 x^3-8 x^4+e (48 x-12 x^2)+(12 e x^2+8 x^4) \log ^2(25))) \, dx\)

Optimal. Leaf size=31 \[ e^{x^2 \left (2 \left (3+e^{\frac {x^2}{e}}\right )-x+x \log ^2(25)\right )^2} \]

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Rubi [A]  time = 6.52, antiderivative size = 59, normalized size of antiderivative = 1.90, number of steps used = 3, number of rules used = 3, integrand size = 201, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {6688, 12, 6706} \begin {gather*} \exp \left (x^4 \left (1-\log ^2(25)\right )^2+4 \left (e^{\frac {x^2}{e}}+3\right )^2 x^2-4 \left (e^{\frac {x^2}{e}}+3\right ) x^3 \left (1-\log ^2(25)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(-1 + 36*x^2 + 4*E^((2*x^2)/E)*x^2 - 12*x^3 + x^4 + (12*x^3 - 2*x^4)*Log[25]^2 + x^4*Log[25]^4 + E^(x^2/
E)*(24*x^2 - 4*x^3 + 4*x^3*Log[25]^2))*(E*(72*x - 36*x^2 + 4*x^3) + E^((2*x^2)/E)*(8*E*x + 16*x^3) + E*(36*x^2
- 8*x^3)*Log[25]^2 + 4*E*x^3*Log[25]^4 + E^(x^2/E)*(48*x^3 - 8*x^4 + E*(48*x - 12*x^2) + (12*E*x^2 + 8*x^4)*L
og[25]^2)),x]

[Out]

E^(4*(3 + E^(x^2/E))^2*x^2 - 4*(3 + E^(x^2/E))*x^3*(1 - Log[25]^2) + x^4*(1 - Log[25]^2)^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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int 4 \exp \left (-1+4 \left (3+e^{\frac {x^2}{e}}\right )^2 x^2+4 \left (3+e^{\frac {x^2}{e}}\right ) x^3 \left (-1+\log ^2(25)\right )+x^4 \left (-1+\log ^2(25)\right )^2\right ) x \left (6+2 e^{\frac {x^2}{e}}+x \left (-1+\log ^2(25)\right )\right ) \left (e^{1+\frac {x^2}{e}}+2 e^{\frac {x^2}{e}} x^2+e \left (3+x \left (-1+\log ^2(25)\right )\right )\right ) \, dx\\ &=4 \int \exp \left (-1+4 \left (3+e^{\frac {x^2}{e}}\right )^2 x^2+4 \left (3+e^{\frac {x^2}{e}}\right ) x^3 \left (-1+\log ^2(25)\right )+x^4 \left (-1+\log ^2(25)\right )^2\right ) x \left (6+2 e^{\frac {x^2}{e}}+x \left (-1+\log ^2(25)\right )\right ) \left (e^{1+\frac {x^2}{e}}+2 e^{\frac {x^2}{e}} x^2+e \left (3+x \left (-1+\log ^2(25)\right )\right )\right ) \, dx\\ &=\exp \left (4 \left (3+e^{\frac {x^2}{e}}\right )^2 x^2-4 \left (3+e^{\frac {x^2}{e}}\right ) x^3 \left (1-\log ^2(25)\right )+x^4 \left (1-\log ^2(25)\right )^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 29, normalized size = 0.94 \begin {gather*} e^{x^2 \left (6+2 e^{\frac {x^2}{e}}+x \left (-1+\log ^2(25)\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(-1 + 36*x^2 + 4*E^((2*x^2)/E)*x^2 - 12*x^3 + x^4 + (12*x^3 - 2*x^4)*Log[25]^2 + x^4*Log[25]^4 + E
^(x^2/E)*(24*x^2 - 4*x^3 + 4*x^3*Log[25]^2))*(E*(72*x - 36*x^2 + 4*x^3) + E^((2*x^2)/E)*(8*E*x + 16*x^3) + E*(
36*x^2 - 8*x^3)*Log[25]^2 + 4*E*x^3*Log[25]^4 + E^(x^2/E)*(48*x^3 - 8*x^4 + E*(48*x - 12*x^2) + (12*E*x^2 + 8*
x^4)*Log[25]^2)),x]

[Out]

E^(x^2*(6 + 2*E^(x^2/E) + x*(-1 + Log[25]^2))^2)

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fricas [B]  time = 1.13, size = 81, normalized size = 2.61 \begin {gather*} e^{\left (16 \, x^{4} \log \relax (5)^{4} + x^{4} - 12 \, x^{3} + 4 \, x^{2} e^{\left (2 \, x^{2} e^{\left (-1\right )}\right )} - 8 \, {\left (x^{4} - 6 \, x^{3}\right )} \log \relax (5)^{2} + 36 \, x^{2} + 4 \, {\left (4 \, x^{3} \log \relax (5)^{2} - x^{3} + 6 \, x^{2}\right )} e^{\left (x^{2} e^{\left (-1\right )}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x*exp(1)+16*x^3)*exp(x^2/exp(1))^2+(4*(12*x^2*exp(1)+8*x^4)*log(5)^2+(-12*x^2+48*x)*exp(1)-8*x^4
+48*x^3)*exp(x^2/exp(1))+64*x^3*exp(1)*log(5)^4+4*(-8*x^3+36*x^2)*exp(1)*log(5)^2+(4*x^3-36*x^2+72*x)*exp(1))*
exp(4*x^2*exp(x^2/exp(1))^2+(16*x^3*log(5)^2-4*x^3+24*x^2)*exp(x^2/exp(1))+16*x^4*log(5)^4+4*(-2*x^4+12*x^3)*l
og(5)^2+x^4-12*x^3+36*x^2)/exp(1),x, algorithm="fricas")

[Out]

e^(16*x^4*log(5)^4 + x^4 - 12*x^3 + 4*x^2*e^(2*x^2*e^(-1)) - 8*(x^4 - 6*x^3)*log(5)^2 + 36*x^2 + 4*(4*x^3*log(
5)^2 - x^3 + 6*x^2)*e^(x^2*e^(-1)))

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giac [B]  time = 1.11, size = 95, normalized size = 3.06 \begin {gather*} e^{\left (16 \, x^{4} \log \relax (5)^{4} - 8 \, x^{4} \log \relax (5)^{2} + 16 \, x^{3} e^{\left (x^{2} e^{\left (-1\right )}\right )} \log \relax (5)^{2} + 48 \, x^{3} \log \relax (5)^{2} + x^{4} - 4 \, x^{3} e^{\left (x^{2} e^{\left (-1\right )}\right )} - 12 \, x^{3} + 4 \, x^{2} e^{\left (2 \, x^{2} e^{\left (-1\right )}\right )} + 24 \, x^{2} e^{\left (x^{2} e^{\left (-1\right )}\right )} + 36 \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x*exp(1)+16*x^3)*exp(x^2/exp(1))^2+(4*(12*x^2*exp(1)+8*x^4)*log(5)^2+(-12*x^2+48*x)*exp(1)-8*x^4
+48*x^3)*exp(x^2/exp(1))+64*x^3*exp(1)*log(5)^4+4*(-8*x^3+36*x^2)*exp(1)*log(5)^2+(4*x^3-36*x^2+72*x)*exp(1))*
exp(4*x^2*exp(x^2/exp(1))^2+(16*x^3*log(5)^2-4*x^3+24*x^2)*exp(x^2/exp(1))+16*x^4*log(5)^4+4*(-2*x^4+12*x^3)*l
og(5)^2+x^4-12*x^3+36*x^2)/exp(1),x, algorithm="giac")

[Out]

e^(16*x^4*log(5)^4 - 8*x^4*log(5)^2 + 16*x^3*e^(x^2*e^(-1))*log(5)^2 + 48*x^3*log(5)^2 + x^4 - 4*x^3*e^(x^2*e^
(-1)) - 12*x^3 + 4*x^2*e^(2*x^2*e^(-1)) + 24*x^2*e^(x^2*e^(-1)) + 36*x^2)

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maple [B]  time = 1.76, size = 82, normalized size = 2.65

method	result	size
risch	\({\mathrm e}^{x^{2} \left (16 \ln \relax (5)^{4} x^{2}+16 \ln \relax (5)^{2} {\mathrm e}^{x^{2} {\mathrm e}^{-1}} x -8 x^{2} \ln \relax (5)^{2}+48 x \ln \relax (5)^{2}-4 \,{\mathrm e}^{x^{2} {\mathrm e}^{-1}} x +x^{2}+4 \,{\mathrm e}^{2 x^{2} {\mathrm e}^{-1}}+24 \,{\mathrm e}^{x^{2} {\mathrm e}^{-1}}-12 x +36\right )}\)	\(82\)
norman	\({\mathrm e}^{4 x^{2} {\mathrm e}^{2 x^{2} {\mathrm e}^{-1}}+\left (16 x^{3} \ln \relax (5)^{2}-4 x^{3}+24 x^{2}\right ) {\mathrm e}^{x^{2} {\mathrm e}^{-1}}+16 x^{4} \ln \relax (5)^{4}+4 \left (-2 x^{4}+12 x^{3}\right ) \ln \relax (5)^{2}+x^{4}-12 x^{3}+36 x^{2}}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x*exp(1)+16*x^3)*exp(x^2/exp(1))^2+(4*(12*x^2*exp(1)+8*x^4)*ln(5)^2+(-12*x^2+48*x)*exp(1)-8*x^4+48*x^3
)*exp(x^2/exp(1))+64*x^3*exp(1)*ln(5)^4+4*(-8*x^3+36*x^2)*exp(1)*ln(5)^2+(4*x^3-36*x^2+72*x)*exp(1))*exp(4*x^2
*exp(x^2/exp(1))^2+(16*x^3*ln(5)^2-4*x^3+24*x^2)*exp(x^2/exp(1))+16*x^4*ln(5)^4+4*(-2*x^4+12*x^3)*ln(5)^2+x^4-
12*x^3+36*x^2)/exp(1),x,method=_RETURNVERBOSE)

[Out]

exp(x^2*(16*ln(5)^4*x^2+16*ln(5)^2*exp(x^2*exp(-1))*x-8*x^2*ln(5)^2+48*x*ln(5)^2-4*exp(x^2*exp(-1))*x+x^2+4*ex
p(2*x^2*exp(-1))+24*exp(x^2*exp(-1))-12*x+36))

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maxima [B]  time = 0.73, size = 95, normalized size = 3.06 \begin {gather*} e^{\left (16 \, x^{4} \log \relax (5)^{4} - 8 \, x^{4} \log \relax (5)^{2} + 16 \, x^{3} e^{\left (x^{2} e^{\left (-1\right )}\right )} \log \relax (5)^{2} + 48 \, x^{3} \log \relax (5)^{2} + x^{4} - 4 \, x^{3} e^{\left (x^{2} e^{\left (-1\right )}\right )} - 12 \, x^{3} + 4 \, x^{2} e^{\left (2 \, x^{2} e^{\left (-1\right )}\right )} + 24 \, x^{2} e^{\left (x^{2} e^{\left (-1\right )}\right )} + 36 \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x*exp(1)+16*x^3)*exp(x^2/exp(1))^2+(4*(12*x^2*exp(1)+8*x^4)*log(5)^2+(-12*x^2+48*x)*exp(1)-8*x^4
+48*x^3)*exp(x^2/exp(1))+64*x^3*exp(1)*log(5)^4+4*(-8*x^3+36*x^2)*exp(1)*log(5)^2+(4*x^3-36*x^2+72*x)*exp(1))*
exp(4*x^2*exp(x^2/exp(1))^2+(16*x^3*log(5)^2-4*x^3+24*x^2)*exp(x^2/exp(1))+16*x^4*log(5)^4+4*(-2*x^4+12*x^3)*l
og(5)^2+x^4-12*x^3+36*x^2)/exp(1),x, algorithm="maxima")

[Out]

e^(16*x^4*log(5)^4 - 8*x^4*log(5)^2 + 16*x^3*e^(x^2*e^(-1))*log(5)^2 + 48*x^3*log(5)^2 + x^4 - 4*x^3*e^(x^2*e^
(-1)) - 12*x^3 + 4*x^2*e^(2*x^2*e^(-1)) + 24*x^2*e^(x^2*e^(-1)) + 36*x^2)

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mupad [B]  time = 5.97, size = 104, normalized size = 3.35 \begin {gather*} {\mathrm {e}}^{x^4}\,{\mathrm {e}}^{-8\,x^4\,{\ln \relax (5)}^2}\,{\mathrm {e}}^{16\,x^4\,{\ln \relax (5)}^4}\,{\mathrm {e}}^{48\,x^3\,{\ln \relax (5)}^2}\,{\mathrm {e}}^{16\,x^3\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^{-1}}\,{\ln \relax (5)}^2}\,{\mathrm {e}}^{-12\,x^3}\,{\mathrm {e}}^{36\,x^2}\,{\mathrm {e}}^{-4\,x^3\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^{-1}}}\,{\mathrm {e}}^{4\,x^2\,{\mathrm {e}}^{2\,x^2\,{\mathrm {e}}^{-1}}}\,{\mathrm {e}}^{24\,x^2\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^{-1}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-1)*exp(16*x^4*log(5)^4 + exp(x^2*exp(-1))*(16*x^3*log(5)^2 + 24*x^2 - 4*x^3) + 4*x^2*exp(2*x^2*exp(-1
)) + 36*x^2 - 12*x^3 + x^4 + 4*log(5)^2*(12*x^3 - 2*x^4))*(exp(2*x^2*exp(-1))*(8*x*exp(1) + 16*x^3) + exp(1)*(
72*x - 36*x^2 + 4*x^3) + exp(x^2*exp(-1))*(exp(1)*(48*x - 12*x^2) + 4*log(5)^2*(12*x^2*exp(1) + 8*x^4) + 48*x^
3 - 8*x^4) + 4*exp(1)*log(5)^2*(36*x^2 - 8*x^3) + 64*x^3*exp(1)*log(5)^4),x)

[Out]

exp(x^4)*exp(-8*x^4*log(5)^2)*exp(16*x^4*log(5)^4)*exp(48*x^3*log(5)^2)*exp(16*x^3*exp(x^2*exp(-1))*log(5)^2)*
exp(-12*x^3)*exp(36*x^2)*exp(-4*x^3*exp(x^2*exp(-1)))*exp(4*x^2*exp(2*x^2*exp(-1)))*exp(24*x^2*exp(x^2*exp(-1)
))

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sympy [B]  time = 0.87, size = 83, normalized size = 2.68 \begin {gather*} e^{x^{4} + 16 x^{4} \log {\relax (5 )}^{4} - 12 x^{3} + 4 x^{2} e^{\frac {2 x^{2}}{e}} + 36 x^{2} + \left (- 8 x^{4} + 48 x^{3}\right ) \log {\relax (5 )}^{2} + \left (- 4 x^{3} + 16 x^{3} \log {\relax (5 )}^{2} + 24 x^{2}\right ) e^{\frac {x^{2}}{e}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x*exp(1)+16*x**3)*exp(x**2/exp(1))**2+(4*(12*x**2*exp(1)+8*x**4)*ln(5)**2+(-12*x**2+48*x)*exp(1)
-8*x**4+48*x**3)*exp(x**2/exp(1))+64*x**3*exp(1)*ln(5)**4+4*(-8*x**3+36*x**2)*exp(1)*ln(5)**2+(4*x**3-36*x**2+
72*x)*exp(1))*exp(4*x**2*exp(x**2/exp(1))**2+(16*x**3*ln(5)**2-4*x**3+24*x**2)*exp(x**2/exp(1))+16*x**4*ln(5)*
*4+4*(-2*x**4+12*x**3)*ln(5)**2+x**4-12*x**3+36*x**2)/exp(1),x)

[Out]

exp(x**4 + 16*x**4*log(5)**4 - 12*x**3 + 4*x**2*exp(2*x**2*exp(-1)) + 36*x**2 + (-8*x**4 + 48*x**3)*log(5)**2
+ (-4*x**3 + 16*x**3*log(5)**2 + 24*x**2)*exp(x**2*exp(-1)))

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