3.77.54 \(\int e^{16 x^2+25 e^{2 x} x^2+e^{2 \log ^4(2)} x^2-8 x^3+x^4+e^{\log ^4(2)} (8 x^2-10 e^x x^2-2 x^3)+e^x (-40 x^2+10 x^3)} (32 x+2 e^{2 \log ^4(2)} x-24 x^2+4 x^3+e^{2 x} (50 x+50 x^2)+e^x (-80 x-10 x^2+10 x^3)+e^{\log ^4(2)} (16 x-6 x^2+e^x (-20 x-10 x^2))) \, dx\)

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

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Rubi [A]  time = 11.23, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 167, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6, 6688, 12, 6706} \begin {gather*} e^{x^2 \left (-x-5 e^x+4+e^{\log ^4(2)}\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(16*x^2 + 25*E^(2*x)*x^2 + E^(2*Log[2]^4)*x^2 - 8*x^3 + x^4 + E^Log[2]^4*(8*x^2 - 10*E^x*x^2 - 2*x^3) +
E^x*(-40*x^2 + 10*x^3))*(32*x + 2*E^(2*Log[2]^4)*x - 24*x^2 + 4*x^3 + E^(2*x)*(50*x + 50*x^2) + E^x*(-80*x - 1
0*x^2 + 10*x^3) + E^Log[2]^4*(16*x - 6*x^2 + E^x*(-20*x - 10*x^2))),x]

[Out]

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

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 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 \exp \left (16 x^2+25 e^{2 x} x^2+e^{2 \log ^4(2)} x^2-8 x^3+x^4+e^{\log ^4(2)} \left (8 x^2-10 e^x x^2-2 x^3\right )+e^x \left (-40 x^2+10 x^3\right )\right ) \left (\left (32+2 e^{2 \log ^4(2)}\right ) x-24 x^2+4 x^3+e^{2 x} \left (50 x+50 x^2\right )+e^x \left (-80 x-10 x^2+10 x^3\right )+e^{\log ^4(2)} \left (16 x-6 x^2+e^x \left (-20 x-10 x^2\right )\right )\right ) \, dx\\ &=\int 2 \exp \left (\left (-5 e^x+4 \left (1+\frac {1}{4} e^{\log ^4(2)}\right )-x\right )^2 x^2\right ) x \left (16 \left (1+\frac {1}{16} e^{2 \log ^4(2)}\right )-12 x+2 x^2+25 e^{2 x} (1+x)+5 e^x \left (-8-x+x^2\right )+e^{\log ^4(2)} \left (8-3 x-5 e^x (2+x)\right )\right ) \, dx\\ &=2 \int \exp \left (\left (-5 e^x+4 \left (1+\frac {1}{4} e^{\log ^4(2)}\right )-x\right )^2 x^2\right ) x \left (16 \left (1+\frac {1}{16} e^{2 \log ^4(2)}\right )-12 x+2 x^2+25 e^{2 x} (1+x)+5 e^x \left (-8-x+x^2\right )+e^{\log ^4(2)} \left (8-3 x-5 e^x (2+x)\right )\right ) \, dx\\ &=e^{\left (4-5 e^x+e^{\log ^4(2)}-x\right )^2 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 24, normalized size = 1.00 \begin {gather*} e^{\left (4-5 e^x+e^{\log ^4(2)}-x\right )^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(16*x^2 + 25*E^(2*x)*x^2 + E^(2*Log[2]^4)*x^2 - 8*x^3 + x^4 + E^Log[2]^4*(8*x^2 - 10*E^x*x^2 - 2*x
^3) + E^x*(-40*x^2 + 10*x^3))*(32*x + 2*E^(2*Log[2]^4)*x - 24*x^2 + 4*x^3 + E^(2*x)*(50*x + 50*x^2) + E^x*(-80
*x - 10*x^2 + 10*x^3) + E^Log[2]^4*(16*x - 6*x^2 + E^x*(-20*x - 10*x^2))),x]

[Out]

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

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fricas [B]  time = 1.45, size = 71, normalized size = 2.96 \begin {gather*} e^{\left (x^{4} - 8 \, x^{3} + x^{2} e^{\left (2 \, \log \relax (2)^{4}\right )} + 25 \, x^{2} e^{\left (2 \, x\right )} + 16 \, x^{2} - 2 \, {\left (x^{3} + 5 \, x^{2} e^{x} - 4 \, x^{2}\right )} e^{\left (\log \relax (2)^{4}\right )} + 10 \, {\left (x^{3} - 4 \, x^{2}\right )} e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(log(2)^4)^2+((-10*x^2-20*x)*exp(x)-6*x^2+16*x)*exp(log(2)^4)+(50*x^2+50*x)*exp(x)^2+(10*x^3
-10*x^2-80*x)*exp(x)+4*x^3-24*x^2+32*x)*exp(x^2*exp(log(2)^4)^2+(-10*exp(x)*x^2-2*x^3+8*x^2)*exp(log(2)^4)+25*
exp(x)^2*x^2+(10*x^3-40*x^2)*exp(x)+x^4-8*x^3+16*x^2),x, algorithm="fricas")

[Out]

e^(x^4 - 8*x^3 + x^2*e^(2*log(2)^4) + 25*x^2*e^(2*x) + 16*x^2 - 2*(x^3 + 5*x^2*e^x - 4*x^2)*e^(log(2)^4) + 10*
(x^3 - 4*x^2)*e^x)

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giac [B]  time = 0.67, size = 81, normalized size = 3.38 \begin {gather*} e^{\left (x^{4} - 2 \, x^{3} e^{\left (\log \relax (2)^{4}\right )} + 10 \, x^{3} e^{x} - 8 \, x^{3} + x^{2} e^{\left (2 \, \log \relax (2)^{4}\right )} - 10 \, x^{2} e^{\left (\log \relax (2)^{4} + x\right )} + 8 \, x^{2} e^{\left (\log \relax (2)^{4}\right )} + 25 \, x^{2} e^{\left (2 \, x\right )} - 40 \, x^{2} e^{x} + 16 \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(log(2)^4)^2+((-10*x^2-20*x)*exp(x)-6*x^2+16*x)*exp(log(2)^4)+(50*x^2+50*x)*exp(x)^2+(10*x^3
-10*x^2-80*x)*exp(x)+4*x^3-24*x^2+32*x)*exp(x^2*exp(log(2)^4)^2+(-10*exp(x)*x^2-2*x^3+8*x^2)*exp(log(2)^4)+25*
exp(x)^2*x^2+(10*x^3-40*x^2)*exp(x)+x^4-8*x^3+16*x^2),x, algorithm="giac")

[Out]

e^(x^4 - 2*x^3*e^(log(2)^4) + 10*x^3*e^x - 8*x^3 + x^2*e^(2*log(2)^4) - 10*x^2*e^(log(2)^4 + x) + 8*x^2*e^(log
(2)^4) + 25*x^2*e^(2*x) - 40*x^2*e^x + 16*x^2)

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maple [B]  time = 0.38, size = 60, normalized size = 2.50

method	result	size
risch	\({\mathrm e}^{x^{2} \left (-10 \,{\mathrm e}^{x +\ln \relax (2)^{4}}+10 \,{\mathrm e}^{x} x -2 \,{\mathrm e}^{\ln \relax (2)^{4}} x +x^{2}-40 \,{\mathrm e}^{x}+{\mathrm e}^{2 \ln \relax (2)^{4}}+8 \,{\mathrm e}^{\ln \relax (2)^{4}}+25 \,{\mathrm e}^{2 x}-8 x +16\right )}\)	\(60\)
norman	\({\mathrm e}^{x^{2} {\mathrm e}^{2 \ln \relax (2)^{4}}+\left (-10 \,{\mathrm e}^{x} x^{2}-2 x^{3}+8 x^{2}\right ) {\mathrm e}^{\ln \relax (2)^{4}}+25 \,{\mathrm e}^{2 x} x^{2}+\left (10 x^{3}-40 x^{2}\right ) {\mathrm e}^{x}+x^{4}-8 x^{3}+16 x^{2}}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*exp(ln(2)^4)^2+((-10*x^2-20*x)*exp(x)-6*x^2+16*x)*exp(ln(2)^4)+(50*x^2+50*x)*exp(x)^2+(10*x^3-10*x^2-
80*x)*exp(x)+4*x^3-24*x^2+32*x)*exp(x^2*exp(ln(2)^4)^2+(-10*exp(x)*x^2-2*x^3+8*x^2)*exp(ln(2)^4)+25*exp(x)^2*x
^2+(10*x^3-40*x^2)*exp(x)+x^4-8*x^3+16*x^2),x,method=_RETURNVERBOSE)

[Out]

exp(x^2*(-10*exp(x+ln(2)^4)+10*exp(x)*x-2*exp(ln(2)^4)*x+x^2-40*exp(x)+exp(2*ln(2)^4)+8*exp(ln(2)^4)+25*exp(2*
x)-8*x+16))

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maxima [B]  time = 0.68, size = 81, normalized size = 3.38 \begin {gather*} e^{\left (x^{4} - 2 \, x^{3} e^{\left (\log \relax (2)^{4}\right )} + 10 \, x^{3} e^{x} - 8 \, x^{3} + x^{2} e^{\left (2 \, \log \relax (2)^{4}\right )} - 10 \, x^{2} e^{\left (\log \relax (2)^{4} + x\right )} + 8 \, x^{2} e^{\left (\log \relax (2)^{4}\right )} + 25 \, x^{2} e^{\left (2 \, x\right )} - 40 \, x^{2} e^{x} + 16 \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(log(2)^4)^2+((-10*x^2-20*x)*exp(x)-6*x^2+16*x)*exp(log(2)^4)+(50*x^2+50*x)*exp(x)^2+(10*x^3
-10*x^2-80*x)*exp(x)+4*x^3-24*x^2+32*x)*exp(x^2*exp(log(2)^4)^2+(-10*exp(x)*x^2-2*x^3+8*x^2)*exp(log(2)^4)+25*
exp(x)^2*x^2+(10*x^3-40*x^2)*exp(x)+x^4-8*x^3+16*x^2),x, algorithm="maxima")

[Out]

e^(x^4 - 2*x^3*e^(log(2)^4) + 10*x^3*e^x - 8*x^3 + x^2*e^(2*log(2)^4) - 10*x^2*e^(log(2)^4 + x) + 8*x^2*e^(log
(2)^4) + 25*x^2*e^(2*x) - 40*x^2*e^x + 16*x^2)

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mupad [B]  time = 5.04, size = 90, normalized size = 3.75 \begin {gather*} {\mathrm {e}}^{x^2\,{\mathrm {e}}^{2\,{\ln \relax (2)}^4}}\,{\mathrm {e}}^{x^4}\,{\mathrm {e}}^{10\,x^3\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-40\,x^2\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-8\,x^3}\,{\mathrm {e}}^{16\,x^2}\,{\mathrm {e}}^{-10\,x^2\,{\mathrm {e}}^{{\ln \relax (2)}^4}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-2\,x^3\,{\mathrm {e}}^{{\ln \relax (2)}^4}}\,{\mathrm {e}}^{8\,x^2\,{\mathrm {e}}^{{\ln \relax (2)}^4}}\,{\mathrm {e}}^{25\,x^2\,{\mathrm {e}}^{2\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(25*x^2*exp(2*x) - exp(log(2)^4)*(10*x^2*exp(x) - 8*x^2 + 2*x^3) - exp(x)*(40*x^2 - 10*x^3) + x^2*exp(2
*log(2)^4) + 16*x^2 - 8*x^3 + x^4)*(32*x + exp(2*x)*(50*x + 50*x^2) + 2*x*exp(2*log(2)^4) - exp(log(2)^4)*(exp
(x)*(20*x + 10*x^2) - 16*x + 6*x^2) - 24*x^2 + 4*x^3 - exp(x)*(80*x + 10*x^2 - 10*x^3)),x)

[Out]

exp(x^2*exp(2*log(2)^4))*exp(x^4)*exp(10*x^3*exp(x))*exp(-40*x^2*exp(x))*exp(-8*x^3)*exp(16*x^2)*exp(-10*x^2*e
xp(log(2)^4)*exp(x))*exp(-2*x^3*exp(log(2)^4))*exp(8*x^2*exp(log(2)^4))*exp(25*x^2*exp(2*x))

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sympy [B]  time = 0.51, size = 75, normalized size = 3.12 \begin {gather*} e^{x^{4} - 8 x^{3} + 25 x^{2} e^{2 x} + x^{2} e^{2 \log {\relax (2 )}^{4}} + 16 x^{2} + \left (10 x^{3} - 40 x^{2}\right ) e^{x} + \left (- 2 x^{3} - 10 x^{2} e^{x} + 8 x^{2}\right ) e^{\log {\relax (2 )}^{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(ln(2)**4)**2+((-10*x**2-20*x)*exp(x)-6*x**2+16*x)*exp(ln(2)**4)+(50*x**2+50*x)*exp(x)**2+(1
0*x**3-10*x**2-80*x)*exp(x)+4*x**3-24*x**2+32*x)*exp(x**2*exp(ln(2)**4)**2+(-10*exp(x)*x**2-2*x**3+8*x**2)*exp
(ln(2)**4)+25*exp(x)**2*x**2+(10*x**3-40*x**2)*exp(x)+x**4-8*x**3+16*x**2),x)

[Out]

exp(x**4 - 8*x**3 + 25*x**2*exp(2*x) + x**2*exp(2*log(2)**4) + 16*x**2 + (10*x**3 - 40*x**2)*exp(x) + (-2*x**3
- 10*x**2*exp(x) + 8*x**2)*exp(log(2)**4))

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