3.76.5 \(\int \frac {1}{8} e^{\frac {1}{16} (144-96 x+112 x^2-32 x^3+16 x^4+e^x (24 x-8 x^2+8 x^3) \log (4)+e^{2 x} x^2 \log ^2(4))} (-48+112 x-48 x^2+32 x^3+e^x (12+4 x+8 x^2+4 x^3) \log (4)+e^{2 x} (x+x^2) \log ^2(4)) \, dx\)

Optimal. Leaf size=24 \[ e^{\left (-3-x-x \left (-2+x+\frac {1}{4} e^x \log (4)\right )\right )^2} \]

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Rubi [A]  time = 5.86, antiderivative size = 25, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, integrand size = 114, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {12, 6741, 6706} \begin {gather*} e^{\frac {1}{16} \left (4 x^2-4 x+e^x x \log (4)+12\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((144 - 96*x + 112*x^2 - 32*x^3 + 16*x^4 + E^x*(24*x - 8*x^2 + 8*x^3)*Log[4] + E^(2*x)*x^2*Log[4]^2)/16
)*(-48 + 112*x - 48*x^2 + 32*x^3 + E^x*(12 + 4*x + 8*x^2 + 4*x^3)*Log[4] + E^(2*x)*(x + x^2)*Log[4]^2))/8,x]

[Out]

E^((12 - 4*x + 4*x^2 + E^x*x*Log[4])^2/16)

Rule 12

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{8} \int \exp \left (\frac {1}{16} \left (144-96 x+112 x^2-32 x^3+16 x^4+e^x \left (24 x-8 x^2+8 x^3\right ) \log (4)+e^{2 x} x^2 \log ^2(4)\right )\right ) \left (-48+112 x-48 x^2+32 x^3+e^x \left (12+4 x+8 x^2+4 x^3\right ) \log (4)+e^{2 x} \left (x+x^2\right ) \log ^2(4)\right ) \, dx\\ &=\frac {1}{8} \int e^{\frac {1}{16} \left (12-4 x+4 x^2+e^x x \log (4)\right )^2} \left (-48+112 x-48 x^2+32 x^3+e^x \left (12+4 x+8 x^2+4 x^3\right ) \log (4)+e^{2 x} \left (x+x^2\right ) \log ^2(4)\right ) \, dx\\ &=e^{\frac {1}{16} \left (12-4 x+4 x^2+e^x x \log (4)\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 5.69, size = 117, normalized size = 4.88 \begin {gather*} \frac {1}{8} \int e^{\frac {1}{16} \left (144-96 x+112 x^2-32 x^3+16 x^4+e^x \left (24 x-8 x^2+8 x^3\right ) \log (4)+e^{2 x} x^2 \log ^2(4)\right )} \left (-48+112 x-48 x^2+32 x^3+e^x \left (12+4 x+8 x^2+4 x^3\right ) \log (4)+e^{2 x} \left (x+x^2\right ) \log ^2(4)\right ) \, dx \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((144 - 96*x + 112*x^2 - 32*x^3 + 16*x^4 + E^x*(24*x - 8*x^2 + 8*x^3)*Log[4] + E^(2*x)*x^2*Log[4]
^2)/16)*(-48 + 112*x - 48*x^2 + 32*x^3 + E^x*(12 + 4*x + 8*x^2 + 4*x^3)*Log[4] + E^(2*x)*(x + x^2)*Log[4]^2))/
8,x]

[Out]

Integrate[E^((144 - 96*x + 112*x^2 - 32*x^3 + 16*x^4 + E^x*(24*x - 8*x^2 + 8*x^3)*Log[4] + E^(2*x)*x^2*Log[4]^
2)/16)*(-48 + 112*x - 48*x^2 + 32*x^3 + E^x*(12 + 4*x + 8*x^2 + 4*x^3)*Log[4] + E^(2*x)*(x + x^2)*Log[4]^2), x
]/8

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fricas [B]  time = 0.57, size = 49, normalized size = 2.04 \begin {gather*} e^{\left (\frac {1}{4} \, x^{2} e^{\left (2 \, x\right )} \log \relax (2)^{2} + x^{4} - 2 \, x^{3} + {\left (x^{3} - x^{2} + 3 \, x\right )} e^{x} \log \relax (2) + 7 \, x^{2} - 6 \, x + 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(4*(x^2+x)*log(2)^2*exp(x)^2+2*(4*x^3+8*x^2+4*x+12)*log(2)*exp(x)+32*x^3-48*x^2+112*x-48)*exp(1/
4*x^2*log(2)^2*exp(x)^2+1/8*(8*x^3-8*x^2+24*x)*log(2)*exp(x)+x^4-2*x^3+7*x^2-6*x+9),x, algorithm="fricas")

[Out]

e^(1/4*x^2*e^(2*x)*log(2)^2 + x^4 - 2*x^3 + (x^3 - x^2 + 3*x)*e^x*log(2) + 7*x^2 - 6*x + 9)

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giac [B]  time = 0.32, size = 56, normalized size = 2.33 \begin {gather*} e^{\left (x^{3} e^{x} \log \relax (2) + \frac {1}{4} \, x^{2} e^{\left (2 \, x\right )} \log \relax (2)^{2} + x^{4} - x^{2} e^{x} \log \relax (2) - 2 \, x^{3} + 3 \, x e^{x} \log \relax (2) + 7 \, x^{2} - 6 \, x + 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(4*(x^2+x)*log(2)^2*exp(x)^2+2*(4*x^3+8*x^2+4*x+12)*log(2)*exp(x)+32*x^3-48*x^2+112*x-48)*exp(1/
4*x^2*log(2)^2*exp(x)^2+1/8*(8*x^3-8*x^2+24*x)*log(2)*exp(x)+x^4-2*x^3+7*x^2-6*x+9),x, algorithm="giac")

[Out]

e^(x^3*e^x*log(2) + 1/4*x^2*e^(2*x)*log(2)^2 + x^4 - x^2*e^x*log(2) - 2*x^3 + 3*x*e^x*log(2) + 7*x^2 - 6*x + 9
)

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maple [B]  time = 0.13, size = 48, normalized size = 2.00

method	result	size
risch	\(2^{x \left (x^{2}-x +3\right ) {\mathrm e}^{x}} {\mathrm e}^{\frac {x^{2} \ln \relax (2)^{2} {\mathrm e}^{2 x}}{4}+9+x^{4}-2 x^{3}+7 x^{2}-6 x}\)	\(48\)
norman	\({\mathrm e}^{\frac {x^{2} \ln \relax (2)^{2} {\mathrm e}^{2 x}}{4}+\frac {\left (8 x^{3}-8 x^{2}+24 x \right ) \ln \relax (2) {\mathrm e}^{x}}{8}+x^{4}-2 x^{3}+7 x^{2}-6 x +9}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/8*(4*(x^2+x)*ln(2)^2*exp(x)^2+2*(4*x^3+8*x^2+4*x+12)*ln(2)*exp(x)+32*x^3-48*x^2+112*x-48)*exp(1/4*x^2*ln
(2)^2*exp(x)^2+1/8*(8*x^3-8*x^2+24*x)*ln(2)*exp(x)+x^4-2*x^3+7*x^2-6*x+9),x,method=_RETURNVERBOSE)

[Out]

2^(x*(x^2-x+3)*exp(x))*exp(1/4*x^2*ln(2)^2*exp(2*x)+9+x^4-2*x^3+7*x^2-6*x)

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maxima [B]  time = 0.69, size = 56, normalized size = 2.33 \begin {gather*} e^{\left (x^{3} e^{x} \log \relax (2) + \frac {1}{4} \, x^{2} e^{\left (2 \, x\right )} \log \relax (2)^{2} + x^{4} - x^{2} e^{x} \log \relax (2) - 2 \, x^{3} + 3 \, x e^{x} \log \relax (2) + 7 \, x^{2} - 6 \, x + 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(4*(x^2+x)*log(2)^2*exp(x)^2+2*(4*x^3+8*x^2+4*x+12)*log(2)*exp(x)+32*x^3-48*x^2+112*x-48)*exp(1/
4*x^2*log(2)^2*exp(x)^2+1/8*(8*x^3-8*x^2+24*x)*log(2)*exp(x)+x^4-2*x^3+7*x^2-6*x+9),x, algorithm="maxima")

[Out]

e^(x^3*e^x*log(2) + 1/4*x^2*e^(2*x)*log(2)^2 + x^4 - x^2*e^x*log(2) - 2*x^3 + 3*x*e^x*log(2) + 7*x^2 - 6*x + 9
)

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mupad [B]  time = 5.17, size = 58, normalized size = 2.42 \begin {gather*} 2^{x^3\,{\mathrm {e}}^x-x^2\,{\mathrm {e}}^x+3\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-6\,x}\,{\mathrm {e}}^{x^4}\,{\mathrm {e}}^9\,{\mathrm {e}}^{-2\,x^3}\,{\mathrm {e}}^{7\,x^2}\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{2\,x}\,{\ln \relax (2)}^2}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(7*x^2 - 6*x - 2*x^3 + x^4 + (x^2*exp(2*x)*log(2)^2)/4 + (exp(x)*log(2)*(24*x - 8*x^2 + 8*x^3))/8 + 9)
*(112*x - 48*x^2 + 32*x^3 + 4*exp(2*x)*log(2)^2*(x + x^2) + 2*exp(x)*log(2)*(4*x + 8*x^2 + 4*x^3 + 12) - 48))/
8,x)

[Out]

2^(x^3*exp(x) - x^2*exp(x) + 3*x*exp(x))*exp(-6*x)*exp(x^4)*exp(9)*exp(-2*x^3)*exp(7*x^2)*exp((x^2*exp(2*x)*lo
g(2)^2)/4)

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sympy [B]  time = 0.45, size = 51, normalized size = 2.12 \begin {gather*} e^{x^{4} - 2 x^{3} + \frac {x^{2} e^{2 x} \log {\relax (2 )}^{2}}{4} + 7 x^{2} - 6 x + \left (x^{3} - x^{2} + 3 x\right ) e^{x} \log {\relax (2 )} + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(4*(x**2+x)*ln(2)**2*exp(x)**2+2*(4*x**3+8*x**2+4*x+12)*ln(2)*exp(x)+32*x**3-48*x**2+112*x-48)*e
xp(1/4*x**2*ln(2)**2*exp(x)**2+1/8*(8*x**3-8*x**2+24*x)*ln(2)*exp(x)+x**4-2*x**3+7*x**2-6*x+9),x)

[Out]

exp(x**4 - 2*x**3 + x**2*exp(2*x)*log(2)**2/4 + 7*x**2 - 6*x + (x**3 - x**2 + 3*x)*exp(x)*log(2) + 9)

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