\(\int \frac {1}{8} e^{\frac {1}{16} (144-96 x+112 x^2-32 x^3+16 x^4+e^x (24 x-8 x^2+8 x^3) \log (4)+e^{2 x} x^2 \log ^2(4))} (-48+112 x-48 x^2+32 x^3+e^x (12+4 x+8 x^2+4 x^3) \log (4)+e^{2 x} (x+x^2) \log ^2(4)) \, dx\) [7505]

Optimal result

Integrand size = 114, antiderivative size = 24 \[ \int \frac {1}{8} e^{\frac {1}{16} \left (144-96 x+112 x^2-32 x^3+16 x^4+e^x \left (24 x-8 x^2+8 x^3\right ) \log (4)+e^{2 x} x^2 \log ^2(4)\right )} \left (-48+112 x-48 x^2+32 x^3+e^x \left (12+4 x+8 x^2+4 x^3\right ) \log (4)+e^{2 x} \left (x+x^2\right ) \log ^2(4)\right ) \, dx=e^{\left (-3-x-x \left (-2+x+\frac {1}{4} e^x \log (4)\right )\right )^2} \]

[Out]

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

Rubi [A] (verified)

Time = 3.59 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {12, 6873, 6838} \[ \int \frac {1}{8} e^{\frac {1}{16} \left (144-96 x+112 x^2-32 x^3+16 x^4+e^x \left (24 x-8 x^2+8 x^3\right ) \log (4)+e^{2 x} x^2 \log ^2(4)\right )} \left (-48+112 x-48 x^2+32 x^3+e^x \left (12+4 x+8 x^2+4 x^3\right ) \log (4)+e^{2 x} \left (x+x^2\right ) \log ^2(4)\right ) \, dx=e^{\frac {1}{16} \left (4 x^2-4 x+e^x x \log (4)+12\right )^2} \]

[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 6838

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 6873

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

Rubi steps \begin{align*} \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{align*}

Mathematica [F]

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

[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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(47\) vs. \(2(20)=40\).

Time = 0.42 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.00

method	result	size
risch	\(2^{x \left (x^{2}-x +3\right ) {\mathrm e}^{x}} {\mathrm e}^{\frac {x^{2} \ln \left (2\right )^{2} {\mathrm e}^{2 x}}{4}+9+x^{4}-2 x^{3}+7 x^{2}-6 x}\)	\(48\)
norman	\({\mathrm e}^{\frac {x^{2} \ln \left (2\right )^{2} {\mathrm e}^{2 x}}{4}+\frac {\left (8 x^{3}-8 x^{2}+24 x \right ) \ln \left (2\right ) {\mathrm e}^{x}}{8}+x^{4}-2 x^{3}+7 x^{2}-6 x +9}\)	\(53\)
parallelrisch	\({\mathrm e}^{9-6 x +x^{4}-2 x^{3}+7 x^{2}+\frac {x^{2} \ln \left (2\right )^{2} {\mathrm e}^{2 x}}{4}+\frac {\ln \left (2\right ) {\mathrm e}^{x} \left (4 x^{3}-4 x^{2}+12 x \right )}{4}}\)	\(53\)

[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)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (22) = 44\).

Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.04 \[ \int \frac {1}{8} e^{\frac {1}{16} \left (144-96 x+112 x^2-32 x^3+16 x^4+e^x \left (24 x-8 x^2+8 x^3\right ) \log (4)+e^{2 x} x^2 \log ^2(4)\right )} \left (-48+112 x-48 x^2+32 x^3+e^x \left (12+4 x+8 x^2+4 x^3\right ) \log (4)+e^{2 x} \left (x+x^2\right ) \log ^2(4)\right ) \, dx=e^{\left (\frac {1}{4} \, x^{2} e^{\left (2 \, x\right )} \log \left (2\right )^{2} + x^{4} - 2 \, x^{3} + {\left (x^{3} - x^{2} + 3 \, x\right )} e^{x} \log \left (2\right ) + 7 \, x^{2} - 6 \, x + 9\right )} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (20) = 40\).

Time = 0.22 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.12 \[ \int \frac {1}{8} e^{\frac {1}{16} \left (144-96 x+112 x^2-32 x^3+16 x^4+e^x \left (24 x-8 x^2+8 x^3\right ) \log (4)+e^{2 x} x^2 \log ^2(4)\right )} \left (-48+112 x-48 x^2+32 x^3+e^x \left (12+4 x+8 x^2+4 x^3\right ) \log (4)+e^{2 x} \left (x+x^2\right ) \log ^2(4)\right ) \, dx=e^{x^{4} - 2 x^{3} + \frac {x^{2} e^{2 x} \log {\left (2 \right )}^{2}}{4} + 7 x^{2} - 6 x + \left (x^{3} - x^{2} + 3 x\right ) e^{x} \log {\left (2 \right )} + 9} \]

[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)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (22) = 44\).

Time = 0.49 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.33 \[ \int \frac {1}{8} e^{\frac {1}{16} \left (144-96 x+112 x^2-32 x^3+16 x^4+e^x \left (24 x-8 x^2+8 x^3\right ) \log (4)+e^{2 x} x^2 \log ^2(4)\right )} \left (-48+112 x-48 x^2+32 x^3+e^x \left (12+4 x+8 x^2+4 x^3\right ) \log (4)+e^{2 x} \left (x+x^2\right ) \log ^2(4)\right ) \, dx=e^{\left (x^{3} e^{x} \log \left (2\right ) + \frac {1}{4} \, x^{2} e^{\left (2 \, x\right )} \log \left (2\right )^{2} + x^{4} - x^{2} e^{x} \log \left (2\right ) - 2 \, x^{3} + 3 \, x e^{x} \log \left (2\right ) + 7 \, x^{2} - 6 \, x + 9\right )} \]

[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
)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (22) = 44\).

Time = 0.36 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.33 \[ \int \frac {1}{8} e^{\frac {1}{16} \left (144-96 x+112 x^2-32 x^3+16 x^4+e^x \left (24 x-8 x^2+8 x^3\right ) \log (4)+e^{2 x} x^2 \log ^2(4)\right )} \left (-48+112 x-48 x^2+32 x^3+e^x \left (12+4 x+8 x^2+4 x^3\right ) \log (4)+e^{2 x} \left (x+x^2\right ) \log ^2(4)\right ) \, dx=e^{\left (x^{3} e^{x} \log \left (2\right ) + \frac {1}{4} \, x^{2} e^{\left (2 \, x\right )} \log \left (2\right )^{2} + x^{4} - x^{2} e^{x} \log \left (2\right ) - 2 \, x^{3} + 3 \, x e^{x} \log \left (2\right ) + 7 \, x^{2} - 6 \, x + 9\right )} \]

[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
)

Mupad [B] (verification not implemented)

Time = 14.05 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.42 \[ \int \frac {1}{8} e^{\frac {1}{16} \left (144-96 x+112 x^2-32 x^3+16 x^4+e^x \left (24 x-8 x^2+8 x^3\right ) \log (4)+e^{2 x} x^2 \log ^2(4)\right )} \left (-48+112 x-48 x^2+32 x^3+e^x \left (12+4 x+8 x^2+4 x^3\right ) \log (4)+e^{2 x} \left (x+x^2\right ) \log ^2(4)\right ) \, dx=2^{x^3\,{\mathrm {e}}^x-x^2\,{\mathrm {e}}^x+3\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-6\,x}\,{\mathrm {e}}^{x^4}\,{\mathrm {e}}^9\,{\mathrm {e}}^{-2\,x^3}\,{\mathrm {e}}^{7\,x^2}\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{2\,x}\,{\ln \left (2\right )}^2}{4}} \]

[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)