3.53.34 \(\int \frac {e^{e^{\frac {2 (4+x \log (2 x+4 x^2+2 x^3))}{\log (2 x+4 x^2+2 x^3)}}+4 e^{\frac {4+x \log (2 x+4 x^2+2 x^3)}{\log (2 x+4 x^2+2 x^3)}} x+4 x^2} ((8 x^2+8 x^3) \log ^2(2 x+4 x^2+2 x^3)+e^{\frac {2 (4+x \log (2 x+4 x^2+2 x^3))}{\log (2 x+4 x^2+2 x^3)}} (-8-24 x+(2 x+2 x^2) \log ^2(2 x+4 x^2+2 x^3))+e^{\frac {4+x \log (2 x+4 x^2+2 x^3)}{\log (2 x+4 x^2+2 x^3)}} (-16 x-48 x^2+(4 x+8 x^2+4 x^3) \log ^2(2 x+4 x^2+2 x^3)))}{(x+x^2) \log ^2(2 x+4 x^2+2 x^3)} \, dx\)

Optimal. Leaf size=28 \[ e^{\left (e^{x+\frac {4}{\log \left ((2+2 x) \left (x+x^2\right )\right )}}+2 x\right )^2} \]

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Rubi [A]  time = 10.64, antiderivative size = 25, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 4, integrand size = 299, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {1593, 6688, 12, 6706} \begin {gather*} e^{\left (2 x+e^{x+\frac {4}{\log \left (2 x (x+1)^2\right )}}\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(E^(x + 4/Log[2*x*(1 + x)^2]) + 2*x)^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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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

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Mathematica [A]  time = 0.39, size = 25, normalized size = 0.89 \begin {gather*} e^{\left (e^{x+\frac {4}{\log \left (2 x (1+x)^2\right )}}+2 x\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(E^(x + 4/Log[2*x*(1 + x)^2]) + 2*x)^2

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fricas [B]  time = 0.56, size = 87, normalized size = 3.11 \begin {gather*} e^{\left (4 \, x^{2} + 4 \, x e^{\left (\frac {x \log \left (2 \, x^{3} + 4 \, x^{2} + 2 \, x\right ) + 4}{\log \left (2 \, x^{3} + 4 \, x^{2} + 2 \, x\right )}\right )} + e^{\left (\frac {2 \, {\left (x \log \left (2 \, x^{3} + 4 \, x^{2} + 2 \, x\right ) + 4\right )}}{\log \left (2 \, x^{3} + 4 \, x^{2} + 2 \, x\right )}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+2*x)*log(2*x^3+4*x^2+2*x)^2-24*x-8)*exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))^2+
((4*x^3+8*x^2+4*x)*log(2*x^3+4*x^2+2*x)^2-48*x^2-16*x)*exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))+(8
*x^3+8*x^2)*log(2*x^3+4*x^2+2*x)^2)*exp(exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))^2+4*x*exp((x*log(
2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))+4*x^2)/(x^2+x)/log(2*x^3+4*x^2+2*x)^2,x, algorithm="fricas")

[Out]

e^(4*x^2 + 4*x*e^((x*log(2*x^3 + 4*x^2 + 2*x) + 4)/log(2*x^3 + 4*x^2 + 2*x)) + e^(2*(x*log(2*x^3 + 4*x^2 + 2*x
) + 4)/log(2*x^3 + 4*x^2 + 2*x)))

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+2*x)*log(2*x^3+4*x^2+2*x)^2-24*x-8)*exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))^2+
((4*x^3+8*x^2+4*x)*log(2*x^3+4*x^2+2*x)^2-48*x^2-16*x)*exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))+(8
*x^3+8*x^2)*log(2*x^3+4*x^2+2*x)^2)*exp(exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))^2+4*x*exp((x*log(
2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))+4*x^2)/(x^2+x)/log(2*x^3+4*x^2+2*x)^2,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.34, size = 88, normalized size = 3.14

method	result	size
risch	\({\mathrm e}^{{\mathrm e}^{\frac {2 x \ln \left (2 x^{3}+4 x^{2}+2 x \right )+8}{\ln \left (2 x^{3}+4 x^{2}+2 x \right )}}+4 x \,{\mathrm e}^{\frac {x \ln \left (2 x^{3}+4 x^{2}+2 x \right )+4}{\ln \left (2 x^{3}+4 x^{2}+2 x \right )}}+4 x^{2}}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^2+2*x)*ln(2*x^3+4*x^2+2*x)^2-24*x-8)*exp((x*ln(2*x^3+4*x^2+2*x)+4)/ln(2*x^3+4*x^2+2*x))^2+((4*x^3+8
*x^2+4*x)*ln(2*x^3+4*x^2+2*x)^2-48*x^2-16*x)*exp((x*ln(2*x^3+4*x^2+2*x)+4)/ln(2*x^3+4*x^2+2*x))+(8*x^3+8*x^2)*
ln(2*x^3+4*x^2+2*x)^2)*exp(exp((x*ln(2*x^3+4*x^2+2*x)+4)/ln(2*x^3+4*x^2+2*x))^2+4*x*exp((x*ln(2*x^3+4*x^2+2*x)
+4)/ln(2*x^3+4*x^2+2*x))+4*x^2)/(x^2+x)/ln(2*x^3+4*x^2+2*x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.80, size = 48, normalized size = 1.71 \begin {gather*} e^{\left (4 \, x^{2} + 4 \, x e^{\left (x + \frac {4}{\log \relax (2) + 2 \, \log \left (x + 1\right ) + \log \relax (x)}\right )} + e^{\left (2 \, x + \frac {8}{\log \relax (2) + 2 \, \log \left (x + 1\right ) + \log \relax (x)}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+2*x)*log(2*x^3+4*x^2+2*x)^2-24*x-8)*exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))^2+
((4*x^3+8*x^2+4*x)*log(2*x^3+4*x^2+2*x)^2-48*x^2-16*x)*exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))+(8
*x^3+8*x^2)*log(2*x^3+4*x^2+2*x)^2)*exp(exp((x*log(2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))^2+4*x*exp((x*log(
2*x^3+4*x^2+2*x)+4)/log(2*x^3+4*x^2+2*x))+4*x^2)/(x^2+x)/log(2*x^3+4*x^2+2*x)^2,x, algorithm="maxima")

[Out]

e^(4*x^2 + 4*x*e^(x + 4/(log(2) + 2*log(x + 1) + log(x))) + e^(2*x + 8/(log(2) + 2*log(x + 1) + log(x))))

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mupad [B]  time = 3.75, size = 59, normalized size = 2.11 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^{\frac {8}{\ln \left (2\,x^3+4\,x^2+2\,x\right )}}\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{4\,x\,{\mathrm {e}}^{\frac {4}{\ln \left (2\,x^3+4\,x^2+2\,x\right )}}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{4\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp((2*(x*log(2*x + 4*x^2 + 2*x^3) + 4))/log(2*x + 4*x^2 + 2*x^3)) + 4*x*exp((x*log(2*x + 4*x^2 + 2*
x^3) + 4)/log(2*x + 4*x^2 + 2*x^3)) + 4*x^2)*(exp((2*(x*log(2*x + 4*x^2 + 2*x^3) + 4))/log(2*x + 4*x^2 + 2*x^3
))*(24*x - log(2*x + 4*x^2 + 2*x^3)^2*(2*x + 2*x^2) + 8) - log(2*x + 4*x^2 + 2*x^3)^2*(8*x^2 + 8*x^3) + exp((x
*log(2*x + 4*x^2 + 2*x^3) + 4)/log(2*x + 4*x^2 + 2*x^3))*(16*x - log(2*x + 4*x^2 + 2*x^3)^2*(4*x + 8*x^2 + 4*x
^3) + 48*x^2)))/(log(2*x + 4*x^2 + 2*x^3)^2*(x + x^2)),x)

[Out]

exp(exp(8/log(2*x + 4*x^2 + 2*x^3))*exp(2*x))*exp(4*x*exp(4/log(2*x + 4*x^2 + 2*x^3))*exp(x))*exp(4*x^2)

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sympy [B]  time = 8.98, size = 82, normalized size = 2.93 \begin {gather*} e^{4 x^{2} + 4 x e^{\frac {x \log {\left (2 x^{3} + 4 x^{2} + 2 x \right )} + 4}{\log {\left (2 x^{3} + 4 x^{2} + 2 x \right )}}} + e^{\frac {2 \left (x \log {\left (2 x^{3} + 4 x^{2} + 2 x \right )} + 4\right )}{\log {\left (2 x^{3} + 4 x^{2} + 2 x \right )}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**2+2*x)*ln(2*x**3+4*x**2+2*x)**2-24*x-8)*exp((x*ln(2*x**3+4*x**2+2*x)+4)/ln(2*x**3+4*x**2+2*x
))**2+((4*x**3+8*x**2+4*x)*ln(2*x**3+4*x**2+2*x)**2-48*x**2-16*x)*exp((x*ln(2*x**3+4*x**2+2*x)+4)/ln(2*x**3+4*
x**2+2*x))+(8*x**3+8*x**2)*ln(2*x**3+4*x**2+2*x)**2)*exp(exp((x*ln(2*x**3+4*x**2+2*x)+4)/ln(2*x**3+4*x**2+2*x)
)**2+4*x*exp((x*ln(2*x**3+4*x**2+2*x)+4)/ln(2*x**3+4*x**2+2*x))+4*x**2)/(x**2+x)/ln(2*x**3+4*x**2+2*x)**2,x)

[Out]

exp(4*x**2 + 4*x*exp((x*log(2*x**3 + 4*x**2 + 2*x) + 4)/log(2*x**3 + 4*x**2 + 2*x)) + exp(2*(x*log(2*x**3 + 4*
x**2 + 2*x) + 4)/log(2*x**3 + 4*x**2 + 2*x)))

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