3.6.91 \(\int \frac {e^{\frac {1}{3} (3 x-4 \log ^2(\log (1+4 x+6 x^2+4 x^3+x^4)))} (e (3+3 x) \log (1+4 x+6 x^2+4 x^3+x^4)-32 e \log (\log (1+4 x+6 x^2+4 x^3+x^4)))}{(3+3 x) \log (1+4 x+6 x^2+4 x^3+x^4)} \, dx\)

Optimal. Leaf size=20 \[ -2+e^{1+x-\frac {4}{3} \log ^2\left (\log \left ((1+x)^4\right )\right )} \]

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Rubi [A]  time = 0.62, antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 3, number of rules used = 3, integrand size = 113, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {6688, 12, 6706} \begin {gather*} e^{x-\frac {4}{3} \log ^2\left (\log \left ((x+1)^4\right )\right )+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((3*x - 4*Log[Log[1 + 4*x + 6*x^2 + 4*x^3 + x^4]]^2)/3)*(E*(3 + 3*x)*Log[1 + 4*x + 6*x^2 + 4*x^3 + x^4]
- 32*E*Log[Log[1 + 4*x + 6*x^2 + 4*x^3 + x^4]]))/((3 + 3*x)*Log[1 + 4*x + 6*x^2 + 4*x^3 + x^4]),x]

[Out]

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

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 \frac {e^{1+x-\frac {4}{3} \log ^2\left (\log \left ((1+x)^4\right )\right )} \left (3 (1+x) \log \left ((1+x)^4\right )-32 \log \left (\log \left ((1+x)^4\right )\right )\right )}{3 (1+x) \log \left ((1+x)^4\right )} \, dx\\ &=\frac {1}{3} \int \frac {e^{1+x-\frac {4}{3} \log ^2\left (\log \left ((1+x)^4\right )\right )} \left (3 (1+x) \log \left ((1+x)^4\right )-32 \log \left (\log \left ((1+x)^4\right )\right )\right )}{(1+x) \log \left ((1+x)^4\right )} \, dx\\ &=e^{1+x-\frac {4}{3} \log ^2\left (\log \left ((1+x)^4\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 18, normalized size = 0.90 \begin {gather*} e^{1+x-\frac {4}{3} \log ^2\left (\log \left ((1+x)^4\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((3*x - 4*Log[Log[1 + 4*x + 6*x^2 + 4*x^3 + x^4]]^2)/3)*(E*(3 + 3*x)*Log[1 + 4*x + 6*x^2 + 4*x^3
+ x^4] - 32*E*Log[Log[1 + 4*x + 6*x^2 + 4*x^3 + x^4]]))/((3 + 3*x)*Log[1 + 4*x + 6*x^2 + 4*x^3 + x^4]),x]

[Out]

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

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fricas [A]  time = 0.68, size = 28, normalized size = 1.40 \begin {gather*} e^{\left (-\frac {4}{3} \, \log \left (\log \left (x^{4} + 4 \, x^{3} + 6 \, x^{2} + 4 \, x + 1\right )\right )^{2} + x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*exp(1)*log(log(x^4+4*x^3+6*x^2+4*x+1))+(3*x+3)*exp(1)*log(x^4+4*x^3+6*x^2+4*x+1))*exp(-4/3*log(
log(x^4+4*x^3+6*x^2+4*x+1))^2+x)/(3*x+3)/log(x^4+4*x^3+6*x^2+4*x+1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 2.30, size = 28, normalized size = 1.40 \begin {gather*} e^{\left (-\frac {4}{3} \, \log \left (\log \left (x^{4} + 4 \, x^{3} + 6 \, x^{2} + 4 \, x + 1\right )\right )^{2} + x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*exp(1)*log(log(x^4+4*x^3+6*x^2+4*x+1))+(3*x+3)*exp(1)*log(x^4+4*x^3+6*x^2+4*x+1))*exp(-4/3*log(
log(x^4+4*x^3+6*x^2+4*x+1))^2+x)/(3*x+3)/log(x^4+4*x^3+6*x^2+4*x+1),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (-32 \,{\mathrm e} \ln \left (\ln \left (x^{4}+4 x^{3}+6 x^{2}+4 x +1\right )\right )+\left (3 x +3\right ) {\mathrm e} \ln \left (x^{4}+4 x^{3}+6 x^{2}+4 x +1\right )\right ) {\mathrm e}^{-\frac {4 \ln \left (\ln \left (x^{4}+4 x^{3}+6 x^{2}+4 x +1\right )\right )^{2}}{3}+x}}{\left (3 x +3\right ) \ln \left (x^{4}+4 x^{3}+6 x^{2}+4 x +1\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-32*exp(1)*ln(ln(x^4+4*x^3+6*x^2+4*x+1))+(3*x+3)*exp(1)*ln(x^4+4*x^3+6*x^2+4*x+1))*exp(-4/3*ln(ln(x^4+4*x
^3+6*x^2+4*x+1))^2+x)/(3*x+3)/ln(x^4+4*x^3+6*x^2+4*x+1),x)

[Out]

int((-32*exp(1)*ln(ln(x^4+4*x^3+6*x^2+4*x+1))+(3*x+3)*exp(1)*ln(x^4+4*x^3+6*x^2+4*x+1))*exp(-4/3*ln(ln(x^4+4*x
^3+6*x^2+4*x+1))^2+x)/(3*x+3)/ln(x^4+4*x^3+6*x^2+4*x+1),x)

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maxima [A]  time = 1.21, size = 28, normalized size = 1.40 \begin {gather*} e^{\left (-\frac {16}{3} \, \log \relax (2)^{2} - \frac {16}{3} \, \log \relax (2) \log \left (\log \left (x + 1\right )\right ) - \frac {4}{3} \, \log \left (\log \left (x + 1\right )\right )^{2} + x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*exp(1)*log(log(x^4+4*x^3+6*x^2+4*x+1))+(3*x+3)*exp(1)*log(x^4+4*x^3+6*x^2+4*x+1))*exp(-4/3*log(
log(x^4+4*x^3+6*x^2+4*x+1))^2+x)/(3*x+3)/log(x^4+4*x^3+6*x^2+4*x+1),x, algorithm="maxima")

[Out]

e^(-16/3*log(2)^2 - 16/3*log(2)*log(log(x + 1)) - 4/3*log(log(x + 1))^2 + x + 1)

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mupad [B]  time = 0.87, size = 30, normalized size = 1.50 \begin {gather*} \mathrm {e}\,{\mathrm {e}}^{-\frac {4\,{\ln \left (\ln \left (x^4+4\,x^3+6\,x^2+4\,x+1\right )\right )}^2}{3}}\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x - (4*log(log(4*x + 6*x^2 + 4*x^3 + x^4 + 1))^2)/3)*(32*log(log(4*x + 6*x^2 + 4*x^3 + x^4 + 1))*exp
(1) - exp(1)*log(4*x + 6*x^2 + 4*x^3 + x^4 + 1)*(3*x + 3)))/(log(4*x + 6*x^2 + 4*x^3 + x^4 + 1)*(3*x + 3)),x)

[Out]

exp(1)*exp(-(4*log(log(4*x + 6*x^2 + 4*x^3 + x^4 + 1))^2)/3)*exp(x)

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sympy [A]  time = 0.57, size = 32, normalized size = 1.60 \begin {gather*} e e^{x - \frac {4 \log {\left (\log {\left (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1 \right )} \right )}^{2}}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*exp(1)*ln(ln(x**4+4*x**3+6*x**2+4*x+1))+(3*x+3)*exp(1)*ln(x**4+4*x**3+6*x**2+4*x+1))*exp(-4/3*l
n(ln(x**4+4*x**3+6*x**2+4*x+1))**2+x)/(3*x+3)/ln(x**4+4*x**3+6*x**2+4*x+1),x)

[Out]

E*exp(x - 4*log(log(x**4 + 4*x**3 + 6*x**2 + 4*x + 1))**2/3)

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