\(\int \frac {e^{\frac {1}{3} (3 x-4 \log ^2(\log (1+4 x+6 x^2+4 x^3+x^4)))} (e (3+3 x) \log (1+4 x+6 x^2+4 x^3+x^4)-32 e \log (\log (1+4 x+6 x^2+4 x^3+x^4)))}{(3+3 x) \log (1+4 x+6 x^2+4 x^3+x^4)} \, dx\) [591]

Optimal result

Integrand size = 113, antiderivative size = 20 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=-2+e^{1+x-\frac {4}{3} \log ^2\left (\log \left ((1+x)^4\right )\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {6820, 12, 6838} \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=e^{x-\frac {4}{3} \log ^2\left (\log \left ((x+1)^4\right )\right )+1} \]

[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 6820

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

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=e^{1+x-\frac {4}{3} \log ^2\left (\log \left ((1+x)^4\right )\right )} \]

[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)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55

method	result	size
parallelrisch	\({\mathrm e} \,{\mathrm e}^{-\frac {4 {\ln \left (\ln \left (x^{4}+4 x^{3}+6 x^{2}+4 x +1\right )\right )}^{2}}{3}+x}\)	\(31\)

[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,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=e^{\left (-\frac {4}{3} \, \log \left (\log \left (x^{4} + 4 \, x^{3} + 6 \, x^{2} + 4 \, x + 1\right )\right )^{2} + x + 1\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=e e^{x - \frac {4 \log {\left (\log {\left (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1 \right )} \right )}^{2}}{3}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=e^{\left (-\frac {16}{3} \, \log \left (2\right )^{2} - \frac {16}{3} \, \log \left (2\right ) \log \left (\log \left (x + 1\right )\right ) - \frac {4}{3} \, \log \left (\log \left (x + 1\right )\right )^{2} + x + 1\right )} \]

[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)

Giac [A] (verification not implemented)

none

Time = 1.34 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=e^{\left (-\frac {4}{3} \, \log \left (\log \left (x^{4} + 4 \, x^{3} + 6 \, x^{2} + 4 \, x + 1\right )\right )^{2} + x + 1\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 8.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=\mathrm {e}\,{\mathrm {e}}^{-\frac {4\,{\ln \left (\ln \left (x^4+4\,x^3+6\,x^2+4\,x+1\right )\right )}^2}{3}}\,{\mathrm {e}}^x \]

[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)