Integrand size = 113, antiderivative size = 20 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=-2+e^{1+x-\frac {4}{3} \log ^2\left (\log \left ((1+x)^4\right )\right )} \] Output:
exp(1)*exp(x-4/3*ln(ln((1+x)^4))^2)-2
Time = 0.13 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=e^{1+x-\frac {4}{3} \log ^2\left (\log \left ((1+x)^4\right )\right )} \] Input:
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]
Output:
E^(1 + x - (4*Log[Log[(1 + x)^4]]^2)/3)
Time = 0.85 (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 = {7239, 27, 7257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (e (3 x+3) \log \left (x^4+4 x^3+6 x^2+4 x+1\right )-32 e \log \left (\log \left (x^4+4 x^3+6 x^2+4 x+1\right )\right )\right ) \exp \left (\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (x^4+4 x^3+6 x^2+4 x+1\right )\right )\right )\right )}{(3 x+3) \log \left (x^4+4 x^3+6 x^2+4 x+1\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{x-\frac {4}{3} \log ^2\left (\log \left ((x+1)^4\right )\right )+1} \left (3 (x+1) \log \left ((x+1)^4\right )-32 \log \left (\log \left ((x+1)^4\right )\right )\right )}{3 (x+1) \log \left ((x+1)^4\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int \frac {e^{-\frac {4}{3} \log ^2\left (\log \left ((x+1)^4\right )\right )+x+1} \left (3 (x+1) \log \left ((x+1)^4\right )-32 \log \left (\log \left ((x+1)^4\right )\right )\right )}{(x+1) \log \left ((x+1)^4\right )}dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle e^{x-\frac {4}{3} \log ^2\left (\log \left ((x+1)^4\right )\right )+1}\) |
Input:
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]
Output:
E^(1 + x - (4*Log[Log[(1 + x)^4]]^2)/3)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 1.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55
method | result | size |
parallelrisch | \({\mathrm e} \,{\mathrm e}^{-\frac {4 {\ln \left (\ln \left (x^{4}+4 x^{3}+6 x^{2}+4 x +1\right )\right )}^{2}}{3}+x}\) | \(31\) |
Input:
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)
Output:
exp(1)*exp(-4/3*ln(ln(x^4+4*x^3+6*x^2+4*x+1))^2+x)
Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=e^{\left (-\frac {4}{3} \, \log \left (\log \left (x^{4} + 4 \, x^{3} + 6 \, x^{2} + 4 \, x + 1\right )\right )^{2} + x + 1\right )} \] Input:
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")
Output:
e^(-4/3*log(log(x^4 + 4*x^3 + 6*x^2 + 4*x + 1))^2 + x + 1)
Time = 0.32 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=e e^{x - \frac {4 \log {\left (\log {\left (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1 \right )} \right )}^{2}}{3}} \] Input:
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*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)
Output:
E*exp(x - 4*log(log(x**4 + 4*x**3 + 6*x**2 + 4*x + 1))**2/3)
Time = 0.32 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=e^{\left (-\frac {16}{3} \, \log \left (2\right )^{2} - \frac {16}{3} \, \log \left (2\right ) \log \left (\log \left (x + 1\right )\right ) - \frac {4}{3} \, \log \left (\log \left (x + 1\right )\right )^{2} + x + 1\right )} \] Input:
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")
Output:
e^(-16/3*log(2)^2 - 16/3*log(2)*log(log(x + 1)) - 4/3*log(log(x + 1))^2 + x + 1)
Time = 0.44 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=e^{\left (-\frac {4}{3} \, \log \left (\log \left (x^{4} + 4 \, x^{3} + 6 \, x^{2} + 4 \, x + 1\right )\right )^{2} + x + 1\right )} \] Input:
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")
Output:
e^(-4/3*log(log(x^4 + 4*x^3 + 6*x^2 + 4*x + 1))^2 + x + 1)
Time = 0.68 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=\mathrm {e}\,{\mathrm {e}}^{-\frac {4\,{\ln \left (\ln \left (x^4+4\,x^3+6\,x^2+4\,x+1\right )\right )}^2}{3}}\,{\mathrm {e}}^x \] Input:
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)
Output:
exp(1)*exp(-(4*log(log(4*x + 6*x^2 + 4*x^3 + x^4 + 1))^2)/3)*exp(x)
Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.65 \[ \int \frac {e^{\frac {1}{3} \left (3 x-4 \log ^2\left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )} \left (e (3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )-32 e \log \left (\log \left (1+4 x+6 x^2+4 x^3+x^4\right )\right )\right )}{(3+3 x) \log \left (1+4 x+6 x^2+4 x^3+x^4\right )} \, dx=\frac {e^{x} e}{e^{\frac {4 {\mathrm {log}\left (\mathrm {log}\left (x^{4}+4 x^{3}+6 x^{2}+4 x +1\right )\right )}^{2}}{3}}} \] Input:
int((-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)
Output:
(e**x*e)/e**((4*log(log(x**4 + 4*x**3 + 6*x**2 + 4*x + 1))**2)/3)