Integrand size = 161, antiderivative size = 36 \[ \int \frac {e^{3-e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x}-x} \left (-4-x+e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x} \left (4+x+e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)} \left (14 x^3+4 x^4+\left (-22 x^2-6 x^3\right ) \log (4+x)+\left (8 x+2 x^2\right ) \log ^2(4+x)\right )\right )\right )}{4+x} \, dx=2+e^{3-e^{3-e^{x^2 (-x+\log (4+x))^2}-x}-x} \] Output:
2+exp(3-exp(3-exp(x^2*(ln(4+x)-x)^2)-x)-x)
Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.19 \[ \int \frac {e^{3-e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x}-x} \left (-4-x+e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x} \left (4+x+e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)} \left (14 x^3+4 x^4+\left (-22 x^2-6 x^3\right ) \log (4+x)+\left (8 x+2 x^2\right ) \log ^2(4+x)\right )\right )\right )}{4+x} \, dx=e^{3-e^{3-x-e^{x^4+x^2 \log ^2(4+x)} (4+x)^{-2 x^3}}-x} \] Input:
Integrate[(E^(3 - E^(3 - E^(x^4 - 2*x^3*Log[4 + x] + x^2*Log[4 + x]^2) - x ) - x)*(-4 - x + E^(3 - E^(x^4 - 2*x^3*Log[4 + x] + x^2*Log[4 + x]^2) - x) *(4 + x + E^(x^4 - 2*x^3*Log[4 + x] + x^2*Log[4 + x]^2)*(14*x^3 + 4*x^4 + (-22*x^2 - 6*x^3)*Log[4 + x] + (8*x + 2*x^2)*Log[4 + x]^2))))/(4 + x),x]
Output:
E^(3 - E^(3 - x - E^(x^4 + x^2*Log[4 + x]^2)/(4 + x)^(2*x^3)) - x)
Time = 4.84 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.19, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.006, Rules used = {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.
\(\displaystyle \int \frac {\exp \left (-\exp \left (-e^{x^4-2 x^3 \log (x+4)+x^2 \log ^2(x+4)}-x+3\right )-x+3\right ) \left (\left (e^{x^4-2 x^3 \log (x+4)+x^2 \log ^2(x+4)} \left (4 x^4+14 x^3+\left (2 x^2+8 x\right ) \log ^2(x+4)+\left (-6 x^3-22 x^2\right ) \log (x+4)\right )+x+4\right ) \exp \left (-e^{x^4-2 x^3 \log (x+4)+x^2 \log ^2(x+4)}-x+3\right )-x-4\right )}{x+4} \, dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle \exp \left (-\exp \left ((x+4)^{-2 x^3} \left (-e^{x^4+x^2 \log ^2(x+4)}\right )-x+3\right )-x+3\right )\) |
Input:
Int[(E^(3 - E^(3 - E^(x^4 - 2*x^3*Log[4 + x] + x^2*Log[4 + x]^2) - x) - x) *(-4 - x + E^(3 - E^(x^4 - 2*x^3*Log[4 + x] + x^2*Log[4 + x]^2) - x)*(4 + x + E^(x^4 - 2*x^3*Log[4 + x] + x^2*Log[4 + x]^2)*(14*x^3 + 4*x^4 + (-22*x ^2 - 6*x^3)*Log[4 + x] + (8*x + 2*x^2)*Log[4 + x]^2))))/(4 + x),x]
Output:
E^(3 - E^(3 - x - E^(x^4 + x^2*Log[4 + x]^2)/(4 + x)^(2*x^3)) - 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 = 4.43 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.14
method | result | size |
risch | \({\mathrm e}^{-{\mathrm e}^{-\left (4+x \right )^{-2 x^{3}} {\mathrm e}^{x^{2} \left (\ln \left (4+x \right )^{2}+x^{2}\right )}+3-x}+3-x}\) | \(41\) |
parallelrisch | \({\mathrm e}^{-{\mathrm e}^{-{\mathrm e}^{x^{2} \ln \left (4+x \right )^{2}-2 x^{3} \ln \left (4+x \right )+x^{4}}+3-x}+3-x}\) | \(41\) |
Input:
int(((((2*x^2+8*x)*ln(4+x)^2+(-6*x^3-22*x^2)*ln(4+x)+4*x^4+14*x^3)*exp(x^2 *ln(4+x)^2-2*x^3*ln(4+x)+x^4)+4+x)*exp(-exp(x^2*ln(4+x)^2-2*x^3*ln(4+x)+x^ 4)+3-x)-x-4)*exp(-exp(-exp(x^2*ln(4+x)^2-2*x^3*ln(4+x)+x^4)+3-x)+3-x)/(4+x ),x,method=_RETURNVERBOSE)
Output:
exp(-exp(-(4+x)^(-2*x^3)*exp(x^2*(ln(4+x)^2+x^2))+3-x)+3-x)
Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.11 \[ \int \frac {e^{3-e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x}-x} \left (-4-x+e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x} \left (4+x+e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)} \left (14 x^3+4 x^4+\left (-22 x^2-6 x^3\right ) \log (4+x)+\left (8 x+2 x^2\right ) \log ^2(4+x)\right )\right )\right )}{4+x} \, dx=e^{\left (-x - e^{\left (-x - e^{\left (x^{4} - 2 \, x^{3} \log \left (x + 4\right ) + x^{2} \log \left (x + 4\right )^{2}\right )} + 3\right )} + 3\right )} \] Input:
integrate(((((2*x^2+8*x)*log(4+x)^2+(-6*x^3-22*x^2)*log(4+x)+4*x^4+14*x^3) *exp(x^2*log(4+x)^2-2*x^3*log(4+x)+x^4)+4+x)*exp(-exp(x^2*log(4+x)^2-2*x^3 *log(4+x)+x^4)+3-x)-x-4)*exp(-exp(-exp(x^2*log(4+x)^2-2*x^3*log(4+x)+x^4)+ 3-x)+3-x)/(4+x),x, algorithm="fricas")
Output:
e^(-x - e^(-x - e^(x^4 - 2*x^3*log(x + 4) + x^2*log(x + 4)^2) + 3) + 3)
Timed out. \[ \int \frac {e^{3-e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x}-x} \left (-4-x+e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x} \left (4+x+e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)} \left (14 x^3+4 x^4+\left (-22 x^2-6 x^3\right ) \log (4+x)+\left (8 x+2 x^2\right ) \log ^2(4+x)\right )\right )\right )}{4+x} \, dx=\text {Timed out} \] Input:
integrate(((((2*x**2+8*x)*ln(4+x)**2+(-6*x**3-22*x**2)*ln(4+x)+4*x**4+14*x **3)*exp(x**2*ln(4+x)**2-2*x**3*ln(4+x)+x**4)+4+x)*exp(-exp(x**2*ln(4+x)** 2-2*x**3*ln(4+x)+x**4)+3-x)-x-4)*exp(-exp(-exp(x**2*ln(4+x)**2-2*x**3*ln(4 +x)+x**4)+3-x)+3-x)/(4+x),x)
Output:
Timed out
Time = 0.33 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.11 \[ \int \frac {e^{3-e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x}-x} \left (-4-x+e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x} \left (4+x+e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)} \left (14 x^3+4 x^4+\left (-22 x^2-6 x^3\right ) \log (4+x)+\left (8 x+2 x^2\right ) \log ^2(4+x)\right )\right )\right )}{4+x} \, dx=e^{\left (-x - e^{\left (-x - e^{\left (x^{4} - 2 \, x^{3} \log \left (x + 4\right ) + x^{2} \log \left (x + 4\right )^{2}\right )} + 3\right )} + 3\right )} \] Input:
integrate(((((2*x^2+8*x)*log(4+x)^2+(-6*x^3-22*x^2)*log(4+x)+4*x^4+14*x^3) *exp(x^2*log(4+x)^2-2*x^3*log(4+x)+x^4)+4+x)*exp(-exp(x^2*log(4+x)^2-2*x^3 *log(4+x)+x^4)+3-x)-x-4)*exp(-exp(-exp(x^2*log(4+x)^2-2*x^3*log(4+x)+x^4)+ 3-x)+3-x)/(4+x),x, algorithm="maxima")
Output:
e^(-x - e^(-x - e^(x^4 - 2*x^3*log(x + 4) + x^2*log(x + 4)^2) + 3) + 3)
Time = 1.64 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.11 \[ \int \frac {e^{3-e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x}-x} \left (-4-x+e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x} \left (4+x+e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)} \left (14 x^3+4 x^4+\left (-22 x^2-6 x^3\right ) \log (4+x)+\left (8 x+2 x^2\right ) \log ^2(4+x)\right )\right )\right )}{4+x} \, dx=e^{\left (-x - e^{\left (-x - e^{\left (x^{4} - 2 \, x^{3} \log \left (x + 4\right ) + x^{2} \log \left (x + 4\right )^{2}\right )} + 3\right )} + 3\right )} \] Input:
integrate(((((2*x^2+8*x)*log(4+x)^2+(-6*x^3-22*x^2)*log(4+x)+4*x^4+14*x^3) *exp(x^2*log(4+x)^2-2*x^3*log(4+x)+x^4)+4+x)*exp(-exp(x^2*log(4+x)^2-2*x^3 *log(4+x)+x^4)+3-x)-x-4)*exp(-exp(-exp(x^2*log(4+x)^2-2*x^3*log(4+x)+x^4)+ 3-x)+3-x)/(4+x),x, algorithm="giac")
Output:
e^(-x - e^(-x - e^(x^4 - 2*x^3*log(x + 4) + x^2*log(x + 4)^2) + 3) + 3)
Time = 4.31 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.25 \[ \int \frac {e^{3-e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x}-x} \left (-4-x+e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x} \left (4+x+e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)} \left (14 x^3+4 x^4+\left (-22 x^2-6 x^3\right ) \log (4+x)+\left (8 x+2 x^2\right ) \log ^2(4+x)\right )\right )\right )}{4+x} \, dx={\mathrm {e}}^{-{\mathrm {e}}^{-x}\,{\mathrm {e}}^3\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{x^4}\,{\mathrm {e}}^{x^2\,{\ln \left (x+4\right )}^2}}{{\left (x+4\right )}^{2\,x^3}}}}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^3 \] Input:
int(-(exp(3 - exp(3 - exp(x^4 - 2*x^3*log(x + 4) + x^2*log(x + 4)^2) - x) - x)*(x - exp(3 - exp(x^4 - 2*x^3*log(x + 4) + x^2*log(x + 4)^2) - x)*(x + exp(x^4 - 2*x^3*log(x + 4) + x^2*log(x + 4)^2)*(log(x + 4)^2*(8*x + 2*x^2 ) - log(x + 4)*(22*x^2 + 6*x^3) + 14*x^3 + 4*x^4) + 4) + 4))/(x + 4),x)
Output:
exp(-exp(-x)*exp(3)*exp(-(exp(x^4)*exp(x^2*log(x + 4)^2))/(x + 4)^(2*x^3)) )*exp(-x)*exp(3)
Time = 0.56 (sec) , antiderivative size = 101, normalized size of antiderivative = 2.81 \[ \int \frac {e^{3-e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x}-x} \left (-4-x+e^{3-e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)}-x} \left (4+x+e^{x^4-2 x^3 \log (4+x)+x^2 \log ^2(4+x)} \left (14 x^3+4 x^4+\left (-22 x^2-6 x^3\right ) \log (4+x)+\left (8 x+2 x^2\right ) \log ^2(4+x)\right )\right )\right )}{4+x} \, dx=\frac {e^{3}}{e^{\frac {e^{\frac {\left (x +4\right )^{2 x^{3}} x +e^{\mathrm {log}\left (x +4\right )^{2} x^{2}+x^{4}}}{\left (x +4\right )^{2 x^{3}}}} x +e^{3}}{e^{\frac {\left (x +4\right )^{2 x^{3}} x +e^{\mathrm {log}\left (x +4\right )^{2} x^{2}+x^{4}}}{\left (x +4\right )^{2 x^{3}}}}}}} \] Input:
int(((((2*x^2+8*x)*log(4+x)^2+(-6*x^3-22*x^2)*log(4+x)+4*x^4+14*x^3)*exp(x ^2*log(4+x)^2-2*x^3*log(4+x)+x^4)+4+x)*exp(-exp(x^2*log(4+x)^2-2*x^3*log(4 +x)+x^4)+3-x)-x-4)*exp(-exp(-exp(x^2*log(4+x)^2-2*x^3*log(4+x)+x^4)+3-x)+3 -x)/(4+x),x)
Output:
e**3/e**((e**(((x + 4)**(2*x**3)*x + e**(log(x + 4)**2*x**2 + x**4))/(x + 4)**(2*x**3))*x + e**3)/e**(((x + 4)**(2*x**3)*x + e**(log(x + 4)**2*x**2 + x**4))/(x + 4)**(2*x**3)))