Integrand size = 107, antiderivative size = 24 \[ \int \frac {x^4+\left (-x^2-2 x^3\right ) \log (4)+x^2 \log ^2(4)+e^x \left (e^4 \left (4 x^2-4 x^3+x^4\right )+e^4 \left (-8 x+8 x^2-2 x^3\right ) \log (4)+e^4 \left (4-4 x+x^2\right ) \log ^2(4)\right )}{x^4-2 x^3 \log (4)+x^2 \log ^2(4)} \, dx=\frac {e^{4+x} (-4+x)}{x}+x+\frac {x}{x-\log (4)} \] Output:
x+(-4+x)*exp(4)*exp(x)/x+x/(x-2*ln(2))
Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {x^4+\left (-x^2-2 x^3\right ) \log (4)+x^2 \log ^2(4)+e^x \left (e^4 \left (4 x^2-4 x^3+x^4\right )+e^4 \left (-8 x+8 x^2-2 x^3\right ) \log (4)+e^4 \left (4-4 x+x^2\right ) \log ^2(4)\right )}{x^4-2 x^3 \log (4)+x^2 \log ^2(4)} \, dx=e^x \left (e^4-\frac {4 e^4}{x}\right )+x+\frac {\log (4)}{x-\log (4)} \] Input:
Integrate[(x^4 + (-x^2 - 2*x^3)*Log[4] + x^2*Log[4]^2 + E^x*(E^4*(4*x^2 - 4*x^3 + x^4) + E^4*(-8*x + 8*x^2 - 2*x^3)*Log[4] + E^4*(4 - 4*x + x^2)*Log [4]^2))/(x^4 - 2*x^3*Log[4] + x^2*Log[4]^2),x]
Output:
E^x*(E^4 - (4*E^4)/x) + x + Log[4]/(x - Log[4])
Time = 0.83 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {2026, 7239, 2009}
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.
\(\displaystyle \int \frac {x^4+x^2 \log ^2(4)+\left (-2 x^3-x^2\right ) \log (4)+e^x \left (e^4 \left (x^2-4 x+4\right ) \log ^2(4)+e^4 \left (-2 x^3+8 x^2-8 x\right ) \log (4)+e^4 \left (x^4-4 x^3+4 x^2\right )\right )}{x^4-2 x^3 \log (4)+x^2 \log ^2(4)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {x^4+x^2 \log ^2(4)+\left (-2 x^3-x^2\right ) \log (4)+e^x \left (e^4 \left (x^2-4 x+4\right ) \log ^2(4)+e^4 \left (-2 x^3+8 x^2-8 x\right ) \log (4)+e^4 \left (x^4-4 x^3+4 x^2\right )\right )}{x^2 \left (x^2-2 x \log (4)+\log ^2(4)\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (\frac {e^{x+4} (x-2)^2}{x^2}+\frac {x^2-2 x \log (4)-(1-\log (4)) \log (4)}{(x-\log (4))^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x+e^{x+4}-\frac {4 e^{x+4}}{x}+\frac {\log (4)}{x-\log (4)}\) |
Input:
Int[(x^4 + (-x^2 - 2*x^3)*Log[4] + x^2*Log[4]^2 + E^x*(E^4*(4*x^2 - 4*x^3 + x^4) + E^4*(-8*x + 8*x^2 - 2*x^3)*Log[4] + E^4*(4 - 4*x + x^2)*Log[4]^2) )/(x^4 - 2*x^3*Log[4] + x^2*Log[4]^2),x]
Output:
E^(4 + x) - (4*E^(4 + x))/x + x + Log[4]/(x - Log[4])
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.73 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08
method | result | size |
risch | \(x -\frac {\ln \left (2\right )}{\ln \left (2\right )-\frac {x}{2}}+\frac {\left (x -4\right ) {\mathrm e}^{4+x}}{x}\) | \(26\) |
parts | \(x +\frac {2 \ln \left (2\right )}{x -2 \ln \left (2\right )}+{\mathrm e}^{4} \left (-\frac {4 \,{\mathrm e}^{x}}{x}+{\mathrm e}^{x}\right )\) | \(28\) |
norman | \(\frac {\left (4 \ln \left (2\right )^{2}-2 \ln \left (2\right )\right ) x +\left (2 \,{\mathrm e}^{4} \ln \left (2\right )+4 \,{\mathrm e}^{4}\right ) x \,{\mathrm e}^{x}-x^{3}-x^{2} {\mathrm e}^{4} {\mathrm e}^{x}-8 \,{\mathrm e}^{4} \ln \left (2\right ) {\mathrm e}^{x}}{x \left (2 \ln \left (2\right )-x \right )}\) | \(66\) |
parallelrisch | \(\frac {2 \,{\mathrm e}^{x} {\mathrm e}^{4} \ln \left (2\right ) x -x^{2} {\mathrm e}^{4} {\mathrm e}^{x}-8 \,{\mathrm e}^{4} \ln \left (2\right ) {\mathrm e}^{x}+4 \,{\mathrm e}^{4} {\mathrm e}^{x} x +4 x \ln \left (2\right )^{2}-x^{3}-2 x \ln \left (2\right )}{x \left (2 \ln \left (2\right )-x \right )}\) | \(66\) |
default | \(x +{\mathrm e}^{4} \left ({\mathrm e}^{x}+4 \ln \left (2\right )^{2} \left (-\frac {{\mathrm e}^{x}}{x -2 \ln \left (2\right )}-4 \,\operatorname {expIntegral}_{1}\left (2 \ln \left (2\right )-x \right )\right )-16 \ln \left (2\right ) \operatorname {expIntegral}_{1}\left (2 \ln \left (2\right )-x \right )\right )+\frac {2 \ln \left (2\right )}{x -2 \ln \left (2\right )}+4 \,{\mathrm e}^{4} \left (-\frac {{\mathrm e}^{x}}{x -2 \ln \left (2\right )}-4 \,\operatorname {expIntegral}_{1}\left (2 \ln \left (2\right )-x \right )\right )-4 \,{\mathrm e}^{4} \left (2 \ln \left (2\right ) \left (-\frac {{\mathrm e}^{x}}{x -2 \ln \left (2\right )}-4 \,\operatorname {expIntegral}_{1}\left (2 \ln \left (2\right )-x \right )\right )-4 \,\operatorname {expIntegral}_{1}\left (2 \ln \left (2\right )-x \right )\right )-\frac {4 \,{\mathrm e}^{4} {\mathrm e}^{x}}{x}+\frac {4 \,{\mathrm e}^{4} {\mathrm e}^{x}}{x -2 \ln \left (2\right )}+16 \,{\mathrm e}^{4} \ln \left (2\right ) \left (-\frac {{\mathrm e}^{x}}{x -2 \ln \left (2\right )}-4 \,\operatorname {expIntegral}_{1}\left (2 \ln \left (2\right )-x \right )\right )-4 \,{\mathrm e}^{4} \ln \left (2\right ) \left (2 \ln \left (2\right ) \left (-\frac {{\mathrm e}^{x}}{x -2 \ln \left (2\right )}-4 \,\operatorname {expIntegral}_{1}\left (2 \ln \left (2\right )-x \right )\right )-4 \,\operatorname {expIntegral}_{1}\left (2 \ln \left (2\right )-x \right )\right )+\frac {8 \,{\mathrm e}^{4} \ln \left (2\right ) {\mathrm e}^{x}}{x -2 \ln \left (2\right )}+32 \,{\mathrm e}^{4} \ln \left (2\right ) \operatorname {expIntegral}_{1}\left (2 \ln \left (2\right )-x \right )+4 \,{\mathrm e}^{4} \ln \left (2\right )^{2} \left (-\frac {{\mathrm e}^{x}}{x -2 \ln \left (2\right )}-4 \,\operatorname {expIntegral}_{1}\left (2 \ln \left (2\right )-x \right )\right )\) | \(308\) |
Input:
int(((4*(x^2-4*x+4)*exp(4)*ln(2)^2+2*(-2*x^3+8*x^2-8*x)*exp(4)*ln(2)+(x^4- 4*x^3+4*x^2)*exp(4))*exp(x)+4*x^2*ln(2)^2+2*(-2*x^3-x^2)*ln(2)+x^4)/(4*x^2 *ln(2)^2-4*x^3*ln(2)+x^4),x,method=_RETURNVERBOSE)
Output:
x-ln(2)/(ln(2)-1/2*x)+(x-4)/x*exp(4+x)
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (23) = 46\).
Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17 \[ \int \frac {x^4+\left (-x^2-2 x^3\right ) \log (4)+x^2 \log ^2(4)+e^x \left (e^4 \left (4 x^2-4 x^3+x^4\right )+e^4 \left (-8 x+8 x^2-2 x^3\right ) \log (4)+e^4 \left (4-4 x+x^2\right ) \log ^2(4)\right )}{x^4-2 x^3 \log (4)+x^2 \log ^2(4)} \, dx=\frac {x^{3} - {\left (2 \, {\left (x - 4\right )} e^{4} \log \left (2\right ) - {\left (x^{2} - 4 \, x\right )} e^{4}\right )} e^{x} - 2 \, {\left (x^{2} - x\right )} \log \left (2\right )}{x^{2} - 2 \, x \log \left (2\right )} \] Input:
integrate(((4*(x^2-4*x+4)*exp(4)*log(2)^2+2*(-2*x^3+8*x^2-8*x)*exp(4)*log( 2)+(x^4-4*x^3+4*x^2)*exp(4))*exp(x)+4*x^2*log(2)^2+2*(-2*x^3-x^2)*log(2)+x ^4)/(4*x^2*log(2)^2-4*x^3*log(2)+x^4),x, algorithm="fricas")
Output:
(x^3 - (2*(x - 4)*e^4*log(2) - (x^2 - 4*x)*e^4)*e^x - 2*(x^2 - x)*log(2))/ (x^2 - 2*x*log(2))
Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {x^4+\left (-x^2-2 x^3\right ) \log (4)+x^2 \log ^2(4)+e^x \left (e^4 \left (4 x^2-4 x^3+x^4\right )+e^4 \left (-8 x+8 x^2-2 x^3\right ) \log (4)+e^4 \left (4-4 x+x^2\right ) \log ^2(4)\right )}{x^4-2 x^3 \log (4)+x^2 \log ^2(4)} \, dx=x + \frac {2 \log {\left (2 \right )}}{x - 2 \log {\left (2 \right )}} + \frac {\left (x e^{4} - 4 e^{4}\right ) e^{x}}{x} \] Input:
integrate(((4*(x**2-4*x+4)*exp(4)*ln(2)**2+2*(-2*x**3+8*x**2-8*x)*exp(4)*l n(2)+(x**4-4*x**3+4*x**2)*exp(4))*exp(x)+4*x**2*ln(2)**2+2*(-2*x**3-x**2)* ln(2)+x**4)/(4*x**2*ln(2)**2-4*x**3*ln(2)+x**4),x)
Output:
x + 2*log(2)/(x - 2*log(2)) + (x*exp(4) - 4*exp(4))*exp(x)/x
Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (23) = 46\).
Time = 0.16 (sec) , antiderivative size = 80, normalized size of antiderivative = 3.33 \[ \int \frac {x^4+\left (-x^2-2 x^3\right ) \log (4)+x^2 \log ^2(4)+e^x \left (e^4 \left (4 x^2-4 x^3+x^4\right )+e^4 \left (-8 x+8 x^2-2 x^3\right ) \log (4)+e^4 \left (4-4 x+x^2\right ) \log ^2(4)\right )}{x^4-2 x^3 \log (4)+x^2 \log ^2(4)} \, dx=4 \, {\left (\frac {2 \, \log \left (2\right )}{x - 2 \, \log \left (2\right )} - \log \left (x - 2 \, \log \left (2\right )\right )\right )} \log \left (2\right ) + 4 \, \log \left (2\right ) \log \left (x - 2 \, \log \left (2\right )\right ) + x + \frac {{\left (x e^{4} - 4 \, e^{4}\right )} e^{x}}{x} - \frac {8 \, \log \left (2\right )^{2}}{x - 2 \, \log \left (2\right )} + \frac {2 \, \log \left (2\right )}{x - 2 \, \log \left (2\right )} \] Input:
integrate(((4*(x^2-4*x+4)*exp(4)*log(2)^2+2*(-2*x^3+8*x^2-8*x)*exp(4)*log( 2)+(x^4-4*x^3+4*x^2)*exp(4))*exp(x)+4*x^2*log(2)^2+2*(-2*x^3-x^2)*log(2)+x ^4)/(4*x^2*log(2)^2-4*x^3*log(2)+x^4),x, algorithm="maxima")
Output:
4*(2*log(2)/(x - 2*log(2)) - log(x - 2*log(2)))*log(2) + 4*log(2)*log(x - 2*log(2)) + x + (x*e^4 - 4*e^4)*e^x/x - 8*log(2)^2/(x - 2*log(2)) + 2*log( 2)/(x - 2*log(2))
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (23) = 46\).
Time = 0.13 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.50 \[ \int \frac {x^4+\left (-x^2-2 x^3\right ) \log (4)+x^2 \log ^2(4)+e^x \left (e^4 \left (4 x^2-4 x^3+x^4\right )+e^4 \left (-8 x+8 x^2-2 x^3\right ) \log (4)+e^4 \left (4-4 x+x^2\right ) \log ^2(4)\right )}{x^4-2 x^3 \log (4)+x^2 \log ^2(4)} \, dx=\frac {x^{3} + x^{2} e^{\left (x + 4\right )} - 2 \, x^{2} \log \left (2\right ) - 2 \, x e^{\left (x + 4\right )} \log \left (2\right ) - 4 \, x e^{\left (x + 4\right )} + 2 \, x \log \left (2\right ) + 8 \, e^{\left (x + 4\right )} \log \left (2\right )}{x^{2} - 2 \, x \log \left (2\right )} \] Input:
integrate(((4*(x^2-4*x+4)*exp(4)*log(2)^2+2*(-2*x^3+8*x^2-8*x)*exp(4)*log( 2)+(x^4-4*x^3+4*x^2)*exp(4))*exp(x)+4*x^2*log(2)^2+2*(-2*x^3-x^2)*log(2)+x ^4)/(4*x^2*log(2)^2-4*x^3*log(2)+x^4),x, algorithm="giac")
Output:
(x^3 + x^2*e^(x + 4) - 2*x^2*log(2) - 2*x*e^(x + 4)*log(2) - 4*x*e^(x + 4) + 2*x*log(2) + 8*e^(x + 4)*log(2))/(x^2 - 2*x*log(2))
Time = 2.44 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.88 \[ \int \frac {x^4+\left (-x^2-2 x^3\right ) \log (4)+x^2 \log ^2(4)+e^x \left (e^4 \left (4 x^2-4 x^3+x^4\right )+e^4 \left (-8 x+8 x^2-2 x^3\right ) \log (4)+e^4 \left (4-4 x+x^2\right ) \log ^2(4)\right )}{x^4-2 x^3 \log (4)+x^2 \log ^2(4)} \, dx=x-\frac {{\mathrm {e}}^x\,\left (4\,{\mathrm {e}}^4-x\,{\mathrm {e}}^4\right )}{x}+\frac {4\,\ln \left (2\right )\,\mathrm {atanh}\left (\frac {2\,x-\ln \left (16\right )}{\sqrt {2\,\ln \left (4\right )+\ln \left (16\right )}\,\sqrt {\ln \left (16\right )-2\,\ln \left (4\right )}}\right )}{\sqrt {2\,\ln \left (4\right )+\ln \left (16\right )}\,\sqrt {\ln \left (16\right )-2\,\ln \left (4\right )}} \] Input:
int((4*x^2*log(2)^2 - 2*log(2)*(x^2 + 2*x^3) + exp(x)*(exp(4)*(4*x^2 - 4*x ^3 + x^4) + 4*exp(4)*log(2)^2*(x^2 - 4*x + 4) - 2*exp(4)*log(2)*(8*x - 8*x ^2 + 2*x^3)) + x^4)/(4*x^2*log(2)^2 - 4*x^3*log(2) + x^4),x)
Output:
x - (exp(x)*(4*exp(4) - x*exp(4)))/x + (4*log(2)*atanh((2*x - log(16))/((2 *log(4) + log(16))^(1/2)*(log(16) - 2*log(4))^(1/2))))/((2*log(4) + log(16 ))^(1/2)*(log(16) - 2*log(4))^(1/2))
Time = 0.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 3.04 \[ \int \frac {x^4+\left (-x^2-2 x^3\right ) \log (4)+x^2 \log ^2(4)+e^x \left (e^4 \left (4 x^2-4 x^3+x^4\right )+e^4 \left (-8 x+8 x^2-2 x^3\right ) \log (4)+e^4 \left (4-4 x+x^2\right ) \log ^2(4)\right )}{x^4-2 x^3 \log (4)+x^2 \log ^2(4)} \, dx=\frac {2 e^{x} \mathrm {log}\left (2\right ) e^{4} x -8 e^{x} \mathrm {log}\left (2\right ) e^{4}-e^{x} e^{4} x^{2}+4 e^{x} e^{4} x +2 \,\mathrm {log}\left (2\right ) x^{2}-x^{3}-x^{2}}{x \left (2 \,\mathrm {log}\left (2\right )-x \right )} \] Input:
int(((4*(x^2-4*x+4)*exp(4)*log(2)^2+2*(-2*x^3+8*x^2-8*x)*exp(4)*log(2)+(x^ 4-4*x^3+4*x^2)*exp(4))*exp(x)+4*x^2*log(2)^2+2*(-2*x^3-x^2)*log(2)+x^4)/(4 *x^2*log(2)^2-4*x^3*log(2)+x^4),x)
Output:
(2*e**x*log(2)*e**4*x - 8*e**x*log(2)*e**4 - e**x*e**4*x**2 + 4*e**x*e**4* x + 2*log(2)*x**2 - x**3 - x**2)/(x*(2*log(2) - x))