\(\int \frac {e^{e^{\frac {1}{4} (1+4 x)}} (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} (4 x-6 x^2+x^3)+(2-2 x+e^{\frac {1}{4} (1+4 x)} (6 x-2 x^2)) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x))}{2 x} \, dx\) [10109]

Optimal result

Integrand size = 100, antiderivative size = 24 \[ \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\frac {1}{2} e^{e^{\frac {1}{4}+x}} \left (-5+(3-x+\log (x))^2\right ) \]

[Out]

1/2*((3-x+ln(x))^2-5)*exp(exp(x+1/4))

Rubi [F]

\[ \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx \]

[In]

Int[(E^E^((1 + 4*x)/4)*(6 - 8*x + 2*x^2 + E^((1 + 4*x)/4)*(4*x - 6*x^2 + x^3) + (2 - 2*x + E^((1 + 4*x)/4)*(6*
x - 2*x^2))*Log[x] + E^((1 + 4*x)/4)*x*Log[x]^2))/(2*x),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{x} \, dx \\ & = \frac {1}{2} \int \frac {e^{e^{\frac {1}{4}+x}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{x} \, dx \\ & = \frac {1}{2} \int \left (\frac {2 e^{e^{\frac {1}{4}+x}} (-1+x) (-3+x-\log (x))}{x}+e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \left (4-6 x+x^2+6 \log (x)-2 x \log (x)+\log ^2(x)\right )\right ) \, dx \\ & = \frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \left (4-6 x+x^2+6 \log (x)-2 x \log (x)+\log ^2(x)\right ) \, dx+\int \frac {e^{e^{\frac {1}{4}+x}} (-1+x) (-3+x-\log (x))}{x} \, dx \\ & = \frac {1}{2} \int \left (4 e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x}-6 e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x+e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x^2+6 e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \log (x)-2 e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \log (x)+e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \log ^2(x)\right ) \, dx+\int \left (\frac {e^{e^{\frac {1}{4}+x}} \left (3-4 x+x^2\right )}{x}-\frac {e^{e^{\frac {1}{4}+x}} (-1+x) \log (x)}{x}\right ) \, dx \\ & = \frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x^2 \, dx+\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \log ^2(x) \, dx+2 \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \, dx-3 \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx+3 \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \log (x) \, dx+\int \frac {e^{e^{\frac {1}{4}+x}} \left (3-4 x+x^2\right )}{x} \, dx-\int \frac {e^{e^{\frac {1}{4}+x}} (-1+x) \log (x)}{x} \, dx-\int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \log (x) \, dx \\ & = 3 e^{e^{\frac {1}{4}+x}} \log (x)-\operatorname {ExpIntegralEi}\left (e^{\frac {1}{4}+x}\right ) \log (x)+\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x^2 \, dx+\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \log ^2(x) \, dx+2 \text {Subst}\left (\int e^{\frac {1}{4}+\sqrt [4]{e} x} \, dx,x,e^x\right )-3 \int \frac {e^{e^{\frac {1}{4}+x}}}{x} \, dx-3 \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx+\log (x) \int \frac {e^{e^{\frac {1}{4}+x}}}{x} \, dx-\log (x) \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx+\int \left (-4 e^{e^{\frac {1}{4}+x}}+\frac {3 e^{e^{\frac {1}{4}+x}}}{x}+e^{e^{\frac {1}{4}+x}} x\right ) \, dx+\int \frac {\operatorname {ExpIntegralEi}\left (e^{\frac {1}{4}+x}\right )-\int \frac {e^{e^{\frac {1}{4}+x}}}{x} \, dx}{x} \, dx+\int \frac {\int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx}{x} \, dx \\ & = 2 e^{e^{\frac {1}{4}+x}}+3 e^{e^{\frac {1}{4}+x}} \log (x)-\operatorname {ExpIntegralEi}\left (e^{\frac {1}{4}+x}\right ) \log (x)+\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x^2 \, dx+\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \log ^2(x) \, dx-3 \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx-4 \int e^{e^{\frac {1}{4}+x}} \, dx+\log (x) \int \frac {e^{e^{\frac {1}{4}+x}}}{x} \, dx-\log (x) \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx+\int e^{e^{\frac {1}{4}+x}} x \, dx+\int \left (\frac {\operatorname {ExpIntegralEi}\left (e^{\frac {1}{4}+x}\right )}{x}-\frac {\int \frac {e^{e^{\frac {1}{4}+x}}}{x} \, dx}{x}\right ) \, dx+\int \frac {\int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx}{x} \, dx \\ & = 2 e^{e^{\frac {1}{4}+x}}+3 e^{e^{\frac {1}{4}+x}} \log (x)-\operatorname {ExpIntegralEi}\left (e^{\frac {1}{4}+x}\right ) \log (x)+\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x^2 \, dx+\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \log ^2(x) \, dx-3 \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx-4 \text {Subst}\left (\int \frac {e^x}{x} \, dx,x,e^{\frac {1}{4}+x}\right )+\log (x) \int \frac {e^{e^{\frac {1}{4}+x}}}{x} \, dx-\log (x) \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx+\int e^{e^{\frac {1}{4}+x}} x \, dx+\int \frac {\operatorname {ExpIntegralEi}\left (e^{\frac {1}{4}+x}\right )}{x} \, dx-\int \frac {\int \frac {e^{e^{\frac {1}{4}+x}}}{x} \, dx}{x} \, dx+\int \frac {\int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx}{x} \, dx \\ & = 2 e^{e^{\frac {1}{4}+x}}-4 \operatorname {ExpIntegralEi}\left (e^{\frac {1}{4}+x}\right )+3 e^{e^{\frac {1}{4}+x}} \log (x)-\operatorname {ExpIntegralEi}\left (e^{\frac {1}{4}+x}\right ) \log (x)+\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x^2 \, dx+\frac {1}{2} \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} \log ^2(x) \, dx-3 \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx+\log (x) \int \frac {e^{e^{\frac {1}{4}+x}}}{x} \, dx-\log (x) \int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx+\int e^{e^{\frac {1}{4}+x}} x \, dx+\int \frac {\operatorname {ExpIntegralEi}\left (e^{\frac {1}{4}+x}\right )}{x} \, dx-\int \frac {\int \frac {e^{e^{\frac {1}{4}+x}}}{x} \, dx}{x} \, dx+\int \frac {\int e^{\frac {1}{4}+e^{\frac {1}{4}+x}+x} x \, dx}{x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\frac {1}{2} e^{e^{\frac {1}{4}+x}} \left (4-6 x+x^2+(6-2 x) \log (x)+\log ^2(x)\right ) \]

[In]

Integrate[(E^E^((1 + 4*x)/4)*(6 - 8*x + 2*x^2 + E^((1 + 4*x)/4)*(4*x - 6*x^2 + x^3) + (2 - 2*x + E^((1 + 4*x)/
4)*(6*x - 2*x^2))*Log[x] + E^((1 + 4*x)/4)*x*Log[x]^2))/(2*x),x]

[Out]

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

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21

method	result	size
risch	\(\frac {\left (x^{2}-2 x \ln \left (x \right )+\ln \left (x \right )^{2}-6 x +6 \ln \left (x \right )+4\right ) {\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}}}{2}\)	\(29\)
parallelrisch	\(\frac {{\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}} x^{2}}{2}-\ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}} x +\frac {{\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}} \ln \left (x \right )^{2}}{2}-3 x \,{\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}}+3 \ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}}+2 \,{\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}}\)	\(57\)

[In]

int(1/2*(x*exp(x+1/4)*ln(x)^2+((-2*x^2+6*x)*exp(x+1/4)-2*x+2)*ln(x)+(x^3-6*x^2+4*x)*exp(x+1/4)+2*x^2-8*x+6)*ex
p(exp(x+1/4))/x,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\frac {1}{2} \, {\left (x^{2} - 2 \, {\left (x - 3\right )} \log \left (x\right ) + \log \left (x\right )^{2} - 6 \, x + 4\right )} e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \]

[In]

integrate(1/2*(x*exp(x+1/4)*log(x)^2+((-2*x^2+6*x)*exp(x+1/4)-2*x+2)*log(x)+(x^3-6*x^2+4*x)*exp(x+1/4)+2*x^2-8
*x+6)*exp(exp(x+1/4))/x,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 31.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\frac {\left (x^{2} - 2 x \log {\left (x \right )} - 6 x + \log {\left (x \right )}^{2} + 6 \log {\left (x \right )} + 4\right ) e^{e^{x + \frac {1}{4}}}}{2} \]

[In]

integrate(1/2*(x*exp(x+1/4)*ln(x)**2+((-2*x**2+6*x)*exp(x+1/4)-2*x+2)*ln(x)+(x**3-6*x**2+4*x)*exp(x+1/4)+2*x**
2-8*x+6)*exp(exp(x+1/4))/x,x)

[Out]

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

Maxima [F]

\[ \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\int { \frac {{\left (x e^{\left (x + \frac {1}{4}\right )} \log \left (x\right )^{2} + 2 \, x^{2} + {\left (x^{3} - 6 \, x^{2} + 4 \, x\right )} e^{\left (x + \frac {1}{4}\right )} - 2 \, {\left ({\left (x^{2} - 3 \, x\right )} e^{\left (x + \frac {1}{4}\right )} + x - 1\right )} \log \left (x\right ) - 8 \, x + 6\right )} e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )}}{2 \, x} \,d x } \]

[In]

integrate(1/2*(x*exp(x+1/4)*log(x)^2+((-2*x^2+6*x)*exp(x+1/4)-2*x+2)*log(x)+(x^3-6*x^2+4*x)*exp(x+1/4)+2*x^2-8
*x+6)*exp(exp(x+1/4))/x,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (18) = 36\).

Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.33 \[ \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\frac {1}{2} \, x^{2} e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} - x e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \log \left (x\right ) + \frac {1}{2} \, e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \log \left (x\right )^{2} - 3 \, x e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} + 3 \, e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \log \left (x\right ) + 2 \, e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \]

[In]

integrate(1/2*(x*exp(x+1/4)*log(x)^2+((-2*x^2+6*x)*exp(x+1/4)-2*x+2)*log(x)+(x^3-6*x^2+4*x)*exp(x+1/4)+2*x^2-8
*x+6)*exp(exp(x+1/4))/x,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 15.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx={\mathrm {e}}^{{\mathrm {e}}^{1/4}\,{\mathrm {e}}^x}\,\left (\frac {x^2}{2}-x\,\ln \left (x\right )-3\,x+\frac {{\ln \left (x\right )}^2}{2}+3\,\ln \left (x\right )+2\right ) \]

[In]

int((exp(exp(x + 1/4))*(log(x)*(exp(x + 1/4)*(6*x - 2*x^2) - 2*x + 2) - 8*x + 2*x^2 + exp(x + 1/4)*(4*x - 6*x^
2 + x^3) + x*exp(x + 1/4)*log(x)^2 + 6))/(2*x),x)

[Out]

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