\(\int \frac {1}{4} e^{e^x} (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} (5+5 e^x x)+e^x (-20-30 x+20 x^2)) \, dx\) [10306]

Optimal result

Integrand size = 51, antiderivative size = 29 \[ \int \frac {1}{4} e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx=-3+5 e^{e^x} x \left (-e^x+\frac {1}{4} \left (-2+e^{\log ^2(5)}\right )+x\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {1}{4} e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx=\int \frac {1}{4} e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx \]

[In]

Int[(E^E^x*(-10 + 40*x - 20*E^(2*x)*x + E^Log[5]^2*(5 + 5*E^x*x) + E^x*(-20 - 30*x + 20*x^2)))/4,x]

[Out]

-5*E^E^x + (5*E^(E^x + Log[5]^2)*x)/4 - (5*ExpIntegralEi[E^x])/2 + 10*Defer[Int][E^E^x*x, x] - (15*Defer[Int][
E^(E^x + x)*x, x])/2 - 5*Defer[Int][E^(E^x + 2*x)*x, x] + 5*Defer[Int][E^(E^x + x)*x^2, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx \\ & = \frac {1}{4} \int \left (-10 e^{e^x}+40 e^{e^x} x-20 e^{e^x+2 x} x+5 e^{e^x+\log ^2(5)} \left (1+e^x x\right )+10 e^{e^x+x} \left (-2-3 x+2 x^2\right )\right ) \, dx \\ & = \frac {5}{4} \int e^{e^x+\log ^2(5)} \left (1+e^x x\right ) \, dx-\frac {5}{2} \int e^{e^x} \, dx+\frac {5}{2} \int e^{e^x+x} \left (-2-3 x+2 x^2\right ) \, dx-5 \int e^{e^x+2 x} x \, dx+10 \int e^{e^x} x \, dx \\ & = \frac {5}{4} e^{e^x+\log ^2(5)} x+\frac {5}{2} \int \left (-2 e^{e^x+x}-3 e^{e^x+x} x+2 e^{e^x+x} x^2\right ) \, dx-\frac {5}{2} \text {Subst}\left (\int \frac {e^x}{x} \, dx,x,e^x\right )-5 \int e^{e^x+2 x} x \, dx+10 \int e^{e^x} x \, dx \\ & = \frac {5}{4} e^{e^x+\log ^2(5)} x-\frac {5 \operatorname {ExpIntegralEi}\left (e^x\right )}{2}-5 \int e^{e^x+x} \, dx-5 \int e^{e^x+2 x} x \, dx+5 \int e^{e^x+x} x^2 \, dx-\frac {15}{2} \int e^{e^x+x} x \, dx+10 \int e^{e^x} x \, dx \\ & = \frac {5}{4} e^{e^x+\log ^2(5)} x-\frac {5 \operatorname {ExpIntegralEi}\left (e^x\right )}{2}-5 \int e^{e^x+2 x} x \, dx+5 \int e^{e^x+x} x^2 \, dx-5 \text {Subst}\left (\int e^x \, dx,x,e^x\right )-\frac {15}{2} \int e^{e^x+x} x \, dx+10 \int e^{e^x} x \, dx \\ & = -5 e^{e^x}+\frac {5}{4} e^{e^x+\log ^2(5)} x-\frac {5 \operatorname {ExpIntegralEi}\left (e^x\right )}{2}-5 \int e^{e^x+2 x} x \, dx+5 \int e^{e^x+x} x^2 \, dx-\frac {15}{2} \int e^{e^x+x} x \, dx+10 \int e^{e^x} x \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.83 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {1}{4} e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx=\frac {5}{4} e^{e^x} x \left (-2-4 e^x+e^{\log ^2(5)}+4 x\right ) \]

[In]

Integrate[(E^E^x*(-10 + 40*x - 20*E^(2*x)*x + E^Log[5]^2*(5 + 5*E^x*x) + E^x*(-20 - 30*x + 20*x^2)))/4,x]

[Out]

(5*E^E^x*x*(-2 - 4*E^x + E^Log[5]^2 + 4*x))/4

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97

method	result	size
risch	\(\frac {\left (5 \,{\mathrm e}^{\ln \left (5\right )^{2}} x +20 x^{2}-20 \,{\mathrm e}^{x} x -10 x \right ) {\mathrm e}^{{\mathrm e}^{x}}}{4}\)	\(28\)
norman	\(\left (\frac {5 \,{\mathrm e}^{\ln \left (5\right )^{2}}}{4}-\frac {5}{2}\right ) x \,{\mathrm e}^{{\mathrm e}^{x}}+5 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}-5 x \,{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x}}\)	\(32\)
parallelrisch	\(\frac {5 x \,{\mathrm e}^{{\mathrm e}^{x}} {\mathrm e}^{\ln \left (5\right )^{2}}}{4}+5 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}-5 x \,{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x}}-\frac {5 x \,{\mathrm e}^{{\mathrm e}^{x}}}{2}\)	\(35\)

[In]

int(1/4*((5*exp(x)*x+5)*exp(ln(5)^2)-20*x*exp(x)^2+(20*x^2-30*x-20)*exp(x)+40*x-10)*exp(exp(x)),x,method=_RETU
RNVERBOSE)

[Out]

1/4*(5*exp(ln(5)^2)*x+20*x^2-20*exp(x)*x-10*x)*exp(exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {1}{4} e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx=\frac {5}{4} \, {\left (4 \, x^{2} + x e^{\left (\log \left (5\right )^{2}\right )} - 4 \, x e^{x} - 2 \, x\right )} e^{\left (e^{x}\right )} \]

[In]

integrate(1/4*((5*exp(x)*x+5)*exp(log(5)^2)-20*x*exp(x)^2+(20*x^2-30*x-20)*exp(x)+40*x-10)*exp(exp(x)),x, algo
rithm="fricas")

[Out]

5/4*(4*x^2 + x*e^(log(5)^2) - 4*x*e^x - 2*x)*e^(e^x)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {1}{4} e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx=\frac {\left (20 x^{2} - 20 x e^{x} - 10 x + 5 x e^{\log {\left (5 \right )}^{2}}\right ) e^{e^{x}}}{4} \]

[In]

integrate(1/4*((5*exp(x)*x+5)*exp(ln(5)**2)-20*x*exp(x)**2+(20*x**2-30*x-20)*exp(x)+40*x-10)*exp(exp(x)),x)

[Out]

(20*x**2 - 20*x*exp(x) - 10*x + 5*x*exp(log(5)**2))*exp(exp(x))/4

Maxima [F]

\[ \int \frac {1}{4} e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx=\int { \frac {5}{4} \, {\left ({\left (x e^{x} + 1\right )} e^{\left (\log \left (5\right )^{2}\right )} - 4 \, x e^{\left (2 \, x\right )} + 2 \, {\left (2 \, x^{2} - 3 \, x - 2\right )} e^{x} + 8 \, x - 2\right )} e^{\left (e^{x}\right )} \,d x } \]

[In]

integrate(1/4*((5*exp(x)*x+5)*exp(log(5)^2)-20*x*exp(x)^2+(20*x^2-30*x-20)*exp(x)+40*x-10)*exp(exp(x)),x, algo
rithm="maxima")

[Out]

5/4*(4*x^2 + x*(e^(log(5)^2) - 2) - 4*x*e^x)*e^(e^x) - 5/2*Ei(e^x) + 5/2*integrate(e^(e^x), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).

Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int \frac {1}{4} e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx=\frac {5}{4} \, {\left (4 \, x^{2} e^{\left (x + e^{x}\right )} + x e^{\left (\log \left (5\right )^{2} + x + e^{x}\right )} - 4 \, x e^{\left (2 \, x + e^{x}\right )} - 2 \, x e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )} \]

[In]

integrate(1/4*((5*exp(x)*x+5)*exp(log(5)^2)-20*x*exp(x)^2+(20*x^2-30*x-20)*exp(x)+40*x-10)*exp(exp(x)),x, algo
rithm="giac")

[Out]

5/4*(4*x^2*e^(x + e^x) + x*e^(log(5)^2 + x + e^x) - 4*x*e^(2*x + e^x) - 2*x*e^(x + e^x))*e^(-x)

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {1}{4} e^{e^x} \left (-10+40 x-20 e^{2 x} x+e^{\log ^2(5)} \left (5+5 e^x x\right )+e^x \left (-20-30 x+20 x^2\right )\right ) \, dx=\frac {5\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (4\,x+{\mathrm {e}}^{{\ln \left (5\right )}^2}-4\,{\mathrm {e}}^x-2\right )}{4} \]

[In]

int(-(exp(exp(x))*(20*x*exp(2*x) - 40*x - exp(log(5)^2)*(5*x*exp(x) + 5) + exp(x)*(30*x - 20*x^2 + 20) + 10))/
4,x)

[Out]

(5*x*exp(exp(x))*(4*x + exp(log(5)^2) - 4*exp(x) - 2))/4