\(\int (-20+e^x (4+4 x)+e^{5+x} (4+4 x)+e^{5 e^2} (-5+e^x (1+x)+e^{5+x} (1+x))) \, dx\) [3996]

Optimal result

Integrand size = 48, antiderivative size = 21 \[ \int \left (-20+e^x (4+4 x)+e^{5+x} (4+4 x)+e^{5 e^2} \left (-5+e^x (1+x)+e^{5+x} (1+x)\right )\right ) \, dx=\left (4+e^{5 e^2}\right ) \left (-5+e^x+e^{5+x}\right ) x \]

[Out]

(exp(5+x)+exp(x)-5)*x*(4+exp(exp(2+ln(5))))

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(96\) vs. \(2(21)=42\).

Time = 0.04 (sec) , antiderivative size = 96, normalized size of antiderivative = 4.57, number of steps used = 10, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {2207, 2225} \[ \int \left (-20+e^x (4+4 x)+e^{5+x} (4+4 x)+e^{5 e^2} \left (-5+e^x (1+x)+e^{5+x} (1+x)\right )\right ) \, dx=-5 e^{5 e^2} x-20 x-4 e^x-4 e^{x+5}-e^{x+5 e^2}-e^{x+5 \left (1+e^2\right )}+4 e^x (x+1)+4 e^{x+5} (x+1)+e^{x+5 e^2} (x+1)+e^{x+5 \left (1+e^2\right )} (x+1) \]

[In]

Int[-20 + E^x*(4 + 4*x) + E^(5 + x)*(4 + 4*x) + E^(5*E^2)*(-5 + E^x*(1 + x) + E^(5 + x)*(1 + x)),x]

[Out]

-4*E^x - 4*E^(5 + x) - E^(5*E^2 + x) - E^(5*(1 + E^2) + x) - 20*x - 5*E^(5*E^2)*x + 4*E^x*(1 + x) + 4*E^(5 + x
)*(1 + x) + E^(5*E^2 + x)*(1 + x) + E^(5*(1 + E^2) + x)*(1 + x)

Rule 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps \begin{align*} \text {integral}& = -20 x+e^{5 e^2} \int \left (-5+e^x (1+x)+e^{5+x} (1+x)\right ) \, dx+\int e^x (4+4 x) \, dx+\int e^{5+x} (4+4 x) \, dx \\ & = -20 x-5 e^{5 e^2} x+4 e^x (1+x)+4 e^{5+x} (1+x)-4 \int e^x \, dx-4 \int e^{5+x} \, dx+e^{5 e^2} \int e^x (1+x) \, dx+e^{5 e^2} \int e^{5+x} (1+x) \, dx \\ & = -4 e^x-4 e^{5+x}-20 x-5 e^{5 e^2} x+4 e^x (1+x)+4 e^{5+x} (1+x)+e^{5 e^2+x} (1+x)+e^{5 \left (1+e^2\right )+x} (1+x)-e^{5 e^2} \int e^x \, dx-e^{5 e^2} \int e^{5+x} \, dx \\ & = -4 e^x-4 e^{5+x}-e^{5 e^2+x}-e^{5 \left (1+e^2\right )+x}-20 x-5 e^{5 e^2} x+4 e^x (1+x)+4 e^{5+x} (1+x)+e^{5 e^2+x} (1+x)+e^{5 \left (1+e^2\right )+x} (1+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \left (-20+e^x (4+4 x)+e^{5+x} (4+4 x)+e^{5 e^2} \left (-5+e^x (1+x)+e^{5+x} (1+x)\right )\right ) \, dx=\left (4+e^{5 e^2}\right ) \left (-5 x+e^x x+e^{5+x} x\right ) \]

[In]

Integrate[-20 + E^x*(4 + 4*x) + E^(5 + x)*(4 + 4*x) + E^(5*E^2)*(-5 + E^x*(1 + x) + E^(5 + x)*(1 + x)),x]

[Out]

(4 + E^(5*E^2))*(-5*x + E^x*x + E^(5 + x)*x)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71

method	result	size
norman	\(\left (-5 \,{\mathrm e}^{5 \,{\mathrm e}^{2}}-20\right ) x +\left ({\mathrm e}^{5} {\mathrm e}^{5 \,{\mathrm e}^{2}}+4 \,{\mathrm e}^{5}+{\mathrm e}^{5 \,{\mathrm e}^{2}}+4\right ) x \,{\mathrm e}^{x}\)	\(36\)
risch	\(-5 \,{\mathrm e}^{5 \,{\mathrm e}^{2}} x +\left ({\mathrm e}^{5}+1\right ) x \,{\mathrm e}^{5 \,{\mathrm e}^{2}+x}+4 x \,{\mathrm e}^{5+x}+4 \,{\mathrm e}^{x} x -20 x\)	\(38\)
parallelrisch	\({\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{2+\ln \left (5\right )}} x +{\mathrm e}^{5+x} {\mathrm e}^{{\mathrm e}^{2+\ln \left (5\right )}} x -5 \,{\mathrm e}^{{\mathrm e}^{2+\ln \left (5\right )}} x +4 \,{\mathrm e}^{x} x +4 x \,{\mathrm e}^{5+x}-20 x\)	\(48\)
default	\(-20 x +4 \,{\mathrm e}^{x} x +4 \left (5+x \right ) {\mathrm e}^{5+x}-20 \,{\mathrm e}^{5+x}+{\mathrm e}^{{\mathrm e}^{2+\ln \left (5\right )}} \left (-5 x +{\mathrm e}^{x} x +\left (5+x \right ) {\mathrm e}^{5+x}-5 \,{\mathrm e}^{5+x}\right )\)	\(54\)
parts	\(-20 x +{\mathrm e}^{x} {\mathrm e}^{5 \,{\mathrm e}^{2}}+{\mathrm e}^{5 \,{\mathrm e}^{2}} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )+4 \,{\mathrm e}^{x} x +{\mathrm e}^{5 \,{\mathrm e}^{2}} \left (\left (5+x \right ) {\mathrm e}^{5+x}-{\mathrm e}^{5+x}\right )+4 \left (5+x \right ) {\mathrm e}^{5+x}-20 \,{\mathrm e}^{5+x}-4 \,{\mathrm e}^{5+x} {\mathrm e}^{5 \,{\mathrm e}^{2}}-5 \,{\mathrm e}^{{\mathrm e}^{2+\ln \left (5\right )}} x\)	\(89\)

[In]

int(((1+x)*exp(5+x)+(1+x)*exp(x)-5)*exp(exp(2+ln(5)))+(4+4*x)*exp(5+x)+(4+4*x)*exp(x)-20,x,method=_RETURNVERBO
SE)

[Out]

(-5*exp(exp(2))^5-20)*x+(exp(5)*exp(exp(2))^5+4*exp(5)+exp(exp(2))^5+4)*x*exp(x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (18) = 36\).

Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.19 \[ \int \left (-20+e^x (4+4 x)+e^{5+x} (4+4 x)+e^{5 e^2} \left (-5+e^x (1+x)+e^{5+x} (1+x)\right )\right ) \, dx=-{\left (20 \, x e^{5} - 4 \, {\left (x e^{5} + x\right )} e^{\left (x + 5\right )} + {\left (5 \, x e^{5} - {\left (x e^{5} + x\right )} e^{\left (x + 5\right )}\right )} e^{\left (5 \, e^{2}\right )}\right )} e^{\left (-5\right )} \]

[In]

integrate(((1+x)*exp(5+x)+(1+x)*exp(x)-5)*exp(exp(2+log(5)))+(4+4*x)*exp(5+x)+(4+4*x)*exp(x)-20,x, algorithm="
fricas")

[Out]

-(20*x*e^5 - 4*(x*e^5 + x)*e^(x + 5) + (5*x*e^5 - (x*e^5 + x)*e^(x + 5))*e^(5*e^2))*e^(-5)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (19) = 38\).

Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.19 \[ \int \left (-20+e^x (4+4 x)+e^{5+x} (4+4 x)+e^{5 e^2} \left (-5+e^x (1+x)+e^{5+x} (1+x)\right )\right ) \, dx=x \left (- 5 e^{5 e^{2}} - 20\right ) + \left (4 x + 4 x e^{5} + x e^{5 e^{2}} + x e^{5} e^{5 e^{2}}\right ) e^{x} \]

[In]

integrate(((1+x)*exp(5+x)+(1+x)*exp(x)-5)*exp(exp(2+ln(5)))+(4+4*x)*exp(5+x)+(4+4*x)*exp(x)-20,x)

[Out]

x*(-5*exp(5*exp(2)) - 20) + (4*x + 4*x*exp(5) + x*exp(5*exp(2)) + x*exp(5)*exp(5*exp(2)))*exp(x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (18) = 36\).

Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.00 \[ \int \left (-20+e^x (4+4 x)+e^{5+x} (4+4 x)+e^{5 e^2} \left (-5+e^x (1+x)+e^{5+x} (1+x)\right )\right ) \, dx=4 \, x e^{\left (x + 5\right )} + 4 \, {\left (x - 1\right )} e^{x} + {\left (x e^{\left (x + 5\right )} + x e^{x} - 5 \, x\right )} e^{\left (5 \, e^{2}\right )} - 20 \, x + 4 \, e^{x} \]

[In]

integrate(((1+x)*exp(5+x)+(1+x)*exp(x)-5)*exp(exp(2+log(5)))+(4+4*x)*exp(5+x)+(4+4*x)*exp(x)-20,x, algorithm="
maxima")

[Out]

4*x*e^(x + 5) + 4*(x - 1)*e^x + (x*e^(x + 5) + x*e^x - 5*x)*e^(5*e^2) - 20*x + 4*e^x

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (18) = 36\).

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.76 \[ \int \left (-20+e^x (4+4 x)+e^{5+x} (4+4 x)+e^{5 e^2} \left (-5+e^x (1+x)+e^{5+x} (1+x)\right )\right ) \, dx=4 \, x e^{\left (x + 5\right )} + 4 \, x e^{x} + {\left (x e^{\left (x + 5\right )} + x e^{x} - 5 \, x\right )} e^{\left (e^{\left (\log \left (5\right ) + 2\right )}\right )} - 20 \, x \]

[In]

integrate(((1+x)*exp(5+x)+(1+x)*exp(x)-5)*exp(exp(2+log(5)))+(4+4*x)*exp(5+x)+(4+4*x)*exp(x)-20,x, algorithm="
giac")

[Out]

4*x*e^(x + 5) + 4*x*e^x + (x*e^(x + 5) + x*e^x - 5*x)*e^(e^(log(5) + 2)) - 20*x

Mupad [B] (verification not implemented)

Time = 9.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.67 \[ \int \left (-20+e^x (4+4 x)+e^{5+x} (4+4 x)+e^{5 e^2} \left (-5+e^x (1+x)+e^{5+x} (1+x)\right )\right ) \, dx=x\,{\mathrm {e}}^x\,\left ({\mathrm {e}}^{5\,{\mathrm {e}}^2}+4\,{\mathrm {e}}^5+{\mathrm {e}}^{5\,{\mathrm {e}}^2+5}+4\right )-x\,\left (5\,{\mathrm {e}}^{5\,{\mathrm {e}}^2}+20\right ) \]

[In]

int(exp(x)*(4*x + 4) + exp(x + 5)*(4*x + 4) + exp(exp(log(5) + 2))*(exp(x)*(x + 1) + exp(x + 5)*(x + 1) - 5) -
20,x)

[Out]

x*exp(x)*(exp(5*exp(2)) + 4*exp(5) + exp(5*exp(2) + 5) + 4) - x*(5*exp(5*exp(2)) + 20)