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

Optimal. Leaf size=21 \[ \left (4+e^{5 e^2}\right ) \left (-5+e^x+e^{5+x}\right ) x \]

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Rubi [B]  time = 0.06, antiderivative size = 96, normalized size of antiderivative = 4.57, number of steps used = 10, number of rules used = 2, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {2176, 2194} \begin {gather*} -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) \end {gather*}

Antiderivative was successfully verified.

[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 2176

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] &&  !$UseGamma === True

Rule 2194

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 1.24 \begin {gather*} \left (4+e^{5 e^2}\right ) \left (-5 x+e^x x+e^{5+x} x\right ) \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [B]  time = 0.87, size = 46, normalized size = 2.19 \begin {gather*} -{\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*exp(5+x)+(x+1)*exp(x)-5)*exp(exp(2+log(5)))+(4*x+4)*exp(5+x)+(4*x+4)*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)

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giac [B]  time = 0.15, size = 37, normalized size = 1.76 \begin {gather*} 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 \relax (5) + 2\right )}\right )} - 20 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*exp(5+x)+(x+1)*exp(x)-5)*exp(exp(2+log(5)))+(4*x+4)*exp(5+x)+(4*x+4)*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

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maple [A]  time = 0.04, size = 36, normalized size = 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	\(-5 \,{\mathrm e}^{5 \,{\mathrm e}^{2}} x +\left ({\mathrm e}^{5}+1\right ) x \,{\mathrm e}^{5 \,{\mathrm e}^{2}+x}+4 x \,{\mathrm e}^{5+x}+4 \,{\mathrm e}^{x} x -20 x\)	\(38\)
default	\(-20 x +4 \,{\mathrm e}^{x} x +4 \,{\mathrm e}^{5+x} \left (5+x \right )-20 \,{\mathrm e}^{5+x}+{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{2+\ln \relax (5)}} x +{\mathrm e}^{5+x} {\mathrm e}^{{\mathrm e}^{2+\ln \relax (5)}} x -5 \,{\mathrm e}^{{\mathrm e}^{2+\ln \relax (5)}} x\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x+1)*exp(5+x)+(x+1)*exp(x)-5)*exp(exp(2+ln(5)))+(4*x+4)*exp(5+x)+(4*x+4)*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)

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maxima [B]  time = 0.37, size = 42, normalized size = 2.00 \begin {gather*} 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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*exp(5+x)+(x+1)*exp(x)-5)*exp(exp(2+log(5)))+(4*x+4)*exp(5+x)+(4*x+4)*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

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mupad [B]  time = 2.53, size = 35, normalized size = 1.67 \begin {gather*} 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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [B]  time = 0.15, size = 46, normalized size = 2.19 \begin {gather*} 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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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