3.36.10 \(\int \frac {1}{5} (5 e^x+e^{\frac {1}{5} (-55-5 e^5+x)} (-10-2 x)) \, dx\)

Optimal. Leaf size=21 \[ e^x-2 e^{-11-e^5+\frac {x}{5}} x \]

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Rubi [A]  time = 0.03, antiderivative size = 41, normalized size of antiderivative = 1.95, number of steps used = 5, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 2194, 2176} \begin {gather*} -2 e^{\frac {1}{5} \left (x-5 \left (11+e^5\right )\right )} (x+5)+e^x+10 e^{\frac {1}{5} \left (x-5 \left (11+e^5\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*E^x + E^((-55 - 5*E^5 + x)/5)*(-10 - 2*x))/5,x]

[Out]

E^x + 10*E^((-5*(11 + E^5) + x)/5) - 2*E^((-5*(11 + E^5) + x)/5)*(5 + x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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} &=\frac {1}{5} \int \left (5 e^x+e^{\frac {1}{5} \left (-55-5 e^5+x\right )} (-10-2 x)\right ) \, dx\\ &=\frac {1}{5} \int e^{\frac {1}{5} \left (-55-5 e^5+x\right )} (-10-2 x) \, dx+\int e^x \, dx\\ &=e^x-2 e^{\frac {1}{5} \left (-5 \left (11+e^5\right )+x\right )} (5+x)+2 \int e^{\frac {1}{5} \left (-55-5 e^5+x\right )} \, dx\\ &=e^x+10 e^{\frac {1}{5} \left (-5 \left (11+e^5\right )+x\right )}-2 e^{\frac {1}{5} \left (-5 \left (11+e^5\right )+x\right )} (5+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 21, normalized size = 1.00 \begin {gather*} e^x-2 e^{-11-e^5+\frac {x}{5}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*E^x + E^((-55 - 5*E^5 + x)/5)*(-10 - 2*x))/5,x]

[Out]

E^x - 2*E^(-11 - E^5 + x/5)*x

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fricas [A]  time = 0.61, size = 16, normalized size = 0.76 \begin {gather*} -2 \, x e^{\left (\frac {1}{5} \, x - e^{5} - 11\right )} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)+1/5*(-2*x-10)*exp(-exp(5)+1/5*x-11),x, algorithm="fricas")

[Out]

-2*x*e^(1/5*x - e^5 - 11) + e^x

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giac [A]  time = 0.16, size = 16, normalized size = 0.76 \begin {gather*} -2 \, x e^{\left (\frac {1}{5} \, x - e^{5} - 11\right )} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)+1/5*(-2*x-10)*exp(-exp(5)+1/5*x-11),x, algorithm="giac")

[Out]

-2*x*e^(1/5*x - e^5 - 11) + e^x

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maple [A]  time = 0.04, size = 17, normalized size = 0.81

method	result	size
risch	\({\mathrm e}^{x}-2 \,{\mathrm e}^{-{\mathrm e}^{5}+\frac {x}{5}-11} x\)	\(17\)
norman	\({\mathrm e}^{x}-2 \,{\mathrm e}^{-{\mathrm e}^{5}} {\mathrm e}^{-11} x \,{\mathrm e}^{\frac {x}{5}}\)	\(22\)
meijerg	\(-1+{\mathrm e}^{x}+10 \,{\mathrm e}^{-{\mathrm e}^{5}-11} \left (1-{\mathrm e}^{\frac {x}{5}}\right )-10 \,{\mathrm e}^{-{\mathrm e}^{5}-11} \left (1-\frac {\left (-\frac {2 x}{5}+2\right ) {\mathrm e}^{\frac {x}{5}}}{2}\right )\)	\(44\)
default	\(-10 \left (-{\mathrm e}^{5}+\frac {x}{5}-11\right ) {\mathrm e}^{-{\mathrm e}^{5}+\frac {x}{5}-11}-110 \,{\mathrm e}^{-{\mathrm e}^{5}+\frac {x}{5}-11}-10 \,{\mathrm e}^{-{\mathrm e}^{5}+\frac {x}{5}-11} {\mathrm e}^{5}+{\mathrm e}^{x}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)+1/5*(-2*x-10)*exp(-exp(5)+1/5*x-11),x,method=_RETURNVERBOSE)

[Out]

exp(x)-2*exp(-exp(5)+1/5*x-11)*x

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maxima [A]  time = 0.37, size = 30, normalized size = 1.43 \begin {gather*} -2 \, {\left (x - 5\right )} e^{\left (\frac {1}{5} \, x - e^{5} - 11\right )} + e^{x} - 10 \, e^{\left (\frac {1}{5} \, x - e^{5} - 11\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)+1/5*(-2*x-10)*exp(-exp(5)+1/5*x-11),x, algorithm="maxima")

[Out]

-2*(x - 5)*e^(1/5*x - e^5 - 11) + e^x - 10*e^(1/5*x - e^5 - 11)

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mupad [B]  time = 0.09, size = 16, normalized size = 0.76 \begin {gather*} {\mathrm {e}}^x-2\,x\,{\mathrm {e}}^{\frac {x}{5}-{\mathrm {e}}^5-11} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x) - (exp(x/5 - exp(5) - 11)*(2*x + 10))/5,x)

[Out]

exp(x) - 2*x*exp(x/5 - exp(5) - 11)

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sympy [A]  time = 0.17, size = 29, normalized size = 1.38 \begin {gather*} \frac {- 2 x \sqrt [5]{e^{x}} + e^{11} e^{x} e^{e^{5}}}{e^{11} e^{e^{5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)+1/5*(-2*x-10)*exp(-exp(5)+1/5*x-11),x)

[Out]

(-2*x*exp(x)**(1/5) + exp(11)*exp(x)*exp(exp(5)))*exp(-11)*exp(-exp(5))

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