3.51.79 \(\int \frac {1}{10} (40+10 e^x+e^{\frac {1}{5} (-5 e^x+x)} (-1+5 e^x)) \, dx\)

Optimal. Leaf size=35 \[ 2 \left (2-\frac {\frac {1}{2} e^{-e^x+\frac {x}{5}}-e^x}{2 x}\right ) x \]

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Rubi [A]  time = 0.05, antiderivative size = 24, normalized size of antiderivative = 0.69, number of steps used = 4, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {12, 2194, 6706} \begin {gather*} 4 x+e^x-\frac {1}{2} e^{\frac {1}{5} \left (x-5 e^x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(40 + 10*E^x + E^((-5*E^x + x)/5)*(-1 + 5*E^x))/10,x]

[Out]

E^x - E^((-5*E^x + x)/5)/2 + 4*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 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]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{10} \int \left (40+10 e^x+e^{\frac {1}{5} \left (-5 e^x+x\right )} \left (-1+5 e^x\right )\right ) \, dx\\ &=4 x+\frac {1}{10} \int e^{\frac {1}{5} \left (-5 e^x+x\right )} \left (-1+5 e^x\right ) \, dx+\int e^x \, dx\\ &=e^x-\frac {1}{2} e^{\frac {1}{5} \left (-5 e^x+x\right )}+4 x\\ \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.06, size = 57, normalized size = 1.63 \begin {gather*} e^x+4 x+\frac {1}{10} e^{-4 x/5} \left (e^x\right )^{4/5} \Gamma \left (\frac {1}{5},e^x\right )-\frac {1}{2} e^{-4 x/5} \left (e^x\right )^{4/5} \Gamma \left (\frac {6}{5},e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(40 + 10*E^x + E^((-5*E^x + x)/5)*(-1 + 5*E^x))/10,x]

[Out]

E^x + 4*x + ((E^x)^(4/5)*Gamma[1/5, E^x])/(10*E^((4*x)/5)) - ((E^x)^(4/5)*Gamma[6/5, E^x])/(2*E^((4*x)/5))

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fricas [A]  time = 0.98, size = 17, normalized size = 0.49 \begin {gather*} 4 \, x + e^{x} - \frac {1}{2} \, e^{\left (\frac {1}{5} \, x - e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(5*exp(x)-1)*exp(1/5*x-exp(x))+exp(x)+4,x, algorithm="fricas")

[Out]

4*x + e^x - 1/2*e^(1/5*x - e^x)

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giac [A]  time = 0.20, size = 17, normalized size = 0.49 \begin {gather*} 4 \, x + e^{x} - \frac {1}{2} \, e^{\left (\frac {1}{5} \, x - e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(5*exp(x)-1)*exp(1/5*x-exp(x))+exp(x)+4,x, algorithm="giac")

[Out]

4*x + e^x - 1/2*e^(1/5*x - e^x)

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maple [A]  time = 0.04, size = 18, normalized size = 0.51

method	result	size
default	\(4 x -\frac {{\mathrm e}^{\frac {x}{5}-{\mathrm e}^{x}}}{2}+{\mathrm e}^{x}\)	\(18\)
norman	\(4 x -\frac {{\mathrm e}^{\frac {x}{5}-{\mathrm e}^{x}}}{2}+{\mathrm e}^{x}\)	\(18\)
risch	\(4 x -\frac {{\mathrm e}^{\frac {x}{5}-{\mathrm e}^{x}}}{2}+{\mathrm e}^{x}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/10*(5*exp(x)-1)*exp(1/5*x-exp(x))+exp(x)+4,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.33, size = 17, normalized size = 0.49 \begin {gather*} 4 \, x + e^{x} - \frac {1}{2} \, e^{\left (\frac {1}{5} \, x - e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(5*exp(x)-1)*exp(1/5*x-exp(x))+exp(x)+4,x, algorithm="maxima")

[Out]

4*x + e^x - 1/2*e^(1/5*x - e^x)

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mupad [B]  time = 0.08, size = 17, normalized size = 0.49 \begin {gather*} 4\,x-\frac {{\mathrm {e}}^{\frac {x}{5}-{\mathrm {e}}^x}}{2}+{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x) + (exp(x/5 - exp(x))*(5*exp(x) - 1))/10 + 4,x)

[Out]

4*x - exp(x/5 - exp(x))/2 + exp(x)

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sympy [A]  time = 0.13, size = 15, normalized size = 0.43 \begin {gather*} 4 x + e^{x} - \frac {e^{\frac {x}{5} - e^{x}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(5*exp(x)-1)*exp(1/5*x-exp(x))+exp(x)+4,x)

[Out]

4*x + exp(x) - exp(x/5 - exp(x))/2

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