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

Optimal. Leaf size=27 \[ \frac {1}{2} \left (3-e^x+\frac {x^2}{4}-10 \left (e^{5+x}+x\right )\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 2194} \begin {gather*} \frac {x^2}{8}-5 x-\frac {e^x}{2}-5 e^{x+5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*E^x - 5*E^(5 + x) - 5*x + x^2/8

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \left (-20-2 e^x-20 e^{5+x}+x\right ) \, dx\\ &=-5 x+\frac {x^2}{8}-\frac {\int e^x \, dx}{2}-5 \int e^{5+x} \, dx\\ &=-\frac {e^x}{2}-5 e^{5+x}-5 x+\frac {x^2}{8}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.00 \begin {gather*} \frac {1}{4} \left (-2 e^x-20 e^{5+x}-20 x+\frac {x^2}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.63, size = 27, normalized size = 1.00 \begin {gather*} \frac {1}{8} \, {\left ({\left (x^{2} - 40 \, x\right )} e^{5} - 4 \, {\left (10 \, e^{5} + 1\right )} e^{\left (x + 5\right )}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((x^2 - 40*x)*e^5 - 4*(10*e^5 + 1)*e^(x + 5))*e^(-5)

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giac [A]  time = 0.16, size = 19, normalized size = 0.70 \begin {gather*} \frac {1}{8} \, x^{2} - 5 \, x - 5 \, e^{\left (x + 5\right )} - \frac {1}{2} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x^2 - 5*x - 5*e^(x + 5) - 1/2*e^x

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maple [A]  time = 0.03, size = 19, normalized size = 0.70

method	result	size
norman	\(\left (-5 \,{\mathrm e}^{5}-\frac {1}{2}\right ) {\mathrm e}^{x}-5 x +\frac {x^{2}}{8}\)	\(19\)
default	\(-5 x +\frac {x^{2}}{8}-\frac {{\mathrm e}^{x}}{2}-5 \,{\mathrm e}^{5+x}\)	\(20\)
risch	\(-5 x +\frac {x^{2}}{8}-\frac {{\mathrm e}^{x}}{2}-5 \,{\mathrm e}^{5+x}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-5*exp(5+x)-1/2*exp(x)+1/4*x-5,x,method=_RETURNVERBOSE)

[Out]

(-5*exp(5)-1/2)*exp(x)-5*x+1/8*x^2

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maxima [A]  time = 0.34, size = 19, normalized size = 0.70 \begin {gather*} \frac {1}{8} \, x^{2} - 5 \, x - 5 \, e^{\left (x + 5\right )} - \frac {1}{2} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x^2 - 5*x - 5*e^(x + 5) - 1/2*e^x

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mupad [B]  time = 1.19, size = 19, normalized size = 0.70 \begin {gather*} \frac {x^2}{8}-{\mathrm {e}}^x\,\left (5\,{\mathrm {e}}^5+\frac {1}{2}\right )-5\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/4 - 5*exp(x + 5) - exp(x)/2 - 5,x)

[Out]

x^2/8 - exp(x)*(5*exp(5) + 1/2) - 5*x

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sympy [A]  time = 0.10, size = 20, normalized size = 0.74 \begin {gather*} \frac {x^{2}}{8} - 5 x + \frac {\left (- 10 e^{5} - 1\right ) e^{x}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2/8 - 5*x + (-10*exp(5) - 1)*exp(x)/2

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