3.54.96 \(\int \frac {1}{3} (100+5 e^x-40 x) \, dx\)

Optimal. Leaf size=18 \[ \frac {5}{3} \left (-1+e^x-(-5+2 x)^2\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 20, normalized size of antiderivative = 1.11, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 2194} \begin {gather*} -\frac {20 x^2}{3}+\frac {100 x}{3}+\frac {5 e^x}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(100 + 5*E^x - 40*x)/3,x]

[Out]

(5*E^x)/3 + (100*x)/3 - (20*x^2)/3

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}{3} \int \left (100+5 e^x-40 x\right ) \, dx\\ &=\frac {100 x}{3}-\frac {20 x^2}{3}+\frac {5 \int e^x \, dx}{3}\\ &=\frac {5 e^x}{3}+\frac {100 x}{3}-\frac {20 x^2}{3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 0.89 \begin {gather*} \frac {5}{3} \left (e^x+20 x-4 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(100 + 5*E^x - 40*x)/3,x]

[Out]

(5*(E^x + 20*x - 4*x^2))/3

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fricas [A]  time = 0.58, size = 13, normalized size = 0.72 \begin {gather*} -\frac {20}{3} \, x^{2} + \frac {100}{3} \, x + \frac {5}{3} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(5/3*exp(x)-40/3*x+100/3,x, algorithm="fricas")

[Out]

-20/3*x^2 + 100/3*x + 5/3*e^x

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giac [A]  time = 0.20, size = 13, normalized size = 0.72 \begin {gather*} -\frac {20}{3} \, x^{2} + \frac {100}{3} \, x + \frac {5}{3} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(5/3*exp(x)-40/3*x+100/3,x, algorithm="giac")

[Out]

-20/3*x^2 + 100/3*x + 5/3*e^x

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maple [A]  time = 0.03, size = 14, normalized size = 0.78

method	result	size
default	\(\frac {100 x}{3}-\frac {20 x^{2}}{3}+\frac {5 \,{\mathrm e}^{x}}{3}\)	\(14\)
norman	\(\frac {100 x}{3}-\frac {20 x^{2}}{3}+\frac {5 \,{\mathrm e}^{x}}{3}\)	\(14\)
risch	\(\frac {100 x}{3}-\frac {20 x^{2}}{3}+\frac {5 \,{\mathrm e}^{x}}{3}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(5/3*exp(x)-40/3*x+100/3,x,method=_RETURNVERBOSE)

[Out]

100/3*x-20/3*x^2+5/3*exp(x)

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maxima [A]  time = 0.42, size = 13, normalized size = 0.72 \begin {gather*} -\frac {20}{3} \, x^{2} + \frac {100}{3} \, x + \frac {5}{3} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(5/3*exp(x)-40/3*x+100/3,x, algorithm="maxima")

[Out]

-20/3*x^2 + 100/3*x + 5/3*e^x

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mupad [B]  time = 3.53, size = 13, normalized size = 0.72 \begin {gather*} \frac {100\,x}{3}+\frac {5\,{\mathrm {e}}^x}{3}-\frac {20\,x^2}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*exp(x))/3 - (40*x)/3 + 100/3,x)

[Out]

(100*x)/3 + (5*exp(x))/3 - (20*x^2)/3

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sympy [A]  time = 0.07, size = 17, normalized size = 0.94 \begin {gather*} - \frac {20 x^{2}}{3} + \frac {100 x}{3} + \frac {5 e^{x}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(5/3*exp(x)-40/3*x+100/3,x)

[Out]

-20*x**2/3 + 100*x/3 + 5*exp(x)/3

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