3.54.78 \(\int \frac {1}{10} (-8+8 e^x-75 x) \, dx\)

Optimal. Leaf size=24 \[ 4-x-\frac {15 x^2}{4}+\frac {1}{5} \left (-3+4 e^x+x\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 20, normalized size of antiderivative = 0.83, 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 {15 x^2}{4}-\frac {4 x}{5}+\frac {4 e^x}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8 + 8*E^x - 75*x)/10,x]

[Out]

(4*E^x)/5 - (4*x)/5 - (15*x^2)/4

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}{10} \int \left (-8+8 e^x-75 x\right ) \, dx\\ &=-\frac {4 x}{5}-\frac {15 x^2}{4}+\frac {4 \int e^x \, dx}{5}\\ &=\frac {4 e^x}{5}-\frac {4 x}{5}-\frac {15 x^2}{4}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 0.83 \begin {gather*} \frac {1}{10} \left (8 e^x-8 x-\frac {75 x^2}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 + 8*E^x - 75*x)/10,x]

[Out]

(8*E^x - 8*x - (75*x^2)/2)/10

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fricas [A]  time = 0.78, size = 13, normalized size = 0.54 \begin {gather*} -\frac {15}{4} \, x^{2} - \frac {4}{5} \, x + \frac {4}{5} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(4/5*exp(x)-15/2*x-4/5,x, algorithm="fricas")

[Out]

-15/4*x^2 - 4/5*x + 4/5*e^x

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giac [A]  time = 0.20, size = 13, normalized size = 0.54 \begin {gather*} -\frac {15}{4} \, x^{2} - \frac {4}{5} \, x + \frac {4}{5} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(4/5*exp(x)-15/2*x-4/5,x, algorithm="giac")

[Out]

-15/4*x^2 - 4/5*x + 4/5*e^x

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maple [A]  time = 0.02, size = 14, normalized size = 0.58

method	result	size
default	\(-\frac {4 x}{5}-\frac {15 x^{2}}{4}+\frac {4 \,{\mathrm e}^{x}}{5}\)	\(14\)
norman	\(-\frac {4 x}{5}-\frac {15 x^{2}}{4}+\frac {4 \,{\mathrm e}^{x}}{5}\)	\(14\)
risch	\(-\frac {4 x}{5}-\frac {15 x^{2}}{4}+\frac {4 \,{\mathrm e}^{x}}{5}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(4/5*exp(x)-15/2*x-4/5,x,method=_RETURNVERBOSE)

[Out]

-4/5*x-15/4*x^2+4/5*exp(x)

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maxima [A]  time = 0.40, size = 13, normalized size = 0.54 \begin {gather*} -\frac {15}{4} \, x^{2} - \frac {4}{5} \, x + \frac {4}{5} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(4/5*exp(x)-15/2*x-4/5,x, algorithm="maxima")

[Out]

-15/4*x^2 - 4/5*x + 4/5*e^x

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mupad [B]  time = 0.04, size = 13, normalized size = 0.54 \begin {gather*} \frac {4\,{\mathrm {e}}^x}{5}-\frac {4\,x}{5}-\frac {15\,x^2}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*exp(x))/5 - (15*x)/2 - 4/5,x)

[Out]

(4*exp(x))/5 - (4*x)/5 - (15*x^2)/4

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sympy [A]  time = 0.07, size = 17, normalized size = 0.71 \begin {gather*} - \frac {15 x^{2}}{4} - \frac {4 x}{5} + \frac {4 e^{x}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(4/5*exp(x)-15/2*x-4/5,x)

[Out]

-15*x**2/4 - 4*x/5 + 4*exp(x)/5

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