3.95.73 \(\int \frac {1}{5} (4-5 e^x-60 x^2) \, dx\)

Optimal. Leaf size=22 \[ 11-e^2-e^x-4 x \left (-\frac {1}{5}+x^2\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 16, normalized size of antiderivative = 0.73, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 2194} \begin {gather*} -4 x^3+\frac {4 x}{5}-e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 - 5*E^x - 60*x^2)/5,x]

[Out]

-E^x + (4*x)/5 - 4*x^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}{5} \int \left (4-5 e^x-60 x^2\right ) \, dx\\ &=\frac {4 x}{5}-4 x^3-\int e^x \, dx\\ &=-e^x+\frac {4 x}{5}-4 x^3\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 0.73 \begin {gather*} -e^x+\frac {4 x}{5}-4 x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 - 5*E^x - 60*x^2)/5,x]

[Out]

-E^x + (4*x)/5 - 4*x^3

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fricas [A]  time = 0.50, size = 13, normalized size = 0.59 \begin {gather*} -4 \, x^{3} + \frac {4}{5} \, x - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(x)-12*x^2+4/5,x, algorithm="fricas")

[Out]

-4*x^3 + 4/5*x - e^x

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giac [A]  time = 0.11, size = 13, normalized size = 0.59 \begin {gather*} -4 \, x^{3} + \frac {4}{5} \, x - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(x)-12*x^2+4/5,x, algorithm="giac")

[Out]

-4*x^3 + 4/5*x - e^x

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maple [A]  time = 0.02, size = 14, normalized size = 0.64

method	result	size
default	\(\frac {4 x}{5}-4 x^{3}-{\mathrm e}^{x}\)	\(14\)
norman	\(\frac {4 x}{5}-4 x^{3}-{\mathrm e}^{x}\)	\(14\)
risch	\(\frac {4 x}{5}-4 x^{3}-{\mathrm e}^{x}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(x)-12*x^2+4/5,x,method=_RETURNVERBOSE)

[Out]

4/5*x-4*x^3-exp(x)

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maxima [A]  time = 0.34, size = 13, normalized size = 0.59 \begin {gather*} -4 \, x^{3} + \frac {4}{5} \, x - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(x)-12*x^2+4/5,x, algorithm="maxima")

[Out]

-4*x^3 + 4/5*x - e^x

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mupad [B]  time = 5.46, size = 13, normalized size = 0.59 \begin {gather*} \frac {4\,x}{5}-{\mathrm {e}}^x-4\,x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(4/5 - 12*x^2 - exp(x),x)

[Out]

(4*x)/5 - exp(x) - 4*x^3

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sympy [A]  time = 0.08, size = 12, normalized size = 0.55 \begin {gather*} - 4 x^{3} + \frac {4 x}{5} - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(x)-12*x**2+4/5,x)

[Out]

-4*x**3 + 4*x/5 - exp(x)

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