3.60.88 \(\int \frac {1}{25} (-8 x+10 e x-15 x^2) \, dx\) [5988]

Optimal. Leaf size=16 \[ \frac {1}{25} (-4+5 (e-x)) x^2 \]

[Out]

x^2*(1/5*exp(1)-1/5*x-4/25)

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Rubi [A]
time = 0.01, antiderivative size = 20, normalized size of antiderivative = 1.25, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6, 12} \begin {gather*} -\frac {x^3}{5}-\frac {1}{25} (4-5 e) x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8*x + 10*E*x - 15*x^2)/25,x]

[Out]

-1/25*((4 - 5*E)*x^2) - x^3/5

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1}{25} \left ((-8+10 e) x-15 x^2\right ) \, dx\\ &=\frac {1}{25} \int \left ((-8+10 e) x-15 x^2\right ) \, dx\\ &=-\frac {1}{25} (4-5 e) x^2-\frac {x^3}{5}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.00, size = 15, normalized size = 0.94 \begin {gather*} \frac {1}{25} (-4+5 e-5 x) x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8*x + 10*E*x - 15*x^2)/25,x]

[Out]

((-4 + 5*E - 5*x)*x^2)/25

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Maple [A]
time = 0.18, size = 18, normalized size = 1.12

method	result	size
gosper	\(\frac {x^{2} \left (-5 x +5 \,{\mathrm e}-4\right )}{25}\)	\(15\)
norman	\(\left (\frac {{\mathrm e}}{5}-\frac {4}{25}\right ) x^{2}-\frac {x^{3}}{5}\)	\(17\)
default	\(-\frac {x^{3}}{5}+\frac {\left (10 \,{\mathrm e}-8\right ) x^{2}}{50}\)	\(18\)
risch	\(\frac {x^{2} {\mathrm e}}{5}-\frac {x^{3}}{5}-\frac {4 x^{2}}{25}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2/5*x*exp(1)-3/5*x^2-8/25*x,x,method=_RETURNVERBOSE)

[Out]

-1/5*x^3+1/50*(10*exp(1)-8)*x^2

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Maxima [A]
time = 0.26, size = 18, normalized size = 1.12 \begin {gather*} -\frac {1}{5} \, x^{3} + \frac {1}{5} \, x^{2} e - \frac {4}{25} \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/5*x*exp(1)-3/5*x^2-8/25*x,x, algorithm="maxima")

[Out]

-1/5*x^3 + 1/5*x^2*e - 4/25*x^2

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Fricas [A]
time = 0.37, size = 18, normalized size = 1.12 \begin {gather*} -\frac {1}{5} \, x^{3} + \frac {1}{5} \, x^{2} e - \frac {4}{25} \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/5*x*exp(1)-3/5*x^2-8/25*x,x, algorithm="fricas")

[Out]

-1/5*x^3 + 1/5*x^2*e - 4/25*x^2

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Sympy [A]
time = 0.01, size = 15, normalized size = 0.94 \begin {gather*} - \frac {x^{3}}{5} + x^{2} \left (- \frac {4}{25} + \frac {e}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/5*x*exp(1)-3/5*x**2-8/25*x,x)

[Out]

-x**3/5 + x**2*(-4/25 + E/5)

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Giac [A]
time = 0.38, size = 18, normalized size = 1.12 \begin {gather*} -\frac {1}{5} \, x^{3} + \frac {1}{5} \, x^{2} e - \frac {4}{25} \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/5*x*exp(1)-3/5*x^2-8/25*x,x, algorithm="giac")

[Out]

-1/5*x^3 + 1/5*x^2*e - 4/25*x^2

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Mupad [B]
time = 0.03, size = 14, normalized size = 0.88 \begin {gather*} -\frac {x^2\,\left (5\,x-5\,\mathrm {e}+4\right )}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*exp(1))/5 - (8*x)/25 - (3*x^2)/5,x)

[Out]

-(x^2*(5*x - 5*exp(1) + 4))/25

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