3.4.89 \(\int \frac {10+e (-1+2 x)}{e} \, dx\)

Optimal. Leaf size=19 \[ -10+\frac {5}{e^3}-x+\left (\frac {5}{e}+x\right )^2 \]

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Rubi [A]  time = 0.01, antiderivative size = 18, normalized size of antiderivative = 0.95, number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12} \begin {gather*} \frac {1}{4} (1-2 x)^2+\frac {10 x}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10 + E*(-1 + 2*x))/E,x]

[Out]

(1 - 2*x)^2/4 + (10*x)/E

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} &=\frac {\int (10+e (-1+2 x)) \, dx}{e}\\ &=\frac {1}{4} (1-2 x)^2+\frac {10 x}{e}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 13, normalized size = 0.68 \begin {gather*} -x+\frac {10 x}{e}+x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10 + E*(-1 + 2*x))/E,x]

[Out]

-x + (10*x)/E + x^2

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fricas [A]  time = 0.60, size = 17, normalized size = 0.89 \begin {gather*} {\left ({\left (x^{2} - x\right )} e + 10 \, x\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-1)*exp(1)+10)/exp(1),x, algorithm="fricas")

[Out]

((x^2 - x)*e + 10*x)*e^(-1)

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giac [A]  time = 0.30, size = 17, normalized size = 0.89 \begin {gather*} {\left ({\left (x^{2} - x\right )} e + 10 \, x\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-1)*exp(1)+10)/exp(1),x, algorithm="giac")

[Out]

((x^2 - x)*e + 10*x)*e^(-1)

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maple [A]  time = 0.02, size = 13, normalized size = 0.68

method	result	size
risch	\(x^{2}-x +10 \,{\mathrm e}^{-1} x\)	\(13\)
norman	\(x^{2}-{\mathrm e}^{-1} \left ({\mathrm e}-10\right ) x\)	\(16\)
gosper	\(x \left (x \,{\mathrm e}-{\mathrm e}+10\right ) {\mathrm e}^{-1}\)	\(17\)
default	\({\mathrm e}^{-1} \left ({\mathrm e} \left (x^{2}-x \right )+10 x \right )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x-1)*exp(1)+10)/exp(1),x,method=_RETURNVERBOSE)

[Out]

x^2-x+10*exp(-1)*x

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maxima [A]  time = 0.35, size = 17, normalized size = 0.89 \begin {gather*} {\left ({\left (x^{2} - x\right )} e + 10 \, x\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-1)*exp(1)+10)/exp(1),x, algorithm="maxima")

[Out]

((x^2 - x)*e + 10*x)*e^(-1)

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mupad [B]  time = 0.27, size = 16, normalized size = 0.84 \begin {gather*} \frac {{\mathrm {e}}^{-2}\,{\left (\mathrm {e}\,\left (2\,x-1\right )+10\right )}^2}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-1)*(exp(1)*(2*x - 1) + 10),x)

[Out]

(exp(-2)*(exp(1)*(2*x - 1) + 10)^2)/4

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sympy [A]  time = 0.05, size = 12, normalized size = 0.63 \begin {gather*} x^{2} + \frac {x \left (10 - e\right )}{e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-1)*exp(1)+10)/exp(1),x)

[Out]

x**2 + x*(10 - E)*exp(-1)

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