3.15.66 \(\int \frac {-25-5 e^4+2 x}{-135+5 e^{e^4}} \, dx\)

Optimal. Leaf size=25 \[ \frac {\left (-4-e^4+\frac {1}{5} (-5+x)\right ) x}{-27+e^{e^4}} \]

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Rubi [A]  time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.12, number of steps used = 1, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {9} \begin {gather*} -\frac {\left (5 \left (5+e^4\right )-2 x\right )^2}{20 \left (27-e^{e^4}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-25 - 5*E^4 + 2*x)/(-135 + 5*E^E^4),x]

[Out]

-1/20*(5*(5 + E^4) - 2*x)^2/(27 - E^E^4)

Rule 9

Int[(a_)*((b_) + (c_.)*(x_)), x_Symbol] :> Simp[(a*(b + c*x)^2)/(2*c), x] /; FreeQ[{a, b, c}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {\left (5 \left (5+e^4\right )-2 x\right )^2}{20 \left (27-e^{e^4}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 27, normalized size = 1.08 \begin {gather*} \frac {25 x+5 e^4 x-x^2}{135-5 e^{e^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25 - 5*E^4 + 2*x)/(-135 + 5*E^E^4),x]

[Out]

(25*x + 5*E^4*x - x^2)/(135 - 5*E^E^4)

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fricas [A]  time = 0.97, size = 21, normalized size = 0.84 \begin {gather*} \frac {x^{2} - 5 \, x e^{4} - 25 \, x}{5 \, {\left (e^{\left (e^{4}\right )} - 27\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*exp(4)+2*x-25)/(5*exp(exp(4))-135),x, algorithm="fricas")

[Out]

1/5*(x^2 - 5*x*e^4 - 25*x)/(e^(e^4) - 27)

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giac [A]  time = 0.15, size = 21, normalized size = 0.84 \begin {gather*} \frac {x^{2} - 5 \, x e^{4} - 25 \, x}{5 \, {\left (e^{\left (e^{4}\right )} - 27\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*exp(4)+2*x-25)/(5*exp(exp(4))-135),x, algorithm="giac")

[Out]

1/5*(x^2 - 5*x*e^4 - 25*x)/(e^(e^4) - 27)

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maple [A]  time = 0.03, size = 20, normalized size = 0.80

method	result	size
gosper	\(-\frac {x \left (-x +5 \,{\mathrm e}^{4}+25\right )}{5 \left ({\mathrm e}^{{\mathrm e}^{4}}-27\right )}\)	\(20\)
default	\(\frac {-5 x \,{\mathrm e}^{4}+x^{2}-25 x}{5 \,{\mathrm e}^{{\mathrm e}^{4}}-135}\)	\(23\)
norman	\(\frac {x^{2}}{5 \,{\mathrm e}^{{\mathrm e}^{4}}-135}-\frac {\left (5+{\mathrm e}^{4}\right ) x}{{\mathrm e}^{{\mathrm e}^{4}}-27}\)	\(28\)
risch	\(-\frac {5 x \,{\mathrm e}^{4}}{5 \,{\mathrm e}^{{\mathrm e}^{4}}-135}+\frac {x^{2}}{5 \,{\mathrm e}^{{\mathrm e}^{4}}-135}-\frac {25 x}{5 \,{\mathrm e}^{{\mathrm e}^{4}}-135}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5*exp(4)+2*x-25)/(5*exp(exp(4))-135),x,method=_RETURNVERBOSE)

[Out]

-1/5*x*(-x+5*exp(4)+25)/(exp(exp(4))-27)

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maxima [A]  time = 0.48, size = 21, normalized size = 0.84 \begin {gather*} \frac {x^{2} - 5 \, x e^{4} - 25 \, x}{5 \, {\left (e^{\left (e^{4}\right )} - 27\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*exp(4)+2*x-25)/(5*exp(exp(4))-135),x, algorithm="maxima")

[Out]

1/5*(x^2 - 5*x*e^4 - 25*x)/(e^(e^4) - 27)

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mupad [B]  time = 1.01, size = 21, normalized size = 0.84 \begin {gather*} \frac {{\left (5\,{\mathrm {e}}^4-2\,x+25\right )}^2}{4\,\left (5\,{\mathrm {e}}^{{\mathrm {e}}^4}-135\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*exp(4) - 2*x + 25)/(5*exp(exp(4)) - 135),x)

[Out]

(5*exp(4) - 2*x + 25)^2/(4*(5*exp(exp(4)) - 135))

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sympy [A]  time = 0.06, size = 26, normalized size = 1.04 \begin {gather*} \frac {x^{2}}{-135 + 5 e^{e^{4}}} + \frac {x \left (- e^{4} - 5\right )}{-27 + e^{e^{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*exp(4)+2*x-25)/(5*exp(exp(4))-135),x)

[Out]

x**2/(-135 + 5*exp(exp(4))) + x*(-exp(4) - 5)/(-27 + exp(exp(4)))

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