3.17.79 \(\int \frac {-1+e^2 x+\frac {2 (-1+e^2 x)^2}{e^4}}{-1+e^2 x} \, dx\)

Optimal. Leaf size=22 \[ -1-e^5+e^{2 \left (-3+\log \left (-1+e^2 x\right )\right )}+x \]

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 0.77, number of steps used = 2, number of rules used = 1, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {1586} \begin {gather*} \frac {x^2}{e^2}+\left (1-\frac {2}{e^4}\right ) x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + E^2*x + (2*(-1 + E^2*x)^2)/E^4)/(-1 + E^2*x),x]

[Out]

(1 - 2/E^4)*x + x^2/E^2

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 15, normalized size = 0.68 \begin {gather*} x-\frac {2 x}{e^4}+\frac {x^2}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + E^2*x + (2*(-1 + E^2*x)^2)/E^4)/(-1 + E^2*x),x]

[Out]

x - (2*x)/E^4 + x^2/E^2

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fricas [A]  time = 0.82, size = 17, normalized size = 0.77 \begin {gather*} {\left (x^{2} e^{2} + x e^{4} - 2 \, x\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(2)*exp(2*log(exp(2)*x-1)-6)+exp(2)*x-1)/(exp(2)*x-1),x, algorithm="fricas")

[Out]

(x^2*e^2 + x*e^4 - 2*x)*e^(-4)

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giac [A]  time = 0.41, size = 17, normalized size = 0.77 \begin {gather*} {\left (x^{2} e^{2} + x e^{4} - 2 \, x\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(2)*exp(2*log(exp(2)*x-1)-6)+exp(2)*x-1)/(exp(2)*x-1),x, algorithm="giac")

[Out]

(x^2*e^2 + x*e^4 - 2*x)*e^(-4)

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maple [A]  time = 0.30, size = 14, normalized size = 0.64

method	result	size
risch	\(-2 x \,{\mathrm e}^{-4}+x^{2} {\mathrm e}^{-2}+x\)	\(14\)
default	\(x +{\mathrm e}^{2 \ln \left ({\mathrm e}^{2} x -1\right )-6}\)	\(15\)
norman	\({\mathrm e}^{4} {\mathrm e}^{-6} x^{2}-\left (2 \,{\mathrm e}^{2}-{\mathrm e}^{6}\right ) {\mathrm e}^{-6} x\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(2)*exp(2*ln(exp(2)*x-1)-6)+exp(2)*x-1)/(exp(2)*x-1),x,method=_RETURNVERBOSE)

[Out]

-2*x*exp(-4)+x^2*exp(-2)+x

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maxima [A]  time = 0.39, size = 16, normalized size = 0.73 \begin {gather*} {\left (x^{2} e^{2} + x {\left (e^{4} - 2\right )}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(2)*exp(2*log(exp(2)*x-1)-6)+exp(2)*x-1)/(exp(2)*x-1),x, algorithm="maxima")

[Out]

(x^2*e^2 + x*(e^4 - 2))*e^(-4)

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mupad [B]  time = 0.07, size = 12, normalized size = 0.55 \begin {gather*} x\,\left (x\,{\mathrm {e}}^{-2}-2\,{\mathrm {e}}^{-4}+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*exp(2) + 2*exp(2*log(x*exp(2) - 1) - 6)*exp(2) - 1)/(x*exp(2) - 1),x)

[Out]

x*(x*exp(-2) - 2*exp(-4) + 1)

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sympy [A]  time = 0.09, size = 15, normalized size = 0.68 \begin {gather*} \frac {x^{2}}{e^{2}} + \frac {x \left (-2 + e^{4}\right )}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(2)*exp(2*ln(exp(2)*x-1)-6)+exp(2)*x-1)/(exp(2)*x-1),x)

[Out]

x**2*exp(-2) + x*(-2 + exp(4))*exp(-4)

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