3.77.57 \(\int \frac {-1-e^2+11 x-5 x^2}{2 e^2-2 x+10 x^2} \, dx\)

Optimal. Leaf size=34 \[ e^{e^3}-x+\frac {1}{2} \left (x-\log \left (\frac {4}{e^2-x+5 x^2}\right )\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 23, normalized size of antiderivative = 0.68, number of steps used = 3, number of rules used = 2, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1657, 628} \begin {gather*} \frac {1}{2} \log \left (5 x^2-x+e^2\right )-\frac {x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 - E^2 + 11*x - 5*x^2)/(2*E^2 - 2*x + 10*x^2),x]

[Out]

-1/2*x + Log[E^2 - x + 5*x^2]/2

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1657

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 21, normalized size = 0.62 \begin {gather*} \frac {1}{2} \left (-x+\log \left (e^2-x+5 x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - E^2 + 11*x - 5*x^2)/(2*E^2 - 2*x + 10*x^2),x]

[Out]

(-x + Log[E^2 - x + 5*x^2])/2

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fricas [A]  time = 1.17, size = 18, normalized size = 0.53 \begin {gather*} -\frac {1}{2} \, x + \frac {1}{2} \, \log \left (5 \, x^{2} - x + e^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(2)-5*x^2+11*x-1)/(2*exp(2)+10*x^2-2*x),x, algorithm="fricas")

[Out]

-1/2*x + 1/2*log(5*x^2 - x + e^2)

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giac [A]  time = 0.16, size = 18, normalized size = 0.53 \begin {gather*} -\frac {1}{2} \, x + \frac {1}{2} \, \log \left (5 \, x^{2} - x + e^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(2)-5*x^2+11*x-1)/(2*exp(2)+10*x^2-2*x),x, algorithm="giac")

[Out]

-1/2*x + 1/2*log(5*x^2 - x + e^2)

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maple [A]  time = 0.52, size = 19, normalized size = 0.56

method	result	size
default	\(-\frac {x}{2}+\frac {\ln \left (5 x^{2}-x +{\mathrm e}^{2}\right )}{2}\)	\(19\)
risch	\(-\frac {x}{2}+\frac {\ln \left (5 x^{2}-x +{\mathrm e}^{2}\right )}{2}\)	\(19\)
norman	\(-\frac {x}{2}+\frac {\ln \left (2 \,{\mathrm e}^{2}+10 x^{2}-2 x \right )}{2}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-exp(2)-5*x^2+11*x-1)/(2*exp(2)+10*x^2-2*x),x,method=_RETURNVERBOSE)

[Out]

-1/2*x+1/2*ln(5*x^2-x+exp(2))

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maxima [A]  time = 0.43, size = 18, normalized size = 0.53 \begin {gather*} -\frac {1}{2} \, x + \frac {1}{2} \, \log \left (5 \, x^{2} - x + e^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(2)-5*x^2+11*x-1)/(2*exp(2)+10*x^2-2*x),x, algorithm="maxima")

[Out]

-1/2*x + 1/2*log(5*x^2 - x + e^2)

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mupad [B]  time = 4.61, size = 18, normalized size = 0.53 \begin {gather*} \frac {\ln \left (5\,x^2-x+{\mathrm {e}}^2\right )}{2}-\frac {x}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2) - 11*x + 5*x^2 + 1)/(2*exp(2) - 2*x + 10*x^2),x)

[Out]

log(exp(2) - x + 5*x^2)/2 - x/2

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sympy [A]  time = 0.14, size = 15, normalized size = 0.44 \begin {gather*} - \frac {x}{2} + \frac {\log {\left (5 x^{2} - x + e^{2} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(2)-5*x**2+11*x-1)/(2*exp(2)+10*x**2-2*x),x)

[Out]

-x/2 + log(5*x**2 - x + exp(2))/2

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