3.51.99 \(\int \frac {2-x^2}{-2 x-15 x^2+e^2 x^2-x^3} \, dx\)

Optimal. Leaf size=14 \[ \log \left (15-e^2+\frac {2}{x}+x\right ) \]

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Rubi [A]  time = 0.07, antiderivative size = 20, normalized size of antiderivative = 1.43, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {6, 1594, 1628, 628} \begin {gather*} \log \left (x^2+\left (15-e^2\right ) x+2\right )-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 - x^2)/(-2*x - 15*x^2 + E^2*x^2 - x^3),x]

[Out]

-Log[x] + Log[2 + (15 - E^2)*x + x^2]

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 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 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 20, normalized size = 1.43 \begin {gather*} -\log (x)+\log \left (2+15 x-e^2 x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - x^2)/(-2*x - 15*x^2 + E^2*x^2 - x^3),x]

[Out]

-Log[x] + Log[2 + 15*x - E^2*x + x^2]

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fricas [A]  time = 0.72, size = 19, normalized size = 1.36 \begin {gather*} \log \left (x^{2} - x e^{2} + 15 \, x + 2\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+2)/(x^2*exp(2)-x^3-15*x^2-2*x),x, algorithm="fricas")

[Out]

log(x^2 - x*e^2 + 15*x + 2) - log(x)

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giac [A]  time = 0.14, size = 21, normalized size = 1.50 \begin {gather*} \log \left ({\left | x^{2} - x e^{2} + 15 \, x + 2 \right |}\right ) - \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+2)/(x^2*exp(2)-x^3-15*x^2-2*x),x, algorithm="giac")

[Out]

log(abs(x^2 - x*e^2 + 15*x + 2)) - log(abs(x))

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maple [A]  time = 0.06, size = 20, normalized size = 1.43

method	result	size
default	\(\ln \left (-{\mathrm e}^{2} x +x^{2}+15 x +2\right )-\ln \relax (x )\)	\(20\)
norman	\(-\ln \relax (x )+\ln \left ({\mathrm e}^{2} x -x^{2}-15 x -2\right )\)	\(21\)
risch	\(-\ln \left (-x \right )+\ln \left (2+x^{2}+\left (-{\mathrm e}^{2}+15\right ) x \right )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^2+2)/(x^2*exp(2)-x^3-15*x^2-2*x),x,method=_RETURNVERBOSE)

[Out]

ln(-exp(2)*x+x^2+15*x+2)-ln(x)

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maxima [A]  time = 0.34, size = 18, normalized size = 1.29 \begin {gather*} \log \left (x^{2} - x {\left (e^{2} - 15\right )} + 2\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+2)/(x^2*exp(2)-x^3-15*x^2-2*x),x, algorithm="maxima")

[Out]

log(x^2 - x*(e^2 - 15) + 2) - log(x)

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mupad [B]  time = 3.34, size = 19, normalized size = 1.36 \begin {gather*} \ln \left (15\,x-x\,{\mathrm {e}}^2+x^2+2\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 - 2)/(2*x - x^2*exp(2) + 15*x^2 + x^3),x)

[Out]

log(15*x - x*exp(2) + x^2 + 2) - log(x)

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sympy [A]  time = 0.29, size = 15, normalized size = 1.07 \begin {gather*} - \log {\relax (x )} + \log {\left (x^{2} + x \left (15 - e^{2}\right ) + 2 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**2+2)/(x**2*exp(2)-x**3-15*x**2-2*x),x)

[Out]

-log(x) + log(x**2 + x*(15 - exp(2)) + 2)

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