\(\int \frac {2-x^2}{-2 x-15 x^2+e^2 x^2-x^3} \, dx\) [5099]

Optimal result

Integrand size = 31, antiderivative size = 14 \[ \int \frac {2-x^2}{-2 x-15 x^2+e^2 x^2-x^3} \, dx=\log \left (15-e^2+\frac {2}{x}+x\right ) \]

[Out]

ln(2/x+x-exp(2)+15)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.43, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {6, 1608, 1642, 642} \[ \int \frac {2-x^2}{-2 x-15 x^2+e^2 x^2-x^3} \, dx=\log \left (x^2+\left (15-e^2\right ) x+2\right )-\log (x) \]

[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 642

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 1608

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 1642

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.43 \[ \int \frac {2-x^2}{-2 x-15 x^2+e^2 x^2-x^3} \, dx=-\log (x)+\log \left (2+15 x-e^2 x+x^2\right ) \]

[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]

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.43

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.36 \[ \int \frac {2-x^2}{-2 x-15 x^2+e^2 x^2-x^3} \, dx=\log \left (x^{2} - x e^{2} + 15 \, x + 2\right ) - \log \left (x\right ) \]

[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)

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {2-x^2}{-2 x-15 x^2+e^2 x^2-x^3} \, dx=- \log {\left (x \right )} + \log {\left (x^{2} + x \left (15 - e^{2}\right ) + 2 \right )} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.29 \[ \int \frac {2-x^2}{-2 x-15 x^2+e^2 x^2-x^3} \, dx=\log \left (x^{2} - x {\left (e^{2} - 15\right )} + 2\right ) - \log \left (x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.50 \[ \int \frac {2-x^2}{-2 x-15 x^2+e^2 x^2-x^3} \, dx=\log \left ({\left | x^{2} - x e^{2} + 15 \, x + 2 \right |}\right ) - \log \left ({\left | x \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.36 \[ \int \frac {2-x^2}{-2 x-15 x^2+e^2 x^2-x^3} \, dx=\ln \left (15\,x-x\,{\mathrm {e}}^2+x^2+2\right )-\ln \left (x\right ) \]

[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)