\(\int \frac {d-e x^2}{d^2-f x^2+e^2 x^4} \, dx\) [33]

Optimal result

Integrand size = 28, antiderivative size = 70 \[ \int \frac {d-e x^2}{d^2-f x^2+e^2 x^4} \, dx=-\frac {\log \left (d-\sqrt {2 d e+f} x+e x^2\right )}{2 \sqrt {2 d e+f}}+\frac {\log \left (d+\sqrt {2 d e+f} x+e x^2\right )}{2 \sqrt {2 d e+f}} \]

[Out]

-1/2*ln(d+e*x^2-x*(2*d*e+f)^(1/2))/(2*d*e+f)^(1/2)+1/2*ln(d+e*x^2+x*(2*d*e+f)^(1/2))/(2*d*e+f)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1178, 642} \[ \int \frac {d-e x^2}{d^2-f x^2+e^2 x^4} \, dx=\frac {\log \left (x \sqrt {2 d e+f}+d+e x^2\right )}{2 \sqrt {2 d e+f}}-\frac {\log \left (-x \sqrt {2 d e+f}+d+e x^2\right )}{2 \sqrt {2 d e+f}} \]

[In]

Int[(d - e*x^2)/(d^2 - f*x^2 + e^2*x^4),x]

[Out]

-1/2*Log[d - Sqrt[2*d*e + f]*x + e*x^2]/Sqrt[2*d*e + f] + Log[d + Sqrt[2*d*e + f]*x + e*x^2]/(2*Sqrt[2*d*e + f
])

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 1178

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e) - b/c, 2]},
Dist[e/(2*c*q), Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x
- x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] &&  !GtQ[b^2
- 4*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\int \frac {\frac {\sqrt {2 d e+f}}{e}+2 x}{-\frac {d}{e}-\frac {\sqrt {2 d e+f} x}{e}-x^2} \, dx}{2 \sqrt {2 d e+f}}-\frac {\int \frac {\frac {\sqrt {2 d e+f}}{e}-2 x}{-\frac {d}{e}+\frac {\sqrt {2 d e+f} x}{e}-x^2} \, dx}{2 \sqrt {2 d e+f}} \\ & = -\frac {\log \left (d-\sqrt {2 d e+f} x+e x^2\right )}{2 \sqrt {2 d e+f}}+\frac {\log \left (d+\sqrt {2 d e+f} x+e x^2\right )}{2 \sqrt {2 d e+f}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(190\) vs. \(2(70)=140\).

Time = 0.08 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.71 \[ \int \frac {d-e x^2}{d^2-f x^2+e^2 x^4} \, dx=\frac {-\frac {\left (-2 d e+f+\sqrt {-4 d^2 e^2+f^2}\right ) \arctan \left (\frac {\sqrt {2} e x}{\sqrt {-f-\sqrt {-4 d^2 e^2+f^2}}}\right )}{\sqrt {-f-\sqrt {-4 d^2 e^2+f^2}}}+\frac {\left (-2 d e+f-\sqrt {-4 d^2 e^2+f^2}\right ) \arctan \left (\frac {\sqrt {2} e x}{\sqrt {-f+\sqrt {-4 d^2 e^2+f^2}}}\right )}{\sqrt {-f+\sqrt {-4 d^2 e^2+f^2}}}}{\sqrt {2} \sqrt {-4 d^2 e^2+f^2}} \]

[In]

Integrate[(d - e*x^2)/(d^2 - f*x^2 + e^2*x^4),x]

[Out]

(-(((-2*d*e + f + Sqrt[-4*d^2*e^2 + f^2])*ArcTan[(Sqrt[2]*e*x)/Sqrt[-f - Sqrt[-4*d^2*e^2 + f^2]]])/Sqrt[-f - S
qrt[-4*d^2*e^2 + f^2]]) + ((-2*d*e + f - Sqrt[-4*d^2*e^2 + f^2])*ArcTan[(Sqrt[2]*e*x)/Sqrt[-f + Sqrt[-4*d^2*e^
2 + f^2]]])/Sqrt[-f + Sqrt[-4*d^2*e^2 + f^2]])/(Sqrt[2]*Sqrt[-4*d^2*e^2 + f^2])

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.87

method	result	size
default	\(\frac {\ln \left (d +e \,x^{2}+x \sqrt {2 e d +f}\right )}{2 \sqrt {2 e d +f}}-\frac {\ln \left (-e \,x^{2}+x \sqrt {2 e d +f}-d \right )}{2 \sqrt {2 e d +f}}\)	\(61\)
risch	\(\frac {\ln \left (\sqrt {2 e d +f}\, e \,x^{2}+\left (2 e d +f \right ) x +\sqrt {2 e d +f}\, d \right )}{2 \sqrt {2 e d +f}}-\frac {\ln \left (\sqrt {2 e d +f}\, e \,x^{2}+\left (-2 e d -f \right ) x +\sqrt {2 e d +f}\, d \right )}{2 \sqrt {2 e d +f}}\)	\(90\)

[In]

int((-e*x^2+d)/(e^2*x^4-f*x^2+d^2),x,method=_RETURNVERBOSE)

[Out]

1/2*ln(d+e*x^2+x*(2*d*e+f)^(1/2))/(2*d*e+f)^(1/2)-1/2/(2*d*e+f)^(1/2)*ln(-e*x^2+x*(2*d*e+f)^(1/2)-d)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.40 \[ \int \frac {d-e x^2}{d^2-f x^2+e^2 x^4} \, dx=\left [\frac {\log \left (\frac {e^{2} x^{4} + {\left (4 \, d e + f\right )} x^{2} + d^{2} + 2 \, {\left (e x^{3} + d x\right )} \sqrt {2 \, d e + f}}{e^{2} x^{4} - f x^{2} + d^{2}}\right )}{2 \, \sqrt {2 \, d e + f}}, -\frac {\sqrt {-2 \, d e - f} \arctan \left (\frac {\sqrt {-2 \, d e - f} e x}{2 \, d e + f}\right ) - \sqrt {-2 \, d e - f} \arctan \left (\frac {{\left (e^{2} x^{3} - {\left (d e + f\right )} x\right )} \sqrt {-2 \, d e - f}}{2 \, d^{2} e + d f}\right )}{2 \, d e + f}\right ] \]

[In]

integrate((-e*x^2+d)/(e^2*x^4-f*x^2+d^2),x, algorithm="fricas")

[Out]

[1/2*log((e^2*x^4 + (4*d*e + f)*x^2 + d^2 + 2*(e*x^3 + d*x)*sqrt(2*d*e + f))/(e^2*x^4 - f*x^2 + d^2))/sqrt(2*d
*e + f), -(sqrt(-2*d*e - f)*arctan(sqrt(-2*d*e - f)*e*x/(2*d*e + f)) - sqrt(-2*d*e - f)*arctan((e^2*x^3 - (d*e
+ f)*x)*sqrt(-2*d*e - f)/(2*d^2*e + d*f)))/(2*d*e + f)]

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.60 \[ \int \frac {d-e x^2}{d^2-f x^2+e^2 x^4} \, dx=- \frac {\sqrt {\frac {1}{2 d e + f}} \log {\left (\frac {d}{e} + x^{2} + \frac {x \left (- 2 d e \sqrt {\frac {1}{2 d e + f}} - f \sqrt {\frac {1}{2 d e + f}}\right )}{e} \right )}}{2} + \frac {\sqrt {\frac {1}{2 d e + f}} \log {\left (\frac {d}{e} + x^{2} + \frac {x \left (2 d e \sqrt {\frac {1}{2 d e + f}} + f \sqrt {\frac {1}{2 d e + f}}\right )}{e} \right )}}{2} \]

[In]

integrate((-e*x**2+d)/(e**2*x**4-f*x**2+d**2),x)

[Out]

-sqrt(1/(2*d*e + f))*log(d/e + x**2 + x*(-2*d*e*sqrt(1/(2*d*e + f)) - f*sqrt(1/(2*d*e + f)))/e)/2 + sqrt(1/(2*
d*e + f))*log(d/e + x**2 + x*(2*d*e*sqrt(1/(2*d*e + f)) + f*sqrt(1/(2*d*e + f)))/e)/2

Maxima [F]

\[ \int \frac {d-e x^2}{d^2-f x^2+e^2 x^4} \, dx=\int { -\frac {e x^{2} - d}{e^{2} x^{4} - f x^{2} + d^{2}} \,d x } \]

[In]

integrate((-e*x^2+d)/(e^2*x^4-f*x^2+d^2),x, algorithm="maxima")

[Out]

-integrate((e*x^2 - d)/(e^2*x^4 - f*x^2 + d^2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (58) = 116\).

Time = 0.69 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.86 \[ \int \frac {d-e x^2}{d^2-f x^2+e^2 x^4} \, dx=\frac {{\left (2 \, d^{2} e^{3} - d e^{4} - d e^{2} f\right )} \sqrt {-2 \, d e - f} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} x}{\sqrt {-\frac {f + \sqrt {-4 \, d^{2} e^{2} + f^{2}}}{e^{2}}}}\right )}{4 \, d^{3} e^{4} - 2 \, d^{2} e^{5} - d e^{4} f - d e^{2} f^{2}} - \frac {{\left (2 \, d^{2} e^{3} - d e^{4} - d e^{2} f\right )} \sqrt {-2 \, d e - f} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} x}{\sqrt {-\frac {f - \sqrt {-4 \, d^{2} e^{2} + f^{2}}}{e^{2}}}}\right )}{4 \, d^{3} e^{4} - 2 \, d^{2} e^{5} - d e^{4} f - d e^{2} f^{2}} \]

[In]

integrate((-e*x^2+d)/(e^2*x^4-f*x^2+d^2),x, algorithm="giac")

[Out]

(2*d^2*e^3 - d*e^4 - d*e^2*f)*sqrt(-2*d*e - f)*arctan(2*sqrt(1/2)*x/sqrt(-(f + sqrt(-4*d^2*e^2 + f^2))/e^2))/(
4*d^3*e^4 - 2*d^2*e^5 - d*e^4*f - d*e^2*f^2) - (2*d^2*e^3 - d*e^4 - d*e^2*f)*sqrt(-2*d*e - f)*arctan(2*sqrt(1/
2)*x/sqrt(-(f - sqrt(-4*d^2*e^2 + f^2))/e^2))/(4*d^3*e^4 - 2*d^2*e^5 - d*e^4*f - d*e^2*f^2)

Mupad [B] (verification not implemented)

Time = 13.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.41 \[ \int \frac {d-e x^2}{d^2-f x^2+e^2 x^4} \, dx=\frac {\mathrm {atanh}\left (\frac {x\,\sqrt {f+2\,d\,e}}{e\,x^2+d}\right )}{\sqrt {f+2\,d\,e}} \]

[In]

int((d - e*x^2)/(d^2 - f*x^2 + e^2*x^4),x)

[Out]

atanh((x*(f + 2*d*e)^(1/2))/(d + e*x^2))/(f + 2*d*e)^(1/2)