\(\int \frac {d-e x^2}{d^2+b x^2+e^2 x^4} \, dx\) [30]

Optimal result

Integrand size = 27, antiderivative size = 78 \[ \int \frac {d-e x^2}{d^2+b x^2+e^2 x^4} \, dx=-\frac {\log \left (d-\sqrt {-b+2 d e} x+e x^2\right )}{2 \sqrt {-b+2 d e}}+\frac {\log \left (d+\sqrt {-b+2 d e} x+e x^2\right )}{2 \sqrt {-b+2 d e}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1178, 642} \[ \int \frac {d-e x^2}{d^2+b x^2+e^2 x^4} \, dx=\frac {\log \left (x \sqrt {2 d e-b}+d+e x^2\right )}{2 \sqrt {2 d e-b}}-\frac {\log \left (-x \sqrt {2 d e-b}+d+e x^2\right )}{2 \sqrt {2 d e-b}} \]

[In]

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

[Out]

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

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 {-b+2 d e}}{e}+2 x}{-\frac {d}{e}-\frac {\sqrt {-b+2 d e} x}{e}-x^2} \, dx}{2 \sqrt {-b+2 d e}}-\frac {\int \frac {\frac {\sqrt {-b+2 d e}}{e}-2 x}{-\frac {d}{e}+\frac {\sqrt {-b+2 d e} x}{e}-x^2} \, dx}{2 \sqrt {-b+2 d e}} \\ & = -\frac {\log \left (d-\sqrt {-b+2 d e} x+e x^2\right )}{2 \sqrt {-b+2 d e}}+\frac {\log \left (d+\sqrt {-b+2 d e} x+e x^2\right )}{2 \sqrt {-b+2 d e}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(182\) vs. \(2(78)=156\).

Time = 0.08 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.33 \[ \int \frac {d-e x^2}{d^2+b x^2+e^2 x^4} \, dx=\frac {\frac {\left (b+2 d e-\sqrt {b^2-4 d^2 e^2}\right ) \arctan \left (\frac {\sqrt {2} e x}{\sqrt {b-\sqrt {b^2-4 d^2 e^2}}}\right )}{\sqrt {b-\sqrt {b^2-4 d^2 e^2}}}-\frac {\left (b+2 d e+\sqrt {b^2-4 d^2 e^2}\right ) \arctan \left (\frac {\sqrt {2} e x}{\sqrt {b+\sqrt {b^2-4 d^2 e^2}}}\right )}{\sqrt {b+\sqrt {b^2-4 d^2 e^2}}}}{\sqrt {2} \sqrt {b^2-4 d^2 e^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.13

method	result	size
default	\(\frac {\sqrt {2 e d -b}\, \ln \left (-e \,x^{2}+x \sqrt {2 e d -b}-d \right )}{-4 e d +2 b}-\frac {\sqrt {2 e d -b}\, \ln \left (d +e \,x^{2}+x \sqrt {2 e d -b}\right )}{-4 e d +2 b}\)	\(88\)
risch	\(-\frac {\ln \left (-e \,x^{2} \sqrt {2 e d -b}+\left (2 e d -b \right ) x -d \sqrt {2 e d -b}\right )}{2 \sqrt {2 e d -b}}+\frac {\ln \left (-e \,x^{2} \sqrt {2 e d -b}+\left (-2 e d +b \right ) x -d \sqrt {2 e d -b}\right )}{2 \sqrt {2 e d -b}}\)	\(106\)

[In]

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

[Out]

1/(-4*d*e+2*b)*(2*d*e-b)^(1/2)*ln(-e*x^2+x*(2*d*e-b)^(1/2)-d)-1/(-4*d*e+2*b)*(2*d*e-b)^(1/2)*ln(d+e*x^2+x*(2*d
*e-b)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.21 \[ \int \frac {d-e x^2}{d^2+b x^2+e^2 x^4} \, dx=\left [\frac {\log \left (\frac {e^{2} x^{4} + {\left (4 \, d e - b\right )} x^{2} + d^{2} + 2 \, {\left (e x^{3} + d x\right )} \sqrt {2 \, d e - b}}{e^{2} x^{4} + b x^{2} + d^{2}}\right )}{2 \, \sqrt {2 \, d e - b}}, -\frac {\sqrt {-2 \, d e + b} \arctan \left (\frac {\sqrt {-2 \, d e + b} e x}{2 \, d e - b}\right ) - \sqrt {-2 \, d e + b} \arctan \left (\frac {{\left (e^{2} x^{3} - {\left (d e - b\right )} x\right )} \sqrt {-2 \, d e + b}}{2 \, d^{2} e - b d}\right )}{2 \, d e - b}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.55 \[ \int \frac {d-e x^2}{d^2+b x^2+e^2 x^4} \, dx=\frac {\sqrt {- \frac {1}{b - 2 d e}} \log {\left (\frac {d}{e} + x^{2} + \frac {x \left (- b \sqrt {- \frac {1}{b - 2 d e}} + 2 d e \sqrt {- \frac {1}{b - 2 d e}}\right )}{e} \right )}}{2} - \frac {\sqrt {- \frac {1}{b - 2 d e}} \log {\left (\frac {d}{e} + x^{2} + \frac {x \left (b \sqrt {- \frac {1}{b - 2 d e}} - 2 d e \sqrt {- \frac {1}{b - 2 d e}}\right )}{e} \right )}}{2} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (66) = 132\).

Time = 0.68 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.42 \[ \int \frac {d-e x^2}{d^2+b x^2+e^2 x^4} \, dx=\frac {{\left (2 \, d^{2} e^{3} - d e^{4} + b d e^{2}\right )} \sqrt {-2 \, d e + b} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} x}{\sqrt {\frac {b + \sqrt {-4 \, d^{2} e^{2} + b^{2}}}{e^{2}}}}\right )}{4 \, d^{3} e^{4} - 2 \, d^{2} e^{5} + b d e^{4} - b^{2} d e^{2}} + \frac {{\left (2 \, d^{2} e^{3} - d e^{4} + b d e^{2}\right )} \sqrt {-2 \, d e + b} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} x}{\sqrt {\frac {b - \sqrt {-4 \, d^{2} e^{2} + b^{2}}}{e^{2}}}}\right )}{4 \, d^{3} e^{4} - 2 \, d^{2} e^{5} + b d e^{4} - b^{2} d e^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.99 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.27 \[ \int \frac {d-e x^2}{d^2+b x^2+e^2 x^4} \, dx=\frac {\mathrm {atan}\left (\frac {b\,x\,\left (b-2\,d\,e\right )+2\,b\,e^2\,x^3+4\,d^2\,e^2\,x-e^2\,x^3\,\left (b-2\,d\,e\right )+3\,d\,e\,x\,\left (b-2\,d\,e\right )}{\left (2\,e\,d^2+b\,d\right )\,\sqrt {b-2\,d\,e}}\right )-\mathrm {atan}\left (\frac {e\,x}{\sqrt {b-2\,d\,e}}\right )}{\sqrt {b-2\,d\,e}} \]

[In]

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

[Out]

(atan((b*x*(b - 2*d*e) + 2*b*e^2*x^3 + 4*d^2*e^2*x - e^2*x^3*(b - 2*d*e) + 3*d*e*x*(b - 2*d*e))/((b*d + 2*d^2*
e)*(b - 2*d*e)^(1/2))) - atan((e*x)/(b - 2*d*e)^(1/2)))/(b - 2*d*e)^(1/2)