\(\int \frac {d-e x^2}{d^2+e^2 x^4} \, dx\) [12]

Optimal result

Integrand size = 22, antiderivative size = 90 \[ \int \frac {d-e x^2}{d^2+e^2 x^4} \, dx=-\frac {\log \left (d-\sqrt {2} \sqrt {d} \sqrt {e} x+e x^2\right )}{2 \sqrt {2} \sqrt {d} \sqrt {e}}+\frac {\log \left (d+\sqrt {2} \sqrt {d} \sqrt {e} x+e x^2\right )}{2 \sqrt {2} \sqrt {d} \sqrt {e}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1179, 642} \[ \int \frac {d-e x^2}{d^2+e^2 x^4} \, dx=\frac {\log \left (\sqrt {2} \sqrt {d} \sqrt {e} x+d+e x^2\right )}{2 \sqrt {2} \sqrt {d} \sqrt {e}}-\frac {\log \left (-\sqrt {2} \sqrt {d} \sqrt {e} x+d+e x^2\right )}{2 \sqrt {2} \sqrt {d} \sqrt {e}} \]

[In]

Int[(d - e*x^2)/(d^2 + e^2*x^4),x]

[Out]

-1/2*Log[d - Sqrt[2]*Sqrt[d]*Sqrt[e]*x + e*x^2]/(Sqrt[2]*Sqrt[d]*Sqrt[e]) + Log[d + Sqrt[2]*Sqrt[d]*Sqrt[e]*x
+ e*x^2]/(2*Sqrt[2]*Sqrt[d]*Sqrt[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 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 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, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.83 \[ \int \frac {d-e x^2}{d^2+e^2 x^4} \, dx=\frac {-\log \left (-d+\sqrt {2} \sqrt {d} \sqrt {e} x-e x^2\right )+\log \left (d+\sqrt {2} \sqrt {d} \sqrt {e} x+e x^2\right )}{2 \sqrt {2} \sqrt {d} \sqrt {e}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.83

method	result	size
risch	\(\frac {\sqrt {2}\, \ln \left (e \,x^{2} \sqrt {e d}+d e x \sqrt {2}+d \sqrt {e d}\right )}{4 \sqrt {e d}}-\frac {\sqrt {2}\, \ln \left (e \,x^{2} \sqrt {e d}-d e x \sqrt {2}+d \sqrt {e d}\right )}{4 \sqrt {e d}}\)	\(75\)
default	\(\frac {\left (\frac {d^{2}}{e^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {d^{2}}{e^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {d^{2}}{e^{2}}}}{x^{2}-\left (\frac {d^{2}}{e^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {d^{2}}{e^{2}}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {d^{2}}{e^{2}}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {d^{2}}{e^{2}}\right )^{\frac {1}{4}}}-1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {d^{2}}{e^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {d^{2}}{e^{2}}}}{x^{2}+\left (\frac {d^{2}}{e^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {d^{2}}{e^{2}}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {d^{2}}{e^{2}}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {d^{2}}{e^{2}}\right )^{\frac {1}{4}}}-1\right )\right )}{8 e \left (\frac {d^{2}}{e^{2}}\right )^{\frac {1}{4}}}\)	\(232\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.56 \[ \int \frac {d-e x^2}{d^2+e^2 x^4} \, dx=\left [\frac {\sqrt {2} \sqrt {d e} \log \left (\frac {e^{2} x^{4} + 4 \, d e x^{2} + 2 \, \sqrt {2} {\left (e x^{3} + d x\right )} \sqrt {d e} + d^{2}}{e^{2} x^{4} + d^{2}}\right )}{4 \, d e}, -\frac {\sqrt {2} \sqrt {-d e} \arctan \left (\frac {\sqrt {2} \sqrt {-d e} x}{2 \, d}\right ) - \sqrt {2} \sqrt {-d e} \arctan \left (\frac {\sqrt {2} {\left (e x^{3} - d x\right )} \sqrt {-d e}}{2 \, d^{2}}\right )}{2 \, d e}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.89 \[ \int \frac {d-e x^2}{d^2+e^2 x^4} \, dx=- \frac {\sqrt {2} \sqrt {\frac {1}{d e}} \log {\left (- \sqrt {2} d x \sqrt {\frac {1}{d e}} + \frac {d}{e} + x^{2} \right )}}{4} + \frac {\sqrt {2} \sqrt {\frac {1}{d e}} \log {\left (\sqrt {2} d x \sqrt {\frac {1}{d e}} + \frac {d}{e} + x^{2} \right )}}{4} \]

[In]

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

[Out]

-sqrt(2)*sqrt(1/(d*e))*log(-sqrt(2)*d*x*sqrt(1/(d*e)) + d/e + x**2)/4 + sqrt(2)*sqrt(1/(d*e))*log(sqrt(2)*d*x*
sqrt(1/(d*e)) + d/e + x**2)/4

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 302 vs. \(2 (62) = 124\).

Time = 0.29 (sec) , antiderivative size = 302, normalized size of antiderivative = 3.36 \[ \int \frac {d-e x^2}{d^2+e^2 x^4} \, dx=-\frac {\sqrt {2} {\left (e - \sqrt {e^{2}}\right )} \log \left (\frac {2 \, \sqrt {e^{2}} x + \sqrt {2} \sqrt {d} {\left (e^{2}\right )}^{\frac {1}{4}} - \sqrt {2} \sqrt {-d \sqrt {e^{2}}}}{2 \, \sqrt {e^{2}} x + \sqrt {2} \sqrt {d} {\left (e^{2}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-d \sqrt {e^{2}}}}\right )}{8 \, \sqrt {e^{2}} \sqrt {-d \sqrt {e^{2}}}} - \frac {\sqrt {2} {\left (e - \sqrt {e^{2}}\right )} \log \left (\frac {2 \, \sqrt {e^{2}} x - \sqrt {2} \sqrt {d} {\left (e^{2}\right )}^{\frac {1}{4}} - \sqrt {2} \sqrt {-d \sqrt {e^{2}}}}{2 \, \sqrt {e^{2}} x - \sqrt {2} \sqrt {d} {\left (e^{2}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-d \sqrt {e^{2}}}}\right )}{8 \, \sqrt {e^{2}} \sqrt {-d \sqrt {e^{2}}}} + \frac {\sqrt {2} {\left (e + \sqrt {e^{2}}\right )} \log \left (\sqrt {e^{2}} x^{2} + \sqrt {2} \sqrt {d} {\left (e^{2}\right )}^{\frac {1}{4}} x + d\right )}{8 \, \sqrt {d} {\left (e^{2}\right )}^{\frac {3}{4}}} - \frac {\sqrt {2} {\left (e + \sqrt {e^{2}}\right )} \log \left (\sqrt {e^{2}} x^{2} - \sqrt {2} \sqrt {d} {\left (e^{2}\right )}^{\frac {1}{4}} x + d\right )}{8 \, \sqrt {d} {\left (e^{2}\right )}^{\frac {3}{4}}} \]

[In]

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

[Out]

-1/8*sqrt(2)*(e - sqrt(e^2))*log((2*sqrt(e^2)*x + sqrt(2)*sqrt(d)*(e^2)^(1/4) - sqrt(2)*sqrt(-d*sqrt(e^2)))/(2
*sqrt(e^2)*x + sqrt(2)*sqrt(d)*(e^2)^(1/4) + sqrt(2)*sqrt(-d*sqrt(e^2))))/(sqrt(e^2)*sqrt(-d*sqrt(e^2))) - 1/8
*sqrt(2)*(e - sqrt(e^2))*log((2*sqrt(e^2)*x - sqrt(2)*sqrt(d)*(e^2)^(1/4) - sqrt(2)*sqrt(-d*sqrt(e^2)))/(2*sqr
t(e^2)*x - sqrt(2)*sqrt(d)*(e^2)^(1/4) + sqrt(2)*sqrt(-d*sqrt(e^2))))/(sqrt(e^2)*sqrt(-d*sqrt(e^2))) + 1/8*sqr
t(2)*(e + sqrt(e^2))*log(sqrt(e^2)*x^2 + sqrt(2)*sqrt(d)*(e^2)^(1/4)*x + d)/(sqrt(d)*(e^2)^(3/4)) - 1/8*sqrt(2
)*(e + sqrt(e^2))*log(sqrt(e^2)*x^2 - sqrt(2)*sqrt(d)*(e^2)^(1/4)*x + d)/(sqrt(d)*(e^2)^(3/4))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.11 \[ \int \frac {d-e x^2}{d^2+e^2 x^4} \, dx=\frac {\sqrt {2} \sqrt {-d e} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {d^{2}}{e^{2}}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {d^{2}}{e^{2}}\right )^{\frac {1}{4}}}\right )}{2 \, d e} + \frac {\sqrt {2} \sqrt {-d e} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {d^{2}}{e^{2}}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {d^{2}}{e^{2}}\right )^{\frac {1}{4}}}\right )}{2 \, d e} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.57 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.46 \[ \int \frac {d-e x^2}{d^2+e^2 x^4} \, dx=\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {2\,\sqrt {2}\,\sqrt {d}\,e^{7/2}\,x}{2\,e^4\,x^2+2\,d\,e^3}\right )}{2\,\sqrt {d}\,\sqrt {e}} \]

[In]

int((d - e*x^2)/(d^2 + e^2*x^4),x)

[Out]

(2^(1/2)*atanh((2*2^(1/2)*d^(1/2)*e^(7/2)*x)/(2*d*e^3 + 2*e^4*x^2)))/(2*d^(1/2)*e^(1/2))