3.1.21 \(\int \frac {d-e x^2}{d^2-f x^2+e^2 x^4} \, dx\)
Optimal. Leaf size=70 \[ \frac {\log \left (x \sqrt {2 d e+f}+d+e x^2\right )}{2 \sqrt {2 d e+f}}-\frac {\log \left (-x \sqrt {2 d e+f}+d+e x^2\right )}{2 \sqrt {2 d e+f}} \]
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Rubi [A] time = 0.05, antiderivative size = 70, normalized size of antiderivative = 1.00,
number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used =
{1164, 628} \begin {gather*} \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}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(d - e*x^2)/(d^2 - f*x^2 + e^2*x^4),x]
[Out]
-Log[d - Sqrt[2*d*e + f]*x + e*x^2]/(2*Sqrt[2*d*e + f]) + Log[d + Sqrt[2*d*e + f]*x + e*x^2]/(2*Sqrt[2*d*e + f
])
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 1164
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*} \int \frac {d-e x^2}{d^2-f x^2+e^2 x^4} \, dx &=-\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*}
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Mathematica [B] time = 0.13, size = 190, normalized size = 2.71 \begin {gather*} \frac {\frac {\left (-\sqrt {f^2-4 d^2 e^2}-2 d e+f\right ) \tan ^{-1}\left (\frac {\sqrt {2} e x}{\sqrt {\sqrt {f^2-4 d^2 e^2}-f}}\right )}{\sqrt {\sqrt {f^2-4 d^2 e^2}-f}}-\frac {\left (\sqrt {f^2-4 d^2 e^2}-2 d e+f\right ) \tan ^{-1}\left (\frac {\sqrt {2} e x}{\sqrt {-\sqrt {f^2-4 d^2 e^2}-f}}\right )}{\sqrt {-\sqrt {f^2-4 d^2 e^2}-f}}}{\sqrt {2} \sqrt {f^2-4 d^2 e^2}} \end {gather*}
Antiderivative was successfully verified.
[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])
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IntegrateAlgebraic [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {d-e x^2}{d^2-f x^2+e^2 x^4} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
IntegrateAlgebraic[(d - e*x^2)/(d^2 - f*x^2 + e^2*x^4),x]
[Out]
IntegrateAlgebraic[(d - e*x^2)/(d^2 - f*x^2 + e^2*x^4), x]
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fricas [A] time = 0.82, size = 168, normalized size = 2.40 \begin {gather*} \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 ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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)]
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giac [B] time = 1.08, size = 1676, normalized size = 23.94
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-e*x^2+d)/(e^2*x^4-f*x^2+d^2),x, algorithm="giac")
[Out]
1/4*(16*sqrt(2)*sqrt(-f*e^2 - sqrt(-4*d^2*e^2 + f^2)*e^2)*d^4*e^4 - 8*sqrt(2)*sqrt(-f*e^2 - sqrt(-4*d^2*e^2 +
f^2)*e^2)*d^2*f^2*e^2 + 4*sqrt(2)*sqrt(-4*d^2*e^2 + f^2)*sqrt(-f*e^2 - sqrt(-4*d^2*e^2 + f^2)*e^2)*d^2*f*e^2 +
sqrt(2)*sqrt(-f*e^2 - sqrt(-4*d^2*e^2 + f^2)*e^2)*f^4 + 32*d^4*e^6 - 8*sqrt(2)*sqrt(-f*e^2 - sqrt(-4*d^2*e^2
+ f^2)*e^2)*d^2*f*e^4 - 16*d^2*f^2*e^4 + 2*sqrt(2)*sqrt(-f*e^2 - sqrt(-4*d^2*e^2 + f^2)*e^2)*f^3*e^2 + 2*f^4*e
^2 - sqrt(2)*sqrt(-4*d^2*e^2 + f^2)*sqrt(-f*e^2 - sqrt(-4*d^2*e^2 + f^2)*e^2)*f^3 - 2*sqrt(2)*sqrt(-4*d^2*e^2
+ f^2)*sqrt(-f*e^2 - sqrt(-4*d^2*e^2 + f^2)*e^2)*f^2*e^2 - 4*sqrt(2)*sqrt(-f*e^2 - sqrt(-4*d^2*e^2 + f^2)*e^2)
*d^2*e^6 - 8*d^2*f*e^6 - 8*(4*d^2*e^2 - f^2)*d^2*e^4 + sqrt(2)*sqrt(-f*e^2 - sqrt(-4*d^2*e^2 + f^2)*e^2)*f^2*e
^4 + 2*f^3*e^4 + 2*(4*d^2*e^2 - f^2)*f^2*e^2 - sqrt(2)*sqrt(-4*d^2*e^2 + f^2)*sqrt(-f*e^2 - sqrt(-4*d^2*e^2 +
f^2)*e^2)*f*e^4 + 2*(4*d^2*e^2 - f^2)*f*e^4 - 2*(4*sqrt(2)*sqrt(-4*d^2*e^2 + f^2)*sqrt(-f*e^2 - sqrt(-4*d^2*e^
2 + f^2)*e^2)*d^3*e^2 - sqrt(2)*sqrt(-4*d^2*e^2 + f^2)*sqrt(-f*e^2 - sqrt(-4*d^2*e^2 + f^2)*e^2)*d*f^2 - 2*sqr
t(2)*sqrt(-4*d^2*e^2 + f^2)*sqrt(-f*e^2 - sqrt(-4*d^2*e^2 + f^2)*e^2)*d*f*e^2 - 8*d^3*e^6 + 2*d*f^2*e^4 - sqrt
(2)*sqrt(-4*d^2*e^2 + f^2)*sqrt(-f*e^2 - sqrt(-4*d^2*e^2 + f^2)*e^2)*d*e^4 + 2*(4*d^2*e^2 - f^2)*d*e^4)*e)*arc
tan(2*sqrt(1/2)*x/sqrt(-(f + sqrt(-4*d^2*e^2 + f^2))*e^(-2)))/(16*d^5*e^6 - 8*d^3*f^2*e^4 + d*f^4*e^2 - 8*d^3*
f*e^6 + 2*d*f^3*e^4 - 4*d^3*e^8 + d*f^2*e^6) + 1/4*(16*sqrt(2)*sqrt(-f*e^2 + sqrt(-4*d^2*e^2 + f^2)*e^2)*d^4*e
^4 - 8*sqrt(2)*sqrt(-f*e^2 + sqrt(-4*d^2*e^2 + f^2)*e^2)*d^2*f^2*e^2 - 4*sqrt(2)*sqrt(-4*d^2*e^2 + f^2)*sqrt(-
f*e^2 + sqrt(-4*d^2*e^2 + f^2)*e^2)*d^2*f*e^2 + sqrt(2)*sqrt(-f*e^2 + sqrt(-4*d^2*e^2 + f^2)*e^2)*f^4 - 32*d^4
*e^6 - 8*sqrt(2)*sqrt(-f*e^2 + sqrt(-4*d^2*e^2 + f^2)*e^2)*d^2*f*e^4 + 16*d^2*f^2*e^4 + 2*sqrt(2)*sqrt(-f*e^2
+ sqrt(-4*d^2*e^2 + f^2)*e^2)*f^3*e^2 - 2*f^4*e^2 + sqrt(2)*sqrt(-4*d^2*e^2 + f^2)*sqrt(-f*e^2 + sqrt(-4*d^2*e
^2 + f^2)*e^2)*f^3 + 2*sqrt(2)*sqrt(-4*d^2*e^2 + f^2)*sqrt(-f*e^2 + sqrt(-4*d^2*e^2 + f^2)*e^2)*f^2*e^2 - 4*sq
rt(2)*sqrt(-f*e^2 + sqrt(-4*d^2*e^2 + f^2)*e^2)*d^2*e^6 + 8*d^2*f*e^6 + 8*(4*d^2*e^2 - f^2)*d^2*e^4 + sqrt(2)*
sqrt(-f*e^2 + sqrt(-4*d^2*e^2 + f^2)*e^2)*f^2*e^4 - 2*f^3*e^4 - 2*(4*d^2*e^2 - f^2)*f^2*e^2 + sqrt(2)*sqrt(-4*
d^2*e^2 + f^2)*sqrt(-f*e^2 + sqrt(-4*d^2*e^2 + f^2)*e^2)*f*e^4 - 2*(4*d^2*e^2 - f^2)*f*e^4 + 2*(4*sqrt(2)*sqrt
(-4*d^2*e^2 + f^2)*sqrt(-f*e^2 + sqrt(-4*d^2*e^2 + f^2)*e^2)*d^3*e^2 - sqrt(2)*sqrt(-4*d^2*e^2 + f^2)*sqrt(-f*
e^2 + sqrt(-4*d^2*e^2 + f^2)*e^2)*d*f^2 - 2*sqrt(2)*sqrt(-4*d^2*e^2 + f^2)*sqrt(-f*e^2 + sqrt(-4*d^2*e^2 + f^2
)*e^2)*d*f*e^2 - 8*d^3*e^6 + 2*d*f^2*e^4 - sqrt(2)*sqrt(-4*d^2*e^2 + f^2)*sqrt(-f*e^2 + sqrt(-4*d^2*e^2 + f^2)
*e^2)*d*e^4 + 2*(4*d^2*e^2 - f^2)*d*e^4)*e)*arctan(2*sqrt(1/2)*x/sqrt(-(f - sqrt(-4*d^2*e^2 + f^2))*e^(-2)))/(
16*d^5*e^6 - 8*d^3*f^2*e^4 + d*f^4*e^2 - 8*d^3*f*e^6 + 2*d*f^3*e^4 - 4*d^3*e^8 + d*f^2*e^6)
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maple [A] time = 0.02, size = 61, normalized size = 0.87 \begin {gather*} \frac {\ln \left (e \,x^{2}+d +\sqrt {2 d e +f}\, x \right )}{2 \sqrt {2 d e +f}}-\frac {\ln \left (-e \,x^{2}-d +\sqrt {2 d e +f}\, x \right )}{2 \sqrt {2 d e +f}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((-e*x^2+d)/(e^2*x^4-f*x^2+d^2),x)
[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)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {e x^{2} - d}{e^{2} x^{4} - f x^{2} + d^{2}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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mupad [B] time = 0.11, size = 29, normalized size = 0.41 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {x\,\sqrt {f+2\,d\,e}}{e\,x^2+d}\right )}{\sqrt {f+2\,d\,e}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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sympy [A] time = 0.61, size = 112, normalized size = 1.60 \begin {gather*} - \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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
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