3.30 \(\int \frac {d-e x^2}{d^2+b x^2+e^2 x^4} \, dx\)
Optimal. Leaf size=78 \[ \frac {\log \left (x \sqrt {2 d e-b}+d+e x^2\right )}{2 \sqrt {2 d e-b}}-\frac {\log \left (-x \sqrt {2 d e-b}+d+e x^2\right )}{2 \sqrt {2 d e-b}} \]
[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)
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Rubi [A] time = 0.05, antiderivative size = 78, normalized size of antiderivative = 1.00,
number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used =
{1164, 628} \[ \frac {\log \left (x \sqrt {2 d e-b}+d+e x^2\right )}{2 \sqrt {2 d e-b}}-\frac {\log \left (-x \sqrt {2 d e-b}+d+e x^2\right )}{2 \sqrt {2 d e-b}} \]
Antiderivative was successfully verified.
[In]
Int[(d - e*x^2)/(d^2 + b*x^2 + e^2*x^4),x]
[Out]
-Log[d - Sqrt[-b + 2*d*e]*x + e*x^2]/(2*Sqrt[-b + 2*d*e]) + Log[d + Sqrt[-b + 2*d*e]*x + e*x^2]/(2*Sqrt[-b + 2
*d*e])
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+b x^2+e^2 x^4} \, dx &=-\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*}
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Mathematica [B] time = 0.12, size = 182, normalized size = 2.33 \[ \frac {\frac {\left (-\sqrt {b^2-4 d^2 e^2}+b+2 d e\right ) \tan ^{-1}\left (\frac {\sqrt {2} e x}{\sqrt {b-\sqrt {b^2-4 d^2 e^2}}}\right )}{\sqrt {b-\sqrt {b^2-4 d^2 e^2}}}-\frac {\left (\sqrt {b^2-4 d^2 e^2}+b+2 d e\right ) \tan ^{-1}\left (\frac {\sqrt {2} e x}{\sqrt {\sqrt {b^2-4 d^2 e^2}+b}}\right )}{\sqrt {\sqrt {b^2-4 d^2 e^2}+b}}}{\sqrt {2} \sqrt {b^2-4 d^2 e^2}} \]
Antiderivative was successfully verified.
[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])
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fricas [A] time = 0.40, size = 172, normalized size = 2.21 \[ \left [\frac {\log \left (\frac {e^{2} x^{4} + {\left (4 \, d e - b\right )} x^{2} + d^{2} + 2 \, {\left (e x^{3} + d x\right )} \sqrt {2 \, d e - b}}{e^{2} x^{4} + b x^{2} + d^{2}}\right )}{2 \, \sqrt {2 \, d e - b}}, -\frac {\sqrt {-2 \, d e + b} \arctan \left (\frac {\sqrt {-2 \, d e + b} e x}{2 \, d e - b}\right ) - \sqrt {-2 \, d e + b} \arctan \left (\frac {{\left (e^{2} x^{3} - {\left (d e - b\right )} x\right )} \sqrt {-2 \, d e + b}}{2 \, d^{2} e - b d}\right )}{2 \, d e - b}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[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)]
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giac [B] time = 1.16, size = 1642, normalized size = 21.05 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-e*x^2+d)/(e^2*x^4+b*x^2+d^2),x, algorithm="giac")
[Out]
1/4*(16*sqrt(2)*sqrt(b*e^2 + sqrt(-4*d^2*e^2 + b^2)*e^2)*d^4*e^4 - 8*sqrt(2)*sqrt(b*e^2 + sqrt(-4*d^2*e^2 + b^
2)*e^2)*b^2*d^2*e^2 + 4*sqrt(2)*sqrt(-4*d^2*e^2 + b^2)*sqrt(b*e^2 + sqrt(-4*d^2*e^2 + b^2)*e^2)*b*d^2*e^2 + sq
rt(2)*sqrt(b*e^2 + sqrt(-4*d^2*e^2 + b^2)*e^2)*b^4 - 32*d^4*e^6 + 8*sqrt(2)*sqrt(b*e^2 + sqrt(-4*d^2*e^2 + b^2
)*e^2)*b*d^2*e^4 + 16*b^2*d^2*e^4 - 2*sqrt(2)*sqrt(b*e^2 + sqrt(-4*d^2*e^2 + b^2)*e^2)*b^3*e^2 - 2*b^4*e^2 - s
qrt(2)*sqrt(-4*d^2*e^2 + b^2)*sqrt(b*e^2 + sqrt(-4*d^2*e^2 + b^2)*e^2)*b^3 + 2*sqrt(2)*sqrt(-4*d^2*e^2 + b^2)*
sqrt(b*e^2 + sqrt(-4*d^2*e^2 + b^2)*e^2)*b^2*e^2 - 4*sqrt(2)*sqrt(b*e^2 + sqrt(-4*d^2*e^2 + b^2)*e^2)*d^2*e^6
- 8*b*d^2*e^6 + sqrt(2)*sqrt(b*e^2 + sqrt(-4*d^2*e^2 + b^2)*e^2)*b^2*e^4 + 2*b^3*e^4 + 8*(4*d^2*e^2 - b^2)*d^2
*e^4 - 2*(4*d^2*e^2 - b^2)*b^2*e^2 - sqrt(2)*sqrt(-4*d^2*e^2 + b^2)*sqrt(b*e^2 + sqrt(-4*d^2*e^2 + b^2)*e^2)*b
*e^4 + 2*(4*d^2*e^2 - b^2)*b*e^4 + 2*(4*sqrt(2)*sqrt(-4*d^2*e^2 + b^2)*sqrt(b*e^2 + sqrt(-4*d^2*e^2 + b^2)*e^2
)*d^3*e^2 - sqrt(2)*sqrt(-4*d^2*e^2 + b^2)*sqrt(b*e^2 + sqrt(-4*d^2*e^2 + b^2)*e^2)*b^2*d + 2*sqrt(2)*sqrt(-4*
d^2*e^2 + b^2)*sqrt(b*e^2 + sqrt(-4*d^2*e^2 + b^2)*e^2)*b*d*e^2 - 8*d^3*e^6 + 2*b^2*d*e^4 - sqrt(2)*sqrt(-4*d^
2*e^2 + b^2)*sqrt(b*e^2 + sqrt(-4*d^2*e^2 + b^2)*e^2)*d*e^4 + 2*(4*d^2*e^2 - b^2)*d*e^4)*e)*arctan(2*sqrt(1/2)
*x*e/sqrt(b + sqrt(-4*d^2*e^2 + b^2)))/(16*d^5*e^6 - 8*b^2*d^3*e^4 + b^4*d*e^2 + 8*b*d^3*e^6 - 2*b^3*d*e^4 - 4
*d^3*e^8 + b^2*d*e^6) + 1/4*(16*sqrt(2)*sqrt(b*e^2 - sqrt(-4*d^2*e^2 + b^2)*e^2)*d^4*e^4 - 8*sqrt(2)*sqrt(b*e^
2 - sqrt(-4*d^2*e^2 + b^2)*e^2)*b^2*d^2*e^2 - 4*sqrt(2)*sqrt(-4*d^2*e^2 + b^2)*sqrt(b*e^2 - sqrt(-4*d^2*e^2 +
b^2)*e^2)*b*d^2*e^2 + sqrt(2)*sqrt(b*e^2 - sqrt(-4*d^2*e^2 + b^2)*e^2)*b^4 + 32*d^4*e^6 + 8*sqrt(2)*sqrt(b*e^2
- sqrt(-4*d^2*e^2 + b^2)*e^2)*b*d^2*e^4 - 16*b^2*d^2*e^4 - 2*sqrt(2)*sqrt(b*e^2 - sqrt(-4*d^2*e^2 + b^2)*e^2)
*b^3*e^2 + 2*b^4*e^2 + sqrt(2)*sqrt(-4*d^2*e^2 + b^2)*sqrt(b*e^2 - sqrt(-4*d^2*e^2 + b^2)*e^2)*b^3 - 2*sqrt(2)
*sqrt(-4*d^2*e^2 + b^2)*sqrt(b*e^2 - sqrt(-4*d^2*e^2 + b^2)*e^2)*b^2*e^2 - 4*sqrt(2)*sqrt(b*e^2 - sqrt(-4*d^2*
e^2 + b^2)*e^2)*d^2*e^6 + 8*b*d^2*e^6 + sqrt(2)*sqrt(b*e^2 - sqrt(-4*d^2*e^2 + b^2)*e^2)*b^2*e^4 - 2*b^3*e^4 -
8*(4*d^2*e^2 - b^2)*d^2*e^4 + 2*(4*d^2*e^2 - b^2)*b^2*e^2 + sqrt(2)*sqrt(-4*d^2*e^2 + b^2)*sqrt(b*e^2 - sqrt(
-4*d^2*e^2 + b^2)*e^2)*b*e^4 - 2*(4*d^2*e^2 - b^2)*b*e^4 - 2*(4*sqrt(2)*sqrt(-4*d^2*e^2 + b^2)*sqrt(b*e^2 - sq
rt(-4*d^2*e^2 + b^2)*e^2)*d^3*e^2 - sqrt(2)*sqrt(-4*d^2*e^2 + b^2)*sqrt(b*e^2 - sqrt(-4*d^2*e^2 + b^2)*e^2)*b^
2*d + 2*sqrt(2)*sqrt(-4*d^2*e^2 + b^2)*sqrt(b*e^2 - sqrt(-4*d^2*e^2 + b^2)*e^2)*b*d*e^2 - 8*d^3*e^6 + 2*b^2*d*
e^4 - sqrt(2)*sqrt(-4*d^2*e^2 + b^2)*sqrt(b*e^2 - sqrt(-4*d^2*e^2 + b^2)*e^2)*d*e^4 + 2*(4*d^2*e^2 - b^2)*d*e^
4)*e)*arctan(2*sqrt(1/2)*x*e/sqrt(b - sqrt(-4*d^2*e^2 + b^2)))/(16*d^5*e^6 - 8*b^2*d^3*e^4 + b^4*d*e^2 + 8*b*d
^3*e^6 - 2*b^3*d*e^4 - 4*d^3*e^8 + b^2*d*e^6)
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maple [A] time = 0.02, size = 88, normalized size = 1.13 \[ -\frac {\sqrt {2 d e -b}\, \ln \left (e \,x^{2}+d +\sqrt {2 d e -b}\, x \right )}{-4 d e +2 b}+\frac {\sqrt {2 d e -b}\, \ln \left (-e \,x^{2}-d +\sqrt {2 d e -b}\, x \right )}{-4 d e +2 b} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((-e*x^2+d)/(e^2*x^4+b*x^2+d^2),x)
[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))
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {e x^{2} - d}{e^{2} x^{4} + b x^{2} + d^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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mupad [B] time = 0.09, size = 99, normalized size = 1.27 \[ \frac {\mathrm {atan}\left (\frac {b\,x\,\left (b-2\,d\,e\right )+2\,b\,e^2\,x^3+4\,d^2\,e^2\,x-e^2\,x^3\,\left (b-2\,d\,e\right )+3\,d\,e\,x\,\left (b-2\,d\,e\right )}{\left (2\,e\,d^2+b\,d\right )\,\sqrt {b-2\,d\,e}}\right )-\mathrm {atan}\left (\frac {e\,x}{\sqrt {b-2\,d\,e}}\right )}{\sqrt {b-2\,d\,e}} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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sympy [A] time = 0.58, size = 121, normalized size = 1.55 \[ \frac {\sqrt {- \frac {1}{b - 2 d e}} \log {\left (\frac {d}{e} + x^{2} + \frac {x \left (- b \sqrt {- \frac {1}{b - 2 d e}} + 2 d e \sqrt {- \frac {1}{b - 2 d e}}\right )}{e} \right )}}{2} - \frac {\sqrt {- \frac {1}{b - 2 d e}} \log {\left (\frac {d}{e} + x^{2} + \frac {x \left (b \sqrt {- \frac {1}{b - 2 d e}} - 2 d e \sqrt {- \frac {1}{b - 2 d e}}\right )}{e} \right )}}{2} \]
Verification of antiderivative is not currently implemented for this CAS.
[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
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