3.29 \(\int \frac {d+e x^2}{d^2-f x^2+e^2 x^4} \, dx\)
Optimal. Leaf size=86 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {2 d e+f}+2 e x}{\sqrt {2 d e-f}}\right )}{\sqrt {2 d e-f}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2 d e+f}-2 e x}{\sqrt {2 d e-f}}\right )}{\sqrt {2 d e-f}} \]
[Out]
-arctan((-2*e*x+(2*d*e+f)^(1/2))/(2*d*e-f)^(1/2))/(2*d*e-f)^(1/2)+arctan((2*e*x+(2*d*e+f)^(1/2))/(2*d*e-f)^(1/
2))/(2*d*e-f)^(1/2)
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Rubi [A] time = 0.10, antiderivative size = 86, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used =
{1161, 618, 204} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {2 d e+f}+2 e x}{\sqrt {2 d e-f}}\right )}{\sqrt {2 d e-f}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2 d e+f}-2 e x}{\sqrt {2 d e-f}}\right )}{\sqrt {2 d e-f}} \]
Antiderivative was successfully verified.
[In]
Int[(d + e*x^2)/(d^2 - f*x^2 + e^2*x^4),x]
[Out]
-(ArcTan[(Sqrt[2*d*e + f] - 2*e*x)/Sqrt[2*d*e - f]]/Sqrt[2*d*e - f]) + ArcTan[(Sqrt[2*d*e + f] + 2*e*x)/Sqrt[2
*d*e - f]]/Sqrt[2*d*e - f]
Rule 204
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])
Rule 618
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]
Rule 1161
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), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/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[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))
Rubi steps
\begin {align*} \int \frac {d+e x^2}{d^2-f x^2+e^2 x^4} \, dx &=\frac {\int \frac {1}{\frac {d}{e}-\frac {\sqrt {2 d e+f} x}{e}+x^2} \, dx}{2 e}+\frac {\int \frac {1}{\frac {d}{e}+\frac {\sqrt {2 d e+f} x}{e}+x^2} \, dx}{2 e}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {1}{-\frac {2 d e-f}{e^2}-x^2} \, dx,x,-\frac {\sqrt {2 d e+f}}{e}+2 x\right )}{e}-\frac {\operatorname {Subst}\left (\int \frac {1}{-\frac {2 d e-f}{e^2}-x^2} \, dx,x,\frac {\sqrt {2 d e+f}}{e}+2 x\right )}{e}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {2 d e+f}-2 e x}{\sqrt {2 d e-f}}\right )}{\sqrt {2 d e-f}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2 d e+f}+2 e x}{\sqrt {2 d e-f}}\right )}{\sqrt {2 d e-f}}\\ \end {align*}
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Mathematica [B] time = 0.11, size = 189, normalized size = 2.20 \[ \frac {\frac {\left (\sqrt {f^2-4 d^2 e^2}+2 d e+f\right ) \tan ^{-1}\left (\frac {\sqrt {2} e x}{\sqrt {-\sqrt {f^2-4 d^2 e^2}-f}}\right )}{\sqrt {-\sqrt {f^2-4 d^2 e^2}-f}}+\frac {\left (\sqrt {f^2-4 d^2 e^2}-2 d e-f\right ) \tan ^{-1}\left (\frac {\sqrt {2} e x}{\sqrt {\sqrt {f^2-4 d^2 e^2}-f}}\right )}{\sqrt {\sqrt {f^2-4 d^2 e^2}-f}}}{\sqrt {2} \sqrt {f^2-4 d^2 e^2}} \]
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 - Sqrt
[-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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fricas [A] time = 0.42, size = 179, normalized size = 2.08 \[ \left [-\frac {\sqrt {-2 \, d e + f} \log \left (\frac {e^{2} x^{4} - {\left (4 \, d e - f\right )} x^{2} + d^{2} - 2 \, {\left (e x^{3} - d x\right )} \sqrt {-2 \, d e + f}}{e^{2} x^{4} - f x^{2} + d^{2}}\right )}{2 \, {\left (2 \, d e - f\right )}}, -\frac {\sqrt {2 \, d e - f} \arctan \left (-\frac {e x}{\sqrt {2 \, d e - f}}\right ) + \sqrt {2 \, d e - f} \arctan \left (-\frac {{\left (e^{2} x^{3} + {\left (d e - f\right )} x\right )} \sqrt {2 \, d e - f}}{2 \, d^{2} e - d f}\right )}{2 \, d e - f}\right ] \]
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*sqrt(-2*d*e + f)*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))/(2*d*e - f), -(sqrt(2*d*e - f)*arctan(-e*x/sqrt(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 = 1.14, size = 1676, normalized size = 19.49 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x^2+d)/(e^2*x^4-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 = 0.03, size = 75, normalized size = 0.87 \[ -\frac {\arctan \left (\frac {-2 e x +\sqrt {2 d e +f}}{\sqrt {2 d e -f}}\right )}{\sqrt {2 d e -f}}+\frac {\arctan \left (\frac {2 e x +\sqrt {2 d e +f}}{\sqrt {2 d e -f}}\right )}{\sqrt {2 d e -f}} \]
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]
-arctan((-2*e*x+(2*d*e+f)^(1/2))/(2*d*e-f)^(1/2))/(2*d*e-f)^(1/2)+arctan((2*e*x+(2*d*e+f)^(1/2))/(2*d*e-f)^(1/
2))/(2*d*e-f)^(1/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e x^{2} + d}{e^{2} x^{4} - f x^{2} + d^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x^2+d)/(e^2*x^4-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 = 4.39, size = 88, normalized size = 1.02 \[ -\frac {\mathrm {atan}\left (\frac {e^2\,x^3\,\sqrt {2\,d\,e-f}-f\,x\,\sqrt {2\,d\,e-f}+d\,e\,x\,\sqrt {2\,d\,e-f}}{d\,\left (f-2\,d\,e\right )}\right )-\mathrm {atan}\left (\frac {e\,x}{\sqrt {2\,d\,e-f}}\right )}{\sqrt {2\,d\,e-f}} \]
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]
-(atan((e^2*x^3*(2*d*e - f)^(1/2) - f*x*(2*d*e - f)^(1/2) + d*e*x*(2*d*e - f)^(1/2))/(d*(f - 2*d*e))) - atan((
e*x)/(2*d*e - f)^(1/2)))/(2*d*e - f)^(1/2)
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sympy [A] time = 0.55, size = 121, normalized size = 1.41 \[ - \frac {\sqrt {- \frac {1}{2 d e - f}} \log {\left (- \frac {d}{e} + x^{2} + \frac {x \left (- 2 d e \sqrt {- \frac {1}{2 d e - f}} + f \sqrt {- \frac {1}{2 d e - f}}\right )}{e} \right )}}{2} + \frac {\sqrt {- \frac {1}{2 d e - f}} \log {\left (- \frac {d}{e} + x^{2} + \frac {x \left (2 d e \sqrt {- \frac {1}{2 d e - f}} - f \sqrt {- \frac {1}{2 d e - f}}\right )}{e} \right )}}{2} \]
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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