3.1.16 \(\int \frac {d+e x^2}{d^2-b x^2+e^2 x^4} \, dx\)
Optimal. Leaf size=78 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b+2 d e}-2 e x}{\sqrt {b-2 d e}}\right )}{\sqrt {b-2 d e}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b+2 d e}+2 e x}{\sqrt {b-2 d e}}\right )}{\sqrt {b-2 d e}} \]
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Rubi [A] time = 0.10, antiderivative size = 78, 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, 206} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {b+2 d e}-2 e x}{\sqrt {b-2 d e}}\right )}{\sqrt {b-2 d e}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b+2 d e}+2 e x}{\sqrt {b-2 d e}}\right )}{\sqrt {b-2 d e}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(d + e*x^2)/(d^2 - b*x^2 + e^2*x^4),x]
[Out]
ArcTanh[(Sqrt[b + 2*d*e] - 2*e*x)/Sqrt[b - 2*d*e]]/Sqrt[b - 2*d*e] - ArcTanh[(Sqrt[b + 2*d*e] + 2*e*x)/Sqrt[b
- 2*d*e]]/Sqrt[b - 2*d*e]
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[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-b x^2+e^2 x^4} \, dx &=\frac {\int \frac {1}{\frac {d}{e}-\frac {\sqrt {b+2 d e} x}{e}+x^2} \, dx}{2 e}+\frac {\int \frac {1}{\frac {d}{e}+\frac {\sqrt {b+2 d e} x}{e}+x^2} \, dx}{2 e}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {1}{\frac {b-2 d e}{e^2}-x^2} \, dx,x,-\frac {\sqrt {b+2 d e}}{e}+2 x\right )}{e}-\frac {\operatorname {Subst}\left (\int \frac {1}{\frac {b-2 d e}{e^2}-x^2} \, dx,x,\frac {\sqrt {b+2 d e}}{e}+2 x\right )}{e}\\ &=\frac {\tanh ^{-1}\left (\frac {\sqrt {b+2 d e}-2 e x}{\sqrt {b-2 d e}}\right )}{\sqrt {b-2 d e}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b+2 d e}+2 e x}{\sqrt {b-2 d e}}\right )}{\sqrt {b-2 d e}}\\ \end {align*}
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Mathematica [B] time = 0.11, size = 189, normalized size = 2.42 \begin {gather*} \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 {-\sqrt {b^2-4 d^2 e^2}-b}}\right )}{\sqrt {-\sqrt {b^2-4 d^2 e^2}-b}}+\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}} \end {gather*}
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]]
])/Sqrt[-b + Sqrt[b^2 - 4*d^2*e^2]])/(Sqrt[2]*Sqrt[b^2 - 4*d^2*e^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-b x^2+e^2 x^4} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
IntegrateAlgebraic[(d + e*x^2)/(d^2 - b*x^2 + e^2*x^4),x]
[Out]
IntegrateAlgebraic[(d + e*x^2)/(d^2 - b*x^2 + e^2*x^4), x]
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fricas [A] time = 0.84, size = 176, normalized size = 2.26 \begin {gather*} \left [-\frac {\sqrt {-2 \, d e + b} \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 \, {\left (2 \, d e - b\right )}}, \frac {\sqrt {2 \, d e - b} \arctan \left (\frac {e x}{\sqrt {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 ] \end {gather*}
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*sqrt(-2*d*e + b)*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))/(2*d*e - b), (sqrt(2*d*e - b)*arctan(e*x/sqrt(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 = 1.12, size = 1676, normalized size = 21.49
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 +
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 - 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*sqr
t(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)*arc
tan(2*sqrt(1/2)*x/sqrt(-(b + sqrt(-4*d^2*e^2 + b^2))*e^(-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*sq
rt(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/sqrt(-(b - sqrt(-4*d^2*e^2 + b^2))*e^(-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 = 0.03, size = 75, normalized size = 0.96 \begin {gather*} -\frac {\arctan \left (\frac {-2 e x +\sqrt {2 d e +b}}{\sqrt {2 d e -b}}\right )}{\sqrt {2 d e -b}}+\frac {\arctan \left (\frac {2 e x +\sqrt {2 d e +b}}{\sqrt {2 d e -b}}\right )}{\sqrt {2 d e -b}} \end {gather*}
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/(2*d*e-b)^(1/2)*arctan((-2*e*x+(2*d*e+b)^(1/2))/(2*d*e-b)^(1/2))+1/(2*d*e-b)^(1/2)*arctan((2*e*x+(2*d*e+b)^
(1/2))/(2*d*e-b)^(1/2))
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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} - b 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-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 = 0.13, size = 30, normalized size = 0.38 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {x\,\sqrt {b-2\,d\,e}}{d-e\,x^2}\right )}{\sqrt {b-2\,d\,e}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d + e*x^2)/(d^2 - b*x^2 + e^2*x^4),x)
[Out]
atanh((x*(b - 2*d*e)^(1/2))/(d - e*x^2))/(b - 2*d*e)^(1/2)
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sympy [A] time = 0.57, size = 110, normalized size = 1.41 \begin {gather*} \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} \end {gather*}
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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