3.26 \(\int \frac {d+e x^2}{d^2+b x^2+e^2 x^4} \, dx\)

Optimal. Leaf size=82 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {2 d e-b}+2 e x}{\sqrt {b+2 d e}}\right )}{\sqrt {b+2 d e}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2 d e-b}-2 e x}{\sqrt {b+2 d e}}\right )}{\sqrt {b+2 d e}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1161, 618, 204} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {2 d e-b}+2 e x}{\sqrt {b+2 d e}}\right )}{\sqrt {b+2 d e}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2 d e-b}-2 e x}{\sqrt {b+2 d e}}\right )}{\sqrt {b+2 d e}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)/(d^2 + b*x^2 + e^2*x^4),x]

[Out]

-(ArcTan[(Sqrt[-b + 2*d*e] - 2*e*x)/Sqrt[b + 2*d*e]]/Sqrt[b + 2*d*e]) + ArcTan[(Sqrt[-b + 2*d*e] + 2*e*x)/Sqrt
[b + 2*d*e]]/Sqrt[b + 2*d*e]

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+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 {\tan ^{-1}\left (\frac {\sqrt {-b+2 d e}-2 e x}{\sqrt {b+2 d e}}\right )}{\sqrt {b+2 d e}}+\frac {\tan ^{-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 = 181, normalized size = 2.21 \[ \frac {\frac {\left (\sqrt {b^2-4 d^2 e^2}-b+2 d e\right ) \tan ^{-1}\left (\frac {\sqrt {2} e x}{\sqrt {b-\sqrt {b^2-4 d^2 e^2}}}\right )}{\sqrt {b-\sqrt {b^2-4 d^2 e^2}}}+\frac {\left (\sqrt {b^2-4 d^2 e^2}+b-2 d e\right ) \tan ^{-1}\left (\frac {\sqrt {2} e x}{\sqrt {\sqrt {b^2-4 d^2 e^2}+b}}\right )}{\sqrt {\sqrt {b^2-4 d^2 e^2}+b}}}{\sqrt {2} \sqrt {b^2-4 d^2 e^2}} \]

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]]])/
Sqrt[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 = 0.43, size = 162, normalized size = 1.98 \[ \left [-\frac {\sqrt {-2 \, d e - b} \log \left (\frac {e^{2} x^{4} - {\left (4 \, d e + b\right )} x^{2} + d^{2} - 2 \, {\left (e x^{3} - d x\right )} \sqrt {-2 \, d e - b}}{e^{2} x^{4} + b x^{2} + d^{2}}\right )}{2 \, {\left (2 \, d e + b\right )}}, \frac {\sqrt {2 \, d e + b} \arctan \left (\frac {e x}{\sqrt {2 \, d e + b}}\right ) + \sqrt {2 \, d e + b} \arctan \left (\frac {{\left (e^{2} x^{3} + {\left (d e + b\right )} x\right )} \sqrt {2 \, d e + b}}{2 \, d^{2} e + b d}\right )}{2 \, d e + b}\right ] \]

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 = 1642, normalized size = 20.02 \[ \text {result too large to display} \]

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 = 0.04, size = 71, normalized size = 0.87 \[ -\frac {\arctan \left (\frac {-2 e x +\sqrt {2 d e -b}}{\sqrt {2 d e +b}}\right )}{\sqrt {2 d e +b}}+\frac {\arctan \left (\frac {2 e x +\sqrt {2 d e -b}}{\sqrt {2 d e +b}}\right )}{\sqrt {2 d e +b}} \]

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]

-arctan((-2*e*x+(2*d*e-b)^(1/2))/(2*d*e+b)^(1/2))/(2*d*e+b)^(1/2)+arctan((2*e*x+(2*d*e-b)^(1/2))/(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 \[ \int \frac {e x^{2} + d}{e^{2} x^{4} + b x^{2} + d^{2}}\,{d x} \]

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 = 4.43, size = 94, normalized size = 1.15 \[ \frac {\mathrm {atan}\left (\frac {e\,x}{\sqrt {b+2\,d\,e}}\right )+\mathrm {atan}\left (\frac {b^2\,x-\frac {x\,{\left (b+2\,d\,e\right )}^2}{2}+\frac {b\,x\,\left (b+2\,d\,e\right )}{2}+2\,b\,e^2\,x^3-e^2\,x^3\,\left (b+2\,d\,e\right )}{\left (b\,d-2\,d^2\,e\right )\,\sqrt {b+2\,d\,e}}\right )}{\sqrt {b+2\,d\,e}} \]

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((e*x)/(b + 2*d*e)^(1/2)) + atan((b^2*x - (x*(b + 2*d*e)^2)/2 + (b*x*(b + 2*d*e))/2 + 2*b*e^2*x^3 - e^2*x
^3*(b + 2*d*e))/((b*d - 2*d^2*e)*(b + 2*d*e)^(1/2))))/(b + 2*d*e)^(1/2)

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sympy [A]  time = 0.54, size = 122, normalized size = 1.49 \[ - \frac {\sqrt {- \frac {1}{b + 2 d e}} \log {\left (- \frac {d}{e} + x^{2} + \frac {x \left (- b \sqrt {- \frac {1}{b + 2 d e}} - 2 d e \sqrt {- \frac {1}{b + 2 d e}}\right )}{e} \right )}}{2} + \frac {\sqrt {- \frac {1}{b + 2 d e}} \log {\left (- \frac {d}{e} + x^{2} + \frac {x \left (b \sqrt {- \frac {1}{b + 2 d e}} + 2 d e \sqrt {- \frac {1}{b + 2 d e}}\right )}{e} \right )}}{2} \]

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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