3.269 \(\int \frac{1}{(d+e x^2)^2 (a+b x^2+c x^4)} \, dx\)

Optimal. Leaf size=429 \[ \frac{\sqrt{c} \left (-2 c e \left (d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (\sqrt{b^2-4 a c}+b\right )+2 c^2 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} \sqrt{b^2-4 a c} \sqrt{b-\sqrt{b^2-4 a c}} \left (a e^2-b d e+c d^2\right )^2}-\frac{\sqrt{c} \left (-2 c e \left (-d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (b-\sqrt{b^2-4 a c}\right )+2 c^2 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} \sqrt{b^2-4 a c} \sqrt{\sqrt{b^2-4 a c}+b} \left (a e^2-b d e+c d^2\right )^2}+\frac{e^2 x}{2 d \left (d+e x^2\right ) \left (a e^2-b d e+c d^2\right )}+\frac{e^{3/2} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{2 d^{3/2} \left (a e^2-b d e+c d^2\right )}+\frac{e^{3/2} (2 c d-b e) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{\sqrt{d} \left (a e^2-b d e+c d^2\right )^2} \]

[Out]

(e^2*x)/(2*d*(c*d^2 - b*d*e + a*e^2)*(d + e*x^2)) + (Sqrt[c]*(2*c^2*d^2 + b*(b + Sqrt[b^2 - 4*a*c])*e^2 - 2*c*
e*(b*d + Sqrt[b^2 - 4*a*c]*d + a*e))*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[b^
2 - 4*a*c]*Sqrt[b - Sqrt[b^2 - 4*a*c]]*(c*d^2 - b*d*e + a*e^2)^2) - (Sqrt[c]*(2*c^2*d^2 + b*(b - Sqrt[b^2 - 4*
a*c])*e^2 - 2*c*e*(b*d - Sqrt[b^2 - 4*a*c]*d + a*e))*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/
(Sqrt[2]*Sqrt[b^2 - 4*a*c]*Sqrt[b + Sqrt[b^2 - 4*a*c]]*(c*d^2 - b*d*e + a*e^2)^2) + (e^(3/2)*(2*c*d - b*e)*Arc
Tan[(Sqrt[e]*x)/Sqrt[d]])/(Sqrt[d]*(c*d^2 - b*d*e + a*e^2)^2) + (e^(3/2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*d^(3/
2)*(c*d^2 - b*d*e + a*e^2))

________________________________________________________________________________________

Rubi [A]  time = 1.4145, antiderivative size = 429, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1170, 199, 205, 1166} \[ \frac{\sqrt{c} \left (-2 c e \left (d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (\sqrt{b^2-4 a c}+b\right )+2 c^2 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} \sqrt{b^2-4 a c} \sqrt{b-\sqrt{b^2-4 a c}} \left (a e^2-b d e+c d^2\right )^2}-\frac{\sqrt{c} \left (-2 c e \left (-d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (b-\sqrt{b^2-4 a c}\right )+2 c^2 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} \sqrt{b^2-4 a c} \sqrt{\sqrt{b^2-4 a c}+b} \left (a e^2-b d e+c d^2\right )^2}+\frac{e^2 x}{2 d \left (d+e x^2\right ) \left (a e^2-b d e+c d^2\right )}+\frac{e^{3/2} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{2 d^{3/2} \left (a e^2-b d e+c d^2\right )}+\frac{e^{3/2} (2 c d-b e) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{\sqrt{d} \left (a e^2-b d e+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x^2)^2*(a + b*x^2 + c*x^4)),x]

[Out]

(e^2*x)/(2*d*(c*d^2 - b*d*e + a*e^2)*(d + e*x^2)) + (Sqrt[c]*(2*c^2*d^2 + b*(b + Sqrt[b^2 - 4*a*c])*e^2 - 2*c*
e*(b*d + Sqrt[b^2 - 4*a*c]*d + a*e))*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[b^
2 - 4*a*c]*Sqrt[b - Sqrt[b^2 - 4*a*c]]*(c*d^2 - b*d*e + a*e^2)^2) - (Sqrt[c]*(2*c^2*d^2 + b*(b - Sqrt[b^2 - 4*
a*c])*e^2 - 2*c*e*(b*d - Sqrt[b^2 - 4*a*c]*d + a*e))*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/
(Sqrt[2]*Sqrt[b^2 - 4*a*c]*Sqrt[b + Sqrt[b^2 - 4*a*c]]*(c*d^2 - b*d*e + a*e^2)^2) + (e^(3/2)*(2*c*d - b*e)*Arc
Tan[(Sqrt[e]*x)/Sqrt[d]])/(Sqrt[d]*(c*d^2 - b*d*e + a*e^2)^2) + (e^(3/2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*d^(3/
2)*(c*d^2 - b*d*e + a*e^2))

Rule 1170

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[(d + e*x
^2)^q/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a
*e^2, 0] && IntegerQ[q]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rubi steps

\begin{align*} \int \frac{1}{\left (d+e x^2\right )^2 \left (a+b x^2+c x^4\right )} \, dx &=\int \left (\frac{e^2}{\left (c d^2-b d e+a e^2\right ) \left (d+e x^2\right )^2}-\frac{e^2 (-2 c d+b e)}{\left (c d^2-b d e+a e^2\right )^2 \left (d+e x^2\right )}+\frac{c^2 d^2+b^2 e^2-c e (2 b d+a e)-c e (2 c d-b e) x^2}{\left (c d^2-b d e+a e^2\right )^2 \left (a+b x^2+c x^4\right )}\right ) \, dx\\ &=\frac{\int \frac{c^2 d^2+b^2 e^2-c e (2 b d+a e)-c e (2 c d-b e) x^2}{a+b x^2+c x^4} \, dx}{\left (c d^2-b d e+a e^2\right )^2}+\frac{\left (e^2 (2 c d-b e)\right ) \int \frac{1}{d+e x^2} \, dx}{\left (c d^2-b d e+a e^2\right )^2}+\frac{e^2 \int \frac{1}{\left (d+e x^2\right )^2} \, dx}{c d^2-b d e+a e^2}\\ &=\frac{e^2 x}{2 d \left (c d^2-b d e+a e^2\right ) \left (d+e x^2\right )}+\frac{e^{3/2} (2 c d-b e) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{\sqrt{d} \left (c d^2-b d e+a e^2\right )^2}+\frac{e^2 \int \frac{1}{d+e x^2} \, dx}{2 d \left (c d^2-b d e+a e^2\right )}+\frac{\left (c \left (2 c^2 d^2+b \left (b+\sqrt{b^2-4 a c}\right ) e^2-2 c e \left (b d+\sqrt{b^2-4 a c} d+a e\right )\right )\right ) \int \frac{1}{\frac{b}{2}-\frac{1}{2} \sqrt{b^2-4 a c}+c x^2} \, dx}{2 \sqrt{b^2-4 a c} \left (c d^2-b d e+a e^2\right )^2}-\frac{\left (c \left (e (2 c d-b e)+\frac{2 c^2 d^2+b^2 e^2-2 c e (b d+a e)}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{1}{\frac{b}{2}+\frac{1}{2} \sqrt{b^2-4 a c}+c x^2} \, dx}{2 \left (c d^2-b d e+a e^2\right )^2}\\ &=\frac{e^2 x}{2 d \left (c d^2-b d e+a e^2\right ) \left (d+e x^2\right )}+\frac{\sqrt{c} \left (2 c^2 d^2+b \left (b+\sqrt{b^2-4 a c}\right ) e^2-2 c e \left (b d+\sqrt{b^2-4 a c} d+a e\right )\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} \sqrt{b^2-4 a c} \sqrt{b-\sqrt{b^2-4 a c}} \left (c d^2-b d e+a e^2\right )^2}-\frac{\sqrt{c} \left (e (2 c d-b e)+\frac{2 c^2 d^2+b^2 e^2-2 c e (b d+a e)}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b+\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} \sqrt{b+\sqrt{b^2-4 a c}} \left (c d^2-b d e+a e^2\right )^2}+\frac{e^{3/2} (2 c d-b e) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{\sqrt{d} \left (c d^2-b d e+a e^2\right )^2}+\frac{e^{3/2} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{2 d^{3/2} \left (c d^2-b d e+a e^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.825196, size = 354, normalized size = 0.83 \[ \frac{\frac{\sqrt{2} \sqrt{c} \left (-2 c e \left (d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (\sqrt{b^2-4 a c}+b\right )+2 c^2 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{b^2-4 a c} \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\sqrt{2} \sqrt{c} \left (2 c e \left (-d \sqrt{b^2-4 a c}+a e+b d\right )+b e^2 \left (\sqrt{b^2-4 a c}-b\right )-2 c^2 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{b^2-4 a c} \sqrt{\sqrt{b^2-4 a c}+b}}+\frac{e^2 x \left (e (a e-b d)+c d^2\right )}{d \left (d+e x^2\right )}+\frac{e^{3/2} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (e (a e-3 b d)+5 c d^2\right )}{d^{3/2}}}{2 \left (e (a e-b d)+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x^2)^2*(a + b*x^2 + c*x^4)),x]

[Out]

((e^2*(c*d^2 + e*(-(b*d) + a*e))*x)/(d*(d + e*x^2)) + (Sqrt[2]*Sqrt[c]*(2*c^2*d^2 + b*(b + Sqrt[b^2 - 4*a*c])*
e^2 - 2*c*e*(b*d + Sqrt[b^2 - 4*a*c]*d + a*e))*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[
b^2 - 4*a*c]*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + (Sqrt[2]*Sqrt[c]*(-2*c^2*d^2 + b*(-b + Sqrt[b^2 - 4*a*c])*e^2 + 2*
c*e*(b*d - Sqrt[b^2 - 4*a*c]*d + a*e))*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[b^2 - 4*
a*c]*Sqrt[b + Sqrt[b^2 - 4*a*c]]) + (e^(3/2)*(5*c*d^2 + e*(-3*b*d + a*e))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/d^(3/2)
)/(2*(c*d^2 + e*(-(b*d) + a*e))^2)

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Maple [B]  time = 0.03, size = 1141, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x^2+d)^2/(c*x^4+b*x^2+a),x)

[Out]

-1/2/(a*e^2-b*d*e+c*d^2)^2*c*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2)*arctanh(c*x*2^(1/2)/(((-4*a*c+b^2)^(1/2)
-b)*c)^(1/2))*e^2*b+1/(a*e^2-b*d*e+c*d^2)^2*c^2*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2)*arctanh(c*x*2^(1/2)/(
((-4*a*c+b^2)^(1/2)-b)*c)^(1/2))*d*e+1/(a*e^2-b*d*e+c*d^2)^2*c^2/(-4*a*c+b^2)^(1/2)*2^(1/2)/(((-4*a*c+b^2)^(1/
2)-b)*c)^(1/2)*arctanh(c*x*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2))*a*e^2-1/2/(a*e^2-b*d*e+c*d^2)^2*c/(-4*a*c
+b^2)^(1/2)*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2)*arctanh(c*x*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2))*b^2
*e^2+1/(a*e^2-b*d*e+c*d^2)^2*c^2/(-4*a*c+b^2)^(1/2)*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2)*arctanh(c*x*2^(1/
2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2))*d*e*b-1/(a*e^2-b*d*e+c*d^2)^2*c^3/(-4*a*c+b^2)^(1/2)*2^(1/2)/(((-4*a*c+b^
2)^(1/2)-b)*c)^(1/2)*arctanh(c*x*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2))*d^2+1/2/(a*e^2-b*d*e+c*d^2)^2*c*2^(
1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*e^2*b-1/(a*e^2-b*d*
e+c*d^2)^2*c^2*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*d
*e+1/(a*e^2-b*d*e+c*d^2)^2*c^2/(-4*a*c+b^2)^(1/2)*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x*2^(1/2)/
((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*a*e^2-1/2/(a*e^2-b*d*e+c*d^2)^2*c/(-4*a*c+b^2)^(1/2)*2^(1/2)/((b+(-4*a*c+b^2
)^(1/2))*c)^(1/2)*arctan(c*x*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*b^2*e^2+1/(a*e^2-b*d*e+c*d^2)^2*c^2/(-4
*a*c+b^2)^(1/2)*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*
d*e*b-1/(a*e^2-b*d*e+c*d^2)^2*c^3/(-4*a*c+b^2)^(1/2)*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x*2^(1/
2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*d^2+1/2*e^4/(a*e^2-b*d*e+c*d^2)^2/d*x/(e*x^2+d)*a-1/2*e^3/(a*e^2-b*d*e+c*
d^2)^2*x/(e*x^2+d)*b+1/2*e^2/(a*e^2-b*d*e+c*d^2)^2*d*x/(e*x^2+d)*c+1/2*e^4/(a*e^2-b*d*e+c*d^2)^2/d/(d*e)^(1/2)
*arctan(x*e/(d*e)^(1/2))*a-3/2*e^3/(a*e^2-b*d*e+c*d^2)^2/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*b+5/2*e^2/(a*e^2-
b*d*e+c*d^2)^2*d/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*c

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x^2+d)^2/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x^2+d)^2/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x**2+d)**2/(c*x**4+b*x**2+a),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x^2+d)^2/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError