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

Optimal. Leaf size=298 \[ -\frac{\sqrt{c} \left (\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} a \sqrt{b-\sqrt{b^2-4 a c}} \left (a e^2-b d e+c d^2\right )}-\frac{\sqrt{c} \left (-\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} a \sqrt{\sqrt{b^2-4 a c}+b} \left (a e^2-b d e+c d^2\right )}-\frac{e^{5/2} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{d^{3/2} \left (a e^2-b d e+c d^2\right )}-\frac{1}{a d x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.960295, antiderivative size = 298, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {1287, 205, 1166} \[ -\frac{\sqrt{c} \left (\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} a \sqrt{b-\sqrt{b^2-4 a c}} \left (a e^2-b d e+c d^2\right )}-\frac{\sqrt{c} \left (-\frac{2 a c e+b^2 (-e)+b c d}{\sqrt{b^2-4 a c}}-b e+c d\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} a \sqrt{\sqrt{b^2-4 a c}+b} \left (a e^2-b d e+c d^2\right )}-\frac{e^{5/2} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{d^{3/2} \left (a e^2-b d e+c d^2\right )}-\frac{1}{a d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1287

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

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}{x^2 \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )} \, dx &=\int \left (\frac{1}{a d x^2}-\frac{e^3}{d \left (c d^2-b d e+a e^2\right ) \left (d+e x^2\right )}+\frac{-b c d+b^2 e-a c e-c (c d-b e) x^2}{a \left (c d^2-b d e+a e^2\right ) \left (a+b x^2+c x^4\right )}\right ) \, dx\\ &=-\frac{1}{a d x}+\frac{\int \frac{-b c d+b^2 e-a c e-c (c d-b e) x^2}{a+b x^2+c x^4} \, dx}{a \left (c d^2-b d e+a e^2\right )}-\frac{e^3 \int \frac{1}{d+e x^2} \, dx}{d \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{1}{a d x}-\frac{e^{5/2} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{d^{3/2} \left (c d^2-b d e+a e^2\right )}-\frac{\left (c \left (c d-b e-\frac{b c d-b^2 e+2 a c 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 a \left (c d^2-b d e+a e^2\right )}-\frac{\left (c \left (c d-b e+\frac{b c d-b^2 e+2 a c 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 a \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{1}{a d x}-\frac{\sqrt{c} \left (c d-b e+\frac{b c d-b^2 e+2 a c 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} a \sqrt{b-\sqrt{b^2-4 a c}} \left (c d^2-b d e+a e^2\right )}-\frac{\sqrt{c} \left (c d-b e-\frac{b c d-b^2 e+2 a c 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} a \sqrt{b+\sqrt{b^2-4 a c}} \left (c d^2-b d e+a e^2\right )}-\frac{e^{5/2} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{d^{3/2} \left (c d^2-b d e+a e^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.433557, size = 340, normalized size = 1.14 \[ -\frac{\sqrt{c} \left (c d \sqrt{b^2-4 a c}-b e \sqrt{b^2-4 a c}+2 a c e+b^2 (-e)+b c d\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} a \sqrt{b^2-4 a c} \sqrt{b-\sqrt{b^2-4 a c}} \left (e (a e-b d)+c d^2\right )}+\frac{\sqrt{c} \left (-c d \sqrt{b^2-4 a c}+b e \sqrt{b^2-4 a c}+2 a c e+b^2 (-e)+b c d\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} a \sqrt{b^2-4 a c} \sqrt{\sqrt{b^2-4 a c}+b} \left (e (a e-b d)+c d^2\right )}-\frac{e^{5/2} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{d^{3/2} \left (a e^2-b d e+c d^2\right )}-\frac{1}{a d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.028, size = 817, 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/x^2/(e*x^2+d)/(c*x^4+b*x^2+a),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Exception raised: NotImplementedError