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

Optimal. Leaf size=254 \[ -\frac{\sqrt{c} \left (e-\frac{2 c d-b e}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} \sqrt{b-\sqrt{b^2-4 a c}} \left (a e^2-b d e+c d^2\right )}-\frac{\sqrt{c} \left (\frac{2 c d-b e}{\sqrt{b^2-4 a c}}+e\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} \sqrt{\sqrt{b^2-4 a c}+b} \left (a e^2-b d e+c d^2\right )}+\frac{e^{3/2} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{\sqrt{d} \left (a e^2-b d e+c d^2\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.585595, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1170, 205, 1166} \[ -\frac{\sqrt{c} \left (e-\frac{2 c d-b e}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} \sqrt{b-\sqrt{b^2-4 a c}} \left (a e^2-b d e+c d^2\right )}-\frac{\sqrt{c} \left (\frac{2 c d-b e}{\sqrt{b^2-4 a c}}+e\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} \sqrt{\sqrt{b^2-4 a c}+b} \left (a e^2-b d e+c d^2\right )}+\frac{e^{3/2} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{\sqrt{d} \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[c]*(e - (2*c*d - b*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqr
t[2]*Sqrt[b - Sqrt[b^2 - 4*a*c]]*(c*d^2 - b*d*e + a*e^2))) - (Sqrt[c]*(e + (2*c*d - b*e)/Sqrt[b^2 - 4*a*c])*Ar
cTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[b + Sqrt[b^2 - 4*a*c]]*(c*d^2 - b*d*e + a
*e^2)) + (e^(3/2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(Sqrt[d]*(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 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 ) \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 )}+\frac{c d-b e-c e x^2}{\left (c d^2-b d e+a e^2\right ) \left (a+b x^2+c x^4\right )}\right ) \, dx\\ &=\frac{\int \frac{c d-b e-c e x^2}{a+b x^2+c x^4} \, dx}{c d^2-b d e+a e^2}+\frac{e^2 \int \frac{1}{d+e x^2} \, dx}{c d^2-b d e+a e^2}\\ &=\frac{e^{3/2} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{\sqrt{d} \left (c d^2-b d e+a e^2\right )}-\frac{\left (c \left (e-\frac{2 c d-b 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 )}-\frac{\left (c \left (e+\frac{2 c d-b 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 )}\\ &=-\frac{\sqrt{c} \left (e-\frac{2 c d-b 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 )}-\frac{\sqrt{c} \left (e+\frac{2 c d-b 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 )}+\frac{e^{3/2} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{\sqrt{d} \left (c d^2-b d e+a e^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.286987, size = 274, normalized size = 1.08 \[ \frac{\sqrt{c} \left (e \sqrt{b^2-4 a c}+b e-2 c d\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} \sqrt{b^2-4 a c} \sqrt{b-\sqrt{b^2-4 a c}} \left (-a e^2+b d e-c d^2\right )}+\frac{\sqrt{c} \left (e \sqrt{b^2-4 a c}-b e+2 c d\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} \sqrt{b^2-4 a c} \sqrt{\sqrt{b^2-4 a c}+b} \left (-a e^2+b d e-c d^2\right )}+\frac{e^{3/2} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{\sqrt{d} \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.023, size = 480, normalized size = 1.9 \begin{align*}{\frac{c\sqrt{2}e}{2\,a{e}^{2}-2\,deb+2\,c{d}^{2}}{\it Artanh} \left ({cx\sqrt{2}{\frac{1}{\sqrt{ \left ( \sqrt{-4\,ac+{b}^{2}}-b \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( \sqrt{-4\,ac+{b}^{2}}-b \right ) c}}}}+{\frac{c\sqrt{2}be}{2\,a{e}^{2}-2\,deb+2\,c{d}^{2}}{\it Artanh} \left ({cx\sqrt{2}{\frac{1}{\sqrt{ \left ( \sqrt{-4\,ac+{b}^{2}}-b \right ) c}}}} \right ){\frac{1}{\sqrt{-4\,ac+{b}^{2}}}}{\frac{1}{\sqrt{ \left ( \sqrt{-4\,ac+{b}^{2}}-b \right ) c}}}}-{\frac{{c}^{2}\sqrt{2}d}{a{e}^{2}-deb+c{d}^{2}}{\it Artanh} \left ({cx\sqrt{2}{\frac{1}{\sqrt{ \left ( \sqrt{-4\,ac+{b}^{2}}-b \right ) c}}}} \right ){\frac{1}{\sqrt{-4\,ac+{b}^{2}}}}{\frac{1}{\sqrt{ \left ( \sqrt{-4\,ac+{b}^{2}}-b \right ) c}}}}-{\frac{c\sqrt{2}e}{2\,a{e}^{2}-2\,deb+2\,c{d}^{2}}\arctan \left ({cx\sqrt{2}{\frac{1}{\sqrt{ \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) c}}}}+{\frac{c\sqrt{2}be}{2\,a{e}^{2}-2\,deb+2\,c{d}^{2}}\arctan \left ({cx\sqrt{2}{\frac{1}{\sqrt{ \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{-4\,ac+{b}^{2}}}}{\frac{1}{\sqrt{ \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) c}}}}-{\frac{{c}^{2}\sqrt{2}d}{a{e}^{2}-deb+c{d}^{2}}\arctan \left ({cx\sqrt{2}{\frac{1}{\sqrt{ \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{-4\,ac+{b}^{2}}}}{\frac{1}{\sqrt{ \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) c}}}}+{\frac{{e}^{2}}{a{e}^{2}-deb+c{d}^{2}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: NotImplementedError