3.180 \(\int \frac{A+B x^2}{x^4 \sqrt{a+b x^2+c x^4}} \, dx\)

Optimal. Leaf size=376 \[ -\frac{\sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\sqrt{a} A \sqrt{c}-3 a B+2 A b\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{6 a^{7/4} \sqrt{a+b x^2+c x^4}}+\frac{(2 A b-3 a B) \sqrt{a+b x^2+c x^4}}{3 a^2 x}-\frac{\sqrt{c} x (2 A b-3 a B) \sqrt{a+b x^2+c x^4}}{3 a^2 \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{\sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x^2\right ) (2 A b-3 a B) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{3 a^{7/4} \sqrt{a+b x^2+c x^4}}-\frac{A \sqrt{a+b x^2+c x^4}}{3 a x^3} \]

[Out]

-(A*Sqrt[a + b*x^2 + c*x^4])/(3*a*x^3) + ((2*A*b - 3*a*B)*Sqrt[a + b*x^2 + c*x^4])/(3*a^2*x) - ((2*A*b - 3*a*B
)*Sqrt[c]*x*Sqrt[a + b*x^2 + c*x^4])/(3*a^2*(Sqrt[a] + Sqrt[c]*x^2)) + ((2*A*b - 3*a*B)*c^(1/4)*(Sqrt[a] + Sqr
t[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/
(Sqrt[a]*Sqrt[c]))/4])/(3*a^(7/4)*Sqrt[a + b*x^2 + c*x^4]) - ((2*A*b - 3*a*B + Sqrt[a]*A*Sqrt[c])*c^(1/4)*(Sqr
t[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)
], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(6*a^(7/4)*Sqrt[a + b*x^2 + c*x^4])

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Rubi [A]  time = 0.227337, antiderivative size = 376, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {1281, 1197, 1103, 1195} \[ \frac{(2 A b-3 a B) \sqrt{a+b x^2+c x^4}}{3 a^2 x}-\frac{\sqrt{c} x (2 A b-3 a B) \sqrt{a+b x^2+c x^4}}{3 a^2 \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{\sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\sqrt{a} A \sqrt{c}-3 a B+2 A b\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{6 a^{7/4} \sqrt{a+b x^2+c x^4}}+\frac{\sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x^2\right ) (2 A b-3 a B) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{3 a^{7/4} \sqrt{a+b x^2+c x^4}}-\frac{A \sqrt{a+b x^2+c x^4}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^4*Sqrt[a + b*x^2 + c*x^4]),x]

[Out]

-(A*Sqrt[a + b*x^2 + c*x^4])/(3*a*x^3) + ((2*A*b - 3*a*B)*Sqrt[a + b*x^2 + c*x^4])/(3*a^2*x) - ((2*A*b - 3*a*B
)*Sqrt[c]*x*Sqrt[a + b*x^2 + c*x^4])/(3*a^2*(Sqrt[a] + Sqrt[c]*x^2)) + ((2*A*b - 3*a*B)*c^(1/4)*(Sqrt[a] + Sqr
t[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/
(Sqrt[a]*Sqrt[c]))/4])/(3*a^(7/4)*Sqrt[a + b*x^2 + c*x^4]) - ((2*A*b - 3*a*B + Sqrt[a]*A*Sqrt[c])*c^(1/4)*(Sqr
t[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)
], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(6*a^(7/4)*Sqrt[a + b*x^2 + c*x^4])

Rule 1281

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*(
f*x)^(m + 1)*(a + b*x^2 + c*x^4)^(p + 1))/(a*f*(m + 1)), x] + Dist[1/(a*f^2*(m + 1)), Int[(f*x)^(m + 2)*(a + b
*x^2 + c*x^4)^p*Simp[a*e*(m + 1) - b*d*(m + 2*p + 3) - c*d*(m + 4*p + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1197

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

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1195

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

Rubi steps

\begin{align*} \int \frac{A+B x^2}{x^4 \sqrt{a+b x^2+c x^4}} \, dx &=-\frac{A \sqrt{a+b x^2+c x^4}}{3 a x^3}-\frac{\int \frac{2 A b-3 a B+A c x^2}{x^2 \sqrt{a+b x^2+c x^4}} \, dx}{3 a}\\ &=-\frac{A \sqrt{a+b x^2+c x^4}}{3 a x^3}+\frac{(2 A b-3 a B) \sqrt{a+b x^2+c x^4}}{3 a^2 x}+\frac{\int \frac{-a A c-(2 A b-3 a B) c x^2}{\sqrt{a+b x^2+c x^4}} \, dx}{3 a^2}\\ &=-\frac{A \sqrt{a+b x^2+c x^4}}{3 a x^3}+\frac{(2 A b-3 a B) \sqrt{a+b x^2+c x^4}}{3 a^2 x}+\frac{\left ((2 A b-3 a B) \sqrt{c}\right ) \int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+b x^2+c x^4}} \, dx}{3 a^{3/2}}-\frac{\left (\left (2 A b-3 a B+\sqrt{a} A \sqrt{c}\right ) \sqrt{c}\right ) \int \frac{1}{\sqrt{a+b x^2+c x^4}} \, dx}{3 a^{3/2}}\\ &=-\frac{A \sqrt{a+b x^2+c x^4}}{3 a x^3}+\frac{(2 A b-3 a B) \sqrt{a+b x^2+c x^4}}{3 a^2 x}-\frac{(2 A b-3 a B) \sqrt{c} x \sqrt{a+b x^2+c x^4}}{3 a^2 \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{(2 A b-3 a B) \sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{3 a^{7/4} \sqrt{a+b x^2+c x^4}}-\frac{\left (2 A b-3 a B+\sqrt{a} A \sqrt{c}\right ) \sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{6 a^{7/4} \sqrt{a+b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.695821, size = 373, normalized size = 0.99 \[ \frac{-\frac{4 \left (a+b x^2+c x^4\right ) \left (a \left (A+3 B x^2\right )-2 A b x^2\right )}{x^3}+\frac{i \sqrt{2} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \left (\left (2 A \left (b \sqrt{b^2-4 a c}+a c-b^2\right )+3 a B \left (b-\sqrt{b^2-4 a c}\right )\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} x \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}}\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right )-\left (\sqrt{b^2-4 a c}-b\right ) (2 A b-3 a B) E\left (i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{b+\sqrt{b^2-4 a c}}} x\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )\right )}{\sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}}}}{12 a^2 \sqrt{a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^4*Sqrt[a + b*x^2 + c*x^4]),x]

[Out]

((-4*(a + b*x^2 + c*x^4)*(-2*A*b*x^2 + a*(A + 3*B*x^2)))/x^3 + (I*Sqrt[2]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^
2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*(-((2*A*b - 3*a*B)*(-b + Sqrt[b^2 - 4*
a*c])*EllipticE[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 -
4*a*c])]) + (3*a*B*(b - Sqrt[b^2 - 4*a*c]) + 2*A*(-b^2 + a*c + b*Sqrt[b^2 - 4*a*c]))*EllipticF[I*ArcSinh[Sqrt[
2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])]))/Sqrt[c/(b + Sqrt[b^2
- 4*a*c])])/(12*a^2*Sqrt[a + b*x^2 + c*x^4])

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Maple [A]  time = 0.018, size = 656, normalized size = 1.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^4/(c*x^4+b*x^2+a)^(1/2),x)

[Out]

B*(-1/a*(c*x^4+b*x^2+a)^(1/2)/x-1/2*c*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(4-2*(-b+(-4*a*c+b^2)^(1/2))/a
*x^2)^(1/2)*(4+2*(b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)/(c*x^4+b*x^2+a)^(1/2)/(b+(-4*a*c+b^2)^(1/2))*(EllipticF(1
/2*x*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2))-EllipticE(1/2*x*
2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2))))+A*(-1/3/a*(c*x^4+b*
x^2+a)^(1/2)/x^3+2/3/a^2*b*(c*x^4+b*x^2+a)^(1/2)/x-1/12/a*c*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(4-2*(-b
+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)*(4+2*(b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)/(c*x^4+b*x^2+a)^(1/2)*EllipticF(1/2
*x*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2))+1/3*b*c/a*2^(1/2)/
((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(4-2*(-b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)*(4+2*(b+(-4*a*c+b^2)^(1/2))/a*x^2)
^(1/2)/(c*x^4+b*x^2+a)^(1/2)/(b+(-4*a*c+b^2)^(1/2))*(EllipticF(1/2*x*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)
,1/2*(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2))-EllipticE(1/2*x*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2),1/2*
(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2))))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{2} + A}{\sqrt{c x^{4} + b x^{2} + a} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x^2 + A)/(sqrt(c*x^4 + b*x^2 + a)*x^4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{4} + b x^{2} + a}{\left (B x^{2} + A\right )}}{c x^{8} + b x^{6} + a x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^4 + b*x^2 + a)*(B*x^2 + A)/(c*x^8 + b*x^6 + a*x^4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B x^{2}}{x^{4} \sqrt{a + b x^{2} + c x^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**4/(c*x**4+b*x**2+a)**(1/2),x)

[Out]

Integral((A + B*x**2)/(x**4*sqrt(a + b*x**2 + c*x**4)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{2} + A}{\sqrt{c x^{4} + b x^{2} + a} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x^2 + A)/(sqrt(c*x^4 + b*x^2 + a)*x^4), x)