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

Optimal. Leaf size=341 \[ \frac{\sqrt [4]{c} \left (2 \sqrt{a} \sqrt{c}+b\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{6 a^{3/4} \sqrt{a+b x^2+c x^4}}-\frac{b \sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{3 a^{3/4} \sqrt{a+b x^2+c x^4}}-\frac{b \sqrt{a+b x^2+c x^4}}{3 a x}+\frac{b \sqrt{c} x \sqrt{a+b x^2+c x^4}}{3 a \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{\sqrt{a+b x^2+c x^4}}{3 x^3} \]

[Out]

-Sqrt[a + b*x^2 + c*x^4]/(3*x^3) - (b*Sqrt[a + b*x^2 + c*x^4])/(3*a*x) + (b*Sqrt[c]*x*Sqrt[a + b*x^2 + c*x^4])
/(3*a*(Sqrt[a] + Sqrt[c]*x^2)) - (b*c^(1/4)*(Sqrt[a] + Sqrt[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^(3/4)*Sqrt[a + b*x^2 + c
*x^4]) + ((b + 2*Sqrt[a]*Sqrt[c])*c^(1/4)*(Sqrt[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^(3/4)*Sqrt[a + b*x^2 + c*x
^4])

________________________________________________________________________________________

Rubi [A]  time = 0.155129, antiderivative size = 341, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1117, 1281, 1197, 1103, 1195} \[ \frac{\sqrt [4]{c} \left (2 \sqrt{a} \sqrt{c}+b\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{6 a^{3/4} \sqrt{a+b x^2+c x^4}}-\frac{b \sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{3 a^{3/4} \sqrt{a+b x^2+c x^4}}-\frac{b \sqrt{a+b x^2+c x^4}}{3 a x}+\frac{b \sqrt{c} x \sqrt{a+b x^2+c x^4}}{3 a \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{\sqrt{a+b x^2+c x^4}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2 + c*x^4]/x^4,x]

[Out]

-Sqrt[a + b*x^2 + c*x^4]/(3*x^3) - (b*Sqrt[a + b*x^2 + c*x^4])/(3*a*x) + (b*Sqrt[c]*x*Sqrt[a + b*x^2 + c*x^4])
/(3*a*(Sqrt[a] + Sqrt[c]*x^2)) - (b*c^(1/4)*(Sqrt[a] + Sqrt[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^(3/4)*Sqrt[a + b*x^2 + c
*x^4]) + ((b + 2*Sqrt[a]*Sqrt[c])*c^(1/4)*(Sqrt[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^(3/4)*Sqrt[a + b*x^2 + c*x
^4])

Rule 1117

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

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

Mathematica [C]  time = 0.895, size = 459, normalized size = 1.35 \[ \frac{-i x^3 \left (b \sqrt{b^2-4 a c}+4 a c-b^2\right ) \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{-2 \sqrt{b^2-4 a c}+2 b+4 c x^2}{b-\sqrt{b^2-4 a c}}} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} x \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}}\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right )-4 \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \left (a+b x^2\right ) \left (a+b x^2+c x^4\right )+i b x^3 \left (\sqrt{b^2-4 a c}-b\right ) \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{-2 \sqrt{b^2-4 a c}+2 b+4 c x^2}{b-\sqrt{b^2-4 a c}}} E\left (i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{b+\sqrt{b^2-4 a c}}} x\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )}{12 a x^3 \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \sqrt{a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^2 + c*x^4]/x^4,x]

[Out]

(-4*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*(a + b*x^2)*(a + b*x^2 + c*x^4) + I*b*(-b + Sqrt[b^2 - 4*a*c])*x^3*Sqrt[(b
+ Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[(2*b - 2*Sqrt[b^2 - 4*a*c] + 4*c*x^2)/(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 - Sq
rt[b^2 - 4*a*c])] - I*(-b^2 + 4*a*c + b*Sqrt[b^2 - 4*a*c])*x^3*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqr
t[b^2 - 4*a*c])]*Sqrt[(2*b - 2*Sqrt[b^2 - 4*a*c] + 4*c*x^2)/(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])])/(12*a*Sqrt[c/(b + Sqr
t[b^2 - 4*a*c])]*x^3*Sqrt[a + b*x^2 + c*x^4])

________________________________________________________________________________________

Maple [A]  time = 0.216, size = 404, normalized size = 1.2 \begin{align*} -{\frac{1}{3\,{x}^{3}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{b}{3\,ax}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{c\sqrt{2}}{6}\sqrt{4-2\,{\frac{ \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ){x}^{2}}{a}}}\sqrt{4+2\,{\frac{ \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ){x}^{2}}{a}}}{\it EllipticF} \left ({\frac{x\sqrt{2}}{2}\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}},{\frac{1}{2}\sqrt{-4+2\,{\frac{b \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) }{ac}}}} \right ){\frac{1}{\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}}}{\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}+a}}}}-{\frac{bc\sqrt{2}}{6}\sqrt{4-2\,{\frac{ \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ){x}^{2}}{a}}}\sqrt{4+2\,{\frac{ \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ){x}^{2}}{a}}} \left ({\it EllipticF} \left ({\frac{x\sqrt{2}}{2}\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}},{\frac{1}{2}\sqrt{-4+2\,{\frac{b \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) }{ac}}}} \right ) -{\it EllipticE} \left ({\frac{x\sqrt{2}}{2}\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}},{\frac{1}{2}\sqrt{-4+2\,{\frac{b \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) }{ac}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}}}{\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}+a}}} \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(c*x^4+b*x^2+a)^(1/2)/x^3-1/3*b*(c*x^4+b*x^2+a)^(1/2)/a/x+1/6*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)*Elli
pticF(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))-1/6*b*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)))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\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((c*x^4+b*x^2+a)^(1/2)/x^4,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^4 + b*x^2 + a)/x^4, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\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((c*x^4+b*x^2+a)^(1/2)/x^4,x, algorithm="giac")

[Out]

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