3.950 \(\int \frac{(a+b x^2+c x^4)^{3/2}}{x^2} \, dx\)

Optimal. Leaf size=361 \[ \frac{\sqrt [4]{a} \left (8 \sqrt{a} b \sqrt{c}+12 a c+b^2\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{10 c^{3/4} \sqrt{a+b x^2+c x^4}}-\frac{\sqrt [4]{a} \left (12 a c+b^2\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{5 c^{3/4} \sqrt{a+b x^2+c x^4}}+\frac{x \left (12 a c+b^2\right ) \sqrt{a+b x^2+c x^4}}{5 \sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{x}+\frac{1}{5} x \left (7 b+6 c x^2\right ) \sqrt{a+b x^2+c x^4} \]

[Out]

((b^2 + 12*a*c)*x*Sqrt[a + b*x^2 + c*x^4])/(5*Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2)) + (x*(7*b + 6*c*x^2)*Sqrt[a + b
*x^2 + c*x^4])/5 - (a + b*x^2 + c*x^4)^(3/2)/x - (a^(1/4)*(b^2 + 12*a*c)*(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])/
(5*c^(3/4)*Sqrt[a + b*x^2 + c*x^4]) + (a^(1/4)*(b^2 + 8*Sqrt[a]*b*Sqrt[c] + 12*a*c)*(Sqrt[a] + Sqrt[c]*x^2)*Sq
rt[(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]*Sqr
t[c]))/4])/(10*c^(3/4)*Sqrt[a + b*x^2 + c*x^4])

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Rubi [A]  time = 0.206346, antiderivative size = 361, 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, 1176, 1197, 1103, 1195} \[ \frac{\sqrt [4]{a} \left (8 \sqrt{a} b \sqrt{c}+12 a c+b^2\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{10 c^{3/4} \sqrt{a+b x^2+c x^4}}-\frac{\sqrt [4]{a} \left (12 a c+b^2\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{5 c^{3/4} \sqrt{a+b x^2+c x^4}}+\frac{x \left (12 a c+b^2\right ) \sqrt{a+b x^2+c x^4}}{5 \sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{x}+\frac{1}{5} x \left (7 b+6 c x^2\right ) \sqrt{a+b x^2+c x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)^(3/2)/x^2,x]

[Out]

((b^2 + 12*a*c)*x*Sqrt[a + b*x^2 + c*x^4])/(5*Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2)) + (x*(7*b + 6*c*x^2)*Sqrt[a + b
*x^2 + c*x^4])/5 - (a + b*x^2 + c*x^4)^(3/2)/x - (a^(1/4)*(b^2 + 12*a*c)*(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])/
(5*c^(3/4)*Sqrt[a + b*x^2 + c*x^4]) + (a^(1/4)*(b^2 + 8*Sqrt[a]*b*Sqrt[c] + 12*a*c)*(Sqrt[a] + Sqrt[c]*x^2)*Sq
rt[(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]*Sqr
t[c]))/4])/(10*c^(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 1176

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(2*b*e*p + c*d*(4*p
+ 3) + c*e*(4*p + 1)*x^2)*(a + b*x^2 + c*x^4)^p)/(c*(4*p + 1)*(4*p + 3)), x] + Dist[(2*p)/(c*(4*p + 1)*(4*p +
3)), Int[Simp[2*a*c*d*(4*p + 3) - a*b*e + (2*a*c*e*(4*p + 1) + b*c*d*(4*p + 3) - b^2*e*(2*p + 1))*x^2, x]*(a +
b*x^2 + c*x^4)^(p - 1), 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] && GtQ[p, 0] && FractionQ[p] && IntegerQ[2*p]

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{\left (a+b x^2+c x^4\right )^{3/2}}{x^2} \, dx &=-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{x}+3 \int \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4} \, dx\\ &=\frac{1}{5} x \left (7 b+6 c x^2\right ) \sqrt{a+b x^2+c x^4}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{x}+\frac{\int \frac{8 a b c+c \left (b^2+12 a c\right ) x^2}{\sqrt{a+b x^2+c x^4}} \, dx}{5 c}\\ &=\frac{1}{5} x \left (7 b+6 c x^2\right ) \sqrt{a+b x^2+c x^4}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{x}-\frac{\left (\sqrt{a} \left (b^2+12 a c\right )\right ) \int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+b x^2+c x^4}} \, dx}{5 \sqrt{c}}+\frac{\left (\sqrt{a} \left (b^2+8 \sqrt{a} b \sqrt{c}+12 a c\right )\right ) \int \frac{1}{\sqrt{a+b x^2+c x^4}} \, dx}{5 \sqrt{c}}\\ &=\frac{\left (b^2+12 a c\right ) x \sqrt{a+b x^2+c x^4}}{5 \sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{1}{5} x \left (7 b+6 c x^2\right ) \sqrt{a+b x^2+c x^4}-\frac{\left (a+b x^2+c x^4\right )^{3/2}}{x}-\frac{\sqrt [4]{a} \left (b^2+12 a c\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}} 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 )}{5 c^{3/4} \sqrt{a+b x^2+c x^4}}+\frac{\sqrt [4]{a} \left (b^2+8 \sqrt{a} b \sqrt{c}+12 a c\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 )}{10 c^{3/4} \sqrt{a+b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 1.26276, size = 505, normalized size = 1.4 \[ \frac{-i x \left (b^2 \sqrt{b^2-4 a c}+12 a c \sqrt{b^2-4 a c}+4 a b c-b^3\right ) \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{-2 \sqrt{b^2-4 a c}+2 b+4 c x^2}{b-\sqrt{b^2-4 a c}}} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} x \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}}\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right )+4 c \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \left (-5 a^2-3 a b x^2-4 a c x^4+2 b^2 x^4+3 b c x^6+c^2 x^8\right )+i x \left (12 a c+b^2\right ) \left (\sqrt{b^2-4 a c}-b\right ) \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{-2 \sqrt{b^2-4 a c}+2 b+4 c x^2}{b-\sqrt{b^2-4 a c}}} E\left (i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{b+\sqrt{b^2-4 a c}}} x\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )}{20 c x \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \sqrt{a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2 + c*x^4)^(3/2)/x^2,x]

[Out]

(4*c*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*(-5*a^2 - 3*a*b*x^2 + 2*b^2*x^4 - 4*a*c*x^4 + 3*b*c*x^6 + c^2*x^8) + I*(b
^2 + 12*a*c)*(-b + Sqrt[b^2 - 4*a*c])*x*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 - Sqrt[b^2 - 4*a*c])] - I*(-b^3 + 4*a*b*c + b^2*Sqrt[b^2 - 4*a*c] +
12*a*c*Sqrt[b^2 - 4*a*c])*x*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[(2*b - 2*Sqr
t[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])])/(20*c*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x*Sqrt[a + b*x^2
+ c*x^4])

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Maple [A]  time = 0.223, size = 430, normalized size = 1.2 \begin{align*} -{\frac{a}{x}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{c{x}^{3}}{5}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{2\,bx}{5}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{2\,ab\sqrt{2}}{5}\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{a\sqrt{2}}{2} \left ({\frac{12\,ac}{5}}+{\frac{{b}^{2}}{5}} \right ) \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)^(3/2)/x^2,x)

[Out]

-a*(c*x^4+b*x^2+a)^(1/2)/x+1/5*c*x^3*(c*x^4+b*x^2+a)^(1/2)+2/5*b*x*(c*x^4+b*x^2+a)^(1/2)+2/5*a*b*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))-1/2*(12/5*a*c+1/5*b^2)*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))*(Ellip
ticF(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{{\left (c x^{4} + b x^{2} + a\right )}^{\frac{3}{2}}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(3/2)/x^2,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2 + a)^(3/2)/x^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c x^{4} + b x^{2} + a\right )}^{\frac{3}{2}}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(3/2)/x^2,x, algorithm="fricas")

[Out]

integral((c*x^4 + b*x^2 + a)^(3/2)/x^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{2} + c x^{4}\right )^{\frac{3}{2}}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2+a)**(3/2)/x**2,x)

[Out]

Integral((a + b*x**2 + c*x**4)**(3/2)/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{4} + b x^{2} + a\right )}^{\frac{3}{2}}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(3/2)/x^2,x, algorithm="giac")

[Out]

integrate((c*x^4 + b*x^2 + a)^(3/2)/x^2, x)