3.114 \(\int \frac{\sqrt{x}}{\sqrt{a x+b x^3+c x^5}} \, dx\)

Optimal. Leaf size=121 \[ \frac{\sqrt{x} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt{a x+b x^3+c x^5}} \]

[Out]

(Sqrt[x]*(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])/(2*a^(1/4)*c^(1/4)*Sqrt[a*x + b*x^3 + c*x^5])

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Rubi [A]  time = 0.0463985, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {1914, 1103} \[ \frac{\sqrt{x} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt{a x+b x^3+c x^5}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/Sqrt[a*x + b*x^3 + c*x^5],x]

[Out]

(Sqrt[x]*(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])/(2*a^(1/4)*c^(1/4)*Sqrt[a*x + b*x^3 + c*x^5])

Rule 1914

Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[(x^(q/2)*Sqrt[a
+ b*x^(n - q) + c*x^(2*(n - q))])/Sqrt[a*x^q + b*x^n + c*x^(2*n - q)], Int[x^(m - q/2)/Sqrt[a + b*x^(n - q) +
c*x^(2*(n - q))], x], x] /; FreeQ[{a, b, c, m, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && ((EqQ[m, 1] &&
EqQ[n, 3] && EqQ[q, 2]) || ((EqQ[m + 1/2] || EqQ[m, 3/2] || EqQ[m, 1/2] || EqQ[m, 5/2]) && EqQ[n, 3] && EqQ[q,
1]))

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]

Rubi steps

\begin{align*} \int \frac{\sqrt{x}}{\sqrt{a x+b x^3+c x^5}} \, dx &=\frac{\left (\sqrt{x} \sqrt{a+b x^2+c x^4}\right ) \int \frac{1}{\sqrt{a+b x^2+c x^4}} \, dx}{\sqrt{a x+b x^3+c x^5}}\\ &=\frac{\sqrt{x} \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 )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt{a x+b x^3+c x^5}}\\ \end{align*}

Mathematica [C]  time = 0.126265, size = 193, normalized size = 1.6 \[ -\frac{i \sqrt{x} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} x \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}}\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right )}{\sqrt{2} \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \sqrt{x \left (a+b x^2+c x^4\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/Sqrt[a*x + b*x^3 + c*x^5],x]

[Out]

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

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Maple [A]  time = 0.016, size = 177, normalized size = 1.5 \begin{align*}{\frac{1}{2\,c{x}^{4}+2\,b{x}^{2}+2\,a}\sqrt{x \left ( c{x}^{4}+b{x}^{2}+a \right ) }\sqrt{-2\,{\frac{{x}^{2}\sqrt{-4\,ac+{b}^{2}}-b{x}^{2}-2\,a}{a}}}\sqrt{{\frac{1}{a} \left ({x}^{2}\sqrt{-4\,ac+{b}^{2}}+b{x}^{2}+2\,a \right ) }}{\it EllipticF} \left ({\frac{x\sqrt{2}}{2}\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}},{\frac{\sqrt{2}}{2}\sqrt{{\frac{1}{ac} \left ( b\sqrt{-4\,ac+{b}^{2}}-2\,ac+{b}^{2} \right ) }}} \right ){\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(c*x^5+b*x^3+a*x)^(1/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x}}{\sqrt{c x^{5} + b x^{3} + a x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(c*x^5+b*x^3+a*x)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(x)/sqrt(c*x^5 + b*x^3 + a*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x}}{\sqrt{c x^{5} + b x^{3} + a x}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(c*x^5+b*x^3+a*x)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(x)/sqrt(c*x^5 + b*x^3 + a*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(1/2)/(c*x**5+b*x**3+a*x)**(1/2),x)

[Out]

Integral(sqrt(x)/sqrt(x*(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{\sqrt{x}}{\sqrt{c x^{5} + b x^{3} + a x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(c*x^5+b*x^3+a*x)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(x)/sqrt(c*x^5 + b*x^3 + a*x), x)