3.63 \(\int \frac{1}{x^2 \sqrt{a x+b x^3}} \, dx\)

Optimal. Leaf size=119 \[ -\frac{b^{3/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{3 a^{5/4} \sqrt{a x+b x^3}}-\frac{2 \sqrt{a x+b x^3}}{3 a x^2} \]

[Out]

(-2*Sqrt[a*x + b*x^3])/(3*a*x^2) - (b^(3/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*
x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(3*a^(5/4)*Sqrt[a*x + b*x^3])

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Rubi [A]  time = 0.0893417, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {2025, 2011, 329, 220} \[ -\frac{b^{3/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{3 a^{5/4} \sqrt{a x+b x^3}}-\frac{2 \sqrt{a x+b x^3}}{3 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*Sqrt[a*x + b*x^3]),x]

[Out]

(-2*Sqrt[a*x + b*x^3])/(3*a*x^2) - (b^(3/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*
x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(3*a^(5/4)*Sqrt[a*x + b*x^3])

Rule 2025

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rule 2011

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 220

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

Rubi steps

\begin{align*} \int \frac{1}{x^2 \sqrt{a x+b x^3}} \, dx &=-\frac{2 \sqrt{a x+b x^3}}{3 a x^2}-\frac{b \int \frac{1}{\sqrt{a x+b x^3}} \, dx}{3 a}\\ &=-\frac{2 \sqrt{a x+b x^3}}{3 a x^2}-\frac{\left (b \sqrt{x} \sqrt{a+b x^2}\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x^2}} \, dx}{3 a \sqrt{a x+b x^3}}\\ &=-\frac{2 \sqrt{a x+b x^3}}{3 a x^2}-\frac{\left (2 b \sqrt{x} \sqrt{a+b x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^4}} \, dx,x,\sqrt{x}\right )}{3 a \sqrt{a x+b x^3}}\\ &=-\frac{2 \sqrt{a x+b x^3}}{3 a x^2}-\frac{b^{3/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{3 a^{5/4} \sqrt{a x+b x^3}}\\ \end{align*}

Mathematica [C]  time = 0.0150918, size = 53, normalized size = 0.45 \[ -\frac{2 \sqrt{\frac{b x^2}{a}+1} \, _2F_1\left (-\frac{3}{4},\frac{1}{2};\frac{1}{4};-\frac{b x^2}{a}\right )}{3 x \sqrt{x \left (a+b x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*Sqrt[a*x + b*x^3]),x]

[Out]

(-2*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[-3/4, 1/2, 1/4, -((b*x^2)/a)])/(3*x*Sqrt[x*(a + b*x^2)])

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Maple [A]  time = 0.014, size = 129, normalized size = 1.1 \begin{align*} -{\frac{2}{3\,a{x}^{2}}\sqrt{b{x}^{3}+ax}}-{\frac{1}{3\,a}\sqrt{-ab}\sqrt{{b \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{-2\,{\frac{b}{\sqrt{-ab}} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }}\sqrt{-{bx{\frac{1}{\sqrt{-ab}}}}}{\it EllipticF} \left ( \sqrt{{b \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ){\frac{1}{\sqrt{-ab}}}}},{\frac{\sqrt{2}}{2}} \right ){\frac{1}{\sqrt{b{x}^{3}+ax}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(b*x^3+a*x)^(1/2)/a/x^2-1/3/a*(-a*b)^(1/2)*((x+1/b*(-a*b)^(1/2))*b/(-a*b)^(1/2))^(1/2)*(-2*(x-1/b*(-a*b)^
(1/2))*b/(-a*b)^(1/2))^(1/2)*(-x*b/(-a*b)^(1/2))^(1/2)/(b*x^3+a*x)^(1/2)*EllipticF(((x+1/b*(-a*b)^(1/2))*b/(-a
*b)^(1/2))^(1/2),1/2*2^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b x^{3} + a x} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(b*x^3 + a*x)*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \sqrt{x \left (a + b x^{2}\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**2*sqrt(x*(a + b*x**2))), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b x^{3} + a x} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(b*x^3 + a*x)*x^2), x)