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

Optimal. Leaf size=114 \[ \frac{\sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{2 a^{5/4} \sqrt [4]{b} \sqrt{a x+b x^3}}+\frac{x}{a \sqrt{a x+b x^3}} \]

[Out]

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

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Rubi [A]  time = 0.083586, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {2023, 2011, 329, 220} \[ \frac{\sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{5/4} \sqrt [4]{b} \sqrt{a x+b x^3}}+\frac{x}{a \sqrt{a x+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a*x + b*x^3)^(3/2),x]

[Out]

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

Rule 2023

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*(n - j)*(p + 1)), x] + Dist[(c^j*(m + n*p + n - j + 1))/(a*(n - j)*(p + 1)),
Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] &
& (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1]

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

Mathematica [C]  time = 0.0192975, size = 54, normalized size = 0.47 \[ \frac{x \sqrt{\frac{b x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{b x^2}{a}\right )+x}{a \sqrt{x \left (a+b x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a*x + b*x^3)^(3/2),x]

[Out]

(x + x*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^2)/a)])/(a*Sqrt[x*(a + b*x^2)])

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Maple [A]  time = 0.016, size = 132, normalized size = 1.2 \begin{align*}{\frac{x}{a}{\frac{1}{\sqrt{ \left ({\frac{a}{b}}+{x}^{2} \right ) bx}}}}+{\frac{1}{2\,ab}\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(x/(b*x^3+a*x)^(3/2),x)

[Out]

x/a/((1/b*a+x^2)*b*x)^(1/2)+1/2/a/b*(-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{x}{{\left (b x^{3} + a x\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(b*x^3 + a*x)^(3/2), 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^{2} x^{5} + 2 \, a b x^{3} + a^{2} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x^3 + a*x)/(b^2*x^5 + 2*a*b*x^3 + a^2*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(b*x^3 + a*x)^(3/2), x)