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

Optimal. Leaf size=116 \[ \frac{x (x+1) \sqrt{\frac{x^2-x+1}{\left (\left (1+\sqrt{3}\right ) x+1\right )^2}} \sqrt{\frac{a}{x}} F\left (\cos ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) x+1}{\left (1+\sqrt{3}\right ) x+1}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{\sqrt [4]{3} \sqrt{\frac{x (x+1)}{\left (\left (1+\sqrt{3}\right ) x+1\right )^2}} \sqrt{x^3+1}} \]

[Out]

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

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Rubi [A]  time = 0.073026, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {15, 329, 225} \[ \frac{x (x+1) \sqrt{\frac{x^2-x+1}{\left (\left (1+\sqrt{3}\right ) x+1\right )^2}} \sqrt{\frac{a}{x}} F\left (\cos ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) x+1}{\left (1+\sqrt{3}\right ) x+1}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{\sqrt [4]{3} \sqrt{\frac{x (x+1)}{\left (\left (1+\sqrt{3}\right ) x+1\right )^2}} \sqrt{x^3+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a/x]/Sqrt[1 + x^3],x]

[Out]

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

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

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 225

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(x*(s
+ r*x^2)*Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2
)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[(r*x^2*(s + r*x^2))/(s + (1
+ Sqrt[3])*r*x^2)^2]), x]] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{\sqrt{\frac{a}{x}}}{\sqrt{1+x^3}} \, dx &=\left (\sqrt{\frac{a}{x}} \sqrt{x}\right ) \int \frac{1}{\sqrt{x} \sqrt{1+x^3}} \, dx\\ &=\left (2 \sqrt{\frac{a}{x}} \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^6}} \, dx,x,\sqrt{x}\right )\\ &=\frac{\sqrt{\frac{a}{x}} x (1+x) \sqrt{\frac{1-x+x^2}{\left (1+\left (1+\sqrt{3}\right ) x\right )^2}} F\left (\cos ^{-1}\left (\frac{1+\left (1-\sqrt{3}\right ) x}{1+\left (1+\sqrt{3}\right ) x}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{\sqrt [4]{3} \sqrt{\frac{x (1+x)}{\left (1+\left (1+\sqrt{3}\right ) x\right )^2}} \sqrt{1+x^3}}\\ \end{align*}

Mathematica [C]  time = 0.005088, size = 27, normalized size = 0.23 \[ 2 x \sqrt{\frac{a}{x}} \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};-x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a/x]/Sqrt[1 + x^3],x]

[Out]

2*Sqrt[a/x]*x*Hypergeometric2F1[1/6, 1/2, 7/6, -x^3]

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Maple [C]  time = 0.105, size = 232, normalized size = 2. \begin{align*} 4\,{\frac{x\sqrt{{x}^{3}+1} \left ( 1+i\sqrt{3} \right ) \left ( 1+x \right ) ^{2}}{\sqrt{x \left ({x}^{3}+1 \right ) } \left ( 3+i\sqrt{3} \right ) \sqrt{-x \left ( 1+x \right ) \left ( i\sqrt{3}+2\,x-1 \right ) \left ( i\sqrt{3}-2\,x+1 \right ) }}\sqrt{{\frac{a}{x}}}\sqrt{{\frac{ \left ( 3+i\sqrt{3} \right ) x}{ \left ( 1+i\sqrt{3} \right ) \left ( 1+x \right ) }}}\sqrt{{\frac{i\sqrt{3}+2\,x-1}{ \left ( i\sqrt{3}-1 \right ) \left ( 1+x \right ) }}}\sqrt{{\frac{i\sqrt{3}-2\,x+1}{ \left ( 1+i\sqrt{3} \right ) \left ( 1+x \right ) }}}{\it EllipticF} \left ( \sqrt{{\frac{ \left ( 3+i\sqrt{3} \right ) x}{ \left ( 1+i\sqrt{3} \right ) \left ( 1+x \right ) }}},\sqrt{{\frac{ \left ( i\sqrt{3}-3 \right ) \left ( 1+i\sqrt{3} \right ) }{ \left ( i\sqrt{3}-1 \right ) \left ( 3+i\sqrt{3} \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a/x)/sqrt(x^3 + 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{\frac{a}{x}}}{\sqrt{x^{3} + 1}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a/x)/sqrt(x^3 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a/x)/sqrt(x^3 + 1), x)