3.4.89 \(\int \frac {\sqrt {\frac {a}{x}}}{\sqrt {1+x^3}} \, dx\) [389]

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

[Out]

1/3*x*(1+x)*((1+x*(1-3^(1/2)))^2/(1+x*(1+3^(1/2)))^2)^(1/2)/(1+x*(1-3^(1/2)))*(1+x*(1+3^(1/2)))*EllipticF((1-(
1+x*(1-3^(1/2)))^2/(1+x*(1+3^(1/2)))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*(a/x)^(1/2)*((x^2-x+1)/(1+x*(1+3^(1/2))
)^2)^(1/2)*3^(3/4)/(x^3+1)^(1/2)/(x*(1+x)/(1+x*(1+3^(1/2)))^2)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 116, normalized size of antiderivative = 1.00, 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, 335, 231} \begin {gather*} \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 (\text {ArcCos}\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}} \end {gather*}

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 231

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]/(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)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^
2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x]

Rule 335

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]

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 ) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.02, size = 27, normalized size = 0.23 \begin {gather*} 2 \sqrt {\frac {a}{x}} x \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};-x^3\right ) \end {gather*}

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] Result contains complex when optimal does not.
time = 0.37, size = 232, normalized size = 2.00

method	result	size
meijerg	\(2 \sqrt {\frac {a}{x}}\, x \hypergeom \left (\left [\frac {1}{6}, \frac {1}{2}\right ], \left [\frac {7}{6}\right ], -x^{3}\right )\)	\(22\)
default	\(\frac {4 \sqrt {\frac {a}{x}}\, x \sqrt {x^{3}+1}\, \left (1+i \sqrt {3}\right ) \sqrt {\frac {\left (i \sqrt {3}+3\right ) x}{\left (1+i \sqrt {3}\right ) \left (1+x \right )}}\, \left (1+x \right )^{2} \sqrt {\frac {i \sqrt {3}+2 x -1}{\left (-1+i \sqrt {3}\right ) \left (1+x \right )}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{\left (1+i \sqrt {3}\right ) \left (1+x \right )}}\, \EllipticF \left (\sqrt {\frac {\left (i \sqrt {3}+3\right ) x}{\left (1+i \sqrt {3}\right ) \left (1+x \right )}}, \sqrt {\frac {\left (-3+i \sqrt {3}\right ) \left (1+i \sqrt {3}\right )}{\left (-1+i \sqrt {3}\right ) \left (i \sqrt {3}+3\right )}}\right )}{\sqrt {x \left (x^{3}+1\right )}\, \left (i \sqrt {3}+3\right ) \sqrt {-x \left (1+x \right ) \left (i \sqrt {3}+2 x -1\right ) \left (i \sqrt {3}-2 x +1\right )}}\)	\(232\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*(a/x)^(1/2)*x*(x^3+1)^(1/2)*(1+I*3^(1/2))*((I*3^(1/2)+3)*x/(1+I*3^(1/2))/(1+x))^(1/2)*(1+x)^2*((I*3^(1/2)+2*
x-1)/(-1+I*3^(1/2))/(1+x))^(1/2)*((I*3^(1/2)-2*x+1)/(1+I*3^(1/2))/(1+x))^(1/2)*EllipticF(((I*3^(1/2)+3)*x/(1+I
*3^(1/2))/(1+x))^(1/2),((-3+I*3^(1/2))*(1+I*3^(1/2))/(-1+I*3^(1/2))/(I*3^(1/2)+3))^(1/2))/(x*(x^3+1))^(1/2)/(I
*3^(1/2)+3)/(-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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.07, size = 11, normalized size = 0.09 \begin {gather*} -2 \, \sqrt {a} {\rm weierstrassPInverse}\left (0, -4, \frac {1}{x}\right ) \end {gather*}

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]

-2*sqrt(a)*weierstrassPInverse(0, -4, 1/x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {\frac {a}{x}}}{\sqrt {\left (x + 1\right ) \left (x^{2} - x + 1\right )}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {\frac {a}{x}}}{\sqrt {x^3+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a/x)^(1/2)/(x^3 + 1)^(1/2),x)

[Out]

int((a/x)^(1/2)/(x^3 + 1)^(1/2), x)

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