Integrand size = 19, antiderivative size = 116 \[ \int \frac {\sqrt {\frac {a}{x}}}{\sqrt {1+x^3}} \, dx=\frac {\sqrt {\frac {a}{x}} x (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1+\left (1-\sqrt {3}\right ) x}{1+\left (1+\sqrt {3}\right ) x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt [4]{3} \sqrt {\frac {x (1+x)}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {1+x^3}} \] Output:
1/3*(a/x)^(1/2)*x*(1+x)*((x^2-x+1)/(1+(1+3^(1/2))*x)^2)^(1/2)*InverseJacob iAM(arccos((1+(1-3^(1/2))*x)/(1+(1+3^(1/2))*x)),1/4*6^(1/2)+1/4*2^(1/2))*3 ^(3/4)/(x*(1+x)/(1+(1+3^(1/2))*x)^2)^(1/2)/(x^3+1)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.23 \[ \int \frac {\sqrt {\frac {a}{x}}}{\sqrt {1+x^3}} \, dx=2 \sqrt {\frac {a}{x}} x \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},-x^3\right ) \] Input:
Integrate[Sqrt[a/x]/Sqrt[1 + x^3],x]
Output:
2*Sqrt[a/x]*x*Hypergeometric2F1[1/6, 1/2, 7/6, -x^3]
Time = 0.39 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {34, 851, 766}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sqrt {\frac {a}{x}}}{\sqrt {x^3+1}} \, dx\) |
\(\Big \downarrow \) 34 |
\(\displaystyle \sqrt {x} \sqrt {\frac {a}{x}} \int \frac {1}{\sqrt {x} \sqrt {x^3+1}}dx\) |
\(\Big \downarrow \) 851 |
\(\displaystyle 2 \sqrt {x} \sqrt {\frac {a}{x}} \int \frac {1}{\sqrt {x^3+1}}d\sqrt {x}\) |
\(\Big \downarrow \) 766 |
\(\displaystyle \frac {x (x+1) \sqrt {\frac {x^2-x+1}{\left (\left (1+\sqrt {3}\right ) x+1\right )^2}} \sqrt {\frac {a}{x}} \operatorname {EllipticF}\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) x+1}{\left (1+\sqrt {3}\right ) x+1}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt [4]{3} \sqrt {\frac {x (x+1)}{\left (\left (1+\sqrt {3}\right ) x+1\right )^2}} \sqrt {x^3+1}}\) |
Input:
Int[Sqrt[a/x]/Sqrt[1 + x^3],x]
Output:
(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])
Int[(u_.)*((a_.)*(x_)^(m_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*x^m)^F racPart[p]/x^(m*FracPart[p])) Int[u*x^(m*p), x], x] /; FreeQ[{a, m, p}, x ] && !IntegerQ[p]
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 ]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/c Subst[Int[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]
Time = 0.75 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.19
method | result | size |
meijerg | \(2 \sqrt {\frac {a}{x}}\, x \operatorname {hypergeom}\left (\left [\frac {1}{6}, \frac {1}{2}\right ], \left [\frac {7}{6}\right ], -x^{3}\right )\) | \(22\) |
default | \(\frac {4 \sqrt {\frac {a}{x}}\, x \sqrt {x^{3}+1}\, \left (1+i \sqrt {3}\right ) \sqrt {\frac {\left (i \sqrt {3}+3\right ) x}{\left (1+i \sqrt {3}\right ) \left (x +1\right )}}\, \left (x +1\right )^{2} \sqrt {\frac {i \sqrt {3}+2 x -1}{\left (-1+i \sqrt {3}\right ) \left (x +1\right )}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{\left (1+i \sqrt {3}\right ) \left (x +1\right )}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (i \sqrt {3}+3\right ) x}{\left (1+i \sqrt {3}\right ) \left (x +1\right )}}, \sqrt {\frac {\left (i \sqrt {3}-3\right ) \left (1+i \sqrt {3}\right )}{\left (-1+i \sqrt {3}\right ) \left (i \sqrt {3}+3\right )}}\right )}{\sqrt {x \left (x^{3}+1\right )}\, \left (i \sqrt {3}+3\right ) \sqrt {-x \left (x +1\right ) \left (i \sqrt {3}+2 x -1\right ) \left (i \sqrt {3}-2 x +1\right )}}\) | \(232\) |
Input:
int((a/x)^(1/2)/(x^3+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
2*(a/x)^(1/2)*x*hypergeom([1/6,1/2],[7/6],-x^3)
Time = 0.10 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.09 \[ \int \frac {\sqrt {\frac {a}{x}}}{\sqrt {1+x^3}} \, dx=-2 \, \sqrt {a} {\rm weierstrassPInverse}\left (0, -4, \frac {1}{x}\right ) \] Input:
integrate((a/x)^(1/2)/(x^3+1)^(1/2),x, algorithm="fricas")
Output:
-2*sqrt(a)*weierstrassPInverse(0, -4, 1/x)
\[ \int \frac {\sqrt {\frac {a}{x}}}{\sqrt {1+x^3}} \, dx=\int \frac {\sqrt {\frac {a}{x}}}{\sqrt {\left (x + 1\right ) \left (x^{2} - x + 1\right )}}\, dx \] Input:
integrate((a/x)**(1/2)/(x**3+1)**(1/2),x)
Output:
Integral(sqrt(a/x)/sqrt((x + 1)*(x**2 - x + 1)), x)
\[ \int \frac {\sqrt {\frac {a}{x}}}{\sqrt {1+x^3}} \, dx=\int { \frac {\sqrt {\frac {a}{x}}}{\sqrt {x^{3} + 1}} \,d x } \] Input:
integrate((a/x)^(1/2)/(x^3+1)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(a/x)/sqrt(x^3 + 1), x)
\[ \int \frac {\sqrt {\frac {a}{x}}}{\sqrt {1+x^3}} \, dx=\int { \frac {\sqrt {\frac {a}{x}}}{\sqrt {x^{3} + 1}} \,d x } \] Input:
integrate((a/x)^(1/2)/(x^3+1)^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(a/x)/sqrt(x^3 + 1), x)
Timed out. \[ \int \frac {\sqrt {\frac {a}{x}}}{\sqrt {1+x^3}} \, dx=\int \frac {\sqrt {\frac {a}{x}}}{\sqrt {x^3+1}} \,d x \] Input:
int((a/x)^(1/2)/(x^3 + 1)^(1/2),x)
Output:
int((a/x)^(1/2)/(x^3 + 1)^(1/2), x)
\[ \int \frac {\sqrt {\frac {a}{x}}}{\sqrt {1+x^3}} \, dx=\sqrt {a}\, \left (\int \frac {\sqrt {x}\, \sqrt {x^{3}+1}}{x^{4}+x}d x \right ) \] Input:
int((a/x)^(1/2)/(x^3+1)^(1/2),x)
Output:
sqrt(a)*int((sqrt(x)*sqrt(x**3 + 1))/(x**4 + x),x)