Integrand size = 15, antiderivative size = 221 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^3}} x^2} \, dx=-\frac {2 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}+\frac {b^{2/3}}{x^2}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt [3]{b} \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}} \] Output:
-2/3*(1/2*6^(1/2)+1/2*2^(1/2))*(a^(1/3)+b^(1/3)/x)*((a^(2/3)+b^(2/3)/x^2-a ^(1/3)*b^(1/3)/x)/((1+3^(1/2))*a^(1/3)+b^(1/3)/x)^2)^(1/2)*EllipticF(((1-3 ^(1/2))*a^(1/3)+b^(1/3)/x)/((1+3^(1/2))*a^(1/3)+b^(1/3)/x),I*3^(1/2)+2*I)* 3^(3/4)/b^(1/3)/(a+b/x^3)^(1/2)/(a^(1/3)*(a^(1/3)+b^(1/3)/x)/((1+3^(1/2))* a^(1/3)+b^(1/3)/x)^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.22 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^3}} x^2} \, dx=\frac {2 \sqrt {1+\frac {a x^3}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},-\frac {a x^3}{b}\right )}{\sqrt {a+\frac {b}{x^3}} x} \] Input:
Integrate[1/(Sqrt[a + b/x^3]*x^2),x]
Output:
(2*Sqrt[1 + (a*x^3)/b]*Hypergeometric2F1[1/6, 1/2, 7/6, -((a*x^3)/b)])/(Sq rt[a + b/x^3]*x)
Time = 0.42 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {858, 759}
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.
\(\displaystyle \int \frac {1}{x^2 \sqrt {a+\frac {b}{x^3}}} \, dx\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\int \frac {1}{\sqrt {a+\frac {b}{x^3}}}d\frac {1}{x}\) |
\(\Big \downarrow \) 759 |
\(\displaystyle -\frac {2 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt [3]{b} \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}\) |
Input:
Int[1/(Sqrt[a + b/x^3]*x^2),x]
Output:
(-2*Sqrt[2 + Sqrt[3]]*(a^(1/3) + b^(1/3)/x)*Sqrt[(a^(2/3) + b^(2/3)/x^2 - (a^(1/3)*b^(1/3))/x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)/x)^2]*EllipticF[ArcS in[((1 - Sqrt[3])*a^(1/3) + b^(1/3)/x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)/x) ], -7 - 4*Sqrt[3]])/(3^(1/4)*b^(1/3)*Sqrt[a + b/x^3]*Sqrt[(a^(1/3)*(a^(1/3 ) + b^(1/3)/x))/((1 + Sqrt[3])*a^(1/3) + b^(1/3)/x)^2])
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[s* ((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] & & PosQ[a]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int egerQ[m]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 436 vs. \(2 (168 ) = 336\).
Time = 0.90 (sec) , antiderivative size = 437, normalized size of antiderivative = 1.98
method | result | size |
default | \(-\frac {4 \left (a \,x^{3}+b \right ) \sqrt {-\frac {\left (i \sqrt {3}-3\right ) x a}{\left (i \sqrt {3}-1\right ) \left (-a x +\left (-a^{2} b \right )^{\frac {1}{3}}\right )}}\, \sqrt {\frac {i \sqrt {3}\, \left (-a^{2} b \right )^{\frac {1}{3}}+2 a x +\left (-a^{2} b \right )^{\frac {1}{3}}}{\left (1+i \sqrt {3}\right ) \left (-a x +\left (-a^{2} b \right )^{\frac {1}{3}}\right )}}\, \sqrt {\frac {i \sqrt {3}\, \left (-a^{2} b \right )^{\frac {1}{3}}-2 a x -\left (-a^{2} b \right )^{\frac {1}{3}}}{\left (i \sqrt {3}-1\right ) \left (-a x +\left (-a^{2} b \right )^{\frac {1}{3}}\right )}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {\left (i \sqrt {3}-3\right ) x a}{\left (i \sqrt {3}-1\right ) \left (-a x +\left (-a^{2} b \right )^{\frac {1}{3}}\right )}}, \sqrt {\frac {\left (i \sqrt {3}+3\right ) \left (i \sqrt {3}-1\right )}{\left (1+i \sqrt {3}\right ) \left (i \sqrt {3}-3\right )}}\right ) \left (i \sqrt {3}\, a^{2} x^{2}-2 i \left (-a^{2} b \right )^{\frac {1}{3}} \sqrt {3}\, a x +i \left (-a^{2} b \right )^{\frac {2}{3}} \sqrt {3}-a^{2} x^{2}+2 \left (-a^{2} b \right )^{\frac {1}{3}} a x -\left (-a^{2} b \right )^{\frac {2}{3}}\right )}{\sqrt {\frac {a \,x^{3}+b}{x^{3}}}\, x a \left (-a^{2} b \right )^{\frac {1}{3}} \sqrt {x \left (a \,x^{3}+b \right )}\, \left (i \sqrt {3}-3\right ) \sqrt {\frac {x \left (-a x +\left (-a^{2} b \right )^{\frac {1}{3}}\right ) \left (i \sqrt {3}\, \left (-a^{2} b \right )^{\frac {1}{3}}+2 a x +\left (-a^{2} b \right )^{\frac {1}{3}}\right ) \left (i \sqrt {3}\, \left (-a^{2} b \right )^{\frac {1}{3}}-2 a x -\left (-a^{2} b \right )^{\frac {1}{3}}\right )}{a^{2}}}}\) | \(437\) |
Input:
int(1/(a+b/x^3)^(1/2)/x^2,x,method=_RETURNVERBOSE)
Output:
-4/((a*x^3+b)/x^3)^(1/2)/x*(a*x^3+b)/a/(-a^2*b)^(1/3)*(-(I*3^(1/2)-3)*x*a/ (I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)+2*a* x+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*( -a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^( 1/2)*EllipticF((-(I*3^(1/2)-3)*x*a/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1 /2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(I*3^ (1/2)*a^2*x^2-2*I*(-a^2*b)^(1/3)*3^(1/2)*a*x+I*(-a^2*b)^(2/3)*3^(1/2)-a^2* x^2+2*(-a^2*b)^(1/3)*a*x-(-a^2*b)^(2/3))/(x*(a*x^3+b))^(1/2)/(I*3^(1/2)-3) /(1/a^2*x*(-a*x+(-a^2*b)^(1/3))*(I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^( 1/3))*(I*3^(1/2)*(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3)))^(1/2)
Time = 0.07 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.07 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^3}} x^2} \, dx=-\frac {2 \, {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, \frac {1}{x}\right )}{\sqrt {b}} \] Input:
integrate(1/(a+b/x^3)^(1/2)/x^2,x, algorithm="fricas")
Output:
-2*weierstrassPInverse(0, -4*a/b, 1/x)/sqrt(b)
Time = 0.55 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.17 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^3}} x^2} \, dx=- \frac {\Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{3}}} \right )}}{3 \sqrt {a} x \Gamma \left (\frac {4}{3}\right )} \] Input:
integrate(1/(a+b/x**3)**(1/2)/x**2,x)
Output:
-gamma(1/3)*hyper((1/3, 1/2), (4/3,), b*exp_polar(I*pi)/(a*x**3))/(3*sqrt( a)*x*gamma(4/3))
\[ \int \frac {1}{\sqrt {a+\frac {b}{x^3}} x^2} \, dx=\int { \frac {1}{\sqrt {a + \frac {b}{x^{3}}} x^{2}} \,d x } \] Input:
integrate(1/(a+b/x^3)^(1/2)/x^2,x, algorithm="maxima")
Output:
integrate(1/(sqrt(a + b/x^3)*x^2), x)
\[ \int \frac {1}{\sqrt {a+\frac {b}{x^3}} x^2} \, dx=\int { \frac {1}{\sqrt {a + \frac {b}{x^{3}}} x^{2}} \,d x } \] Input:
integrate(1/(a+b/x^3)^(1/2)/x^2,x, algorithm="giac")
Output:
integrate(1/(sqrt(a + b/x^3)*x^2), x)
Time = 0.50 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.18 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^3}} x^2} \, dx=-\frac {\sqrt {\frac {b}{a\,x^3}+1}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{2};\ \frac {4}{3};\ -\frac {b}{a\,x^3}\right )}{x\,\sqrt {a+\frac {b}{x^3}}} \] Input:
int(1/(x^2*(a + b/x^3)^(1/2)),x)
Output:
-((b/(a*x^3) + 1)^(1/2)*hypergeom([1/3, 1/2], 4/3, -b/(a*x^3)))/(x*(a + b/ x^3)^(1/2))
\[ \int \frac {1}{\sqrt {a+\frac {b}{x^3}} x^2} \, dx=\int \frac {\sqrt {x}\, \sqrt {a \,x^{3}+b}}{a \,x^{4}+b x}d x \] Input:
int(1/(a+b/x^3)^(1/2)/x^2,x)
Output:
int((sqrt(x)*sqrt(a*x**3 + b))/(a*x**4 + b*x),x)