Integrand size = 15, antiderivative size = 248 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^2} \, dx=-\frac {2}{3 a \sqrt {a+\frac {b}{x^3}} x}-\frac {2 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}+\frac {b^{2/3}}{x^2}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} a \sqrt [3]{b} \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}} \] Output:
-2/3/a/(a+b/x^3)^(1/2)/x-2/9*(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)/a/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 = 57, normalized size of antiderivative = 0.23 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^2} \, dx=\frac {2 \left (-1+\sqrt {1+\frac {a x^3}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},-\frac {a x^3}{b}\right )\right )}{3 a \sqrt {a+\frac {b}{x^3}} x} \] Input:
Integrate[1/((a + b/x^3)^(3/2)*x^2),x]
Output:
(2*(-1 + Sqrt[1 + (a*x^3)/b]*Hypergeometric2F1[1/6, 1/2, 7/6, -((a*x^3)/b) ]))/(3*a*Sqrt[a + b/x^3]*x)
Time = 0.45 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {858, 749, 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.
\(\displaystyle \int \frac {1}{x^2 \left (a+\frac {b}{x^3}\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2}}d\frac {1}{x}\) |
\(\Big \downarrow \) 749 |
\(\displaystyle -\frac {\int \frac {1}{\sqrt {a+\frac {b}{x^3}}}d\frac {1}{x}}{3 a}-\frac {2}{3 a x \sqrt {a+\frac {b}{x^3}}}\) |
\(\Big \downarrow \) 759 |
\(\displaystyle -\frac {2 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} a \sqrt [3]{b} \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}-\frac {2}{3 a x \sqrt {a+\frac {b}{x^3}}}\) |
Input:
Int[1/((a + b/x^3)^(3/2)*x^2),x]
Output:
-2/(3*a*Sqrt[a + b/x^3]*x) - (2*Sqrt[2 + Sqrt[3]]*(a^(1/3) + b^(1/3)/x)*Sq rt[(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[ArcSin[((1 - Sqrt[3])*a^(1/3) + b^(1/3)/x)/((1 + S qrt[3])*a^(1/3) + b^(1/3)/x)], -7 - 4*Sqrt[3]])/(3*3^(1/4)*a*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[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Simp[(n*(p + 1) + 1)/(a*n*(p + 1)) Int[(a + b*x^ n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (Inte gerQ[2*p] || Denominator[p + 1/n] < Denominator[p])
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. 1821 vs. \(2 (189 ) = 378\).
Time = 0.52 (sec) , antiderivative size = 1822, normalized size of antiderivative = 7.35
Input:
int(1/(a+b/x^3)^(3/2)/x^2,x,method=_RETURNVERBOSE)
Output:
-2/3/((a*x^3+b)/x^3)^(3/2)/x^5*(a*x^3+b)/a^2/(-a^2*b)^(1/3)*(2*I*(-(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))*(x*(a*x^3+b))^(1/2)*3^(1/2)*a^2*x^2-4*I*(-(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)*Ell ipticF((-(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))*(x*(a*x^3+b)) ^(1/2)*(-a^2*b)^(1/3)*3^(1/2)*a*x+2*I*(-(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))*(x*(a*x^3+b))^(1/2)*( -a^2*b)^(2/3)*3^(1/2)-2*(-(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...
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.24 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^2} \, dx=-\frac {2 \, {\left (b x^{2} \sqrt {\frac {a x^{3} + b}{x^{3}}} + {\left (a x^{3} + b\right )} \sqrt {b} {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, \frac {1}{x}\right )\right )}}{3 \, {\left (a^{2} b x^{3} + a b^{2}\right )}} \] Input:
integrate(1/(a+b/x^3)^(3/2)/x^2,x, algorithm="fricas")
Output:
-2/3*(b*x^2*sqrt((a*x^3 + b)/x^3) + (a*x^3 + b)*sqrt(b)*weierstrassPInvers e(0, -4*a/b, 1/x))/(a^2*b*x^3 + a*b^2)
Time = 0.75 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.15 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^2} \, dx=- \frac {\Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {3}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{3}}} \right )}}{3 a^{\frac {3}{2}} x \Gamma \left (\frac {4}{3}\right )} \] Input:
integrate(1/(a+b/x**3)**(3/2)/x**2,x)
Output:
-gamma(1/3)*hyper((1/3, 3/2), (4/3,), b*exp_polar(I*pi)/(a*x**3))/(3*a**(3 /2)*x*gamma(4/3))
\[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^2} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{2}} x^{2}} \,d x } \] Input:
integrate(1/(a+b/x^3)^(3/2)/x^2,x, algorithm="maxima")
Output:
integrate(1/((a + b/x^3)^(3/2)*x^2), x)
\[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^2} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{2}} x^{2}} \,d x } \] Input:
integrate(1/(a+b/x^3)^(3/2)/x^2,x, algorithm="giac")
Output:
integrate(1/((a + b/x^3)^(3/2)*x^2), x)
Time = 0.67 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.16 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^2} \, dx=-\frac {{\left (\frac {b}{a\,x^3}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {3}{2};\ \frac {4}{3};\ -\frac {b}{a\,x^3}\right )}{x\,{\left (a+\frac {b}{x^3}\right )}^{3/2}} \] Input:
int(1/(x^2*(a + b/x^3)^(3/2)),x)
Output:
-((b/(a*x^3) + 1)^(3/2)*hypergeom([1/3, 3/2], 4/3, -b/(a*x^3)))/(x*(a + b/ x^3)^(3/2))
\[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^2} \, dx=\frac {-2 \sqrt {x}\, \sqrt {a \,x^{3}+b}+\left (\int \frac {\sqrt {x}\, \sqrt {a \,x^{3}+b}}{a^{2} x^{7}+2 a b \,x^{4}+b^{2} x}d x \right ) a b \,x^{3}+\left (\int \frac {\sqrt {x}\, \sqrt {a \,x^{3}+b}}{a^{2} x^{7}+2 a b \,x^{4}+b^{2} x}d x \right ) b^{2}}{2 a \left (a \,x^{3}+b \right )} \] Input:
int(1/(a+b/x^3)^(3/2)/x^2,x)
Output:
( - 2*sqrt(x)*sqrt(a*x**3 + b) + int((sqrt(x)*sqrt(a*x**3 + b))/(a**2*x**7 + 2*a*b*x**4 + b**2*x),x)*a*b*x**3 + int((sqrt(x)*sqrt(a*x**3 + b))/(a**2 *x**7 + 2*a*b*x**4 + b**2*x),x)*b**2)/(2*a*(a*x**3 + b))