Integrand size = 15, antiderivative size = 245 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^5} \, dx=\frac {2}{3 b \sqrt {a+\frac {b}{x^3}} x}-\frac {4 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}+\frac {b^{2/3}}{x^2}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} b^{4/3} \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}} \] Output:
2/3/b/(a+b/x^3)^(1/2)/x-4/9*(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)/b^(4/3)/(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 = 10.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.24 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^5} \, dx=\frac {2+4 \sqrt {1+\frac {a x^3}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},-\frac {a x^3}{b}\right )}{3 b \sqrt {a+\frac {b}{x^3}} x} \] Input:
Integrate[1/((a + b/x^3)^(3/2)*x^5),x]
Output:
(2 + 4*Sqrt[1 + (a*x^3)/b]*Hypergeometric2F1[1/6, 1/2, 7/6, -((a*x^3)/b)]) /(3*b*Sqrt[a + b/x^3]*x)
Time = 0.46 (sec) , antiderivative size = 245, 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, 817, 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.
\(\displaystyle \int \frac {1}{x^5 \left (a+\frac {b}{x^3}\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^3}d\frac {1}{x}\) |
\(\Big \downarrow \) 817 |
\(\displaystyle \frac {2}{3 b x \sqrt {a+\frac {b}{x^3}}}-\frac {2 \int \frac {1}{\sqrt {a+\frac {b}{x^3}}}d\frac {1}{x}}{3 b}\) |
\(\Big \downarrow \) 759 |
\(\displaystyle \frac {2}{3 b x \sqrt {a+\frac {b}{x^3}}}-\frac {4 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} b^{4/3} \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}\) |
Input:
Int[1/((a + b/x^3)^(3/2)*x^5),x]
Output:
2/(3*b*Sqrt[a + b/x^3]*x) - (4*Sqrt[2 + Sqrt[3]]*(a^(1/3) + b^(1/3)/x)*Sqr t[(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 + Sq rt[3])*a^(1/3) + b^(1/3)/x)], -7 - 4*Sqrt[3]])/(3*3^(1/4)*b^(4/3)*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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n *((m - n + 1)/(b*n*(p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x ] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && ! ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
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. 1824 vs. \(2 (186 ) = 372\).
Time = 1.82 (sec) , antiderivative size = 1825, normalized size of antiderivative = 7.45
Input:
int(1/(a+b/x^3)^(3/2)/x^5,x,method=_RETURNVERBOSE)
Output:
-2/3/((a*x^3+b)/x^3)^(3/2)/x^5*(a*x^3+b)/a/(-a^2*b)^(1/3)/b*(4*I*(x*(a*x^3 +b))^(1/2)*3^(1/2)*(-(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^(1/2))/(-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))*a^2*x^2-8*I*(x*(a*x^3+b))^(1/2)*(-a^2*b) ^(1/3)*3^(1/2)*(-(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^(1/2))/(-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))*a*x+4*I*(x*(a*x^3+b))^(1/2)*(-a^2*b)^(2/3)*3 ^(1/2)*(-(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^(1/2))/(-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))-4*(-(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 = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.24 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^5} \, dx=\frac {2 \, {\left (b x^{2} \sqrt {\frac {a x^{3} + b}{x^{3}}} - 2 \, {\left (a x^{3} + b\right )} \sqrt {b} {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, \frac {1}{x}\right )\right )}}{3 \, {\left (a b^{2} x^{3} + b^{3}\right )}} \] Input:
integrate(1/(a+b/x^3)^(3/2)/x^5,x, algorithm="fricas")
Output:
2/3*(b*x^2*sqrt((a*x^3 + b)/x^3) - 2*(a*x^3 + b)*sqrt(b)*weierstrassPInver se(0, -4*a/b, 1/x))/(a*b^2*x^3 + b^3)
Time = 0.77 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.16 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^5} \, dx=- \frac {\Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, \frac {3}{2} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{3}}} \right )}}{3 a^{\frac {3}{2}} x^{4} \Gamma \left (\frac {7}{3}\right )} \] Input:
integrate(1/(a+b/x**3)**(3/2)/x**5,x)
Output:
-gamma(4/3)*hyper((4/3, 3/2), (7/3,), b*exp_polar(I*pi)/(a*x**3))/(3*a**(3 /2)*x**4*gamma(7/3))
\[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^5} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{2}} x^{5}} \,d x } \] Input:
integrate(1/(a+b/x^3)^(3/2)/x^5,x, algorithm="maxima")
Output:
integrate(1/((a + b/x^3)^(3/2)*x^5), x)
\[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^5} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{2}} x^{5}} \,d x } \] Input:
integrate(1/(a+b/x^3)^(3/2)/x^5,x, algorithm="giac")
Output:
integrate(1/((a + b/x^3)^(3/2)*x^5), x)
Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^5} \, dx=\int \frac {1}{x^5\,{\left (a+\frac {b}{x^3}\right )}^{3/2}} \,d x \] Input:
int(1/(x^5*(a + b/x^3)^(3/2)),x)
Output:
int(1/(x^5*(a + b/x^3)^(3/2)), x)
\[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^5} \, dx=\int \frac {\sqrt {x}\, \sqrt {a \,x^{3}+b}}{a^{2} x^{7}+2 a b \,x^{4}+b^{2} x}d x \] Input:
int(1/(a+b/x^3)^(3/2)/x^5,x)
Output:
int((sqrt(x)*sqrt(a*x**3 + b))/(a**2*x**7 + 2*a*b*x**4 + b**2*x),x)