Integrand size = 15, antiderivative size = 267 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^8} \, dx=\frac {2}{3 b \sqrt {a+\frac {b}{x^3}} x^4}-\frac {16 \sqrt {a+\frac {b}{x^3}}}{15 b^2 x}+\frac {32 \sqrt {2+\sqrt {3}} a \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}+\frac {b^{2/3}}{x^2}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right ),-7-4 \sqrt {3}\right )}{15 \sqrt [4]{3} b^{7/3} \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}} \] Output:
2/3/b/(a+b/x^3)^(1/2)/x^4-16/15*(a+b/x^3)^(1/2)/b^2/x+32/45*(1/2*6^(1/2)+1 /2*2^(1/2))*a*(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^(7/3)/( 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 = 54, normalized size of antiderivative = 0.20 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^8} \, dx=-\frac {2 \sqrt {1+\frac {a x^3}{b}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {3}{2},\frac {1}{6},-\frac {a x^3}{b}\right )}{5 b \sqrt {a+\frac {b}{x^3}} x^4} \] Input:
Integrate[1/((a + b/x^3)^(3/2)*x^8),x]
Output:
(-2*Sqrt[1 + (a*x^3)/b]*Hypergeometric2F1[-5/6, 3/2, 1/6, -((a*x^3)/b)])/( 5*b*Sqrt[a + b/x^3]*x^4)
Time = 0.50 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {858, 817, 843, 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.
\(\displaystyle \int \frac {1}{x^8 \left (a+\frac {b}{x^3}\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^6}d\frac {1}{x}\) |
\(\Big \downarrow \) 817 |
\(\displaystyle \frac {2}{3 b x^4 \sqrt {a+\frac {b}{x^3}}}-\frac {8 \int \frac {1}{\sqrt {a+\frac {b}{x^3}} x^3}d\frac {1}{x}}{3 b}\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {2}{3 b x^4 \sqrt {a+\frac {b}{x^3}}}-\frac {8 \left (\frac {2 \sqrt {a+\frac {b}{x^3}}}{5 b x}-\frac {2 a \int \frac {1}{\sqrt {a+\frac {b}{x^3}}}d\frac {1}{x}}{5 b}\right )}{3 b}\) |
\(\Big \downarrow \) 759 |
\(\displaystyle \frac {2}{3 b x^4 \sqrt {a+\frac {b}{x^3}}}-\frac {8 \left (\frac {2 \sqrt {a+\frac {b}{x^3}}}{5 b x}-\frac {4 \sqrt {2+\sqrt {3}} a \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} b^{4/3} \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}\right )}{3 b}\) |
Input:
Int[1/((a + b/x^3)^(3/2)*x^8),x]
Output:
2/(3*b*Sqrt[a + b/x^3]*x^4) - (8*((2*Sqrt[a + b/x^3])/(5*b*x) - (4*Sqrt[2 + Sqrt[3]]*a*(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[ArcSin[((1 - Sqrt[3])*a^(1/3) + b^(1/3)/x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)/x)], -7 - 4 *Sqrt[3]])/(5*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])))/(3*b)
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[((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*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 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. 1430 vs. \(2 (204 ) = 408\).
Time = 3.39 (sec) , antiderivative size = 1431, normalized size of antiderivative = 5.36
method | result | size |
risch | \(\text {Expression too large to display}\) | \(1431\) |
default | \(\text {Expression too large to display}\) | \(2056\) |
Input:
int(1/(a+b/x^3)^(3/2)/x^8,x,method=_RETURNVERBOSE)
Output:
-2/5/b^2*(a*x^3+b)/x^4/((a*x^3+b)/x^3)^(1/2)-1/5/b^2*a*(4*(1/2/a*(-a^2*b)^ (1/3)-1/2*I*3^(1/2)/a*(-a^2*b)^(1/3))*((-3/2/a*(-a^2*b)^(1/3)+1/2*I*3^(1/2 )/a*(-a^2*b)^(1/3))*x/(-1/2/a*(-a^2*b)^(1/3)+1/2*I*3^(1/2)/a*(-a^2*b)^(1/3 ))/(x-1/a*(-a^2*b)^(1/3)))^(1/2)*(x-1/a*(-a^2*b)^(1/3))^2*(1/a*(-a^2*b)^(1 /3)*(x+1/2/a*(-a^2*b)^(1/3)+1/2*I*3^(1/2)/a*(-a^2*b)^(1/3))/(-1/2/a*(-a^2* b)^(1/3)-1/2*I*3^(1/2)/a*(-a^2*b)^(1/3))/(x-1/a*(-a^2*b)^(1/3)))^(1/2)*(1/ a*(-a^2*b)^(1/3)*(x+1/2/a*(-a^2*b)^(1/3)-1/2*I*3^(1/2)/a*(-a^2*b)^(1/3))/( -1/2/a*(-a^2*b)^(1/3)+1/2*I*3^(1/2)/a*(-a^2*b)^(1/3))/(x-1/a*(-a^2*b)^(1/3 )))^(1/2)/(-3/2/a*(-a^2*b)^(1/3)+1/2*I*3^(1/2)/a*(-a^2*b)^(1/3))*a/(-a^2*b )^(1/3)/(a*x*(x-1/a*(-a^2*b)^(1/3))*(x+1/2/a*(-a^2*b)^(1/3)+1/2*I*3^(1/2)/ a*(-a^2*b)^(1/3))*(x+1/2/a*(-a^2*b)^(1/3)-1/2*I*3^(1/2)/a*(-a^2*b)^(1/3))) ^(1/2)*EllipticF(((-3/2/a*(-a^2*b)^(1/3)+1/2*I*3^(1/2)/a*(-a^2*b)^(1/3))*x /(-1/2/a*(-a^2*b)^(1/3)+1/2*I*3^(1/2)/a*(-a^2*b)^(1/3))/(x-1/a*(-a^2*b)^(1 /3)))^(1/2),((3/2/a*(-a^2*b)^(1/3)+1/2*I*3^(1/2)/a*(-a^2*b)^(1/3))*(1/2/a* (-a^2*b)^(1/3)-1/2*I*3^(1/2)/a*(-a^2*b)^(1/3))/(1/2/a*(-a^2*b)^(1/3)+1/2*I *3^(1/2)/a*(-a^2*b)^(1/3))/(3/2/a*(-a^2*b)^(1/3)-1/2*I*3^(1/2)/a*(-a^2*b)^ (1/3)))^(1/2))+5*b*(2/3*x/b/((x^3+b/a)*a*x)^(1/2)+4/3/b*(1/2/a*(-a^2*b)^(1 /3)-1/2*I*3^(1/2)/a*(-a^2*b)^(1/3))*((-3/2/a*(-a^2*b)^(1/3)+1/2*I*3^(1/2)/ a*(-a^2*b)^(1/3))*x/(-1/2/a*(-a^2*b)^(1/3)+1/2*I*3^(1/2)/a*(-a^2*b)^(1/3)) /(x-1/a*(-a^2*b)^(1/3)))^(1/2)*(x-1/a*(-a^2*b)^(1/3))^2*(1/a*(-a^2*b)^(...
Time = 0.10 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.28 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^8} \, dx=\frac {2 \, {\left (16 \, {\left (a^{2} x^{4} + a b x\right )} \sqrt {b} {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, \frac {1}{x}\right ) - {\left (8 \, a b x^{3} + 3 \, b^{2}\right )} \sqrt {\frac {a x^{3} + b}{x^{3}}}\right )}}{15 \, {\left (a b^{3} x^{4} + b^{4} x\right )}} \] Input:
integrate(1/(a+b/x^3)^(3/2)/x^8,x, algorithm="fricas")
Output:
2/15*(16*(a^2*x^4 + a*b*x)*sqrt(b)*weierstrassPInverse(0, -4*a/b, 1/x) - ( 8*a*b*x^3 + 3*b^2)*sqrt((a*x^3 + b)/x^3))/(a*b^3*x^4 + b^4*x)
Time = 0.98 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.15 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^8} \, dx=- \frac {\Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{3}}} \right )}}{3 a^{\frac {3}{2}} x^{7} \Gamma \left (\frac {10}{3}\right )} \] Input:
integrate(1/(a+b/x**3)**(3/2)/x**8,x)
Output:
-gamma(7/3)*hyper((3/2, 7/3), (10/3,), b*exp_polar(I*pi)/(a*x**3))/(3*a**( 3/2)*x**7*gamma(10/3))
\[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^8} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{2}} x^{8}} \,d x } \] Input:
integrate(1/(a+b/x^3)^(3/2)/x^8,x, algorithm="maxima")
Output:
integrate(1/((a + b/x^3)^(3/2)*x^8), x)
\[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^8} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{2}} x^{8}} \,d x } \] Input:
integrate(1/(a+b/x^3)^(3/2)/x^8,x, algorithm="giac")
Output:
integrate(1/((a + b/x^3)^(3/2)*x^8), x)
Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^8} \, dx=\int \frac {1}{x^8\,{\left (a+\frac {b}{x^3}\right )}^{3/2}} \,d x \] Input:
int(1/(x^8*(a + b/x^3)^(3/2)),x)
Output:
int(1/(x^8*(a + b/x^3)^(3/2)), x)
\[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^{3/2} x^8} \, dx=\int \frac {\sqrt {x}\, \sqrt {a \,x^{3}+b}}{a^{2} x^{10}+2 a b \,x^{7}+b^{2} x^{4}}d x \] Input:
int(1/(a+b/x^3)^(3/2)/x^8,x)
Output:
int((sqrt(x)*sqrt(a*x**3 + b))/(a**2*x**10 + 2*a*b*x**7 + b**2*x**4),x)