Integrand size = 13, antiderivative size = 64 \[ \int \frac {1}{\left (a+b x^{2/3}\right )^{9/2}} \, dx=\frac {3 x}{7 a \left (a+b x^{2/3}\right )^{7/2}}+\frac {12 x}{35 a^2 \left (a+b x^{2/3}\right )^{5/2}}+\frac {8 x}{35 a^3 \left (a+b x^{2/3}\right )^{3/2}} \] Output:
3/7*x/a/(a+b*x^(2/3))^(7/2)+12/35*x/a^2/(a+b*x^(2/3))^(5/2)+8/35*x/a^3/(a+ b*x^(2/3))^(3/2)
Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.72 \[ \int \frac {1}{\left (a+b x^{2/3}\right )^{9/2}} \, dx=\frac {35 a^2 x+28 a b x^{5/3}+8 b^2 x^{7/3}}{35 a^3 \left (a+b x^{2/3}\right )^{7/2}} \] Input:
Integrate[(a + b*x^(2/3))^(-9/2),x]
Output:
(35*a^2*x + 28*a*b*x^(5/3) + 8*b^2*x^(7/3))/(35*a^3*(a + b*x^(2/3))^(7/2))
Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.31, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {774, 245, 245, 242}
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.
\(\displaystyle \int \frac {1}{\left (a+b x^{2/3}\right )^{9/2}} \, dx\) |
\(\Big \downarrow \) 774 |
\(\displaystyle 3 \int \frac {x^{2/3}}{\left (a+b x^{2/3}\right )^{9/2}}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 245 |
\(\displaystyle 3 \left (\frac {4 b \int \frac {x^{4/3}}{\left (a+b x^{2/3}\right )^{9/2}}d\sqrt [3]{x}}{3 a}+\frac {x}{3 a \left (a+b x^{2/3}\right )^{7/2}}\right )\) |
\(\Big \downarrow \) 245 |
\(\displaystyle 3 \left (\frac {4 b \left (\frac {2 b \int \frac {x^2}{\left (a+b x^{2/3}\right )^{9/2}}d\sqrt [3]{x}}{5 a}+\frac {x^{5/3}}{5 a \left (a+b x^{2/3}\right )^{7/2}}\right )}{3 a}+\frac {x}{3 a \left (a+b x^{2/3}\right )^{7/2}}\right )\) |
\(\Big \downarrow \) 242 |
\(\displaystyle 3 \left (\frac {4 b \left (\frac {2 b x^{7/3}}{35 a^2 \left (a+b x^{2/3}\right )^{7/2}}+\frac {x^{5/3}}{5 a \left (a+b x^{2/3}\right )^{7/2}}\right )}{3 a}+\frac {x}{3 a \left (a+b x^{2/3}\right )^{7/2}}\right )\) |
Input:
Int[(a + b*x^(2/3))^(-9/2),x]
Output:
3*(x/(3*a*(a + b*x^(2/3))^(7/2)) + (4*b*(x^(5/3)/(5*a*(a + b*x^(2/3))^(7/2 )) + (2*b*x^(7/3))/(35*a^2*(a + b*x^(2/3))^(7/2))))/(3*a))
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, p}, x ] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Simp[k Subst[Int[x^(k - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; Fre eQ[{a, b, p}, x] && FractionQ[n]
Leaf count of result is larger than twice the leaf count of optimal. \(105\) vs. \(2(46)=92\).
Time = 0.36 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.66
method | result | size |
derivativedivides | \(-\frac {x^{\frac {1}{3}}}{2 \left (a +b \,x^{\frac {2}{3}}\right )^{\frac {7}{2}} b}+\frac {a \left (\frac {x^{\frac {1}{3}}}{7 a \left (a +b \,x^{\frac {2}{3}}\right )^{\frac {7}{2}}}+\frac {\frac {6 x^{\frac {1}{3}}}{35 a \left (a +b \,x^{\frac {2}{3}}\right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x^{\frac {1}{3}}}{15 a \left (a +b \,x^{\frac {2}{3}}\right )^{\frac {3}{2}}}+\frac {8 x^{\frac {1}{3}}}{15 a^{2} \sqrt {a +b \,x^{\frac {2}{3}}}}\right )}{7 a}}{a}\right )}{2 b}\) | \(106\) |
default | \(-\frac {x^{\frac {1}{3}}}{2 \left (a +b \,x^{\frac {2}{3}}\right )^{\frac {7}{2}} b}+\frac {a \left (\frac {x^{\frac {1}{3}}}{7 a \left (a +b \,x^{\frac {2}{3}}\right )^{\frac {7}{2}}}+\frac {\frac {6 x^{\frac {1}{3}}}{35 a \left (a +b \,x^{\frac {2}{3}}\right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x^{\frac {1}{3}}}{15 a \left (a +b \,x^{\frac {2}{3}}\right )^{\frac {3}{2}}}+\frac {8 x^{\frac {1}{3}}}{15 a^{2} \sqrt {a +b \,x^{\frac {2}{3}}}}\right )}{7 a}}{a}\right )}{2 b}\) | \(106\) |
Input:
int(1/(a+b*x^(2/3))^(9/2),x,method=_RETURNVERBOSE)
Output:
-1/2/(a+b*x^(2/3))^(7/2)/b*x^(1/3)+1/2*a/b*(1/7*x^(1/3)/a/(a+b*x^(2/3))^(7 /2)+6/7/a*(1/5*x^(1/3)/a/(a+b*x^(2/3))^(5/2)+4/5/a*(1/3*x^(1/3)/a/(a+b*x^( 2/3))^(3/2)+2/3/a^2*x^(1/3)/(a+b*x^(2/3))^(1/2))))
Timed out. \[ \int \frac {1}{\left (a+b x^{2/3}\right )^{9/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(a+b*x^(2/3))^(9/2),x, algorithm="fricas")
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 573 vs. \(2 (58) = 116\).
Time = 6.33 (sec) , antiderivative size = 573, normalized size of antiderivative = 8.95 \[ \int \frac {1}{\left (a+b x^{2/3}\right )^{9/2}} \, dx=\frac {35 a^{5} x}{35 a^{\frac {19}{2}} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}} + 140 a^{\frac {17}{2}} b x^{\frac {2}{3}} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}} + 210 a^{\frac {15}{2}} b^{2} x^{\frac {4}{3}} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}} + 140 a^{\frac {13}{2}} b^{3} x^{2} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}} + 35 a^{\frac {11}{2}} b^{4} x^{\frac {8}{3}} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}}} + \frac {63 a^{4} b x^{\frac {5}{3}}}{35 a^{\frac {19}{2}} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}} + 140 a^{\frac {17}{2}} b x^{\frac {2}{3}} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}} + 210 a^{\frac {15}{2}} b^{2} x^{\frac {4}{3}} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}} + 140 a^{\frac {13}{2}} b^{3} x^{2} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}} + 35 a^{\frac {11}{2}} b^{4} x^{\frac {8}{3}} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}}} + \frac {36 a^{3} b^{2} x^{\frac {7}{3}}}{35 a^{\frac {19}{2}} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}} + 140 a^{\frac {17}{2}} b x^{\frac {2}{3}} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}} + 210 a^{\frac {15}{2}} b^{2} x^{\frac {4}{3}} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}} + 140 a^{\frac {13}{2}} b^{3} x^{2} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}} + 35 a^{\frac {11}{2}} b^{4} x^{\frac {8}{3}} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}}} + \frac {8 a^{2} b^{3} x^{3}}{35 a^{\frac {19}{2}} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}} + 140 a^{\frac {17}{2}} b x^{\frac {2}{3}} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}} + 210 a^{\frac {15}{2}} b^{2} x^{\frac {4}{3}} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}} + 140 a^{\frac {13}{2}} b^{3} x^{2} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}} + 35 a^{\frac {11}{2}} b^{4} x^{\frac {8}{3}} \sqrt {1 + \frac {b x^{\frac {2}{3}}}{a}}} \] Input:
integrate(1/(a+b*x**(2/3))**(9/2),x)
Output:
35*a**5*x/(35*a**(19/2)*sqrt(1 + b*x**(2/3)/a) + 140*a**(17/2)*b*x**(2/3)* sqrt(1 + b*x**(2/3)/a) + 210*a**(15/2)*b**2*x**(4/3)*sqrt(1 + b*x**(2/3)/a ) + 140*a**(13/2)*b**3*x**2*sqrt(1 + b*x**(2/3)/a) + 35*a**(11/2)*b**4*x** (8/3)*sqrt(1 + b*x**(2/3)/a)) + 63*a**4*b*x**(5/3)/(35*a**(19/2)*sqrt(1 + b*x**(2/3)/a) + 140*a**(17/2)*b*x**(2/3)*sqrt(1 + b*x**(2/3)/a) + 210*a**( 15/2)*b**2*x**(4/3)*sqrt(1 + b*x**(2/3)/a) + 140*a**(13/2)*b**3*x**2*sqrt( 1 + b*x**(2/3)/a) + 35*a**(11/2)*b**4*x**(8/3)*sqrt(1 + b*x**(2/3)/a)) + 3 6*a**3*b**2*x**(7/3)/(35*a**(19/2)*sqrt(1 + b*x**(2/3)/a) + 140*a**(17/2)* b*x**(2/3)*sqrt(1 + b*x**(2/3)/a) + 210*a**(15/2)*b**2*x**(4/3)*sqrt(1 + b *x**(2/3)/a) + 140*a**(13/2)*b**3*x**2*sqrt(1 + b*x**(2/3)/a) + 35*a**(11/ 2)*b**4*x**(8/3)*sqrt(1 + b*x**(2/3)/a)) + 8*a**2*b**3*x**3/(35*a**(19/2)* sqrt(1 + b*x**(2/3)/a) + 140*a**(17/2)*b*x**(2/3)*sqrt(1 + b*x**(2/3)/a) + 210*a**(15/2)*b**2*x**(4/3)*sqrt(1 + b*x**(2/3)/a) + 140*a**(13/2)*b**3*x **2*sqrt(1 + b*x**(2/3)/a) + 35*a**(11/2)*b**4*x**(8/3)*sqrt(1 + b*x**(2/3 )/a))
Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (a+b x^{2/3}\right )^{9/2}} \, dx=\frac {{\left (15 \, b^{2} - \frac {42 \, {\left (b x^{\frac {2}{3}} + a\right )} b}{x^{\frac {2}{3}}} + \frac {35 \, {\left (b x^{\frac {2}{3}} + a\right )}^{2}}{x^{\frac {4}{3}}}\right )} x^{\frac {7}{3}}}{35 \, {\left (b x^{\frac {2}{3}} + a\right )}^{\frac {7}{2}} a^{3}} \] Input:
integrate(1/(a+b*x^(2/3))^(9/2),x, algorithm="maxima")
Output:
1/35*(15*b^2 - 42*(b*x^(2/3) + a)*b/x^(2/3) + 35*(b*x^(2/3) + a)^2/x^(4/3) )*x^(7/3)/((b*x^(2/3) + a)^(7/2)*a^3)
Time = 0.14 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.64 \[ \int \frac {1}{\left (a+b x^{2/3}\right )^{9/2}} \, dx=\frac {{\left (4 \, x^{\frac {2}{3}} {\left (\frac {2 \, b^{2} x^{\frac {2}{3}}}{a^{3}} + \frac {7 \, b}{a^{2}}\right )} + \frac {35}{a}\right )} x}{35 \, {\left (b x^{\frac {2}{3}} + a\right )}^{\frac {7}{2}}} \] Input:
integrate(1/(a+b*x^(2/3))^(9/2),x, algorithm="giac")
Output:
1/35*(4*x^(2/3)*(2*b^2*x^(2/3)/a^3 + 7*b/a^2) + 35/a)*x/(b*x^(2/3) + a)^(7 /2)
Time = 0.94 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.59 \[ \int \frac {1}{\left (a+b x^{2/3}\right )^{9/2}} \, dx=-\frac {x\,{\left (\frac {a}{b\,x^{2/3}}+1\right )}^{9/2}\,{{}}_2{\mathrm {F}}_1\left (3,\frac {9}{2};\ 4;\ -\frac {a}{b\,x^{2/3}}\right )}{2\,{\left (a+b\,x^{2/3}\right )}^{9/2}} \] Input:
int(1/(a + b*x^(2/3))^(9/2),x)
Output:
-(x*(a/(b*x^(2/3)) + 1)^(9/2)*hypergeom([3, 9/2], 4, -a/(b*x^(2/3))))/(2*( a + b*x^(2/3))^(9/2))
Time = 0.21 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.39 \[ \int \frac {1}{\left (a+b x^{2/3}\right )^{9/2}} \, dx=\frac {28 x^{\frac {5}{3}} \sqrt {x^{\frac {2}{3}} b +a}\, a \,b^{3}+8 x^{\frac {7}{3}} \sqrt {x^{\frac {2}{3}} b +a}\, b^{4}+35 \sqrt {x^{\frac {2}{3}} b +a}\, a^{2} b^{2} x -32 x^{\frac {2}{3}} \sqrt {b}\, a^{3} b -8 x^{\frac {8}{3}} \sqrt {b}\, b^{4}-48 x^{\frac {4}{3}} \sqrt {b}\, a^{2} b^{2}-8 \sqrt {b}\, a^{4}-32 \sqrt {b}\, a \,b^{3} x^{2}}{35 a^{3} b^{2} \left (4 x^{\frac {2}{3}} a^{3} b +x^{\frac {8}{3}} b^{4}+6 x^{\frac {4}{3}} a^{2} b^{2}+a^{4}+4 a \,b^{3} x^{2}\right )} \] Input:
int(1/(a+b*x^(2/3))^(9/2),x)
Output:
(28*x**(2/3)*sqrt(x**(2/3)*b + a)*a*b**3*x + 8*x**(1/3)*sqrt(x**(2/3)*b + a)*b**4*x**2 + 35*sqrt(x**(2/3)*b + a)*a**2*b**2*x - 32*x**(2/3)*sqrt(b)*a **3*b - 8*x**(2/3)*sqrt(b)*b**4*x**2 - 48*x**(1/3)*sqrt(b)*a**2*b**2*x - 8 *sqrt(b)*a**4 - 32*sqrt(b)*a*b**3*x**2)/(35*a**3*b**2*(4*x**(2/3)*a**3*b + x**(2/3)*b**4*x**2 + 6*x**(1/3)*a**2*b**2*x + a**4 + 4*a*b**3*x**2))