Integrand size = 16, antiderivative size = 99 \[ \int \frac {x^8}{\left (a-b x^4\right )^{3/2}} \, dx=\frac {x^5}{2 b \sqrt {a-b x^4}}+\frac {5 x \sqrt {a-b x^4}}{6 b^2}-\frac {5 a^{5/4} \sqrt {1-\frac {b x^4}{a}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),-1\right )}{6 b^{9/4} \sqrt {a-b x^4}} \] Output:
1/2*x^5/b/(-b*x^4+a)^(1/2)+5/6*x*(-b*x^4+a)^(1/2)/b^2-5/6*a^(5/4)*(1-b*x^4 /a)^(1/2)*EllipticF(b^(1/4)*x/a^(1/4),I)/b^(9/4)/(-b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.68 \[ \int \frac {x^8}{\left (a-b x^4\right )^{3/2}} \, dx=\frac {5 a x-2 b x^5-5 a x \sqrt {1-\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {b x^4}{a}\right )}{6 b^2 \sqrt {a-b x^4}} \] Input:
Integrate[x^8/(a - b*x^4)^(3/2),x]
Output:
(5*a*x - 2*b*x^5 - 5*a*x*Sqrt[1 - (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5 /4, (b*x^4)/a])/(6*b^2*Sqrt[a - b*x^4])
Time = 0.35 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {817, 843, 765, 762}
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.
\(\displaystyle \int \frac {x^8}{\left (a-b x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 817 |
\(\displaystyle \frac {x^5}{2 b \sqrt {a-b x^4}}-\frac {5 \int \frac {x^4}{\sqrt {a-b x^4}}dx}{2 b}\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {x^5}{2 b \sqrt {a-b x^4}}-\frac {5 \left (\frac {a \int \frac {1}{\sqrt {a-b x^4}}dx}{3 b}-\frac {x \sqrt {a-b x^4}}{3 b}\right )}{2 b}\) |
\(\Big \downarrow \) 765 |
\(\displaystyle \frac {x^5}{2 b \sqrt {a-b x^4}}-\frac {5 \left (\frac {a \sqrt {1-\frac {b x^4}{a}} \int \frac {1}{\sqrt {1-\frac {b x^4}{a}}}dx}{3 b \sqrt {a-b x^4}}-\frac {x \sqrt {a-b x^4}}{3 b}\right )}{2 b}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {x^5}{2 b \sqrt {a-b x^4}}-\frac {5 \left (\frac {a^{5/4} \sqrt {1-\frac {b x^4}{a}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),-1\right )}{3 b^{5/4} \sqrt {a-b x^4}}-\frac {x \sqrt {a-b x^4}}{3 b}\right )}{2 b}\) |
Input:
Int[x^8/(a - b*x^4)^(3/2),x]
Output:
x^5/(2*b*Sqrt[a - b*x^4]) - (5*(-1/3*(x*Sqrt[a - b*x^4])/b + (a^(5/4)*Sqrt [1 - (b*x^4)/a]*EllipticF[ArcSin[(b^(1/4)*x)/a^(1/4)], -1])/(3*b^(5/4)*Sqr t[a - b*x^4])))/(2*b)
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) )*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[Sqrt[1 + b*(x^4/a)]/Sqrt [a + b*x^4] Int[1/Sqrt[1 + b*(x^4/a)], x], x] /; FreeQ[{a, b}, x] && NegQ [b/a] && !GtQ[a, 0]
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]
Time = 1.65 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.09
method | result | size |
default | \(\frac {x a}{2 b^{2} \sqrt {-\left (x^{4}-\frac {a}{b}\right ) b}}+\frac {x \sqrt {-b \,x^{4}+a}}{3 b^{2}}-\frac {5 a \sqrt {1-\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )}{6 b^{2} \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}\) | \(108\) |
elliptic | \(\frac {x a}{2 b^{2} \sqrt {-\left (x^{4}-\frac {a}{b}\right ) b}}+\frac {x \sqrt {-b \,x^{4}+a}}{3 b^{2}}-\frac {5 a \sqrt {1-\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )}{6 b^{2} \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}\) | \(108\) |
risch | \(\frac {x \sqrt {-b \,x^{4}+a}}{3 b^{2}}-\frac {a \left (\frac {4 \sqrt {1-\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}+3 a \left (-\frac {x}{2 a \sqrt {-\left (x^{4}-\frac {a}{b}\right ) b}}-\frac {\sqrt {1-\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}\right )\right )}{3 b^{2}}\) | \(181\) |
Input:
int(x^8/(-b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/2/b^2*x*a/(-(x^4-a/b)*b)^(1/2)+1/3*x*(-b*x^4+a)^(1/2)/b^2-5/6*a/b^2/(1/a ^(1/2)*b^(1/2))^(1/2)*(1-b^(1/2)*x^2/a^(1/2))^(1/2)*(1+b^(1/2)*x^2/a^(1/2) )^(1/2)/(-b*x^4+a)^(1/2)*EllipticF(x*(1/a^(1/2)*b^(1/2))^(1/2),I)
Time = 0.10 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.80 \[ \int \frac {x^8}{\left (a-b x^4\right )^{3/2}} \, dx=-\frac {5 \, {\left (b x^{4} - a\right )} \sqrt {-b} \left (\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - {\left (2 \, b x^{5} - 5 \, a x\right )} \sqrt {-b x^{4} + a}}{6 \, {\left (b^{3} x^{4} - a b^{2}\right )}} \] Input:
integrate(x^8/(-b*x^4+a)^(3/2),x, algorithm="fricas")
Output:
-1/6*(5*(b*x^4 - a)*sqrt(-b)*(a/b)^(3/4)*elliptic_f(arcsin((a/b)^(1/4)/x), -1) - (2*b*x^5 - 5*a*x)*sqrt(-b*x^4 + a))/(b^3*x^4 - a*b^2)
Time = 0.54 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.39 \[ \int \frac {x^8}{\left (a-b x^4\right )^{3/2}} \, dx=\frac {x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {13}{4}\right )} \] Input:
integrate(x**8/(-b*x**4+a)**(3/2),x)
Output:
x**9*gamma(9/4)*hyper((3/2, 9/4), (13/4,), b*x**4*exp_polar(2*I*pi)/a)/(4* a**(3/2)*gamma(13/4))
\[ \int \frac {x^8}{\left (a-b x^4\right )^{3/2}} \, dx=\int { \frac {x^{8}}{{\left (-b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^8/(-b*x^4+a)^(3/2),x, algorithm="maxima")
Output:
integrate(x^8/(-b*x^4 + a)^(3/2), x)
\[ \int \frac {x^8}{\left (a-b x^4\right )^{3/2}} \, dx=\int { \frac {x^{8}}{{\left (-b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^8/(-b*x^4+a)^(3/2),x, algorithm="giac")
Output:
integrate(x^8/(-b*x^4 + a)^(3/2), x)
Timed out. \[ \int \frac {x^8}{\left (a-b x^4\right )^{3/2}} \, dx=\int \frac {x^8}{{\left (a-b\,x^4\right )}^{3/2}} \,d x \] Input:
int(x^8/(a - b*x^4)^(3/2),x)
Output:
int(x^8/(a - b*x^4)^(3/2), x)
\[ \int \frac {x^8}{\left (a-b x^4\right )^{3/2}} \, dx=\frac {5 \sqrt {-b \,x^{4}+a}\, a x -\sqrt {-b \,x^{4}+a}\, b \,x^{5}-5 \left (\int \frac {\sqrt {-b \,x^{4}+a}}{b^{2} x^{8}-2 a b \,x^{4}+a^{2}}d x \right ) a^{3}+5 \left (\int \frac {\sqrt {-b \,x^{4}+a}}{b^{2} x^{8}-2 a b \,x^{4}+a^{2}}d x \right ) a^{2} b \,x^{4}}{3 b^{2} \left (-b \,x^{4}+a \right )} \] Input:
int(x^8/(-b*x^4+a)^(3/2),x)
Output:
(5*sqrt(a - b*x**4)*a*x - sqrt(a - b*x**4)*b*x**5 - 5*int(sqrt(a - b*x**4) /(a**2 - 2*a*b*x**4 + b**2*x**8),x)*a**3 + 5*int(sqrt(a - b*x**4)/(a**2 - 2*a*b*x**4 + b**2*x**8),x)*a**2*b*x**4)/(3*b**2*(a - b*x**4))