Integrand size = 16, antiderivative size = 119 \[ \int x^4 \left (a-b x^4\right )^{3/2} \, dx=-\frac {4 a^2 x \sqrt {a-b x^4}}{77 b}+\frac {6}{77} a x^5 \sqrt {a-b x^4}+\frac {1}{11} x^5 \left (a-b x^4\right )^{3/2}+\frac {4 a^{13/4} \sqrt {1-\frac {b x^4}{a}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),-1\right )}{77 b^{5/4} \sqrt {a-b x^4}} \] Output:
-4/77*a^2*x*(-b*x^4+a)^(1/2)/b+6/77*a*x^5*(-b*x^4+a)^(1/2)+1/11*x^5*(-b*x^ 4+a)^(3/2)+4/77*a^(13/4)*(1-b*x^4/a)^(1/2)*EllipticF(b^(1/4)*x/a^(1/4),I)/ b^(5/4)/(-b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.88 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.59 \[ \int x^4 \left (a-b x^4\right )^{3/2} \, dx=\frac {x \sqrt {a-b x^4} \left (-\left (a-b x^4\right )^2+\frac {a^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},\frac {5}{4},\frac {b x^4}{a}\right )}{\sqrt {1-\frac {b x^4}{a}}}\right )}{11 b} \] Input:
Integrate[x^4*(a - b*x^4)^(3/2),x]
Output:
(x*Sqrt[a - b*x^4]*(-(a - b*x^4)^2 + (a^2*Hypergeometric2F1[-3/2, 1/4, 5/4 , (b*x^4)/a])/Sqrt[1 - (b*x^4)/a]))/(11*b)
Time = 0.39 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {811, 811, 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.
\(\displaystyle \int x^4 \left (a-b x^4\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {6}{11} a \int x^4 \sqrt {a-b x^4}dx+\frac {1}{11} x^5 \left (a-b x^4\right )^{3/2}\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {6}{11} a \left (\frac {2}{7} a \int \frac {x^4}{\sqrt {a-b x^4}}dx+\frac {1}{7} x^5 \sqrt {a-b x^4}\right )+\frac {1}{11} x^5 \left (a-b x^4\right )^{3/2}\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {6}{11} a \left (\frac {2}{7} a \left (\frac {a \int \frac {1}{\sqrt {a-b x^4}}dx}{3 b}-\frac {x \sqrt {a-b x^4}}{3 b}\right )+\frac {1}{7} x^5 \sqrt {a-b x^4}\right )+\frac {1}{11} x^5 \left (a-b x^4\right )^{3/2}\) |
\(\Big \downarrow \) 765 |
\(\displaystyle \frac {6}{11} a \left (\frac {2}{7} a \left (\frac {a \sqrt {1-\frac {b x^4}{a}} \int \frac {1}{\sqrt {1-\frac {b x^4}{a}}}dx}{3 b \sqrt {a-b x^4}}-\frac {x \sqrt {a-b x^4}}{3 b}\right )+\frac {1}{7} x^5 \sqrt {a-b x^4}\right )+\frac {1}{11} x^5 \left (a-b x^4\right )^{3/2}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {6}{11} a \left (\frac {2}{7} a \left (\frac {a^{5/4} \sqrt {1-\frac {b x^4}{a}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),-1\right )}{3 b^{5/4} \sqrt {a-b x^4}}-\frac {x \sqrt {a-b x^4}}{3 b}\right )+\frac {1}{7} x^5 \sqrt {a-b x^4}\right )+\frac {1}{11} x^5 \left (a-b x^4\right )^{3/2}\) |
Input:
Int[x^4*(a - b*x^4)^(3/2),x]
Output:
(x^5*(a - b*x^4)^(3/2))/11 + (6*a*((x^5*Sqrt[a - b*x^4])/7 + (2*a*(-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)*Sqrt[a - b*x^4])))/7))/11
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* x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 )) Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 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 = 0.63 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.92
method | result | size |
risch | \(-\frac {x \left (7 b^{2} x^{8}-13 a b \,x^{4}+4 a^{2}\right ) \sqrt {-b \,x^{4}+a}}{77 b}+\frac {4 a^{3} \sqrt {1-\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )}{77 b \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}\) | \(109\) |
default | \(-\frac {b \,x^{9} \sqrt {-b \,x^{4}+a}}{11}+\frac {13 a \,x^{5} \sqrt {-b \,x^{4}+a}}{77}-\frac {4 a^{2} x \sqrt {-b \,x^{4}+a}}{77 b}+\frac {4 a^{3} \sqrt {1-\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )}{77 b \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}\) | \(123\) |
elliptic | \(-\frac {b \,x^{9} \sqrt {-b \,x^{4}+a}}{11}+\frac {13 a \,x^{5} \sqrt {-b \,x^{4}+a}}{77}-\frac {4 a^{2} x \sqrt {-b \,x^{4}+a}}{77 b}+\frac {4 a^{3} \sqrt {1-\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )}{77 b \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}\) | \(123\) |
Input:
int(x^4*(-b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/77*x*(7*b^2*x^8-13*a*b*x^4+4*a^2)*(-b*x^4+a)^(1/2)/b+4/77/b*a^3/(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 = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.60 \[ \int x^4 \left (a-b x^4\right )^{3/2} \, dx=\frac {4 \, a^{2} \sqrt {-b} \left (\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - {\left (7 \, b^{2} x^{9} - 13 \, a b x^{5} + 4 \, a^{2} x\right )} \sqrt {-b x^{4} + a}}{77 \, b} \] Input:
integrate(x^4*(-b*x^4+a)^(3/2),x, algorithm="fricas")
Output:
1/77*(4*a^2*sqrt(-b)*(a/b)^(3/4)*elliptic_f(arcsin((a/b)^(1/4)/x), -1) - ( 7*b^2*x^9 - 13*a*b*x^5 + 4*a^2*x)*sqrt(-b*x^4 + a))/b
Time = 0.58 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.34 \[ \int x^4 \left (a-b x^4\right )^{3/2} \, dx=\frac {a^{\frac {3}{2}} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} \] Input:
integrate(x**4*(-b*x**4+a)**(3/2),x)
Output:
a**(3/2)*x**5*gamma(5/4)*hyper((-3/2, 5/4), (9/4,), b*x**4*exp_polar(2*I*p i)/a)/(4*gamma(9/4))
\[ \int x^4 \left (a-b x^4\right )^{3/2} \, dx=\int { {\left (-b x^{4} + a\right )}^{\frac {3}{2}} x^{4} \,d x } \] Input:
integrate(x^4*(-b*x^4+a)^(3/2),x, algorithm="maxima")
Output:
integrate((-b*x^4 + a)^(3/2)*x^4, x)
\[ \int x^4 \left (a-b x^4\right )^{3/2} \, dx=\int { {\left (-b x^{4} + a\right )}^{\frac {3}{2}} x^{4} \,d x } \] Input:
integrate(x^4*(-b*x^4+a)^(3/2),x, algorithm="giac")
Output:
integrate((-b*x^4 + a)^(3/2)*x^4, x)
Timed out. \[ \int x^4 \left (a-b x^4\right )^{3/2} \, dx=\int x^4\,{\left (a-b\,x^4\right )}^{3/2} \,d x \] Input:
int(x^4*(a - b*x^4)^(3/2),x)
Output:
int(x^4*(a - b*x^4)^(3/2), x)
\[ \int x^4 \left (a-b x^4\right )^{3/2} \, dx=\frac {-4 \sqrt {-b \,x^{4}+a}\, a^{2} x +13 \sqrt {-b \,x^{4}+a}\, a b \,x^{5}-7 \sqrt {-b \,x^{4}+a}\, b^{2} x^{9}+4 \left (\int \frac {\sqrt {-b \,x^{4}+a}}{-b \,x^{4}+a}d x \right ) a^{3}}{77 b} \] Input:
int(x^4*(-b*x^4+a)^(3/2),x)
Output:
( - 4*sqrt(a - b*x**4)*a**2*x + 13*sqrt(a - b*x**4)*a*b*x**5 - 7*sqrt(a - b*x**4)*b**2*x**9 + 4*int(sqrt(a - b*x**4)/(a - b*x**4),x)*a**3)/(77*b)