Integrand size = 22, antiderivative size = 173 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {a (b c-a d) x}{6 b^3 \left (a+b x^4\right )^{3/2}}-\frac {(7 b c-13 a d) x}{12 b^3 \sqrt {a+b x^4}}+\frac {d x \sqrt {a+b x^4}}{3 b^3}+\frac {5 (b c-3 a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{24 \sqrt [4]{a} b^{13/4} \sqrt {a+b x^4}} \] Output:
1/6*a*(-a*d+b*c)*x/b^3/(b*x^4+a)^(3/2)-1/12*(-13*a*d+7*b*c)*x/b^3/(b*x^4+a )^(1/2)+1/3*d*x*(b*x^4+a)^(1/2)/b^3+5/24*(-3*a*d+b*c)*(a^(1/2)+b^(1/2)*x^2 )*((b*x^4+a)/(a^(1/2)+b^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/ 4)*x/a^(1/4)),1/2*2^(1/2))/a^(1/4)/b^(13/4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.10 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.61 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {x \left (15 a^2 d+b^2 x^4 \left (-7 c+4 d x^4\right )+a b \left (-5 c+21 d x^4\right )+5 (b c-3 a d) \left (a+b x^4\right ) \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )\right )}{12 b^3 \left (a+b x^4\right )^{3/2}} \] Input:
Integrate[(x^8*(c + d*x^4))/(a + b*x^4)^(5/2),x]
Output:
(x*(15*a^2*d + b^2*x^4*(-7*c + 4*d*x^4) + a*b*(-5*c + 21*d*x^4) + 5*(b*c - 3*a*d)*(a + b*x^4)*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, - ((b*x^4)/a)]))/(12*b^3*(a + b*x^4)^(3/2))
Time = 0.44 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {959, 817, 817, 761}
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.
| \(\displaystyle \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 959 |
| \(\displaystyle \frac {(b c-3 a d) \int \frac {x^8}{\left (b x^4+a\right )^{5/2}}dx}{b}+\frac {d x^9}{3 b \left (a+b x^4\right )^{3/2}}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {(b c-3 a d) \left (\frac {5 \int \frac {x^4}{\left (b x^4+a\right )^{3/2}}dx}{6 b}-\frac {x^5}{6 b \left (a+b x^4\right )^{3/2}}\right )}{b}+\frac {d x^9}{3 b \left (a+b x^4\right )^{3/2}}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {(b c-3 a d) \left (\frac {5 \left (\frac {\int \frac {1}{\sqrt {b x^4+a}}dx}{2 b}-\frac {x}{2 b \sqrt {a+b x^4}}\right )}{6 b}-\frac {x^5}{6 b \left (a+b x^4\right )^{3/2}}\right )}{b}+\frac {d x^9}{3 b \left (a+b x^4\right )^{3/2}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {(b c-3 a d) \left (\frac {5 \left (\frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 \sqrt [4]{a} b^{5/4} \sqrt {a+b x^4}}-\frac {x}{2 b \sqrt {a+b x^4}}\right )}{6 b}-\frac {x^5}{6 b \left (a+b x^4\right )^{3/2}}\right )}{b}+\frac {d x^9}{3 b \left (a+b x^4\right )^{3/2}}\) |
Input:
Int[(x^8*(c + d*x^4))/(a + b*x^4)^(5/2),x]
Output:
(d*x^9)/(3*b*(a + b*x^4)^(3/2)) + ((b*c - 3*a*d)*(-1/6*x^5/(b*(a + b*x^4)^ (3/2)) + (5*(-1/2*x/(b*Sqrt[a + b*x^4]) + ((Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4 )], 1/2])/(4*a^(1/4)*b^(5/4)*Sqrt[a + b*x^4])))/(6*b)))/b
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m + n*(p + 1) + 1)) Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + n*(p + 1) + 1, 0]
Result contains complex when optimal does not.
Time = 4.59 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.08
| method | result | size |
| elliptic | \(-\frac {a x \left (a d -c b \right ) \sqrt {b \,x^{4}+a}}{6 b^{5} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {x \left (13 a d -7 c b \right )}{12 b^{3} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {d x \sqrt {b \,x^{4}+a}}{3 b^{3}}+\frac {\left (-\frac {2 a d -c b}{b^{3}}+\frac {13 a d -7 c b}{12 b^{3}}-\frac {d a}{3 b^{3}}\right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(186\) |
| default | \(c \left (\frac {a x \sqrt {b \,x^{4}+a}}{6 b^{4} \left (x^{4}+\frac {a}{b}\right )^{2}}-\frac {7 x}{12 b^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {5 \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{12 b^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+d \left (-\frac {a^{2} x \sqrt {b \,x^{4}+a}}{6 b^{5} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {13 a x}{12 b^{3} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {x \sqrt {b \,x^{4}+a}}{3 b^{3}}-\frac {5 a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{4 b^{3} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\) | \(265\) |
| risch | \(\frac {d x \sqrt {b \,x^{4}+a}}{3 b^{3}}-\frac {\frac {7 a d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-3 a \left (3 a d -2 c b \right ) \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+3 a^{2} \left (a d -c b \right ) \left (\frac {x \sqrt {b \,x^{4}+a}}{6 a \,b^{2} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {5 x}{12 a^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {5 \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{12 a^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )-\frac {3 c b \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}}{3 b^{3}}\) | \(408\) |
Input:
int(x^8*(d*x^4+c)/(b*x^4+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/6*a*x/b^5*(a*d-b*c)*(b*x^4+a)^(1/2)/(x^4+a/b)^2+1/12/b^3*x*(13*a*d-7*b* c)/((x^4+a/b)*b)^(1/2)+1/3*d*x*(b*x^4+a)^(1/2)/b^3+(-1/b^3*(2*a*d-b*c)+1/1 2/b^3*(13*a*d-7*b*c)-1/3*d/b^3*a)/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b ^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)*Elliptic F(x*(I/a^(1/2)*b^(1/2))^(1/2),I)
Time = 0.09 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.95 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {5 \, {\left ({\left (b^{3} c - 3 \, a b^{2} d\right )} x^{8} + 2 \, {\left (a b^{2} c - 3 \, a^{2} b d\right )} x^{4} + a^{2} b c - 3 \, a^{3} d\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (4 \, a b^{2} d x^{9} - 7 \, {\left (a b^{2} c - 3 \, a^{2} b d\right )} x^{5} - 5 \, {\left (a^{2} b c - 3 \, a^{3} d\right )} x\right )} \sqrt {b x^{4} + a}}{12 \, {\left (a b^{5} x^{8} + 2 \, a^{2} b^{4} x^{4} + a^{3} b^{3}\right )}} \] Input:
integrate(x^8*(d*x^4+c)/(b*x^4+a)^(5/2),x, algorithm="fricas")
Output:
1/12*(5*((b^3*c - 3*a*b^2*d)*x^8 + 2*(a*b^2*c - 3*a^2*b*d)*x^4 + a^2*b*c - 3*a^3*d)*sqrt(b)*(-a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/4)/x), -1) + (4 *a*b^2*d*x^9 - 7*(a*b^2*c - 3*a^2*b*d)*x^5 - 5*(a^2*b*c - 3*a^3*d)*x)*sqrt (b*x^4 + a))/(a*b^5*x^8 + 2*a^2*b^4*x^4 + a^3*b^3)
Result contains complex when optimal does not.
Time = 38.87 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.46 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {c x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, \frac {5}{2} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{2}} \Gamma \left (\frac {13}{4}\right )} + \frac {d x^{13} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{2}} \Gamma \left (\frac {17}{4}\right )} \] Input:
integrate(x**8*(d*x**4+c)/(b*x**4+a)**(5/2),x)
Output:
c*x**9*gamma(9/4)*hyper((9/4, 5/2), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4* a**(5/2)*gamma(13/4)) + d*x**13*gamma(13/4)*hyper((5/2, 13/4), (17/4,), b* x**4*exp_polar(I*pi)/a)/(4*a**(5/2)*gamma(17/4))
\[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{8}}{{\left (b x^{4} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(x^8*(d*x^4+c)/(b*x^4+a)^(5/2),x, algorithm="maxima")
Output:
integrate((d*x^4 + c)*x^8/(b*x^4 + a)^(5/2), x)
\[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{8}}{{\left (b x^{4} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(x^8*(d*x^4+c)/(b*x^4+a)^(5/2),x, algorithm="giac")
Output:
integrate((d*x^4 + c)*x^8/(b*x^4 + a)^(5/2), x)
Timed out. \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\int \frac {x^8\,\left (d\,x^4+c\right )}{{\left (b\,x^4+a\right )}^{5/2}} \,d x \] Input:
int((x^8*(c + d*x^4))/(a + b*x^4)^(5/2),x)
Output:
int((x^8*(c + d*x^4))/(a + b*x^4)^(5/2), x)
\[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {9 \sqrt {b \,x^{4}+a}\, a^{2} d x -3 \sqrt {b \,x^{4}+a}\, a b c x +9 \sqrt {b \,x^{4}+a}\, a b d \,x^{5}-3 \sqrt {b \,x^{4}+a}\, b^{2} c \,x^{5}+\sqrt {b \,x^{4}+a}\, b^{2} d \,x^{9}-9 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{5} d +3 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{4} b c -18 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{4} b d \,x^{4}+6 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{3} b^{2} c \,x^{4}-9 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{3} b^{2} d \,x^{8}+3 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{2} b^{3} c \,x^{8}}{3 b^{3} \left (b^{2} x^{8}+2 a b \,x^{4}+a^{2}\right )} \] Input:
int(x^8*(d*x^4+c)/(b*x^4+a)^(5/2),x)
Output:
(9*sqrt(a + b*x**4)*a**2*d*x - 3*sqrt(a + b*x**4)*a*b*c*x + 9*sqrt(a + b*x **4)*a*b*d*x**5 - 3*sqrt(a + b*x**4)*b**2*c*x**5 + sqrt(a + b*x**4)*b**2*d *x**9 - 9*int(sqrt(a + b*x**4)/(a**3 + 3*a**2*b*x**4 + 3*a*b**2*x**8 + b** 3*x**12),x)*a**5*d + 3*int(sqrt(a + b*x**4)/(a**3 + 3*a**2*b*x**4 + 3*a*b* *2*x**8 + b**3*x**12),x)*a**4*b*c - 18*int(sqrt(a + b*x**4)/(a**3 + 3*a**2 *b*x**4 + 3*a*b**2*x**8 + b**3*x**12),x)*a**4*b*d*x**4 + 6*int(sqrt(a + b* x**4)/(a**3 + 3*a**2*b*x**4 + 3*a*b**2*x**8 + b**3*x**12),x)*a**3*b**2*c*x **4 - 9*int(sqrt(a + b*x**4)/(a**3 + 3*a**2*b*x**4 + 3*a*b**2*x**8 + b**3* x**12),x)*a**3*b**2*d*x**8 + 3*int(sqrt(a + b*x**4)/(a**3 + 3*a**2*b*x**4 + 3*a*b**2*x**8 + b**3*x**12),x)*a**2*b**3*c*x**8)/(3*b**3*(a**2 + 2*a*b*x **4 + b**2*x**8))