Integrand size = 22, antiderivative size = 176 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {a (b c-a d) x}{2 b^3 \sqrt {a+b x^4}}+\frac {(7 b c-12 a d) x \sqrt {a+b x^4}}{21 b^3}+\frac {d x^5 \sqrt {a+b x^4}}{7 b^2}-\frac {5 a^{3/4} (7 b c-9 a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{84 b^{13/4} \sqrt {a+b x^4}} \] Output:
1/2*a*(-a*d+b*c)*x/b^3/(b*x^4+a)^(1/2)+1/21*(-12*a*d+7*b*c)*x*(b*x^4+a)^(1 /2)/b^3+1/7*d*x^5*(b*x^4+a)^(1/2)/b^2-5/84*a^(3/4)*(-9*a*d+7*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))/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 = 10.08 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.58 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {x \left (-45 a^2 d+a b \left (35 c-18 d x^4\right )+2 b^2 x^4 \left (7 c+3 d x^4\right )+5 a (-7 b c+9 a d) \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )\right )}{42 b^3 \sqrt {a+b x^4}} \] Input:
Integrate[(x^8*(c + d*x^4))/(a + b*x^4)^(3/2),x]
Output:
(x*(-45*a^2*d + a*b*(35*c - 18*d*x^4) + 2*b^2*x^4*(7*c + 3*d*x^4) + 5*a*(- 7*b*c + 9*a*d)*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x ^4)/a)]))/(42*b^3*Sqrt[a + b*x^4])
Time = 0.45 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {959, 817, 843, 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.
| \(\displaystyle \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 959 |
| \(\displaystyle \frac {(7 b c-9 a d) \int \frac {x^8}{\left (b x^4+a\right )^{3/2}}dx}{7 b}+\frac {d x^9}{7 b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {(7 b c-9 a d) \left (\frac {5 \int \frac {x^4}{\sqrt {b x^4+a}}dx}{2 b}-\frac {x^5}{2 b \sqrt {a+b x^4}}\right )}{7 b}+\frac {d x^9}{7 b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 843 |
| \(\displaystyle \frac {(7 b c-9 a d) \left (\frac {5 \left (\frac {x \sqrt {a+b x^4}}{3 b}-\frac {a \int \frac {1}{\sqrt {b x^4+a}}dx}{3 b}\right )}{2 b}-\frac {x^5}{2 b \sqrt {a+b x^4}}\right )}{7 b}+\frac {d x^9}{7 b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {(7 b c-9 a d) \left (\frac {5 \left (\frac {x \sqrt {a+b x^4}}{3 b}-\frac {a^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{6 b^{5/4} \sqrt {a+b x^4}}\right )}{2 b}-\frac {x^5}{2 b \sqrt {a+b x^4}}\right )}{7 b}+\frac {d x^9}{7 b \sqrt {a+b x^4}}\) |
Input:
Int[(x^8*(c + d*x^4))/(a + b*x^4)^(3/2),x]
Output:
(d*x^9)/(7*b*Sqrt[a + b*x^4]) + ((7*b*c - 9*a*d)*(-1/2*x^5/(b*Sqrt[a + b*x ^4]) + (5*((x*Sqrt[a + b*x^4])/(3*b) - (a^(3/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sq rt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a ^(1/4)], 1/2])/(6*b^(5/4)*Sqrt[a + b*x^4])))/(2*b)))/(7*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[((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[((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 = 4.79 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.11
| method | result | size |
| elliptic | \(-\frac {a x \left (a d -c b \right )}{2 b^{3} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {d \,x^{5} \sqrt {b \,x^{4}+a}}{7 b^{2}}+\frac {\left (-\frac {a d -c b}{b^{2}}-\frac {5 d a}{7 b^{2}}\right ) x \sqrt {b \,x^{4}+a}}{3 b}+\frac {\left (\frac {a \left (a d -c b \right )}{2 b^{3}}-\frac {\left (-\frac {a d -c b}{b^{2}}-\frac {5 d a}{7 b^{2}}\right ) a}{3 b}\right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(195\) |
| default | \(c \left (\frac {a x}{2 b^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {x \sqrt {b \,x^{4}+a}}{3 b^{2}}-\frac {5 a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{6 b^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+d \left (-\frac {a^{2} x}{2 b^{3} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {x^{5} \sqrt {b \,x^{4}+a}}{7 b^{2}}-\frac {4 a x \sqrt {b \,x^{4}+a}}{7 b^{3}}+\frac {15 a^{2} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{14 b^{3} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\) | \(248\) |
| risch | \(-\frac {x \left (-3 d b \,x^{4}+12 a d -7 c b \right ) \sqrt {b \,x^{4}+a}}{21 b^{3}}+\frac {a \left (b \left (33 a d -28 c b \right ) \left (-\frac {x}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+12 a^{2} d \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )-7 a b c \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\right )}{21 b^{3}}\) | \(341\) |
Input:
int(x^8*(d*x^4+c)/(b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/2/b^3*a*x*(a*d-b*c)/((x^4+a/b)*b)^(1/2)+1/7*d*x^5*(b*x^4+a)^(1/2)/b^2+1 /3*(-1/b^2*(a*d-b*c)-5/7/b^2*d*a)/b*x*(b*x^4+a)^(1/2)+(1/2*a*(a*d-b*c)/b^3 -1/3*(-1/b^2*(a*d-b*c)-5/7/b^2*d*a)/b*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)*E llipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)
Time = 0.10 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.72 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/2}} \, dx=-\frac {5 \, {\left ({\left (7 \, b^{2} c - 9 \, a b d\right )} x^{4} + 7 \, a b c - 9 \, a^{2} d\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - {\left (6 \, b^{2} d x^{9} + 2 \, {\left (7 \, b^{2} c - 9 \, a b d\right )} x^{5} + 5 \, {\left (7 \, a b c - 9 \, a^{2} d\right )} x\right )} \sqrt {b x^{4} + a}}{42 \, {\left (b^{4} x^{4} + a b^{3}\right )}} \] Input:
integrate(x^8*(d*x^4+c)/(b*x^4+a)^(3/2),x, algorithm="fricas")
Output:
-1/42*(5*((7*b^2*c - 9*a*b*d)*x^4 + 7*a*b*c - 9*a^2*d)*sqrt(b)*(-a/b)^(3/4 )*elliptic_f(arcsin((-a/b)^(1/4)/x), -1) - (6*b^2*d*x^9 + 2*(7*b^2*c - 9*a *b*d)*x^5 + 5*(7*a*b*c - 9*a^2*d)*x)*sqrt(b*x^4 + a))/(b^4*x^4 + a*b^3)
Result contains complex when optimal does not.
Time = 11.36 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.45 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {c x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {13}{4}\right )} + \frac {d x^{13} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {17}{4}\right )} \] Input:
integrate(x**8*(d*x**4+c)/(b*x**4+a)**(3/2),x)
Output:
c*x**9*gamma(9/4)*hyper((3/2, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4* a**(3/2)*gamma(13/4)) + d*x**13*gamma(13/4)*hyper((3/2, 13/4), (17/4,), b* x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(17/4))
\[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{8}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^8*(d*x^4+c)/(b*x^4+a)^(3/2),x, algorithm="maxima")
Output:
integrate((d*x^4 + c)*x^8/(b*x^4 + a)^(3/2), x)
\[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{8}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^8*(d*x^4+c)/(b*x^4+a)^(3/2),x, algorithm="giac")
Output:
integrate((d*x^4 + c)*x^8/(b*x^4 + a)^(3/2), x)
Timed out. \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\int \frac {x^8\,\left (d\,x^4+c\right )}{{\left (b\,x^4+a\right )}^{3/2}} \,d x \] Input:
int((x^8*(c + d*x^4))/(a + b*x^4)^(3/2),x)
Output:
int((x^8*(c + d*x^4))/(a + b*x^4)^(3/2), x)
\[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {-45 \sqrt {b \,x^{4}+a}\, a^{2} d x +35 \sqrt {b \,x^{4}+a}\, a b c x -9 \sqrt {b \,x^{4}+a}\, a b d \,x^{5}+7 \sqrt {b \,x^{4}+a}\, b^{2} c \,x^{5}+3 \sqrt {b \,x^{4}+a}\, b^{2} d \,x^{9}+45 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{4} d -35 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{3} b c +45 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{3} b d \,x^{4}-35 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{2} b^{2} c \,x^{4}}{21 b^{3} \left (b \,x^{4}+a \right )} \] Input:
int(x^8*(d*x^4+c)/(b*x^4+a)^(3/2),x)
Output:
( - 45*sqrt(a + b*x**4)*a**2*d*x + 35*sqrt(a + b*x**4)*a*b*c*x - 9*sqrt(a + b*x**4)*a*b*d*x**5 + 7*sqrt(a + b*x**4)*b**2*c*x**5 + 3*sqrt(a + b*x**4) *b**2*d*x**9 + 45*int(sqrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)* a**4*d - 35*int(sqrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a**3*b *c + 45*int(sqrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a**3*b*d*x **4 - 35*int(sqrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a**2*b**2 *c*x**4)/(21*b**3*(a + b*x**4))