\(\int \frac {x^8 (c+d x^4)}{\sqrt {a+b x^4}} \, dx\) [25]

Optimal result

Integrand size = 22, antiderivative size = 179 \[ \int \frac {x^8 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=-\frac {5 a (11 b c-9 a d) x \sqrt {a+b x^4}}{231 b^3}+\frac {(11 b c-9 a d) x^5 \sqrt {a+b x^4}}{77 b^2}+\frac {d x^9 \sqrt {a+b x^4}}{11 b}+\frac {5 a^{7/4} (11 b c-9 a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{462 b^{13/4} \sqrt {a+b x^4}} \] Output:

-5/231*a*(-9*a*d+11*b*c)*x*(b*x^4+a)^(1/2)/b^3+1/77*(-9*a*d+11*b*c)*x^5*(b 
*x^4+a)^(1/2)/b^2+1/11*d*x^9*(b*x^4+a)^(1/2)/b+5/462*a^(7/4)*(-9*a*d+11*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)*Inverse 
JacobiAM(2*arctan(b^(1/4)*x/a^(1/4)),1/2*2^(1/2))/b^(13/4)/(b*x^4+a)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.11 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.64 \[ \int \frac {x^8 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=\frac {x \left (\left (a+b x^4\right ) \left (45 a^2 d+3 b^2 x^4 \left (11 c+7 d x^4\right )-a b \left (55 c+27 d x^4\right )\right )+5 a^2 (11 b c-9 a d) \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )\right )}{231 b^3 \sqrt {a+b x^4}} \] Input:

Integrate[(x^8*(c + d*x^4))/Sqrt[a + b*x^4],x]
 

Output:

(x*((a + b*x^4)*(45*a^2*d + 3*b^2*x^4*(11*c + 7*d*x^4) - a*b*(55*c + 27*d* 
x^4)) + 5*a^2*(11*b*c - 9*a*d)*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/4, 
1/2, 5/4, -((b*x^4)/a)]))/(231*b^3*Sqrt[a + b*x^4])
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 177, 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, 843, 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.

\(\displaystyle \int \frac {x^8 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx\)

\(\Big \downarrow \) 959

\(\displaystyle \frac {(11 b c-9 a d) \int \frac {x^8}{\sqrt {b x^4+a}}dx}{11 b}+\frac {d x^9 \sqrt {a+b x^4}}{11 b}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {(11 b c-9 a d) \left (\frac {x^5 \sqrt {a+b x^4}}{7 b}-\frac {5 a \int \frac {x^4}{\sqrt {b x^4+a}}dx}{7 b}\right )}{11 b}+\frac {d x^9 \sqrt {a+b x^4}}{11 b}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {(11 b c-9 a d) \left (\frac {x^5 \sqrt {a+b x^4}}{7 b}-\frac {5 a \left (\frac {x \sqrt {a+b x^4}}{3 b}-\frac {a \int \frac {1}{\sqrt {b x^4+a}}dx}{3 b}\right )}{7 b}\right )}{11 b}+\frac {d x^9 \sqrt {a+b x^4}}{11 b}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {(11 b c-9 a d) \left (\frac {x^5 \sqrt {a+b x^4}}{7 b}-\frac {5 a \left (\frac {x \sqrt {a+b x^4}}{3 b}-\frac {a^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{6 b^{5/4} \sqrt {a+b x^4}}\right )}{7 b}\right )}{11 b}+\frac {d x^9 \sqrt {a+b x^4}}{11 b}\)

Input:

Int[(x^8*(c + d*x^4))/Sqrt[a + b*x^4],x]
 

Output:

(d*x^9*Sqrt[a + b*x^4])/(11*b) + ((11*b*c - 9*a*d)*((x^5*Sqrt[a + b*x^4])/ 
(7*b) - (5*a*((x*Sqrt[a + b*x^4])/(3*b) - (a^(3/4)*(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])/(6*b^(5/4)*Sqrt[a + b*x^4])))/(7*b)))/(11*b)
 

Defintions of rubi rules used

rule 761

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]
 

rule 843

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]
 

rule 959

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]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 3.84 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {x \left (21 d \,b^{2} x^{8}-27 a b d \,x^{4}+33 b^{2} c \,x^{4}+45 a^{2} d -55 a b c \right ) \sqrt {b \,x^{4}+a}}{231 b^{3}}-\frac {5 a^{2} \left (9 a d -11 c b \right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{231 b^{3} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(140\)
elliptic	\(\frac {d \,x^{9} \sqrt {b \,x^{4}+a}}{11 b}+\frac {\left (c -\frac {9 a d}{11 b}\right ) x^{5} \sqrt {b \,x^{4}+a}}{7 b}-\frac {5 a \left (c -\frac {9 a d}{11 b}\right ) x \sqrt {b \,x^{4}+a}}{21 b^{2}}+\frac {5 a^{2} \left (c -\frac {9 a d}{11 b}\right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{21 b^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(156\)
default	\(c \left (\frac {x^{5} \sqrt {b \,x^{4}+a}}{7 b}-\frac {5 a x \sqrt {b \,x^{4}+a}}{21 b^{2}}+\frac {5 a^{2} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{21 b^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+d \left (\frac {x^{9} \sqrt {b \,x^{4}+a}}{11 b}-\frac {9 a \,x^{5} \sqrt {b \,x^{4}+a}}{77 b^{2}}+\frac {15 a^{2} x \sqrt {b \,x^{4}+a}}{77 b^{3}}-\frac {15 a^{3} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{77 b^{3} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\)	\(246\)

Input:

int(x^8*(d*x^4+c)/(b*x^4+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/231*x*(21*b^2*d*x^8-27*a*b*d*x^4+33*b^2*c*x^4+45*a^2*d-55*a*b*c)/b^3*(b* 
x^4+a)^(1/2)-5/231*a^2*(9*a*d-11*b*c)/b^3/(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)* 
EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.55 \[ \int \frac {x^8 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=\frac {5 \, {\left (11 \, a b c - 9 \, a^{2} d\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (21 \, b^{2} d x^{9} + 3 \, {\left (11 \, b^{2} c - 9 \, a b d\right )} x^{5} - 5 \, {\left (11 \, a b c - 9 \, a^{2} d\right )} x\right )} \sqrt {b x^{4} + a}}{231 \, b^{3}} \] Input:

integrate(x^8*(d*x^4+c)/(b*x^4+a)^(1/2),x, algorithm="fricas")
 

Output:

1/231*(5*(11*a*b*c - 9*a^2*d)*sqrt(b)*(-a/b)^(3/4)*elliptic_f(arcsin((-a/b 
)^(1/4)/x), -1) + (21*b^2*d*x^9 + 3*(11*b^2*c - 9*a*b*d)*x^5 - 5*(11*a*b*c 
 - 9*a^2*d)*x)*sqrt(b*x^4 + a))/b^3
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.64 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.45 \[ \int \frac {x^8 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=\frac {c x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {13}{4}\right )} + \frac {d x^{13} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {17}{4}\right )} \] Input:

integrate(x**8*(d*x**4+c)/(b*x**4+a)**(1/2),x)
 

Output:

c*x**9*gamma(9/4)*hyper((1/2, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4* 
sqrt(a)*gamma(13/4)) + d*x**13*gamma(13/4)*hyper((1/2, 13/4), (17/4,), b*x 
**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(17/4))
 

Maxima [F]

\[ \int \frac {x^8 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{8}}{\sqrt {b x^{4} + a}} \,d x } \] Input:

integrate(x^8*(d*x^4+c)/(b*x^4+a)^(1/2),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)*x^8/sqrt(b*x^4 + a), x)
 

Giac [F]

\[ \int \frac {x^8 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{8}}{\sqrt {b x^{4} + a}} \,d x } \] Input:

integrate(x^8*(d*x^4+c)/(b*x^4+a)^(1/2),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)*x^8/sqrt(b*x^4 + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^8 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=\int \frac {x^8\,\left (d\,x^4+c\right )}{\sqrt {b\,x^4+a}} \,d x \] Input:

int((x^8*(c + d*x^4))/(a + b*x^4)^(1/2),x)
 

Output:

int((x^8*(c + d*x^4))/(a + b*x^4)^(1/2), x)
 

Reduce [F]

\[ \int \frac {x^8 \left (c+d x^4\right )}{\sqrt {a+b x^4}} \, dx=\frac {45 \sqrt {b \,x^{4}+a}\, a^{2} d x -55 \sqrt {b \,x^{4}+a}\, a b c x -27 \sqrt {b \,x^{4}+a}\, a b d \,x^{5}+33 \sqrt {b \,x^{4}+a}\, b^{2} c \,x^{5}+21 \sqrt {b \,x^{4}+a}\, b^{2} d \,x^{9}-45 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{4}+a}d x \right ) a^{3} d +55 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{4}+a}d x \right ) a^{2} b c}{231 b^{3}} \] Input:

int(x^8*(d*x^4+c)/(b*x^4+a)^(1/2),x)
 

Output:

(45*sqrt(a + b*x**4)*a**2*d*x - 55*sqrt(a + b*x**4)*a*b*c*x - 27*sqrt(a + 
b*x**4)*a*b*d*x**5 + 33*sqrt(a + b*x**4)*b**2*c*x**5 + 21*sqrt(a + b*x**4) 
*b**2*d*x**9 - 45*int(sqrt(a + b*x**4)/(a + b*x**4),x)*a**3*d + 55*int(sqr 
t(a + b*x**4)/(a + b*x**4),x)*a**2*b*c)/(231*b**3)