\(\int \frac {(c+d x^4)^2}{\sqrt {a+b x^4}} \, dx\) [32]

Optimal result

Integrand size = 21, antiderivative size = 165 \[ \int \frac {\left (c+d x^4\right )^2}{\sqrt {a+b x^4}} \, dx=\frac {d (14 b c-5 a d) x \sqrt {a+b x^4}}{21 b^2}+\frac {d^2 x^5 \sqrt {a+b x^4}}{7 b}+\frac {\left (21 b^2 c^2-14 a b c d+5 a^2 d^2\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{42 \sqrt [4]{a} b^{9/4} \sqrt {a+b x^4}} \] Output:

1/21*d*(-5*a*d+14*b*c)*x*(b*x^4+a)^(1/2)/b^2+1/7*d^2*x^5*(b*x^4+a)^(1/2)/b 
+1/42*(5*a^2*d^2-14*a*b*c*d+21*b^2*c^2)*(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^(9/4)/(b*x^4+a)^(1/2)
 

Mathematica [C] (warning: unable to verify)

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

Time = 13.40 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.98 \[ \int \frac {\left (c+d x^4\right )^2}{\sqrt {a+b x^4}} \, dx=\frac {x \sqrt {1+\frac {b x^4}{a}} \left (13 a \left (45 c^2+18 c d x^4+5 d^2 x^8\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {13}{4},-\frac {b x^4}{a}\right )-4 b x^4 \left (7 c^2+10 c d x^4+3 d^2 x^8\right ) \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {3}{2},\frac {17}{4},-\frac {b x^4}{a}\right )-8 b x^4 \left (c+d x^4\right )^2 \, _3F_2\left (\frac {5}{4},\frac {3}{2},2;1,\frac {17}{4};-\frac {b x^4}{a}\right )\right )}{585 a \sqrt {a+b x^4}} \] Input:

Integrate[(c + d*x^4)^2/Sqrt[a + b*x^4],x]
 

Output:

(x*Sqrt[1 + (b*x^4)/a]*(13*a*(45*c^2 + 18*c*d*x^4 + 5*d^2*x^8)*Hypergeomet 
ric2F1[1/4, 1/2, 13/4, -((b*x^4)/a)] - 4*b*x^4*(7*c^2 + 10*c*d*x^4 + 3*d^2 
*x^8)*Hypergeometric2F1[5/4, 3/2, 17/4, -((b*x^4)/a)] - 8*b*x^4*(c + d*x^4 
)^2*HypergeometricPFQ[{5/4, 3/2, 2}, {1, 17/4}, -((b*x^4)/a)]))/(585*a*Sqr 
t[a + b*x^4])
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.07, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {933, 913, 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.

\(\displaystyle \int \frac {\left (c+d x^4\right )^2}{\sqrt {a+b x^4}} \, dx\)

\(\Big \downarrow \) 933

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

\(\Big \downarrow \) 913

\(\displaystyle \frac {\frac {\left (5 a^2 d^2-14 a b c d+21 b^2 c^2\right ) \int \frac {1}{\sqrt {b x^4+a}}dx}{3 b}+\frac {d x \sqrt {a+b x^4} (11 b c-5 a d)}{3 b}}{7 b}+\frac {d x \sqrt {a+b x^4} \left (c+d x^4\right )}{7 b}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {\frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (5 a^2 d^2-14 a b c d+21 b^2 c^2\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{6 \sqrt [4]{a} b^{5/4} \sqrt {a+b x^4}}+\frac {d x \sqrt {a+b x^4} (11 b c-5 a d)}{3 b}}{7 b}+\frac {d x \sqrt {a+b x^4} \left (c+d x^4\right )}{7 b}\)

Input:

Int[(c + d*x^4)^2/Sqrt[a + b*x^4],x]
 

Output:

(d*x*Sqrt[a + b*x^4]*(c + d*x^4))/(7*b) + ((d*(11*b*c - 5*a*d)*x*Sqrt[a + 
b*x^4])/(3*b) + ((21*b^2*c^2 - 14*a*b*c*d + 5*a^2*d^2)*(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*a^(1/4)*b^(5/4)*Sqrt[a + b*x^4]))/(7*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 913

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si 
mp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(p + 1) + 1))), x] - Simp[(a*d - b*c*(n*( 
p + 1) + 1))/(b*(n*(p + 1) + 1))   Int[(a + b*x^n)^p, x], x] /; FreeQ[{a, b 
, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]
 

rule 933

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[d*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), 
x] + Simp[1/(b*(n*(p + q) + 1))   Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Sim 
p[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q 
- 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d 
, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[ 
a, b, c, d, n, p, q, x]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 3.79 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.79

method	result	size
risch	\(-\frac {x d \left (-3 b d \,x^{4}+5 a d -14 b c \right ) \sqrt {b \,x^{4}+a}}{21 b^{2}}+\frac {\left (5 a^{2} d^{2}-14 a b c d +21 b^{2} c^{2}\right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{21 b^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(130\)
elliptic	\(\frac {d^{2} x^{5} \sqrt {b \,x^{4}+a}}{7 b}+\frac {\left (2 c d -\frac {5 d^{2} a}{7 b}\right ) x \sqrt {b \,x^{4}+a}}{3 b}+\frac {\left (c^{2}-\frac {a \left (2 c d -\frac {5 d^{2} a}{7 b}\right )}{3 b}\right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(144\)
default	\(\frac {c^{2} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+d^{2} \left (\frac {x^{5} \sqrt {b \,x^{4}+a}}{7 b}-\frac {5 a x \sqrt {b \,x^{4}+a}}{21 b^{2}}+\frac {5 a^{2} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{21 b^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+2 c d \left (\frac {x \sqrt {b \,x^{4}+a}}{3 b}-\frac {a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\)	\(282\)

Input:

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

Output:

-1/21*x*d*(-3*b*d*x^4+5*a*d-14*b*c)/b^2*(b*x^4+a)^(1/2)+1/21*(5*a^2*d^2-14 
*a*b*c*d+21*b^2*c^2)/b^2/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I*b^(1/2)*x^2/a^(1/2 
))^(1/2)*(1+I*b^(1/2)*x^2/a^(1/2))^(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 = 0.13 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.59 \[ \int \frac {\left (c+d x^4\right )^2}{\sqrt {a+b x^4}} \, dx=\frac {{\left (21 \, b^{2} c^{2} - 14 \, a b c d + 5 \, a^{2} d^{2}\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (3 \, a b d^{2} x^{5} + {\left (14 \, a b c d - 5 \, a^{2} d^{2}\right )} x\right )} \sqrt {b x^{4} + a}}{21 \, a b^{2}} \] Input:

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

Output:

1/21*((21*b^2*c^2 - 14*a*b*c*d + 5*a^2*d^2)*sqrt(b)*(-a/b)^(3/4)*elliptic_ 
f(arcsin((-a/b)^(1/4)/x), -1) + (3*a*b*d^2*x^5 + (14*a*b*c*d - 5*a^2*d^2)* 
x)*sqrt(b*x^4 + a))/(a*b^2)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.62 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.75 \[ \int \frac {\left (c+d x^4\right )^2}{\sqrt {a+b x^4}} \, dx=\frac {c^{2} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} + \frac {c d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {9}{4}\right )} + \frac {d^{2} x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {13}{4}\right )} \] Input:

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

Output:

c**2*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*s 
qrt(a)*gamma(5/4)) + c*d*x**5*gamma(5/4)*hyper((1/2, 5/4), (9/4,), b*x**4* 
exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(9/4)) + d**2*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))
 

Maxima [F]

\[ \int \frac {\left (c+d x^4\right )^2}{\sqrt {a+b x^4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )}^{2}}{\sqrt {b x^{4} + a}} \,d x } \] Input:

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

Output:

integrate((d*x^4 + c)^2/sqrt(b*x^4 + a), x)
 

Giac [F]

\[ \int \frac {\left (c+d x^4\right )^2}{\sqrt {a+b x^4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )}^{2}}{\sqrt {b x^{4} + a}} \,d x } \] Input:

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

Output:

integrate((d*x^4 + c)^2/sqrt(b*x^4 + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^4\right )^2}{\sqrt {a+b x^4}} \, dx=\int \frac {{\left (d\,x^4+c\right )}^2}{\sqrt {b\,x^4+a}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\left (c+d x^4\right )^2}{\sqrt {a+b x^4}} \, dx=\frac {-5 \sqrt {b \,x^{4}+a}\, a \,d^{2} x +14 \sqrt {b \,x^{4}+a}\, b c d x +3 \sqrt {b \,x^{4}+a}\, b \,d^{2} x^{5}+5 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{4}+a}d x \right ) a^{2} d^{2}-14 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{4}+a}d x \right ) a b c d +21 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{4}+a}d x \right ) b^{2} c^{2}}{21 b^{2}} \] Input:

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

Output:

( - 5*sqrt(a + b*x**4)*a*d**2*x + 14*sqrt(a + b*x**4)*b*c*d*x + 3*sqrt(a + 
 b*x**4)*b*d**2*x**5 + 5*int(sqrt(a + b*x**4)/(a + b*x**4),x)*a**2*d**2 - 
14*int(sqrt(a + b*x**4)/(a + b*x**4),x)*a*b*c*d + 21*int(sqrt(a + b*x**4)/ 
(a + b*x**4),x)*b**2*c**2)/(21*b**2)