\(\int \sqrt [4]{a+b x^4} (c+d x^4)^2 \, dx\) [98]

Optimal result

Integrand size = 21, antiderivative size = 180 \[ \int \sqrt [4]{a+b x^4} \left (c+d x^4\right )^2 \, dx=\frac {\left (12 b^2 c^2-4 a b c d+a^2 d^2\right ) x \sqrt [4]{a+b x^4}}{24 b^2}+\frac {d (4 b c-a d) x \left (a+b x^4\right )^{5/4}}{12 b^2}+\frac {d^2 x^5 \left (a+b x^4\right )^{5/4}}{10 b}-\frac {\sqrt {a} \left (12 b^2 c^2-4 a b c d+a^2 d^2\right ) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{24 b^{3/2} \left (a+b x^4\right )^{3/4}} \] Output:

1/24*(a^2*d^2-4*a*b*c*d+12*b^2*c^2)*x*(b*x^4+a)^(1/4)/b^2+1/12*d*(-a*d+4*b 
*c)*x*(b*x^4+a)^(5/4)/b^2+1/10*d^2*x^5*(b*x^4+a)^(5/4)/b-1/24*a^(1/2)*(a^2 
*d^2-4*a*b*c*d+12*b^2*c^2)*(1+a/b/x^4)^(3/4)*x^3*InverseJacobiAM(1/2*arcco 
t(b^(1/2)*x^2/a^(1/2)),2^(1/2))/b^(3/2)/(b*x^4+a)^(3/4)
 

Mathematica [C] (verified)

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

Time = 11.80 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.99 \[ \int \sqrt [4]{a+b x^4} \left (c+d x^4\right )^2 \, dx=\frac {x \sqrt [4]{a+b x^4} \left (13 a \left (45 c^2+18 c d x^4+5 d^2 x^8\right ) \operatorname {Gamma}\left (-\frac {1}{4}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{4},\frac {13}{4},-\frac {b x^4}{a}\right )-8 b x^4 \left (7 c^2+10 c d x^4+3 d^2 x^8\right ) \operatorname {Gamma}\left (\frac {3}{4}\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {17}{4},-\frac {b x^4}{a}\right )-16 b x^4 \left (c+d x^4\right )^2 \operatorname {Gamma}\left (\frac {3}{4}\right ) \, _3F_2\left (\frac {3}{4},\frac {5}{4},2;1,\frac {17}{4};-\frac {b x^4}{a}\right )\right )}{585 a \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Gamma}\left (-\frac {1}{4}\right )} \] Input:

Integrate[(a + b*x^4)^(1/4)*(c + d*x^4)^2,x]
 

Output:

(x*(a + b*x^4)^(1/4)*(13*a*(45*c^2 + 18*c*d*x^4 + 5*d^2*x^8)*Gamma[-1/4]*H 
ypergeometric2F1[-1/4, 1/4, 13/4, -((b*x^4)/a)] - 8*b*x^4*(7*c^2 + 10*c*d* 
x^4 + 3*d^2*x^8)*Gamma[3/4]*Hypergeometric2F1[3/4, 5/4, 17/4, -((b*x^4)/a) 
] - 16*b*x^4*(c + d*x^4)^2*Gamma[3/4]*HypergeometricPFQ[{3/4, 5/4, 2}, {1, 
 17/4}, -((b*x^4)/a)]))/(585*a*(1 + (b*x^4)/a)^(1/4)*Gamma[-1/4])
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {933, 913, 748, 768, 858, 807, 229}

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.

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

\(\Big \downarrow \) 933

\(\displaystyle \frac {\int \sqrt [4]{b x^4+a} \left (d (14 b c-5 a d) x^4+c (10 b c-a d)\right )dx}{10 b}+\frac {d x \left (a+b x^4\right )^{5/4} \left (c+d x^4\right )}{10 b}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {\frac {5 \left (a^2 d^2-4 a b c d+12 b^2 c^2\right ) \int \sqrt [4]{b x^4+a}dx}{6 b}+\frac {d x \left (a+b x^4\right )^{5/4} (14 b c-5 a d)}{6 b}}{10 b}+\frac {d x \left (a+b x^4\right )^{5/4} \left (c+d x^4\right )}{10 b}\)

\(\Big \downarrow \) 748

\(\displaystyle \frac {\frac {5 \left (a^2 d^2-4 a b c d+12 b^2 c^2\right ) \left (\frac {1}{2} a \int \frac {1}{\left (b x^4+a\right )^{3/4}}dx+\frac {1}{2} x \sqrt [4]{a+b x^4}\right )}{6 b}+\frac {d x \left (a+b x^4\right )^{5/4} (14 b c-5 a d)}{6 b}}{10 b}+\frac {d x \left (a+b x^4\right )^{5/4} \left (c+d x^4\right )}{10 b}\)

\(\Big \downarrow \) 768

\(\displaystyle \frac {\frac {5 \left (a^2 d^2-4 a b c d+12 b^2 c^2\right ) \left (\frac {a x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x^3}dx}{2 \left (a+b x^4\right )^{3/4}}+\frac {1}{2} x \sqrt [4]{a+b x^4}\right )}{6 b}+\frac {d x \left (a+b x^4\right )^{5/4} (14 b c-5 a d)}{6 b}}{10 b}+\frac {d x \left (a+b x^4\right )^{5/4} \left (c+d x^4\right )}{10 b}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {\frac {5 \left (a^2 d^2-4 a b c d+12 b^2 c^2\right ) \left (\frac {1}{2} x \sqrt [4]{a+b x^4}-\frac {a x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x}d\frac {1}{x}}{2 \left (a+b x^4\right )^{3/4}}\right )}{6 b}+\frac {d x \left (a+b x^4\right )^{5/4} (14 b c-5 a d)}{6 b}}{10 b}+\frac {d x \left (a+b x^4\right )^{5/4} \left (c+d x^4\right )}{10 b}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {\frac {5 \left (a^2 d^2-4 a b c d+12 b^2 c^2\right ) \left (\frac {1}{2} x \sqrt [4]{a+b x^4}-\frac {a x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4}}d\frac {1}{x^2}}{4 \left (a+b x^4\right )^{3/4}}\right )}{6 b}+\frac {d x \left (a+b x^4\right )^{5/4} (14 b c-5 a d)}{6 b}}{10 b}+\frac {d x \left (a+b x^4\right )^{5/4} \left (c+d x^4\right )}{10 b}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {\frac {5 \left (a^2 d^2-4 a b c d+12 b^2 c^2\right ) \left (\frac {1}{2} x \sqrt [4]{a+b x^4}-\frac {\sqrt {a} \sqrt {b} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{2 \left (a+b x^4\right )^{3/4}}\right )}{6 b}+\frac {d x \left (a+b x^4\right )^{5/4} (14 b c-5 a d)}{6 b}}{10 b}+\frac {d x \left (a+b x^4\right )^{5/4} \left (c+d x^4\right )}{10 b}\)

Input:

Int[(a + b*x^4)^(1/4)*(c + d*x^4)^2,x]
 

Output:

(d*x*(a + b*x^4)^(5/4)*(c + d*x^4))/(10*b) + ((d*(14*b*c - 5*a*d)*x*(a + b 
*x^4)^(5/4))/(6*b) + (5*(12*b^2*c^2 - 4*a*b*c*d + a^2*d^2)*((x*(a + b*x^4) 
^(1/4))/2 - (Sqrt[a]*Sqrt[b]*(1 + a/(b*x^4))^(3/4)*x^3*EllipticF[ArcTan[Sq 
rt[a]/(Sqrt[b]*x^2)]/2, 2])/(2*(a + b*x^4)^(3/4))))/(6*b))/(10*b)
 

Defintions of rubi rules used

rule 229

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) 
)*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 748

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p 
+ 1)), x] + Simp[a*n*(p/(n*p + 1))   Int[(a + b*x^n)^(p - 1), x], x] /; Fre 
eQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || LtQ[Denominat 
or[p + 1/n], Denominator[p]])
 

rule 768

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 
/4)/(a + b*x^4)^(3/4))   Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ 
[{a, b}, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

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 [F]

\[\int \left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (d \,x^{4}+c \right )^{2}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \sqrt [4]{a+b x^4} \left (c+d x^4\right )^2 \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {1}{4}} {\left (d x^{4} + c\right )}^{2} \,d x } \] Input:

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.64 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.72 \[ \int \sqrt [4]{a+b x^4} \left (c+d x^4\right )^2 \, dx=\frac {\sqrt [4]{a} c^{2} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {\sqrt [4]{a} c d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {9}{4}\right )} + \frac {\sqrt [4]{a} d^{2} x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {13}{4}\right )} \] Input:

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

Output:

a**(1/4)*c**2*x*gamma(1/4)*hyper((-1/4, 1/4), (5/4,), b*x**4*exp_polar(I*p 
i)/a)/(4*gamma(5/4)) + a**(1/4)*c*d*x**5*gamma(5/4)*hyper((-1/4, 5/4), (9/ 
4,), b*x**4*exp_polar(I*pi)/a)/(2*gamma(9/4)) + a**(1/4)*d**2*x**9*gamma(9 
/4)*hyper((-1/4, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(13/4))
 

Maxima [F]

\[ \int \sqrt [4]{a+b x^4} \left (c+d x^4\right )^2 \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {1}{4}} {\left (d x^{4} + c\right )}^{2} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \sqrt [4]{a+b x^4} \left (c+d x^4\right )^2 \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {1}{4}} {\left (d x^{4} + c\right )}^{2} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \sqrt [4]{a+b x^4} \left (c+d x^4\right )^2 \, dx=\int {\left (b\,x^4+a\right )}^{1/4}\,{\left (d\,x^4+c\right )}^2 \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \sqrt [4]{a+b x^4} \left (c+d x^4\right )^2 \, dx=\frac {-5 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} d^{2} x +20 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a b c d x +2 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a b \,d^{2} x^{5}+60 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2} c^{2} x +40 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2} c d \,x^{5}+12 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2} d^{2} x^{9}+5 \left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}d x \right ) a^{3} d^{2}-20 \left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}d x \right ) a^{2} b c d +60 \left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}d x \right ) a \,b^{2} c^{2}}{120 b^{2}} \] Input:

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

Output:

( - 5*(a + b*x**4)**(1/4)*a**2*d**2*x + 20*(a + b*x**4)**(1/4)*a*b*c*d*x + 
 2*(a + b*x**4)**(1/4)*a*b*d**2*x**5 + 60*(a + b*x**4)**(1/4)*b**2*c**2*x 
+ 40*(a + b*x**4)**(1/4)*b**2*c*d*x**5 + 12*(a + b*x**4)**(1/4)*b**2*d**2* 
x**9 + 5*int((a + b*x**4)**(1/4)/(a + b*x**4),x)*a**3*d**2 - 20*int((a + b 
*x**4)**(1/4)/(a + b*x**4),x)*a**2*b*c*d + 60*int((a + b*x**4)**(1/4)/(a + 
 b*x**4),x)*a*b**2*c**2)/(120*b**2)