\(\int \sqrt [3]{a+b x^4} (c+d x^4)^2 \, dx\) [63]

Optimal result

Integrand size = 21, antiderivative size = 125 \[ \int \sqrt [3]{a+b x^4} \left (c+d x^4\right )^2 \, dx=\frac {3 d (62 b c-15 a d) x \left (a+b x^4\right )^{4/3}}{589 b^2}+\frac {3 d^2 x^5 \left (a+b x^4\right )^{4/3}}{31 b}+\frac {\left (589 c^2-\frac {3 a d (62 b c-15 a d)}{b^2}\right ) x \sqrt [3]{a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{4},\frac {5}{4},-\frac {b x^4}{a}\right )}{589 \sqrt [3]{1+\frac {b x^4}{a}}} \] Output:

3/589*d*(-15*a*d+62*b*c)*x*(b*x^4+a)^(4/3)/b^2+3/31*d^2*x^5*(b*x^4+a)^(4/3 
)/b+1/589*(589*c^2-3*a*d*(-15*a*d+62*b*c)/b^2)*x*(b*x^4+a)^(1/3)*hypergeom 
([-1/3, 1/4],[5/4],-b*x^4/a)/(1+b*x^4/a)^(1/3)
 

Mathematica [A] (verified)

Time = 13.65 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.43 \[ \int \sqrt [3]{a+b x^4} \left (c+d x^4\right )^2 \, dx=\frac {x \sqrt [3]{a+b x^4} \left (13 a \left (45 c^2+18 c d x^4+5 d^2 x^8\right ) \operatorname {Gamma}\left (-\frac {1}{3}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{4},\frac {13}{4},-\frac {b x^4}{a}\right )-8 b x^4 \left (7 c^2+10 c d x^4+3 d^2 x^8\right ) \operatorname {Gamma}\left (\frac {2}{3}\right ) \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {5}{4},\frac {17}{4},-\frac {b x^4}{a}\right )-16 b x^4 \left (c+d x^4\right )^2 \operatorname {Gamma}\left (\frac {2}{3}\right ) \, _3F_2\left (\frac {2}{3},\frac {5}{4},2;1,\frac {17}{4};-\frac {b x^4}{a}\right )\right )}{585 a \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {Gamma}\left (-\frac {1}{3}\right )} \] Input:

Integrate[(a + b*x^4)^(1/3)*(c + d*x^4)^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {933, 27, 913, 779, 778}

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.

\(\displaystyle \int \sqrt [3]{a+b x^4} \left (c+d x^4\right )^2 \, dx\)

\(\Big \downarrow \) 933

\(\displaystyle \frac {3 \int \frac {1}{3} \sqrt [3]{b x^4+a} \left (d (43 b c-15 a d) x^4+c (31 b c-3 a d)\right )dx}{31 b}+\frac {3 d x \left (a+b x^4\right )^{4/3} \left (c+d x^4\right )}{31 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sqrt [3]{b x^4+a} \left (d (43 b c-15 a d) x^4+c (31 b c-3 a d)\right )dx}{31 b}+\frac {3 d x \left (a+b x^4\right )^{4/3} \left (c+d x^4\right )}{31 b}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {\frac {\left (45 a^2 d^2-186 a b c d+589 b^2 c^2\right ) \int \sqrt [3]{b x^4+a}dx}{19 b}+\frac {3 d x \left (a+b x^4\right )^{4/3} (43 b c-15 a d)}{19 b}}{31 b}+\frac {3 d x \left (a+b x^4\right )^{4/3} \left (c+d x^4\right )}{31 b}\)

\(\Big \downarrow \) 779

\(\displaystyle \frac {\frac {\sqrt [3]{a+b x^4} \left (45 a^2 d^2-186 a b c d+589 b^2 c^2\right ) \int \sqrt [3]{\frac {b x^4}{a}+1}dx}{19 b \sqrt [3]{\frac {b x^4}{a}+1}}+\frac {3 d x \left (a+b x^4\right )^{4/3} (43 b c-15 a d)}{19 b}}{31 b}+\frac {3 d x \left (a+b x^4\right )^{4/3} \left (c+d x^4\right )}{31 b}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {\frac {x \sqrt [3]{a+b x^4} \left (45 a^2 d^2-186 a b c d+589 b^2 c^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{4},\frac {5}{4},-\frac {b x^4}{a}\right )}{19 b \sqrt [3]{\frac {b x^4}{a}+1}}+\frac {3 d x \left (a+b x^4\right )^{4/3} (43 b c-15 a d)}{19 b}}{31 b}+\frac {3 d x \left (a+b x^4\right )^{4/3} \left (c+d x^4\right )}{31 b}\)

Input:

Int[(a + b*x^4)^(1/3)*(c + d*x^4)^2,x]
 

Output:

(3*d*x*(a + b*x^4)^(4/3)*(c + d*x^4))/(31*b) + ((3*d*(43*b*c - 15*a*d)*x*( 
a + b*x^4)^(4/3))/(19*b) + ((589*b^2*c^2 - 186*a*b*c*d + 45*a^2*d^2)*x*(a 
+ b*x^4)^(1/3)*Hypergeometric2F1[-1/3, 1/4, 5/4, -((b*x^4)/a)])/(19*b*(1 + 
 (b*x^4)/a)^(1/3)))/(31*b)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 778

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 
1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p 
, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || 
GtQ[a, 0])
 

rule 779

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x 
^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p])   Int[(1 + b*(x^n/a))^p, x], x 
] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Si 
mplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

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]

\[\int \left (b \,x^{4}+a \right )^{\frac {1}{3}} \left (d \,x^{4}+c \right )^{2}d x\]

Input:

int((b*x^4+a)^(1/3)*(d*x^4+c)^2,x)
 

Output:

int((b*x^4+a)^(1/3)*(d*x^4+c)^2,x)
 

Fricas [F]

\[ \int \sqrt [3]{a+b x^4} \left (c+d x^4\right )^2 \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {1}{3}} {\left (d x^{4} + c\right )}^{2} \,d x } \] Input:

integrate((b*x^4+a)^(1/3)*(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)^(1/3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.58 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.03 \[ \int \sqrt [3]{a+b x^4} \left (c+d x^4\right )^2 \, dx=\frac {\sqrt [3]{a} c^{2} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {\sqrt [3]{a} c d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {9}{4}\right )} + \frac {\sqrt [3]{a} d^{2} x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {13}{4}\right )} \] Input:

integrate((b*x**4+a)**(1/3)*(d*x**4+c)**2,x)
 

Output:

a**(1/3)*c**2*x*gamma(1/4)*hyper((-1/3, 1/4), (5/4,), b*x**4*exp_polar(I*p 
i)/a)/(4*gamma(5/4)) + a**(1/3)*c*d*x**5*gamma(5/4)*hyper((-1/3, 5/4), (9/ 
4,), b*x**4*exp_polar(I*pi)/a)/(2*gamma(9/4)) + a**(1/3)*d**2*x**9*gamma(9 
/4)*hyper((-1/3, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(13/4))
 

Maxima [F]

\[ \int \sqrt [3]{a+b x^4} \left (c+d x^4\right )^2 \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {1}{3}} {\left (d x^{4} + c\right )}^{2} \,d x } \] Input:

integrate((b*x^4+a)^(1/3)*(d*x^4+c)^2,x, algorithm="maxima")
 

Output:

integrate((b*x^4 + a)^(1/3)*(d*x^4 + c)^2, x)
 

Giac [F]

\[ \int \sqrt [3]{a+b x^4} \left (c+d x^4\right )^2 \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {1}{3}} {\left (d x^{4} + c\right )}^{2} \,d x } \] Input:

integrate((b*x^4+a)^(1/3)*(d*x^4+c)^2,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((b*x^4 + a)^(1/3)*(d*x^4 + c)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt [3]{a+b x^4} \left (c+d x^4\right )^2 \, dx=\int {\left (b\,x^4+a\right )}^{1/3}\,{\left (d\,x^4+c\right )}^2 \,d x \] Input:

int((a + b*x^4)^(1/3)*(c + d*x^4)^2,x)
 

Output:

int((a + b*x^4)^(1/3)*(c + d*x^4)^2, x)
 

Reduce [F]

\[ \int \sqrt [3]{a+b x^4} \left (c+d x^4\right )^2 \, dx=\frac {-180 \left (b \,x^{4}+a \right )^{\frac {1}{3}} a^{2} d^{2} x +744 \left (b \,x^{4}+a \right )^{\frac {1}{3}} a b c d x +84 \left (b \,x^{4}+a \right )^{\frac {1}{3}} a b \,d^{2} x^{5}+1767 \left (b \,x^{4}+a \right )^{\frac {1}{3}} b^{2} c^{2} x +1302 \left (b \,x^{4}+a \right )^{\frac {1}{3}} b^{2} c d \,x^{5}+399 \left (b \,x^{4}+a \right )^{\frac {1}{3}} b^{2} d^{2} x^{9}+180 \left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {2}{3}}}d x \right ) a^{3} d^{2}-744 \left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {2}{3}}}d x \right ) a^{2} b c d +2356 \left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {2}{3}}}d x \right ) a \,b^{2} c^{2}}{4123 b^{2}} \] Input:

int((b*x^4+a)^(1/3)*(d*x^4+c)^2,x)
 

Output:

( - 180*(a + b*x**4)**(1/3)*a**2*d**2*x + 744*(a + b*x**4)**(1/3)*a*b*c*d* 
x + 84*(a + b*x**4)**(1/3)*a*b*d**2*x**5 + 1767*(a + b*x**4)**(1/3)*b**2*c 
**2*x + 1302*(a + b*x**4)**(1/3)*b**2*c*d*x**5 + 399*(a + b*x**4)**(1/3)*b 
**2*d**2*x**9 + 180*int((a + b*x**4)**(1/3)/(a + b*x**4),x)*a**3*d**2 - 74 
4*int((a + b*x**4)**(1/3)/(a + b*x**4),x)*a**2*b*c*d + 2356*int((a + b*x** 
4)**(1/3)/(a + b*x**4),x)*a*b**2*c**2)/(4123*b**2)