\(\int x^4 \sqrt [3]{a+b x^4} (A+B x^2+C x^4) \, dx\) [93]

Optimal result

Integrand size = 27, antiderivative size = 138 \[ \int x^4 \sqrt [3]{a+b x^4} \left (A+B x^2+C x^4\right ) \, dx=\frac {3 C x^5 \left (a+b x^4\right )^{4/3}}{31 b}+\frac {(31 A b-15 a C) x^5 \sqrt [3]{a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {5}{4},\frac {9}{4},-\frac {b x^4}{a}\right )}{155 b \sqrt [3]{1+\frac {b x^4}{a}}}+\frac {B x^7 \sqrt [3]{a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {7}{4},\frac {11}{4},-\frac {b x^4}{a}\right )}{7 \sqrt [3]{1+\frac {b x^4}{a}}} \] Output:

3/31*C*x^5*(b*x^4+a)^(4/3)/b+1/155*(31*A*b-15*C*a)*x^5*(b*x^4+a)^(1/3)*hyp 
ergeom([-1/3, 5/4],[9/4],-b*x^4/a)/b/(1+b*x^4/a)^(1/3)+1/7*B*x^7*(b*x^4+a) 
^(1/3)*hypergeom([-1/3, 7/4],[11/4],-b*x^4/a)/(1+b*x^4/a)^(1/3)
 

Mathematica [A] (verified)

Time = 10.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.76 \[ \int x^4 \sqrt [3]{a+b x^4} \left (A+B x^2+C x^4\right ) \, dx=\frac {\sqrt [3]{a+b x^4} \left (63 A x^5 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {5}{4},\frac {9}{4},-\frac {b x^4}{a}\right )+45 B x^7 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {7}{4},\frac {11}{4},-\frac {b x^4}{a}\right )+35 C x^9 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {9}{4},\frac {13}{4},-\frac {b x^4}{a}\right )\right )}{315 \sqrt [3]{1+\frac {b x^4}{a}}} \] Input:

Integrate[x^4*(a + b*x^4)^(1/3)*(A + B*x^2 + C*x^4),x]
 

Output:

((a + b*x^4)^(1/3)*(63*A*x^5*Hypergeometric2F1[-1/3, 5/4, 9/4, -((b*x^4)/a 
)] + 45*B*x^7*Hypergeometric2F1[-1/3, 7/4, 11/4, -((b*x^4)/a)] + 35*C*x^9* 
Hypergeometric2F1[-1/3, 9/4, 13/4, -((b*x^4)/a)]))/(315*(1 + (b*x^4)/a)^(1 
/3))
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.14, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2432, 2009}

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.

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

\(\Big \downarrow \) 2432

\(\displaystyle \int \left (A x^4 \sqrt [3]{a+b x^4}+B x^6 \sqrt [3]{a+b x^4}+C x^8 \sqrt [3]{a+b x^4}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {A x^5 \sqrt [3]{a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {5}{4},\frac {9}{4},-\frac {b x^4}{a}\right )}{5 \sqrt [3]{\frac {b x^4}{a}+1}}+\frac {B x^7 \sqrt [3]{a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {7}{4},\frac {11}{4},-\frac {b x^4}{a}\right )}{7 \sqrt [3]{\frac {b x^4}{a}+1}}+\frac {C x^9 \sqrt [3]{a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {9}{4},\frac {13}{4},-\frac {b x^4}{a}\right )}{9 \sqrt [3]{\frac {b x^4}{a}+1}}\)

Input:

Int[x^4*(a + b*x^4)^(1/3)*(A + B*x^2 + C*x^4),x]
 

Output:

(A*x^5*(a + b*x^4)^(1/3)*Hypergeometric2F1[-1/3, 5/4, 9/4, -((b*x^4)/a)])/ 
(5*(1 + (b*x^4)/a)^(1/3)) + (B*x^7*(a + b*x^4)^(1/3)*Hypergeometric2F1[-1/ 
3, 7/4, 11/4, -((b*x^4)/a)])/(7*(1 + (b*x^4)/a)^(1/3)) + (C*x^9*(a + b*x^4 
)^(1/3)*Hypergeometric2F1[-1/3, 9/4, 13/4, -((b*x^4)/a)])/(9*(1 + (b*x^4)/ 
a)^(1/3))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2432

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[ 
Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n, p}, x] && (PolyQ[Pq, x] || Poly 
Q[Pq, x^n])
 

Maple [F]

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

Input:

int(x^4*(b*x^4+a)^(1/3)*(C*x^4+B*x^2+A),x)
 

Output:

int(x^4*(b*x^4+a)^(1/3)*(C*x^4+B*x^2+A),x)
 

Fricas [F]

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

integrate(x^4*(b*x^4+a)^(1/3)*(C*x^4+B*x^2+A),x, algorithm="fricas")
 

Output:

integral((C*x^8 + B*x^6 + A*x^4)*(b*x^4 + a)^(1/3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.59 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.91 \[ \int x^4 \sqrt [3]{a+b x^4} \left (A+B x^2+C x^4\right ) \, dx=\frac {A \sqrt [3]{a} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} + \frac {B \sqrt [3]{a} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} + \frac {C \sqrt [3]{a} x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {13}{4}\right )} \] Input:

integrate(x**4*(b*x**4+a)**(1/3)*(C*x**4+B*x**2+A),x)
 

Output:

A*a**(1/3)*x**5*gamma(5/4)*hyper((-1/3, 5/4), (9/4,), b*x**4*exp_polar(I*p 
i)/a)/(4*gamma(9/4)) + B*a**(1/3)*x**7*gamma(7/4)*hyper((-1/3, 7/4), (11/4 
,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(11/4)) + C*a**(1/3)*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]

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

integrate(x^4*(b*x^4+a)^(1/3)*(C*x^4+B*x^2+A),x, algorithm="maxima")
 

Output:

integrate((C*x^4 + B*x^2 + A)*(b*x^4 + a)^(1/3)*x^4, x)
 

Giac [F]

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

integrate(x^4*(b*x^4+a)^(1/3)*(C*x^4+B*x^2+A),x, algorithm="giac")
 

Output:

integrate((C*x^4 + B*x^2 + A)*(b*x^4 + a)^(1/3)*x^4, x)
 

Mupad [F(-1)]

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

int(x^4*(a + b*x^4)^(1/3)*(A + B*x^2 + C*x^4),x)
 

Output:

int(x^4*(a + b*x^4)^(1/3)*(A + B*x^2 + C*x^4), x)
 

Reduce [F]

\[ \int x^4 \sqrt [3]{a+b x^4} \left (A+B x^2+C x^4\right ) \, dx=\frac {\frac {12 \left (b \,x^{4}+a \right )^{\frac {1}{3}} a^{2} b x}{133}-\frac {180 \left (b \,x^{4}+a \right )^{\frac {1}{3}} a^{2} c x}{4123}+\frac {3 \left (b \,x^{4}+a \right )^{\frac {1}{3}} a \,b^{2} x^{5}}{19}+\frac {12 \left (b \,x^{4}+a \right )^{\frac {1}{3}} a \,b^{2} x^{3}}{325}+\frac {12 \left (b \,x^{4}+a \right )^{\frac {1}{3}} a b c \,x^{5}}{589}+\frac {3 \left (b \,x^{4}+a \right )^{\frac {1}{3}} b^{3} x^{7}}{25}+\frac {3 \left (b \,x^{4}+a \right )^{\frac {1}{3}} b^{2} c \,x^{9}}{31}-\frac {12 \left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {2}{3}}}d x \right ) a^{3} b}{133}+\frac {180 \left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {2}{3}}}d x \right ) a^{3} c}{4123}-\frac {36 \left (\int \frac {x^{2}}{\left (b \,x^{4}+a \right )^{\frac {2}{3}}}d x \right ) a^{2} b^{2}}{325}}{b^{2}} \] Input:

int(x^4*(b*x^4+a)^(1/3)*(C*x^4+B*x^2+A),x)
 

Output:

(3*(40300*(a + b*x**4)**(1/3)*a**2*b*x - 19500*(a + b*x**4)**(1/3)*a**2*c* 
x + 70525*(a + b*x**4)**(1/3)*a*b**2*x**5 + 16492*(a + b*x**4)**(1/3)*a*b* 
*2*x**3 + 9100*(a + b*x**4)**(1/3)*a*b*c*x**5 + 53599*(a + b*x**4)**(1/3)* 
b**3*x**7 + 43225*(a + b*x**4)**(1/3)*b**2*c*x**9 - 40300*int((a + b*x**4) 
**(1/3)/(a + b*x**4),x)*a**3*b + 19500*int((a + b*x**4)**(1/3)/(a + b*x**4 
),x)*a**3*c - 49476*int(((a + b*x**4)**(1/3)*x**2)/(a + b*x**4),x)*a**2*b* 
*2))/(1339975*b**2)