\(\int x^2 \sqrt [3]{a+b x^4} (A+B x^2+C x^4) \, dx\) [94]

Optimal result

Integrand size = 27, antiderivative size = 138 \[ \int x^2 \sqrt [3]{a+b x^4} \left (A+B x^2+C x^4\right ) \, dx=\frac {3 C x^3 \left (a+b x^4\right )^{4/3}}{25 b}+\frac {(25 A b-9 a C) x^3 \sqrt [3]{a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{75 b \sqrt [3]{1+\frac {b x^4}{a}}}+\frac {B x^5 \sqrt [3]{a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {5}{4},\frac {9}{4},-\frac {b x^4}{a}\right )}{5 \sqrt [3]{1+\frac {b x^4}{a}}} \] Output:

3/25*C*x^3*(b*x^4+a)^(4/3)/b+1/75*(25*A*b-9*C*a)*x^3*(b*x^4+a)^(1/3)*hyper 
geom([-1/3, 3/4],[7/4],-b*x^4/a)/b/(1+b*x^4/a)^(1/3)+1/5*B*x^5*(b*x^4+a)^( 
1/3)*hypergeom([-1/3, 5/4],[9/4],-b*x^4/a)/(1+b*x^4/a)^(1/3)
 

Mathematica [A] (verified)

Time = 10.05 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.76 \[ \int x^2 \sqrt [3]{a+b x^4} \left (A+B x^2+C x^4\right ) \, dx=\frac {\sqrt [3]{a+b x^4} \left (35 A x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )+21 B x^5 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {5}{4},\frac {9}{4},-\frac {b x^4}{a}\right )+15 C x^7 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {7}{4},\frac {11}{4},-\frac {b x^4}{a}\right )\right )}{105 \sqrt [3]{1+\frac {b x^4}{a}}} \] Input:

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

Output:

((a + b*x^4)^(1/3)*(35*A*x^3*Hypergeometric2F1[-1/3, 3/4, 7/4, -((b*x^4)/a 
)] + 21*B*x^5*Hypergeometric2F1[-1/3, 5/4, 9/4, -((b*x^4)/a)] + 15*C*x^7*H 
ypergeometric2F1[-1/3, 7/4, 11/4, -((b*x^4)/a)]))/(105*(1 + (b*x^4)/a)^(1/ 
3))
 

Rubi [A] (verified)

Time = 0.57 (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.

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

\(\Big \downarrow \) 2432

\(\displaystyle \int \left (A x^2 \sqrt [3]{a+b x^4}+B x^4 \sqrt [3]{a+b x^4}+C x^6 \sqrt [3]{a+b x^4}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((C*x^6 + B*x^4 + A*x^2)*(b*x^4 + a)^(1/3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.53 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.91 \[ \int x^2 \sqrt [3]{a+b x^4} \left (A+B x^2+C x^4\right ) \, dx=\frac {A \sqrt [3]{a} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {B \sqrt [3]{a} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} + \frac {C \sqrt [3]{a} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} \] Input:

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

Output:

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

Maxima [F]

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

integrate(x^2*(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^2, x)
 

Giac [F]

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

integrate(x^2*(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^2, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x^2 \sqrt [3]{a+b x^4} \left (A+B x^2+C x^4\right ) \, dx=\frac {9975 \left (b \,x^{4}+a \right )^{\frac {1}{3}} a b \,x^{3}+3900 \left (b \,x^{4}+a \right )^{\frac {1}{3}} a b x +1596 \left (b \,x^{4}+a \right )^{\frac {1}{3}} a c \,x^{3}+6825 \left (b \,x^{4}+a \right )^{\frac {1}{3}} b^{2} x^{5}+5187 \left (b \,x^{4}+a \right )^{\frac {1}{3}} b c \,x^{7}-3900 \left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {2}{3}}}d x \right ) a^{2} b +13300 \left (\int \frac {x^{2}}{\left (b \,x^{4}+a \right )^{\frac {2}{3}}}d x \right ) a^{2} b -4788 \left (\int \frac {x^{2}}{\left (b \,x^{4}+a \right )^{\frac {2}{3}}}d x \right ) a^{2} c}{43225 b} \] Input:

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

Output:

(9975*(a + b*x**4)**(1/3)*a*b*x**3 + 3900*(a + b*x**4)**(1/3)*a*b*x + 1596 
*(a + b*x**4)**(1/3)*a*c*x**3 + 6825*(a + b*x**4)**(1/3)*b**2*x**5 + 5187* 
(a + b*x**4)**(1/3)*b*c*x**7 - 3900*int((a + b*x**4)**(1/3)/(a + b*x**4),x 
)*a**2*b + 13300*int(((a + b*x**4)**(1/3)*x**2)/(a + b*x**4),x)*a**2*b - 4 
788*int(((a + b*x**4)**(1/3)*x**2)/(a + b*x**4),x)*a**2*c)/(43225*b)