\(\int \frac {x^2 (A+B x^2+C x^4)}{(a+b x^4)^{4/3}} \, dx\) [105]

Optimal result

Integrand size = 27, antiderivative size = 144 \[ \int \frac {x^2 \left (A+B x^2+C x^4\right )}{\left (a+b x^4\right )^{4/3}} \, dx=\frac {3 C x^3}{5 b \sqrt [3]{a+b x^4}}+\frac {(5 A b-9 a C) x^3 \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {4}{3},\frac {7}{4},-\frac {b x^4}{a}\right )}{15 a b \sqrt [3]{a+b x^4}}+\frac {B x^5 \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {4}{3},\frac {9}{4},-\frac {b x^4}{a}\right )}{5 a \sqrt [3]{a+b x^4}} \] Output:

3/5*C*x^3/b/(b*x^4+a)^(1/3)+1/15*(5*A*b-9*C*a)*x^3*(1+b*x^4/a)^(1/3)*hyper 
geom([3/4, 4/3],[7/4],-b*x^4/a)/a/b/(b*x^4+a)^(1/3)+1/5*B*x^5*(1+b*x^4/a)^ 
(1/3)*hypergeom([5/4, 4/3],[9/4],-b*x^4/a)/a/(b*x^4+a)^(1/3)
 

Mathematica [A] (verified)

Time = 10.04 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.75 \[ \int \frac {x^2 \left (A+B x^2+C x^4\right )}{\left (a+b x^4\right )^{4/3}} \, dx=\frac {\sqrt [3]{1+\frac {b x^4}{a}} \left (35 A x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {4}{3},\frac {7}{4},-\frac {b x^4}{a}\right )+21 B x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {4}{3},\frac {9}{4},-\frac {b x^4}{a}\right )+15 C x^7 \operatorname {Hypergeometric2F1}\left (\frac {4}{3},\frac {7}{4},\frac {11}{4},-\frac {b x^4}{a}\right )\right )}{105 a \sqrt [3]{a+b x^4}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2367, 27, 1516, 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.

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

\(\Big \downarrow \) 2367

\(\displaystyle -\frac {3 \int -\frac {b \left (3 a B-(5 A b-9 a C) x^2\right )}{3 \sqrt [3]{b x^4+a}}dx}{4 a b^2}-\frac {3 x \left (a B-x^2 (A b-a C)\right )}{4 a b \sqrt [3]{a+b x^4}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {3 a B-(5 A b-9 a C) x^2}{\sqrt [3]{b x^4+a}}dx}{4 a b}-\frac {3 x \left (a B-x^2 (A b-a C)\right )}{4 a b \sqrt [3]{a+b x^4}}\)

\(\Big \downarrow \) 1516

\(\displaystyle \frac {\int \left (\frac {3 a B}{\sqrt [3]{b x^4+a}}-\frac {(5 A b-9 a C) x^2}{\sqrt [3]{b x^4+a}}\right )dx}{4 a b}-\frac {3 x \left (a B-x^2 (A b-a C)\right )}{4 a b \sqrt [3]{a+b x^4}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {3 a B x \sqrt [3]{\frac {b x^4}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {5}{4},-\frac {b x^4}{a}\right )}{\sqrt [3]{a+b x^4}}-\frac {x^3 \sqrt [3]{\frac {b x^4}{a}+1} (5 A b-9 a C) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{3 \sqrt [3]{a+b x^4}}}{4 a b}-\frac {3 x \left (a B-x^2 (A b-a C)\right )}{4 a b \sqrt [3]{a+b x^4}}\)

Input:

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

Output:

(-3*x*(a*B - (A*b - a*C)*x^2))/(4*a*b*(a + b*x^4)^(1/3)) + ((3*a*B*x*(1 + 
(b*x^4)/a)^(1/3)*Hypergeometric2F1[1/4, 1/3, 5/4, -((b*x^4)/a)])/(a + b*x^ 
4)^(1/3) - ((5*A*b - 9*a*C)*x^3*(1 + (b*x^4)/a)^(1/3)*Hypergeometric2F1[1/ 
3, 3/4, 7/4, -((b*x^4)/a)])/(3*(a + b*x^4)^(1/3)))/(4*a*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 1516

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[Expa 
ndIntegrand[(d + e*x^2)*(a + c*x^4)^p, x], x] /; FreeQ[{a, c, d, e}, x] && 
NeQ[c*d^2 + a*e^2, 0]
 

rule 2009

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

rule 2367

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = 
 m + Expon[Pq, x]}, Module[{Q = PolynomialQuotient[b^(Floor[(q - 1)/n] + 1) 
*x^m*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] + 1)*x^ 
m*Pq, a + b*x^n, x]}, Simp[(-x)*R*((a + b*x^n)^(p + 1)/(a*n*(p + 1)*b^(Floo 
r[(q - 1)/n] + 1))), x] + Simp[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1))   I 
nt[(a + b*x^n)^(p + 1)*ExpandToSum[a*n*(p + 1)*Q + n*(p + 1)*R + D[x*R, x], 
 x], x], x]] /; GeQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0 
] && LtQ[p, -1] && IGtQ[m, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((C*x^6 + B*x^4 + A*x^2)*(b*x^4 + a)^(2/3)/(b^2*x^8 + 2*a*b*x^4 + 
a^2), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.34 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.84 \[ \int \frac {x^2 \left (A+B x^2+C x^4\right )}{\left (a+b x^4\right )^{4/3}} \, dx=\frac {A x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {4}{3} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {4}{3}} \Gamma \left (\frac {7}{4}\right )} + \frac {B x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {4}{3} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {4}{3}} \Gamma \left (\frac {9}{4}\right )} + \frac {C x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {4}{3}} \Gamma \left (\frac {11}{4}\right )} \] Input:

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

Output:

A*x**3*gamma(3/4)*hyper((3/4, 4/3), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*a 
**(4/3)*gamma(7/4)) + B*x**5*gamma(5/4)*hyper((5/4, 4/3), (9/4,), b*x**4*e 
xp_polar(I*pi)/a)/(4*a**(4/3)*gamma(9/4)) + C*x**7*gamma(7/4)*hyper((4/3, 
7/4), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(4/3)*gamma(11/4))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^2 \left (A+B x^2+C x^4\right )}{\left (a+b x^4\right )^{4/3}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{4}+a \right )^{\frac {1}{3}} a +\left (b \,x^{4}+a \right )^{\frac {1}{3}} b \,x^{4}}d x \right ) c +\left (\int \frac {x^{4}}{\left (b \,x^{4}+a \right )^{\frac {1}{3}} a +\left (b \,x^{4}+a \right )^{\frac {1}{3}} b \,x^{4}}d x \right ) b +\left (\int \frac {x^{2}}{\left (b \,x^{4}+a \right )^{\frac {1}{3}} a +\left (b \,x^{4}+a \right )^{\frac {1}{3}} b \,x^{4}}d x \right ) a \] Input:

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

Output:

int(x**6/((a + b*x**4)**(1/3)*a + (a + b*x**4)**(1/3)*b*x**4),x)*c + int(x 
**4/((a + b*x**4)**(1/3)*a + (a + b*x**4)**(1/3)*b*x**4),x)*b + int(x**2/( 
(a + b*x**4)**(1/3)*a + (a + b*x**4)**(1/3)*b*x**4),x)*a