\(\int \frac {A+B x^2+C x^4}{(a+b x^4)^{4/3}} \, dx\) [106]

Optimal result

Integrand size = 24, antiderivative size = 134 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^4\right )^{4/3}} \, dx=-\frac {3 C x}{b \sqrt [3]{a+b x^4}}+\frac {(A b+3 a C) x \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {4}{3},\frac {5}{4},-\frac {b x^4}{a}\right )}{a b \sqrt [3]{a+b x^4}}+\frac {B x^3 \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {4}{3},\frac {7}{4},-\frac {b x^4}{a}\right )}{3 a \sqrt [3]{a+b x^4}} \] Output:

-3*C*x/b/(b*x^4+a)^(1/3)+(A*b+3*C*a)*x*(1+b*x^4/a)^(1/3)*hypergeom([1/4, 4 
/3],[5/4],-b*x^4/a)/a/b/(b*x^4+a)^(1/3)+1/3*B*x^3*(1+b*x^4/a)^(1/3)*hyperg 
eom([3/4, 4/3],[7/4],-b*x^4/a)/a/(b*x^4+a)^(1/3)
 

Mathematica [A] (verified)

Time = 10.04 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^4\right )^{4/3}} \, dx=\frac {\sqrt [3]{1+\frac {b x^4}{a}} \left (15 A x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {4}{3},\frac {5}{4},-\frac {b x^4}{a}\right )+5 B x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {4}{3},\frac {7}{4},-\frac {b x^4}{a}\right )+3 C x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {4}{3},\frac {9}{4},-\frac {b x^4}{a}\right )\right )}{15 a \sqrt [3]{a+b x^4}} \] Input:

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

Output:

((1 + (b*x^4)/a)^(1/3)*(15*A*x*Hypergeometric2F1[1/4, 4/3, 5/4, -((b*x^4)/ 
a)] + 5*B*x^3*Hypergeometric2F1[3/4, 4/3, 7/4, -((b*x^4)/a)] + 3*C*x^5*Hyp 
ergeometric2F1[5/4, 4/3, 9/4, -((b*x^4)/a)]))/(15*a*(a + b*x^4)^(1/3))
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.16, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2397, 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.

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

\(\Big \downarrow \) 2397

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1516

\(\displaystyle \frac {\int \left (\frac {A b \left (\frac {3 a C}{A b}+1\right )}{\sqrt [3]{b x^4+a}}-\frac {5 b B x^2}{\sqrt [3]{b x^4+a}}\right )dx}{4 a b}+\frac {3 x \left (-a C+A b+b B x^2\right )}{4 a b \sqrt [3]{a+b x^4}}\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(3*x*(A*b - a*C + b*B*x^2))/(4*a*b*(a + b*x^4)^(1/3)) + (((A*b + 3*a*C)*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*b*B*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 2397

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, 
x]}, Module[{Q = PolynomialQuotient[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, 
 x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x]}, S 
imp[(-x)*R*((a + b*x^n)^(p + 1)/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1))), x] 
 + Simp[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1))   Int[(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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((C*x^4 + B*x^2 + A)*(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 = 3.78 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^4\right )^{4/3}} \, dx=\frac {A x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {4}{3} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {4}{3}} \Gamma \left (\frac {5}{4}\right )} + \frac {B x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {4}{3} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {4}{3}} \Gamma \left (\frac {7}{4}\right )} + \frac {C x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {4}{3} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {4}{3}} \Gamma \left (\frac {9}{4}\right )} \] Input:

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

Output:

A*x*gamma(1/4)*hyper((1/4, 4/3), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**( 
4/3)*gamma(5/4)) + B*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)) + C*x**5*gamma(5/4)*hyper((5/4, 4/3 
), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(4/3)*gamma(9/4))
 

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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