\(\int \frac {A+B x^2+C x^4}{x^4 \sqrt [3]{a+b x^4}} \, dx\) [102]

Optimal result

Integrand size = 27, antiderivative size = 133 \[ \int \frac {A+B x^2+C x^4}{x^4 \sqrt [3]{a+b x^4}} \, dx=-\frac {A \left (a+b x^4\right )^{2/3}}{3 a x^3}-\frac {B \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{3},\frac {3}{4},-\frac {b x^4}{a}\right )}{x \sqrt [3]{a+b x^4}}-\frac {(A b-9 a C) x \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {5}{4},-\frac {b x^4}{a}\right )}{9 a \sqrt [3]{a+b x^4}} \] Output:

-1/3*A*(b*x^4+a)^(2/3)/a/x^3-B*(1+b*x^4/a)^(1/3)*hypergeom([-1/4, 1/3],[3/ 
4],-b*x^4/a)/x/(b*x^4+a)^(1/3)-1/9*(A*b-9*C*a)*x*(1+b*x^4/a)^(1/3)*hyperge 
om([1/4, 1/3],[5/4],-b*x^4/a)/a/(b*x^4+a)^(1/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 10.05 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x^2+C x^4}{x^4 \sqrt [3]{a+b x^4}} \, dx=\frac {\sqrt [3]{1+\frac {b x^4}{a}} \left (-A \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{3},\frac {1}{4},-\frac {b x^4}{a}\right )-3 B x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{3},\frac {3}{4},-\frac {b x^4}{a}\right )+3 C x^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {5}{4},-\frac {b x^4}{a}\right )\right )}{3 x^3 \sqrt [3]{a+b x^4}} \] Input:

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

Output:

((1 + (b*x^4)/a)^(1/3)*(-(A*Hypergeometric2F1[-3/4, 1/3, 1/4, -((b*x^4)/a) 
]) - 3*B*x^2*Hypergeometric2F1[-1/4, 1/3, 3/4, -((b*x^4)/a)] + 3*C*x^4*Hyp 
ergeometric2F1[1/4, 1/3, 5/4, -((b*x^4)/a)]))/(3*x^3*(a + b*x^4)^(1/3))
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2374, 9, 27, 1675, 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.

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

\(\Big \downarrow \) 2374

\(\displaystyle -\frac {\int -\frac {2 \left (9 a B x-(A b-9 a C) x^3\right )}{3 x^3 \sqrt [3]{b x^4+a}}dx}{6 a}-\frac {A \left (a+b x^4\right )^{2/3}}{3 a x^3}\)

\(\Big \downarrow \) 9

\(\displaystyle -\frac {\int -\frac {2 \left (9 a B-(A b-9 a C) x^2\right )}{3 x^2 \sqrt [3]{b x^4+a}}dx}{6 a}-\frac {A \left (a+b x^4\right )^{2/3}}{3 a x^3}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1675

\(\displaystyle \frac {\int \left (\frac {9 a B}{x^2 \sqrt [3]{b x^4+a}}-\frac {A b \left (1-\frac {9 a C}{A b}\right )}{\sqrt [3]{b x^4+a}}\right )dx}{9 a}-\frac {A \left (a+b x^4\right )^{2/3}}{3 a x^3}\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-1/3*(A*(a + b*x^4)^(2/3))/(a*x^3) + ((-9*a*B*(1 + (b*x^4)/a)^(1/3)*Hyperg 
eometric2F1[-1/4, 1/3, 3/4, -((b*x^4)/a)])/(x*(a + b*x^4)^(1/3)) - ((A*b - 
 9*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))/(9*a)
 

Defintions of rubi rules used

rule 9

Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, 
x, Min]}, Simp[1/e^(p*r)   Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, 
x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && 
  !MonomialQ[Px, x]
 

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 1675

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p 
_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + c*x^4)^p, 
x], x] /; FreeQ[{a, c, d, e, f, m, p, q}, x] && (IGtQ[p, 0] || IGtQ[q, 0] | 
| IntegersQ[m, q])
 

rule 2009

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

rule 2374

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Wit 
h[{Pq0 = Coeff[Pq, x, 0]}, Simp[Pq0*(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c 
*(m + 1))), x] + Simp[1/(2*a*c*(m + 1))   Int[(c*x)^(m + 1)*ExpandToSum[2*a 
*(m + 1)*((Pq - Pq0)/x) - 2*b*Pq0*(m + n*(p + 1) + 1)*x^(n - 1), x]*(a + b* 
x^n)^p, x], x] /; NeQ[Pq0, 0]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && 
 IGtQ[n, 0] && LtQ[m, -1] && LeQ[n - 1, Expon[Pq, x]]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.31 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.93 \[ \int \frac {A+B x^2+C x^4}{x^4 \sqrt [3]{a+b x^4}} \, dx=\frac {A \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{3} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [3]{a} x^{3} \Gamma \left (\frac {1}{4}\right )} + \frac {B \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{3} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [3]{a} x \Gamma \left (\frac {3}{4}\right )} + \frac {C x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{3} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [3]{a} \Gamma \left (\frac {5}{4}\right )} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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