\(\int \frac {A+B x^2+C x^4}{x^2 \sqrt [3]{a+b x^4}} \, dx\) [101]

Optimal result

Integrand size = 27, antiderivative size = 131 \[ \int \frac {A+B x^2+C x^4}{x^2 \sqrt [3]{a+b x^4}} \, dx=-\frac {A \left (a+b x^4\right )^{2/3}}{a x}+\frac {B x \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {5}{4},-\frac {b x^4}{a}\right )}{\sqrt [3]{a+b x^4}}+\frac {(5 A b+3 a C) x^3 \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{9 a \sqrt [3]{a+b x^4}} \] Output:

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

Mathematica [A] (verified)

Time = 10.06 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x^2+C x^4}{x^2 \sqrt [3]{a+b x^4}} \, dx=\frac {\sqrt [3]{1+\frac {b x^4}{a}} \left (-3 A \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{3},\frac {3}{4},-\frac {b x^4}{a}\right )+3 B x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {5}{4},-\frac {b x^4}{a}\right )+C x^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{3 x \sqrt [3]{a+b x^4}} \] Input:

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

Output:

((1 + (b*x^4)/a)^(1/3)*(-3*A*Hypergeometric2F1[-1/4, 1/3, 3/4, -((b*x^4)/a 
)] + 3*B*x^2*Hypergeometric2F1[1/4, 1/3, 5/4, -((b*x^4)/a)] + C*x^4*Hyperg 
eometric2F1[1/3, 3/4, 7/4, -((b*x^4)/a)]))/(3*x*(a + b*x^4)^(1/3))
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.05, 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, 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.

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

\(\Big \downarrow \) 2374

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

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1516

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-((A*(a + b*x^4)^(2/3))/(a*x)) + ((3*a*B*x*(1 + (b*x^4)/a)^(1/3)*Hypergeom 
etric2F1[1/4, 1/3, 5/4, -((b*x^4)/a)])/(a + b*x^4)^(1/3) + ((5*A*b + 3*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)))/(3*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 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 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]

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

Input:

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

Output:

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

Fricas [F]

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

integrate((C*x^4+B*x^2+A)/x^2/(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^6 + a*x^2), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.26 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x^2+C x^4}{x^2 \sqrt [3]{a+b x^4}} \, dx=\frac {A \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{3} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [3]{a} x \Gamma \left (\frac {3}{4}\right )} + \frac {B x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{3} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [3]{a} \Gamma \left (\frac {5}{4}\right )} + \frac {C x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [3]{a} \Gamma \left (\frac {7}{4}\right )} \] Input:

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

Output:

A*gamma(-1/4)*hyper((-1/4, 1/3), (3/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**( 
1/3)*x*gamma(3/4)) + B*x*gamma(1/4)*hyper((1/4, 1/3), (5/4,), b*x**4*exp_p 
olar(I*pi)/a)/(4*a**(1/3)*gamma(5/4)) + C*x**3*gamma(3/4)*hyper((1/3, 3/4) 
, (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(1/3)*gamma(7/4))
 

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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