\(\int \frac {A+B x^2+C x^4}{x^4 (a+b x^4)^{4/3}} \, dx\) [108]

Optimal result

Integrand size = 27, antiderivative size = 137 \[ \int \frac {A+B x^2+C x^4}{x^4 \left (a+b x^4\right )^{4/3}} \, dx=-\frac {A}{3 a x^3 \sqrt [3]{a+b x^4}}-\frac {B \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {4}{3},\frac {3}{4},-\frac {b x^4}{a}\right )}{a x \sqrt [3]{a+b x^4}}-\frac {(13 A b-9 a C) x \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {4}{3},\frac {5}{4},-\frac {b x^4}{a}\right )}{9 a^2 \sqrt [3]{a+b x^4}} \] Output:

-1/3*A/a/x^3/(b*x^4+a)^(1/3)-B*(1+b*x^4/a)^(1/3)*hypergeom([-1/4, 4/3],[3/ 
4],-b*x^4/a)/a/x/(b*x^4+a)^(1/3)-1/9*(13*A*b-9*C*a)*x*(1+b*x^4/a)^(1/3)*hy 
pergeom([1/4, 4/3],[5/4],-b*x^4/a)/a^2/(b*x^4+a)^(1/3)
 

Mathematica [A] (verified)

Time = 10.06 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x^2+C x^4}{x^4 \left (a+b x^4\right )^{4/3}} \, dx=\frac {\sqrt [3]{1+\frac {b x^4}{a}} \left (-A \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {4}{3},\frac {1}{4},-\frac {b x^4}{a}\right )-3 B x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {4}{3},\frac {3}{4},-\frac {b x^4}{a}\right )+3 C x^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {4}{3},\frac {5}{4},-\frac {b x^4}{a}\right )\right )}{3 a x^3 \sqrt [3]{a+b x^4}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.20 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.49, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {2368, 27, 2374, 9, 27, 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.

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

\(\Big \downarrow \) 2368

\(\displaystyle -\frac {3 \int -\frac {\frac {5 b^2 B x^6}{a}-b \left (\frac {A b}{a}-C\right ) x^4+4 b B x^2+4 A b}{3 x^4 \sqrt [3]{b x^4+a}}dx}{4 a b}-\frac {3 x \left (-a C+A b+b B x^2\right )}{4 a^2 \sqrt [3]{a+b x^4}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\frac {5 b^2 B x^6}{a}-b \left (\frac {A b}{a}-C\right ) x^4+4 b B x^2+4 A b}{x^4 \sqrt [3]{b x^4+a}}dx}{4 a b}-\frac {3 x \left (-a C+A b+b B x^2\right )}{4 a^2 \sqrt [3]{a+b x^4}}\)

\(\Big \downarrow \) 2374

\(\displaystyle \frac {-\frac {\int -\frac {2 \left (45 b^2 B x^5-b (13 A b-9 a C) x^3+36 a b B x\right )}{3 x^3 \sqrt [3]{b x^4+a}}dx}{6 a}-\frac {4 A b \left (a+b x^4\right )^{2/3}}{3 a x^3}}{4 a b}-\frac {3 x \left (-a C+A b+b B x^2\right )}{4 a^2 \sqrt [3]{a+b x^4}}\)

\(\Big \downarrow \) 9

\(\displaystyle \frac {-\frac {\int -\frac {2 \left (45 b^2 B x^4-b (13 A b-9 a C) x^2+36 a b B\right )}{3 x^2 \sqrt [3]{b x^4+a}}dx}{6 a}-\frac {4 A b \left (a+b x^4\right )^{2/3}}{3 a x^3}}{4 a b}-\frac {3 x \left (-a C+A b+b B x^2\right )}{4 a^2 \sqrt [3]{a+b x^4}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {45 b^2 B x^4-b (13 A b-9 a C) x^2+36 a b B}{x^2 \sqrt [3]{b x^4+a}}dx}{9 a}-\frac {4 A b \left (a+b x^4\right )^{2/3}}{3 a x^3}}{4 a b}-\frac {3 x \left (-a C+A b+b B x^2\right )}{4 a^2 \sqrt [3]{a+b x^4}}\)

\(\Big \downarrow \) 2374

\(\displaystyle \frac {\frac {-\frac {\int \frac {2 \left (a b (13 A b-9 a C) x-105 a b^2 B x^3\right )}{x \sqrt [3]{b x^4+a}}dx}{2 a}-\frac {36 b B \left (a+b x^4\right )^{2/3}}{x}}{9 a}-\frac {4 A b \left (a+b x^4\right )^{2/3}}{3 a x^3}}{4 a b}-\frac {3 x \left (-a C+A b+b B x^2\right )}{4 a^2 \sqrt [3]{a+b x^4}}\)

\(\Big \downarrow \) 9

\(\displaystyle \frac {\frac {-\frac {\int \frac {2 a b \left (-105 b B x^2+13 A b-9 a C\right )}{\sqrt [3]{b x^4+a}}dx}{2 a}-\frac {36 b B \left (a+b x^4\right )^{2/3}}{x}}{9 a}-\frac {4 A b \left (a+b x^4\right )^{2/3}}{3 a x^3}}{4 a b}-\frac {3 x \left (-a C+A b+b B x^2\right )}{4 a^2 \sqrt [3]{a+b x^4}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-b \int \frac {-105 b B x^2+13 A b-9 a C}{\sqrt [3]{b x^4+a}}dx-\frac {36 b B \left (a+b x^4\right )^{2/3}}{x}}{9 a}-\frac {4 A b \left (a+b x^4\right )^{2/3}}{3 a x^3}}{4 a b}-\frac {3 x \left (-a C+A b+b B x^2\right )}{4 a^2 \sqrt [3]{a+b x^4}}\)

\(\Big \downarrow \) 1516

\(\displaystyle \frac {\frac {-b \int \left (\frac {13 A b \left (1-\frac {9 a C}{13 A b}\right )}{\sqrt [3]{b x^4+a}}-\frac {105 b B x^2}{\sqrt [3]{b x^4+a}}\right )dx-\frac {36 b B \left (a+b x^4\right )^{2/3}}{x}}{9 a}-\frac {4 A b \left (a+b x^4\right )^{2/3}}{3 a x^3}}{4 a b}-\frac {3 x \left (-a C+A b+b B x^2\right )}{4 a^2 \sqrt [3]{a+b x^4}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {-b \left (\frac {x \sqrt [3]{\frac {b x^4}{a}+1} (13 A b-9 a C) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {5}{4},-\frac {b x^4}{a}\right )}{\sqrt [3]{a+b x^4}}-\frac {35 b B x^3 \sqrt [3]{\frac {b x^4}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{\sqrt [3]{a+b x^4}}\right )-\frac {36 b B \left (a+b x^4\right )^{2/3}}{x}}{9 a}-\frac {4 A b \left (a+b x^4\right )^{2/3}}{3 a x^3}}{4 a b}-\frac {3 x \left (-a C+A b+b B x^2\right )}{4 a^2 \sqrt [3]{a+b x^4}}\)

Input:

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

Output:

(-3*x*(A*b - a*C + b*B*x^2))/(4*a^2*(a + b*x^4)^(1/3)) + ((-4*A*b*(a + b*x 
^4)^(2/3))/(3*a*x^3) + ((-36*b*B*(a + b*x^4)^(2/3))/x - b*(((13*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) - (35*b*B*x^3*(1 + (b*x^4)/a)^(1/3)*Hypergeometric2F1[1 
/3, 3/4, 7/4, -((b*x^4)/a)])/(a + b*x^4)^(1/3)))/(9*a))/(4*a*b)
 

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 2368

Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = 
Expon[Pq, x]}, Module[{Q = PolynomialQuotient[a*b^(Floor[(q - 1)/n] + 1)*x^ 
m*Pq, a + b*x^n, x], R = PolynomialRemainder[a*b^(Floor[(q - 1)/n] + 1)*x^m 
*Pq, a + b*x^n, x], i}, Simp[(-x)*R*((a + b*x^n)^(p + 1)/(a^2*n*(p + 1)*b^( 
Floor[(q - 1)/n] + 1))), x] + Simp[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)) 
   Int[x^m*(a + b*x^n)^(p + 1)*ExpandToSum[(n*(p + 1)*Q)/x^m + Sum[((n*(p + 
 1) + i + 1)/a)*Coeff[R, x, i]*x^(i - m), {i, 0, n - 1}], x], x], x]]] /; F 
reeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && ILtQ[m, 0]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

integrate((C*x^4+B*x^2+A)/x^4/(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^12 + 2*a*b*x^8 + a^2 
*x^4), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 8.49 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x^2+C x^4}{x^4 \left (a+b x^4\right )^{4/3}} \, dx=\frac {A \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {4}{3} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {4}{3}} x^{3} \Gamma \left (\frac {1}{4}\right )} + \frac {B \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {4}{3} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {4}{3}} x \Gamma \left (\frac {3}{4}\right )} + \frac {C x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {4}{3} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {4}{3}} \Gamma \left (\frac {5}{4}\right )} \] Input:

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

Output:

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

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {A+B x^2+C x^4}{x^4 \left (a+b x^4\right )^{4/3}} \, dx=\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{3}} a \,x^{4}+\left (b \,x^{4}+a \right )^{\frac {1}{3}} b \,x^{8}}d x \right ) a +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{3}} a \,x^{2}+\left (b \,x^{4}+a \right )^{\frac {1}{3}} b \,x^{6}}d x \right ) b +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{3}} a +\left (b \,x^{4}+a \right )^{\frac {1}{3}} b \,x^{4}}d x \right ) c \] Input:

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

Output:

int(1/((a + b*x**4)**(1/3)*a*x**4 + (a + b*x**4)**(1/3)*b*x**8),x)*a + int 
(1/((a + b*x**4)**(1/3)*a*x**2 + (a + b*x**4)**(1/3)*b*x**6),x)*b + int(1/ 
((a + b*x**4)**(1/3)*a + (a + b*x**4)**(1/3)*b*x**4),x)*c