\(\int \frac {A+B x^2+C x^4}{(a+b x^2)^{2/3}} \, dx\) [123]

Optimal result

Integrand size = 24, antiderivative size = 317 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {3 (11 b B-9 a C) x \sqrt [3]{a+b x^2}}{55 b^2}+\frac {3 C x^3 \sqrt [3]{a+b x^2}}{11 b}-\frac {3^{3/4} \sqrt {2-\sqrt {3}} \left (55 A b^2-33 a b B+27 a^2 C\right ) \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),-7+4 \sqrt {3}\right )}{55 b^3 x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}} \] Output:

3/55*(11*B*b-9*C*a)*x*(b*x^2+a)^(1/3)/b^2+3/11*C*x^3*(b*x^2+a)^(1/3)/b-1/5 
5*3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2))*(55*A*b^2-33*B*a*b+27*C*a^2)*(a^(1/3)- 
(b*x^2+a)^(1/3))*((a^(2/3)+a^(1/3)*(b*x^2+a)^(1/3)+(b*x^2+a)^(2/3))/((1-3^ 
(1/2))*a^(1/3)-(b*x^2+a)^(1/3))^2)^(1/2)*EllipticF(((1+3^(1/2))*a^(1/3)-(b 
*x^2+a)^(1/3))/((1-3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3)),2*I-I*3^(1/2))/b^3/x/ 
(-a^(1/3)*(a^(1/3)-(b*x^2+a)^(1/3))/((1-3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3))^ 
2)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.07 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.31 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {x \left (-3 \left (a+b x^2\right ) \left (-11 b B+9 a C-5 b C x^2\right )+\left (55 A b^2+3 a (-11 b B+9 a C)\right ) \left (1+\frac {b x^2}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{55 b^2 \left (a+b x^2\right )^{2/3}} \] Input:

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

Output:

(x*(-3*(a + b*x^2)*(-11*b*B + 9*a*C - 5*b*C*x^2) + (55*A*b^2 + 3*a*(-11*b* 
B + 9*a*C))*(1 + (b*x^2)/a)^(2/3)*Hypergeometric2F1[1/2, 2/3, 3/2, -((b*x^ 
2)/a)]))/(55*b^2*(a + b*x^2)^(2/3))
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1473, 27, 299, 234, 760}

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.

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

\(\Big \downarrow \) 1473

\(\displaystyle \frac {3 \int \frac {(11 b B-9 a C) x^2+11 A b}{3 \left (b x^2+a\right )^{2/3}}dx}{11 b}+\frac {3 C x^3 \sqrt [3]{a+b x^2}}{11 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(11 b B-9 a C) x^2+11 A b}{\left (b x^2+a\right )^{2/3}}dx}{11 b}+\frac {3 C x^3 \sqrt [3]{a+b x^2}}{11 b}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\left (55 A b^2-3 a (11 b B-9 a C)\right ) \int \frac {1}{\left (b x^2+a\right )^{2/3}}dx}{5 b}+\frac {3 x \sqrt [3]{a+b x^2} (11 b B-9 a C)}{5 b}}{11 b}+\frac {3 C x^3 \sqrt [3]{a+b x^2}}{11 b}\)

\(\Big \downarrow \) 234

\(\displaystyle \frac {\frac {3 \sqrt {b x^2} \left (55 A b^2-3 a (11 b B-9 a C)\right ) \int \frac {1}{\sqrt {b x^2}}d\sqrt [3]{b x^2+a}}{10 b^2 x}+\frac {3 x \sqrt [3]{a+b x^2} (11 b B-9 a C)}{5 b}}{11 b}+\frac {3 C x^3 \sqrt [3]{a+b x^2}}{11 b}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {\frac {3 x \sqrt [3]{a+b x^2} (11 b B-9 a C)}{5 b}-\frac {3^{3/4} \sqrt {2-\sqrt {3}} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \left (55 A b^2-3 a (11 b B-9 a C)\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right ),-7+4 \sqrt {3}\right )}{5 b^2 x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}}{11 b}+\frac {3 C x^3 \sqrt [3]{a+b x^2}}{11 b}\)

Input:

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

Output:

(3*C*x^3*(a + b*x^2)^(1/3))/(11*b) + ((3*(11*b*B - 9*a*C)*x*(a + b*x^2)^(1 
/3))/(5*b) - (3^(3/4)*Sqrt[2 - Sqrt[3]]*(55*A*b^2 - 3*a*(11*b*B - 9*a*C))* 
(a^(1/3) - (a + b*x^2)^(1/3))*Sqrt[(a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3) + 
(a + b*x^2)^(2/3))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))^2]*Elliptic 
F[ArcSin[((1 + Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))/((1 - Sqrt[3])*a^(1/3 
) - (a + b*x^2)^(1/3))], -7 + 4*Sqrt[3]])/(5*b^2*x*Sqrt[-((a^(1/3)*(a^(1/3 
) - (a + b*x^2)^(1/3)))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))^2)]))/ 
(11*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 234

Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Simp[3*(Sqrt[b*x^2]/(2*b*x)) 
   Subst[Int[1/Sqrt[-a + x^3], x], x, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b 
}, x]
 

rule 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 760

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s 
*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- 
s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) 
*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x 
] && NegQ[a]
 

rule 1473

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> Simp[c^p*x^(4*p - 1)*((d + e*x^2)^(q + 1)/(e*(4*p + 2*q + 1))) 
, x] + Simp[1/(e*(4*p + 2*q + 1))   Int[(d + e*x^2)^q*ExpandToSum[e*(4*p + 
2*q + 1)*(a + b*x^2 + c*x^4)^p - d*c^p*(4*p - 1)*x^(4*p - 2) - e*c^p*(4*p + 
 2*q + 1)*x^(4*p), x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b^2 - 
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] &&  !LtQ[q, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [A] (verification not implemented)

Time = 1.17 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.27 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {A x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {2}{3} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {2}{3}}} + \frac {B x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}}} + \frac {C x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5 a^{\frac {2}{3}}} \] Input:

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

Output:

A*x*hyper((1/2, 2/3), (3/2,), b*x**2*exp_polar(I*pi)/a)/a**(2/3) + B*x**3* 
hyper((2/3, 3/2), (5/2,), b*x**2*exp_polar(I*pi)/a)/(3*a**(2/3)) + C*x**5* 
hyper((2/3, 5/2), (7/2,), b*x**2*exp_polar(I*pi)/a)/(5*a**(2/3))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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