\(\int \sqrt [3]{a+b x^2} (A+B x^2+C x^4) \, dx\) [122]

Optimal result

Integrand size = 24, antiderivative size = 353 \[ \int \sqrt [3]{a+b x^2} \left (A+B x^2+C x^4\right ) \, dx=\frac {3}{935} \left (187 A-\frac {3 a (17 b B-9 a C)}{b^2}\right ) x \sqrt [3]{a+b x^2}+\frac {3 (17 b B-9 a C) x \left (a+b x^2\right )^{4/3}}{187 b^2}+\frac {3 C x^3 \left (a+b x^2\right )^{4/3}}{17 b}-\frac {2\ 3^{3/4} \sqrt {2-\sqrt {3}} a \left (187 A b^2-51 a b B+27 a^2 C\right ) \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),-7+4 \sqrt {3}\right )}{935 b^3 x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}} \] Output:

3/935*(187*A-3*a*(17*B*b-9*C*a)/b^2)*x*(b*x^2+a)^(1/3)+3/187*(17*B*b-9*C*a 
)*x*(b*x^2+a)^(4/3)/b^2+3/17*C*x^3*(b*x^2+a)^(4/3)/b-2/935*3^(3/4)*(1/2*6^ 
(1/2)-1/2*2^(1/2))*a*(187*A*b^2-51*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 = 5.63 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.28 \[ \int \sqrt [3]{a+b x^2} \left (A+B x^2+C x^4\right ) \, dx=\frac {x \sqrt [3]{a+b x^2} \left (-3 \left (a+b x^2\right ) \left (-17 b B+9 a C-11 b C x^2\right )+\frac {\left (187 A b^2+3 a (-17 b B+9 a C)\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )}{\sqrt [3]{1+\frac {b x^2}{a}}}\right )}{187 b^2} \] Input:

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

Output:

(x*(a + b*x^2)^(1/3)*(-3*(a + b*x^2)*(-17*b*B + 9*a*C - 11*b*C*x^2) + ((18 
7*A*b^2 + 3*a*(-17*b*B + 9*a*C))*Hypergeometric2F1[-1/3, 1/2, 3/2, -((b*x^ 
2)/a)])/(1 + (b*x^2)/a)^(1/3)))/(187*b^2)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 351, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1473, 27, 299, 211, 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.

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

\(\Big \downarrow \) 1473

\(\displaystyle \frac {3 \int \frac {1}{3} \sqrt [3]{b x^2+a} \left ((17 b B-9 a C) x^2+17 A b\right )dx}{17 b}+\frac {3 C x^3 \left (a+b x^2\right )^{4/3}}{17 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sqrt [3]{b x^2+a} \left ((17 b B-9 a C) x^2+17 A b\right )dx}{17 b}+\frac {3 C x^3 \left (a+b x^2\right )^{4/3}}{17 b}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\left (187 A b^2-3 a (17 b B-9 a C)\right ) \int \sqrt [3]{b x^2+a}dx}{11 b}+\frac {3 x \left (a+b x^2\right )^{4/3} (17 b B-9 a C)}{11 b}}{17 b}+\frac {3 C x^3 \left (a+b x^2\right )^{4/3}}{17 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {\left (187 A b^2-3 a (17 b B-9 a C)\right ) \left (\frac {2}{5} a \int \frac {1}{\left (b x^2+a\right )^{2/3}}dx+\frac {3}{5} x \sqrt [3]{a+b x^2}\right )}{11 b}+\frac {3 x \left (a+b x^2\right )^{4/3} (17 b B-9 a C)}{11 b}}{17 b}+\frac {3 C x^3 \left (a+b x^2\right )^{4/3}}{17 b}\)

\(\Big \downarrow \) 234

\(\displaystyle \frac {\frac {\left (187 A b^2-3 a (17 b B-9 a C)\right ) \left (\frac {3 a \sqrt {b x^2} \int \frac {1}{\sqrt {b x^2}}d\sqrt [3]{b x^2+a}}{5 b x}+\frac {3}{5} x \sqrt [3]{a+b x^2}\right )}{11 b}+\frac {3 x \left (a+b x^2\right )^{4/3} (17 b B-9 a C)}{11 b}}{17 b}+\frac {3 C x^3 \left (a+b x^2\right )^{4/3}}{17 b}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {\frac {\left (187 A b^2-3 a (17 b B-9 a C)\right ) \left (\frac {3}{5} x \sqrt [3]{a+b x^2}-\frac {2\ 3^{3/4} \sqrt {2-\sqrt {3}} a \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right ),-7+4 \sqrt {3}\right )}{5 b x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}\right )}{11 b}+\frac {3 x \left (a+b x^2\right )^{4/3} (17 b B-9 a C)}{11 b}}{17 b}+\frac {3 C x^3 \left (a+b x^2\right )^{4/3}}{17 b}\)

Input:

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

Output:

(3*C*x^3*(a + b*x^2)^(4/3))/(17*b) + ((3*(17*b*B - 9*a*C)*x*(a + b*x^2)^(4 
/3))/(11*b) + ((187*A*b^2 - 3*a*(17*b*B - 9*a*C))*((3*x*(a + b*x^2)^(1/3)) 
/5 - (2*3^(3/4)*Sqrt[2 - Sqrt[3]]*a*(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]*EllipticF[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*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))/(17*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 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [A] (verification not implemented)

Time = 1.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.26 \[ \int \sqrt [3]{a+b x^2} \left (A+B x^2+C x^4\right ) \, dx=A \sqrt [3]{a} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {B \sqrt [3]{a} x^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + \frac {C \sqrt [3]{a} x^{5} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt [3]{a+b x^2} \left (A+B x^2+C x^4\right ) \, dx=\frac {-54 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a^{2} c x +663 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a \,b^{2} x +30 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a b c \,x^{3}+255 \left (b \,x^{2}+a \right )^{\frac {1}{3}} b^{3} x^{3}+165 \left (b \,x^{2}+a \right )^{\frac {1}{3}} b^{2} c \,x^{5}+54 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x \right ) a^{3} c +272 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x \right ) a^{2} b^{2}}{935 b^{2}} \] Input:

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

Output:

( - 54*(a + b*x**2)**(1/3)*a**2*c*x + 663*(a + b*x**2)**(1/3)*a*b**2*x + 3 
0*(a + b*x**2)**(1/3)*a*b*c*x**3 + 255*(a + b*x**2)**(1/3)*b**3*x**3 + 165 
*(a + b*x**2)**(1/3)*b**2*c*x**5 + 54*int((a + b*x**2)**(1/3)/(a + b*x**2) 
,x)*a**3*c + 272*int((a + b*x**2)**(1/3)/(a + b*x**2),x)*a**2*b**2)/(935*b 
**2)