\(\int \sqrt [3]{a+b x^2} (A+B x+C x^2) \, dx\) [38]

Optimal result

Integrand size = 22, antiderivative size = 326 \[ \int \sqrt [3]{a+b x^2} \left (A+B x+C x^2\right ) \, dx=\frac {3 (11 A b-3 a C) x \sqrt [3]{a+b x^2}}{55 b}+\frac {3 B \left (a+b x^2\right )^{4/3}}{8 b}+\frac {3 C x \left (a+b x^2\right )^{4/3}}{11 b}-\frac {2\ 3^{3/4} \sqrt {2-\sqrt {3}} a (11 A b-3 a C) \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),-7+4 \sqrt {3}\right )}{55 b^2 x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}} \] Output:

3/55*(11*A*b-3*C*a)*x*(b*x^2+a)^(1/3)/b+3/8*B*(b*x^2+a)^(4/3)/b+3/11*C*x*( 
b*x^2+a)^(4/3)/b-2/55*3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2))*a*(11*A*b-3*C*a)*( 
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^2/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 = 6.05 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.25 \[ \int \sqrt [3]{a+b x^2} \left (A+B x+C x^2\right ) \, dx=\frac {\sqrt [3]{a+b x^2} \left (3 (11 B+8 C x) \left (a+b x^2\right )+\frac {8 (11 A b-3 a C) x \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )}{\sqrt [3]{1+\frac {b x^2}{a}}}\right )}{88 b} \] Input:

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

Output:

((a + b*x^2)^(1/3)*(3*(11*B + 8*C*x)*(a + b*x^2) + (8*(11*A*b - 3*a*C)*x*H 
ypergeometric2F1[-1/3, 1/2, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^(1/3)))/(8 
8*b)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 321, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2346, 27, 455, 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.

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

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 455

\(\displaystyle \frac {(11 A b-3 a C) \int \sqrt [3]{b x^2+a}dx+\frac {33}{8} B \left (a+b x^2\right )^{4/3}}{11 b}+\frac {3 C x \left (a+b x^2\right )^{4/3}}{11 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {(11 A b-3 a C) \left (\frac {2}{5} a \int \frac {1}{\left (b x^2+a\right )^{2/3}}dx+\frac {3}{5} x \sqrt [3]{a+b x^2}\right )+\frac {33}{8} B \left (a+b x^2\right )^{4/3}}{11 b}+\frac {3 C x \left (a+b x^2\right )^{4/3}}{11 b}\)

\(\Big \downarrow \) 234

\(\displaystyle \frac {(11 A b-3 a C) \left (\frac {3 a \sqrt {b x^2} \int \frac {1}{\sqrt {b x^2}}d\sqrt [3]{b x^2+a}}{5 b x}+\frac {3}{5} x \sqrt [3]{a+b x^2}\right )+\frac {33}{8} B \left (a+b x^2\right )^{4/3}}{11 b}+\frac {3 C x \left (a+b x^2\right )^{4/3}}{11 b}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {(11 A b-3 a C) \left (\frac {3}{5} x \sqrt [3]{a+b x^2}-\frac {2\ 3^{3/4} \sqrt {2-\sqrt {3}} a \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right ),-7+4 \sqrt {3}\right )}{5 b x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}\right )+\frac {33}{8} B \left (a+b x^2\right )^{4/3}}{11 b}+\frac {3 C x \left (a+b x^2\right )^{4/3}}{11 b}\)

Input:

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

Output:

(3*C*x*(a + b*x^2)^(4/3))/(11*b) + ((33*B*(a + b*x^2)^(4/3))/8 + (11*A*b - 
 3*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[ArcS 
in[((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)
 

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 455

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

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 2346

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], 
e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( 
q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1))   Int[(a + b*x^2)^p*ExpandToS 
um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], 
x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !LeQ[p, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [A] (verification not implemented)

Time = 1.09 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.27 \[ \int \sqrt [3]{a+b x^2} \left (A+B x+C x^2\right ) \, dx=A \sqrt [3]{a} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + B \left (\begin {cases} \frac {\sqrt [3]{a} x^{2}}{2} & \text {for}\: b = 0 \\\frac {3 \left (a + b x^{2}\right )^{\frac {4}{3}}}{8 b} & \text {otherwise} \end {cases}\right ) + \frac {C \sqrt [3]{a} x^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} \] Input:

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

Output:

A*a**(1/3)*x*hyper((-1/3, 1/2), (3/2,), b*x**2*exp_polar(I*pi)/a) + B*Piec 
ewise((a**(1/3)*x**2/2, Eq(b, 0)), (3*(a + b*x**2)**(4/3)/(8*b), True)) + 
C*a**(1/3)*x**3*hyper((-1/3, 3/2), (5/2,), b*x**2*exp_polar(I*pi)/a)/3
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt [3]{a+b x^2} \left (A+B x+C x^2\right ) \, dx=\frac {264 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a b x +165 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a b +48 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a c x +165 \left (b \,x^{2}+a \right )^{\frac {1}{3}} b^{2} x^{2}+120 \left (b \,x^{2}+a \right )^{\frac {1}{3}} b c \,x^{3}+176 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x \right ) a^{2} b -48 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x \right ) a^{2} c}{440 b} \] Input:

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

Output:

(264*(a + b*x**2)**(1/3)*a*b*x + 165*(a + b*x**2)**(1/3)*a*b + 48*(a + b*x 
**2)**(1/3)*a*c*x + 165*(a + b*x**2)**(1/3)*b**2*x**2 + 120*(a + b*x**2)** 
(1/3)*b*c*x**3 + 176*int((a + b*x**2)**(1/3)/(a + b*x**2),x)*a**2*b - 48*i 
nt((a + b*x**2)**(1/3)/(a + b*x**2),x)*a**2*c)/(440*b)