\(\int \frac {A+B x+C x^2+D x^3}{(a+b x^2)^{2/3}} \, dx\) [63]

Optimal result

Integrand size = 27, antiderivative size = 323 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {3 (b B-a D) \sqrt [3]{a+b x^2}}{2 b^2}+\frac {3 C x \sqrt [3]{a+b x^2}}{5 b}+\frac {3 D \left (a+b x^2\right )^{4/3}}{8 b^2}-\frac {3^{3/4} \sqrt {2-\sqrt {3}} (5 A b-3 a C) \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),-7+4 \sqrt {3}\right )}{5 b^2 x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}} \] Output:

3/2*(B*b-D*a)*(b*x^2+a)^(1/3)/b^2+3/5*C*x*(b*x^2+a)^(1/3)/b+3/8*D*(b*x^2+a 
)^(4/3)/b^2-1/5*3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2))*(5*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 = 5.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.30 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {-3 \left (a+b x^2\right ) \left (15 a D-b \left (20 B+8 C x+5 D x^2\right )\right )+8 b (5 A b-3 a C) x \left (1+\frac {b x^2}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},-\frac {b x^2}{a}\right )}{40 b^2 \left (a+b x^2\right )^{2/3}} \] Input:

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

Output:

(-3*(a + b*x^2)*(15*a*D - b*(20*B + 8*C*x + 5*D*x^2)) + 8*b*(5*A*b - 3*a*C 
)*x*(1 + (b*x^2)/a)^(2/3)*Hypergeometric2F1[1/2, 2/3, 3/2, -((b*x^2)/a)])/ 
(40*b^2*(a + b*x^2)^(2/3))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2346, 27, 2346, 27, 455, 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.

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

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2346

\(\displaystyle \frac {\frac {3 \int \frac {b (4 (5 A b-3 a C)+5 (4 b B-3 a D) x)}{3 \left (b x^2+a\right )^{2/3}}dx}{5 b}+\frac {12}{5} C x \sqrt [3]{a+b x^2}}{4 b}+\frac {3 D x^2 \sqrt [3]{a+b x^2}}{8 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \int \frac {4 (5 A b-3 a C)+5 (4 b B-3 a D) x}{\left (b x^2+a\right )^{2/3}}dx+\frac {12}{5} C x \sqrt [3]{a+b x^2}}{4 b}+\frac {3 D x^2 \sqrt [3]{a+b x^2}}{8 b}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {\frac {1}{5} \left (4 (5 A b-3 a C) \int \frac {1}{\left (b x^2+a\right )^{2/3}}dx+\frac {15 \sqrt [3]{a+b x^2} (4 b B-3 a D)}{2 b}\right )+\frac {12}{5} C x \sqrt [3]{a+b x^2}}{4 b}+\frac {3 D x^2 \sqrt [3]{a+b x^2}}{8 b}\)

\(\Big \downarrow \) 234

\(\displaystyle \frac {\frac {1}{5} \left (\frac {6 \sqrt {b x^2} (5 A b-3 a C) \int \frac {1}{\sqrt {b x^2}}d\sqrt [3]{b x^2+a}}{b x}+\frac {15 \sqrt [3]{a+b x^2} (4 b B-3 a D)}{2 b}\right )+\frac {12}{5} C x \sqrt [3]{a+b x^2}}{4 b}+\frac {3 D x^2 \sqrt [3]{a+b x^2}}{8 b}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {\frac {1}{5} \left (\frac {15 \sqrt [3]{a+b x^2} (4 b B-3 a D)}{2 b}-\frac {4\ 3^{3/4} \sqrt {2-\sqrt {3}} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} (5 A b-3 a C) \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right ),-7+4 \sqrt {3}\right )}{b x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}\right )+\frac {12}{5} C x \sqrt [3]{a+b x^2}}{4 b}+\frac {3 D x^2 \sqrt [3]{a+b x^2}}{8 b}\)

Input:

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

Output:

(3*D*x^2*(a + b*x^2)^(1/3))/(8*b) + ((12*C*x*(a + b*x^2)^(1/3))/5 + ((15*( 
4*b*B - 3*a*D)*(a + b*x^2)^(1/3))/(2*b) - (4*3^(3/4)*Sqrt[2 - Sqrt[3]]*(5* 
A*b - 3*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]*EllipticF[ArcSin[((1 + Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))/((1 - Sq 
rt[3])*a^(1/3) - (a + b*x^2)^(1/3))], -7 + 4*Sqrt[3]])/(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)]))/5)/(4*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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [A] (verification not implemented)

Time = 1.43 (sec) , antiderivative size = 272, normalized size of antiderivative = 0.84 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {A x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {2}{3} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {2}{3}}} + B \left (\begin {cases} \frac {x^{2}}{2 a^{\frac {2}{3}}} & \text {for}\: b = 0 \\\frac {3 \sqrt [3]{a + b x^{2}}}{2 b} & \text {otherwise} \end {cases}\right ) + \frac {C x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}}} - \frac {9 D a^{\frac {10}{3}} \sqrt [3]{1 + \frac {b x^{2}}{a}}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} + \frac {9 D a^{\frac {10}{3}}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} - \frac {6 D a^{\frac {7}{3}} b x^{2} \sqrt [3]{1 + \frac {b x^{2}}{a}}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} + \frac {9 D a^{\frac {7}{3}} b x^{2}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} + \frac {3 D a^{\frac {4}{3}} b^{2} x^{4} \sqrt [3]{1 + \frac {b x^{2}}{a}}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} \] Input:

integrate((D*x**3+C*x**2+B*x+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*Piece 
wise((x**2/(2*a**(2/3)), Eq(b, 0)), (3*(a + b*x**2)**(1/3)/(2*b), True)) + 
 C*x**3*hyper((2/3, 3/2), (5/2,), b*x**2*exp_polar(I*pi)/a)/(3*a**(2/3)) - 
 9*D*a**(10/3)*(1 + b*x**2/a)**(1/3)/(8*a**2*b**2 + 8*a*b**3*x**2) + 9*D*a 
**(10/3)/(8*a**2*b**2 + 8*a*b**3*x**2) - 6*D*a**(7/3)*b*x**2*(1 + b*x**2/a 
)**(1/3)/(8*a**2*b**2 + 8*a*b**3*x**2) + 9*D*a**(7/3)*b*x**2/(8*a**2*b**2 
+ 8*a*b**3*x**2) + 3*D*a**(4/3)*b**2*x**4*(1 + b*x**2/a)**(1/3)/(8*a**2*b* 
*2 + 8*a*b**3*x**2)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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