\(\int \frac {A+B x+C x^2+D x^3}{(a+b x^2)^{5/3}} \, dx\) [64]

Optimal result

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

1/4*(-3*a*(B*b-D*a)+3*b*(A*b-C*a)*x)/a/b^2/(b*x^2+a)^(2/3)+3/2*D*(b*x^2+a) 
^(1/3)/b^2-1/4*3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2))*(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))/a/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 = 10.06 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.29 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{5/3}} \, dx=\frac {9 a^2 D+3 A b^2 x-3 a b (B+x (C-2 D x))+b (A b+3 a C) x \left (1+\frac {b x^2}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},-\frac {b x^2}{a}\right )}{4 a b^2 \left (a+b x^2\right )^{2/3}} \] Input:

Integrate[(A + B*x + C*x^2 + D*x^3)/(a + b*x^2)^(5/3),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2345, 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.

\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{5/3}} \, dx\)

\(\Big \downarrow \) 2345

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 234

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

\(\Big \downarrow \) 760

\(\displaystyle \frac {\frac {6 a D \sqrt [3]{a+b x^2}}{b}-\frac {3^{3/4} \sqrt {2-\sqrt {3}} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} (3 a C+A b) \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right ),-7+4 \sqrt {3}\right )}{b x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}}{4 a b}-\frac {3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^{2/3}}\)

Input:

Int[(A + B*x + C*x^2 + D*x^3)/(a + b*x^2)^(5/3),x]
 

Output:

(-3*(a*(B - (a*D)/b) - (A*b - a*C)*x))/(4*a*b*(a + b*x^2)^(2/3)) + ((6*a*D 
*(a + b*x^2)^(1/3))/b - (3^(3/4)*Sqrt[2 - Sqrt[3]]*(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 - Sqrt[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)]))/(4*a*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 2345

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot 
ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b 
*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   In 
t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] 
/; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
 

Maple [F]

\[\int \frac {D x^{3}+C \,x^{2}+B x +A}{\left (b \,x^{2}+a \right )^{\frac {5}{3}}}d x\]

Input:

int((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/3),x)
 

Output:

int((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/3),x)
 

Fricas [F]

\[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{5/3}} \, dx=\int { \frac {D x^{3} + C x^{2} + B x + A}{{\left (b x^{2} + a\right )}^{\frac {5}{3}}} \,d x } \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/3),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 4.03 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.41 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{5/3}} \, dx=\frac {A x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{3} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {5}{3}}} + B \left (\begin {cases} - \frac {3}{4 b \left (a + b x^{2}\right )^{\frac {2}{3}}} & \text {for}\: b \neq 0 \\\frac {x^{2}}{2 a^{\frac {5}{3}}} & \text {otherwise} \end {cases}\right ) + \frac {C x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {5}{3} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 a^{\frac {5}{3}}} + D \left (\begin {cases} \frac {9 a}{4 b^{2} \left (a + b x^{2}\right )^{\frac {2}{3}}} + \frac {3 x^{2}}{2 b \left (a + b x^{2}\right )^{\frac {2}{3}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {5}{3}}} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((D*x**3+C*x**2+B*x+A)/(b*x**2+a)**(5/3),x)
 

Output:

A*x*hyper((1/2, 5/3), (3/2,), b*x**2*exp_polar(I*pi)/a)/a**(5/3) + B*Piece 
wise((-3/(4*b*(a + b*x**2)**(2/3)), Ne(b, 0)), (x**2/(2*a**(5/3)), True)) 
+ C*x**3*hyper((3/2, 5/3), (5/2,), b*x**2*exp_polar(I*pi)/a)/(3*a**(5/3)) 
+ D*Piecewise((9*a/(4*b**2*(a + b*x**2)**(2/3)) + 3*x**2/(2*b*(a + b*x**2) 
**(2/3)), Ne(b, 0)), (x**4/(4*a**(5/3)), True))
 

Maxima [F]

\[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{5/3}} \, dx=\int { \frac {D x^{3} + C x^{2} + B x + A}{{\left (b x^{2} + a\right )}^{\frac {5}{3}}} \,d x } \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/3),x, algorithm="maxima")
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)/(b*x^2 + a)^(5/3), x)
 

Giac [F]

\[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{5/3}} \, dx=\int { \frac {D x^{3} + C x^{2} + B x + A}{{\left (b x^{2} + a\right )}^{\frac {5}{3}}} \,d x } \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/3),x, algorithm="giac")
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)/(b*x^2 + a)^(5/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{5/3}} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{{\left (b\,x^2+a\right )}^{5/3}} \,d x \] Input:

int((A + B*x + C*x^2 + x^3*D)/(a + b*x^2)^(5/3),x)
 

Output:

int((A + B*x + C*x^2 + x^3*D)/(a + b*x^2)^(5/3), x)
 

Reduce [F]

\[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{5/3}} \, dx=\left (\int \frac {x^{3}}{\left (b \,x^{2}+a \right )^{\frac {2}{3}} a +\left (b \,x^{2}+a \right )^{\frac {2}{3}} b \,x^{2}}d x \right ) d +\left (\int \frac {x^{2}}{\left (b \,x^{2}+a \right )^{\frac {2}{3}} a +\left (b \,x^{2}+a \right )^{\frac {2}{3}} b \,x^{2}}d x \right ) c +\left (\int \frac {x}{\left (b \,x^{2}+a \right )^{\frac {2}{3}} a +\left (b \,x^{2}+a \right )^{\frac {2}{3}} b \,x^{2}}d x \right ) b +\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {2}{3}} a +\left (b \,x^{2}+a \right )^{\frac {2}{3}} b \,x^{2}}d x \right ) a \] Input:

int((D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/3),x)
 

Output:

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