\(\int \frac {A+B x^2+C x^4+D x^6}{(a+b x^2)^{3/4}} \, dx\) [144]

Optimal result

Integrand size = 29, antiderivative size = 176 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {2 \left (77 b^2 B-66 a b C+60 a^2 D\right ) x \sqrt [4]{a+b x^2}}{231 b^3}+\frac {2 (11 b C-10 a D) x^3 \sqrt [4]{a+b x^2}}{77 b^2}+\frac {2 D x^5 \sqrt [4]{a+b x^2}}{11 b}+\frac {2 \sqrt {a} \left (231 A b^3-2 a \left (77 b^2 B-66 a b C+60 a^2 D\right )\right ) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{231 b^{7/2} \left (a+b x^2\right )^{3/4}} \] Output:

2/231*(77*B*b^2-66*C*a*b+60*D*a^2)*x*(b*x^2+a)^(1/4)/b^3+2/77*(11*C*b-10*D 
*a)*x^3*(b*x^2+a)^(1/4)/b^2+2/11*D*x^5*(b*x^2+a)^(1/4)/b+2/231*a^(1/2)*(23 
1*A*b^3-2*a*(77*B*b^2-66*C*a*b+60*D*a^2))*(1+b*x^2/a)^(3/4)*InverseJacobiA 
M(1/2*arctan(b^(1/2)*x/a^(1/2)),2^(1/2))/b^(7/2)/(b*x^2+a)^(3/4)
 

Mathematica [C] (verified)

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

Time = 10.12 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.75 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {x \left (2 \left (a+b x^2\right ) \left (60 a^2 D-6 a b \left (11 C+5 D x^2\right )+b^2 \left (77 B+33 C x^2+21 D x^4\right )\right )+\left (231 A b^3-2 a \left (77 b^2 B-66 a b C+60 a^2 D\right )\right ) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{231 b^3 \left (a+b x^2\right )^{3/4}} \] Input:

Integrate[(A + B*x^2 + C*x^4 + D*x^6)/(a + b*x^2)^(3/4),x]
 

Output:

(x*(2*(a + b*x^2)*(60*a^2*D - 6*a*b*(11*C + 5*D*x^2) + b^2*(77*B + 33*C*x^ 
2 + 21*D*x^4)) + (231*A*b^3 - 2*a*(77*b^2*B - 66*a*b*C + 60*a^2*D))*(1 + ( 
b*x^2)/a)^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, -((b*x^2)/a)]))/(231*b^3* 
(a + b*x^2)^(3/4))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2346, 27, 1473, 27, 299, 231, 229}

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.

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

\(\Big \downarrow \) 2346

\(\displaystyle \frac {2 \int \frac {(11 b C-10 a D) x^4+11 b B x^2+11 A b}{2 \left (b x^2+a\right )^{3/4}}dx}{11 b}+\frac {2 D x^5 \sqrt [4]{a+b x^2}}{11 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(11 b C-10 a D) x^4+11 b B x^2+11 A b}{\left (b x^2+a\right )^{3/4}}dx}{11 b}+\frac {2 D x^5 \sqrt [4]{a+b x^2}}{11 b}\)

\(\Big \downarrow \) 1473

\(\displaystyle \frac {\frac {2 \int \frac {77 A b^2+\left (60 D a^2-66 b C a+77 b^2 B\right ) x^2}{2 \left (b x^2+a\right )^{3/4}}dx}{7 b}+\frac {2 x^3 \sqrt [4]{a+b x^2} (11 b C-10 a D)}{7 b}}{11 b}+\frac {2 D x^5 \sqrt [4]{a+b x^2}}{11 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {77 A b^2+\left (60 D a^2-66 b C a+77 b^2 B\right ) x^2}{\left (b x^2+a\right )^{3/4}}dx}{7 b}+\frac {2 x^3 \sqrt [4]{a+b x^2} (11 b C-10 a D)}{7 b}}{11 b}+\frac {2 D x^5 \sqrt [4]{a+b x^2}}{11 b}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\frac {\left (231 A b^3-2 a \left (60 a^2 D-66 a b C+77 b^2 B\right )\right ) \int \frac {1}{\left (b x^2+a\right )^{3/4}}dx}{3 b}+\frac {2 x \sqrt [4]{a+b x^2} \left (60 a^2 D-66 a b C+77 b^2 B\right )}{3 b}}{7 b}+\frac {2 x^3 \sqrt [4]{a+b x^2} (11 b C-10 a D)}{7 b}}{11 b}+\frac {2 D x^5 \sqrt [4]{a+b x^2}}{11 b}\)

\(\Big \downarrow \) 231

\(\displaystyle \frac {\frac {\frac {\left (\frac {b x^2}{a}+1\right )^{3/4} \left (231 A b^3-2 a \left (60 a^2 D-66 a b C+77 b^2 B\right )\right ) \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{3 b \left (a+b x^2\right )^{3/4}}+\frac {2 x \sqrt [4]{a+b x^2} \left (60 a^2 D-66 a b C+77 b^2 B\right )}{3 b}}{7 b}+\frac {2 x^3 \sqrt [4]{a+b x^2} (11 b C-10 a D)}{7 b}}{11 b}+\frac {2 D x^5 \sqrt [4]{a+b x^2}}{11 b}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {\frac {\frac {2 \sqrt {a} \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right ) \left (231 A b^3-2 a \left (60 a^2 D-66 a b C+77 b^2 B\right )\right )}{3 b^{3/2} \left (a+b x^2\right )^{3/4}}+\frac {2 x \sqrt [4]{a+b x^2} \left (60 a^2 D-66 a b C+77 b^2 B\right )}{3 b}}{7 b}+\frac {2 x^3 \sqrt [4]{a+b x^2} (11 b C-10 a D)}{7 b}}{11 b}+\frac {2 D x^5 \sqrt [4]{a+b x^2}}{11 b}\)

Input:

Int[(A + B*x^2 + C*x^4 + D*x^6)/(a + b*x^2)^(3/4),x]
 

Output:

(2*D*x^5*(a + b*x^2)^(1/4))/(11*b) + ((2*(11*b*C - 10*a*D)*x^3*(a + b*x^2) 
^(1/4))/(7*b) + ((2*(77*b^2*B - 66*a*b*C + 60*a^2*D)*x*(a + b*x^2)^(1/4))/ 
(3*b) + (2*Sqrt[a]*(231*A*b^3 - 2*a*(77*b^2*B - 66*a*b*C + 60*a^2*D))*(1 + 
 (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(3*b^(3/2)* 
(a + b*x^2)^(3/4)))/(7*b))/(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 229

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) 
)*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(3/4)/( 
a + b*x^2)^(3/4)   Int[1/(1 + b*(x^2/a))^(3/4), x], x] /; FreeQ[{a, b}, x] 
&& PosQ[a]
 

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 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]
 

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]

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

Input:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(3/4),x)
 

Output:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(3/4),x)
 

Fricas [F]

\[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {D x^{6} + C x^{4} + B x^{2} + A}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(3/4),x, algorithm="fricas")
 

Output:

integral((D*x^6 + C*x^4 + B*x^2 + A)/(b*x^2 + a)^(3/4), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.54 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.66 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {A x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {3}{4}}} + \frac {B x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 a^{\frac {3}{4}}} + \frac {C x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5 a^{\frac {3}{4}}} + \frac {D x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7 a^{\frac {3}{4}}} \] Input:

integrate((D*x**6+C*x**4+B*x**2+A)/(b*x**2+a)**(3/4),x)
 

Output:

A*x*hyper((1/2, 3/4), (3/2,), b*x**2*exp_polar(I*pi)/a)/a**(3/4) + B*x**3* 
hyper((3/4, 3/2), (5/2,), b*x**2*exp_polar(I*pi)/a)/(3*a**(3/4)) + C*x**5* 
hyper((3/4, 5/2), (7/2,), b*x**2*exp_polar(I*pi)/a)/(5*a**(3/4)) + D*x**7* 
hyper((3/4, 7/2), (9/2,), b*x**2*exp_polar(I*pi)/a)/(7*a**(3/4))
 

Maxima [F]

\[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {D x^{6} + C x^{4} + B x^{2} + A}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(3/4),x, algorithm="maxima")
 

Output:

integrate((D*x^6 + C*x^4 + B*x^2 + A)/(b*x^2 + a)^(3/4), x)
 

Giac [F]

\[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {D x^{6} + C x^{4} + B x^{2} + A}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(3/4),x, algorithm="giac")
 

Output:

integrate((D*x^6 + C*x^4 + B*x^2 + A)/(b*x^2 + a)^(3/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{3/4}} \, dx=\int \frac {A+B\,x^2+C\,x^4+x^6\,D}{{\left (b\,x^2+a\right )}^{3/4}} \,d x \] Input:

int((A + B*x^2 + C*x^4 + x^6*D)/(a + b*x^2)^(3/4),x)
 

Output:

int((A + B*x^2 + C*x^4 + x^6*D)/(a + b*x^2)^(3/4), x)
 

Reduce [F]

\[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{3/4}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) d +\left (\int \frac {x^{4}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) c +\left (\int \frac {x^{2}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) b +\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a \] Input:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(3/4),x)
 

Output:

int(x**6/(a + b*x**2)**(3/4),x)*d + int(x**4/(a + b*x**2)**(3/4),x)*c + in 
t(x**2/(a + b*x**2)**(3/4),x)*b + int(1/(a + b*x**2)**(3/4),x)*a