\(\int \frac {A+B x^2+C x^4+D x^6}{(a+b x^2)^{11/4}} \, dx\) [148]

Optimal result

Integrand size = 29, antiderivative size = 198 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{11/4}} \, dx=\frac {2 \left (\frac {A}{a}-\frac {b^2 B-a b C+a^2 D}{b^3}\right ) x}{7 \left (a+b x^2\right )^{7/4}}+\frac {2 \left (5 A b^3+a \left (2 b^2 B-9 a b C+16 a^2 D\right )\right ) x}{21 a^2 b^3 \left (a+b x^2\right )^{3/4}}+\frac {2 D x \sqrt [4]{a+b x^2}}{3 b^3}+\frac {2 \left (5 A b^3+2 a \left (b^2 B+6 a b C-20 a^2 D\right )\right ) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{21 a^{3/2} b^{7/2} \left (a+b x^2\right )^{3/4}} \] Output:

2/7*(A/a-(B*b^2-C*a*b+D*a^2)/b^3)*x/(b*x^2+a)^(7/4)+2/21*(5*A*b^3+a*(2*B*b 
^2-9*C*a*b+16*D*a^2))*x/a^2/b^3/(b*x^2+a)^(3/4)+2/3*D*x*(b*x^2+a)^(1/4)/b^ 
3+2/21*(5*A*b^3+2*a*(B*b^2+6*C*a*b-20*D*a^2))*(1+b*x^2/a)^(3/4)*InverseJac 
obiAM(1/2*arctan(b^(1/2)*x/a^(1/2)),2^(1/2))/a^(3/2)/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 = 10.13 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.81 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{11/4}} \, dx=\frac {2 x \left (20 a^4 D+5 A b^4 x^2+2 a b^3 \left (4 A+B x^2\right )-6 a^3 b \left (C-5 D x^2\right )-a^2 b^2 \left (B+9 C x^2-7 D x^4\right )\right )+\left (5 A b^3+2 a \left (b^2 B+6 a b C-20 a^2 D\right )\right ) x \left (a+b x^2\right ) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\frac {b x^2}{a}\right )}{21 a^2 b^3 \left (a+b x^2\right )^{7/4}} \] Input:

Integrate[(A + B*x^2 + C*x^4 + D*x^6)/(a + b*x^2)^(11/4),x]
 

Output:

(2*x*(20*a^4*D + 5*A*b^4*x^2 + 2*a*b^3*(4*A + B*x^2) - 6*a^3*b*(C - 5*D*x^ 
2) - a^2*b^2*(B + 9*C*x^2 - 7*D*x^4)) + (5*A*b^3 + 2*a*(b^2*B + 6*a*b*C - 
20*a^2*D))*x*(a + b*x^2)*(1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[1/2, 3/4, 
 3/2, -((b*x^2)/a)])/(21*a^2*b^3*(a + b*x^2)^(7/4))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2345, 27, 1471, 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.

\(\displaystyle \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{11/4}} \, dx\)

\(\Big \downarrow \) 2345

\(\displaystyle \frac {2 x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{7 a \left (a+b x^2\right )^{7/4}}-\frac {2 \int -\frac {\frac {7 a D x^4}{b}+\frac {7 a (b C-a D) x^2}{b^2}+5 A+\frac {2 a \left (D a^2-b C a+b^2 B\right )}{b^3}}{2 \left (b x^2+a\right )^{7/4}}dx}{7 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\frac {7 a D x^4}{b}+\frac {7 a (b C-a D) x^2}{b^2}+5 A+\frac {2 a \left (D a^2-b C a+b^2 B\right )}{b^3}}{\left (b x^2+a\right )^{7/4}}dx}{7 a}+\frac {2 x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{7 a \left (a+b x^2\right )^{7/4}}\)

\(\Big \downarrow \) 1471

\(\displaystyle \frac {\frac {2 x \left (\frac {16 a^2 D-9 a b C+2 b^2 B}{b^3}+\frac {5 A}{a}\right )}{3 \left (a+b x^2\right )^{3/4}}-\frac {2 \int -\frac {\left (5 A+\frac {2 a \left (-13 D a^2+6 b C a+b^2 B\right )}{b^3}\right ) b^2+21 a^2 D x^2}{2 b^2 \left (b x^2+a\right )^{3/4}}dx}{3 a}}{7 a}+\frac {2 x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{7 a \left (a+b x^2\right )^{7/4}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {5 A b^2+21 a^2 D x^2+2 a \left (-\frac {13 D a^2}{b}+6 C a+b B\right )}{\left (b x^2+a\right )^{3/4}}dx}{3 a b^2}+\frac {2 x \left (\frac {16 a^2 D-9 a b C+2 b^2 B}{b^3}+\frac {5 A}{a}\right )}{3 \left (a+b x^2\right )^{3/4}}}{7 a}+\frac {2 x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{7 a \left (a+b x^2\right )^{7/4}}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\frac {\left (2 a \left (-20 a^2 D+6 a b C+b^2 B\right )+5 A b^3\right ) \int \frac {1}{\left (b x^2+a\right )^{3/4}}dx}{b}+\frac {14 a^2 D x \sqrt [4]{a+b x^2}}{b}}{3 a b^2}+\frac {2 x \left (\frac {16 a^2 D-9 a b C+2 b^2 B}{b^3}+\frac {5 A}{a}\right )}{3 \left (a+b x^2\right )^{3/4}}}{7 a}+\frac {2 x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{7 a \left (a+b x^2\right )^{7/4}}\)

\(\Big \downarrow \) 231

\(\displaystyle \frac {\frac {\frac {\left (\frac {b x^2}{a}+1\right )^{3/4} \left (2 a \left (-20 a^2 D+6 a b C+b^2 B\right )+5 A b^3\right ) \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{b \left (a+b x^2\right )^{3/4}}+\frac {14 a^2 D x \sqrt [4]{a+b x^2}}{b}}{3 a b^2}+\frac {2 x \left (\frac {16 a^2 D-9 a b C+2 b^2 B}{b^3}+\frac {5 A}{a}\right )}{3 \left (a+b x^2\right )^{3/4}}}{7 a}+\frac {2 x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{7 a \left (a+b x^2\right )^{7/4}}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {\frac {\frac {2 \sqrt {a} \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right ) \left (2 a \left (-20 a^2 D+6 a b C+b^2 B\right )+5 A b^3\right )}{b^{3/2} \left (a+b x^2\right )^{3/4}}+\frac {14 a^2 D x \sqrt [4]{a+b x^2}}{b}}{3 a b^2}+\frac {2 x \left (\frac {16 a^2 D-9 a b C+2 b^2 B}{b^3}+\frac {5 A}{a}\right )}{3 \left (a+b x^2\right )^{3/4}}}{7 a}+\frac {2 x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{7 a \left (a+b x^2\right )^{7/4}}\)

Input:

Int[(A + B*x^2 + C*x^4 + D*x^6)/(a + b*x^2)^(11/4),x]
 

Output:

(2*(A - (a*(b^2*B - a*b*C + a^2*D))/b^3)*x)/(7*a*(a + b*x^2)^(7/4)) + ((2* 
((5*A)/a + (2*b^2*B - 9*a*b*C + 16*a^2*D)/b^3)*x)/(3*(a + b*x^2)^(3/4)) + 
((14*a^2*D*x*(a + b*x^2)^(1/4))/b + (2*Sqrt[a]*(5*A*b^3 + 2*a*(b^2*B + 6*a 
*b*C - 20*a^2*D))*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/Sqrt[ 
a]]/2, 2])/(b^(3/2)*(a + b*x^2)^(3/4)))/(3*a*b^2))/(7*a)
 

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 1471

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 
, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x 
, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q 
 + 1))   Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), 
x], x], x]] /; FreeQ[{a, b, c, d, e}, 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 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]

\[\int \frac {D x^{6}+C \,x^{4}+x^{2} B +A}{\left (b \,x^{2}+a \right )^{\frac {11}{4}}}d x\]

Input:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(11/4),x)
 

Output:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(11/4),x)
 

Fricas [F]

\[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{11/4}} \, dx=\int { \frac {D x^{6} + C x^{4} + B x^{2} + A}{{\left (b x^{2} + a\right )}^{\frac {11}{4}}} \,d x } \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(11/4),x, algorithm="fricas")
 

Output:

integral((D*x^6 + C*x^4 + B*x^2 + A)*(b*x^2 + a)^(1/4)/(b^3*x^6 + 3*a*b^2* 
x^4 + 3*a^2*b*x^2 + a^3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 12.95 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.59 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{11/4}} \, dx=\frac {A x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {11}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {11}{4}}} + \frac {B x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {11}{4} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 a^{\frac {11}{4}}} + \frac {C x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, \frac {11}{4} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5 a^{\frac {11}{4}}} + \frac {D x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {11}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7 a^{\frac {11}{4}}} \] Input:

integrate((D*x**6+C*x**4+B*x**2+A)/(b*x**2+a)**(11/4),x)
 

Output:

A*x*hyper((1/2, 11/4), (3/2,), b*x**2*exp_polar(I*pi)/a)/a**(11/4) + B*x** 
3*hyper((3/2, 11/4), (5/2,), b*x**2*exp_polar(I*pi)/a)/(3*a**(11/4)) + C*x 
**5*hyper((5/2, 11/4), (7/2,), b*x**2*exp_polar(I*pi)/a)/(5*a**(11/4)) + D 
*x**7*hyper((11/4, 7/2), (9/2,), b*x**2*exp_polar(I*pi)/a)/(7*a**(11/4))
 

Maxima [F]

\[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{11/4}} \, dx=\int { \frac {D x^{6} + C x^{4} + B x^{2} + A}{{\left (b x^{2} + a\right )}^{\frac {11}{4}}} \,d x } \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(11/4),x, algorithm="maxima")
 

Output:

integrate((D*x^6 + C*x^4 + B*x^2 + A)/(b*x^2 + a)^(11/4), x)
 

Giac [F]

\[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{11/4}} \, dx=\int { \frac {D x^{6} + C x^{4} + B x^{2} + A}{{\left (b x^{2} + a\right )}^{\frac {11}{4}}} \,d x } \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(11/4),x, algorithm="giac")
 

Output:

integrate((D*x^6 + C*x^4 + B*x^2 + A)/(b*x^2 + a)^(11/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{11/4}} \, dx=\int \frac {A+B\,x^2+C\,x^4+x^6\,D}{{\left (b\,x^2+a\right )}^{11/4}} \,d x \] Input:

int((A + B*x^2 + C*x^4 + x^6*D)/(a + b*x^2)^(11/4),x)
 

Output:

int((A + B*x^2 + C*x^4 + x^6*D)/(a + b*x^2)^(11/4), x)
 

Reduce [F]

\[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{11/4}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} a^{2}+2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a b \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {3}{4}} b^{2} x^{4}}d x \right ) d +\left (\int \frac {x^{4}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} a^{2}+2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a b \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {3}{4}} b^{2} x^{4}}d x \right ) c +\left (\int \frac {x^{2}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} a^{2}+2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a b \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {3}{4}} b^{2} x^{4}}d x \right ) b +\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} a^{2}+2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a b \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {3}{4}} b^{2} x^{4}}d x \right ) a \] Input:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(11/4),x)
 

Output:

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