\(\int \frac {A+B x^2+C x^4+D x^6}{(a+b x^2)^{15/4}} \, dx\) [150]

Optimal result

Integrand size = 29, antiderivative size = 227 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{15/4}} \, dx=\frac {2 \left (\frac {A}{a}-\frac {b^2 B-a b C+a^2 D}{b^3}\right ) x}{11 \left (a+b x^2\right )^{11/4}}+\frac {2 \left (9 A b^3+a \left (2 b^2 B-13 a b C+24 a^2 D\right )\right ) x}{77 a^2 b^3 \left (a+b x^2\right )^{7/4}}+\frac {2 \left (45 A b^3+a \left (10 b^2 B+12 a b C-111 a^2 D\right )\right ) x}{231 a^3 b^3 \left (a+b x^2\right )^{3/4}}+\frac {2 \left (45 A b^3+10 a b^2 B+12 a^2 b C+120 a^3 D\right ) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{231 a^{5/2} b^{7/2} \left (a+b x^2\right )^{3/4}} \] Output:

2/11*(A/a-(B*b^2-C*a*b+D*a^2)/b^3)*x/(b*x^2+a)^(11/4)+2/77*(9*A*b^3+a*(2*B 
*b^2-13*C*a*b+24*D*a^2))*x/a^2/b^3/(b*x^2+a)^(7/4)+2/231*(45*A*b^3+a*(10*B 
*b^2+12*C*a*b-111*D*a^2))*x/a^3/b^3/(b*x^2+a)^(3/4)+2/231*(45*A*b^3+10*B*a 
*b^2+12*C*a^2*b+120*D*a^3)*(1+b*x^2/a)^(3/4)*InverseJacobiAM(1/2*arctan(b^ 
(1/2)*x/a^(1/2)),2^(1/2))/a^(5/2)/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 = 10.19 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.85 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{15/4}} \, dx=\frac {-2 x \left (60 a^5 D-45 A b^5 x^4-a b^4 x^2 \left (117 A+10 B x^2\right )+6 a^4 b \left (C+25 D x^2\right )-a^2 b^3 \left (93 A+26 B x^2+12 C x^4\right )+a^3 b^2 \left (5 B+15 C x^2+111 D x^4\right )\right )+\left (45 A b^3+2 a \left (5 b^2 B+6 a b C+60 a^2 D\right )\right ) x \left (a+b x^2\right )^2 \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\frac {b x^2}{a}\right )}{231 a^3 b^3 \left (a+b x^2\right )^{11/4}} \] Input:

Integrate[(A + B*x^2 + C*x^4 + D*x^6)/(a + b*x^2)^(15/4),x]
 

Output:

(-2*x*(60*a^5*D - 45*A*b^5*x^4 - a*b^4*x^2*(117*A + 10*B*x^2) + 6*a^4*b*(C 
 + 25*D*x^2) - a^2*b^3*(93*A + 26*B*x^2 + 12*C*x^4) + a^3*b^2*(5*B + 15*C* 
x^2 + 111*D*x^4)) + (45*A*b^3 + 2*a*(5*b^2*B + 6*a*b*C + 60*a^2*D))*x*(a + 
 b*x^2)^2*(1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, -((b*x^2) 
/a)])/(231*a^3*b^3*(a + b*x^2)^(11/4))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 242, 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, 298, 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.

\(\displaystyle \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{15/4}} \, dx\)

\(\Big \downarrow \) 2345

\(\displaystyle \frac {2 x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{11 a \left (a+b x^2\right )^{11/4}}-\frac {2 \int -\frac {\frac {11 a D x^4}{b}+\frac {11 a (b C-a D) x^2}{b^2}+9 A+\frac {2 a \left (D a^2-b C a+b^2 B\right )}{b^3}}{2 \left (b x^2+a\right )^{11/4}}dx}{11 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\frac {11 a D x^4}{b}+\frac {11 a (b C-a D) x^2}{b^2}+9 A+\frac {2 a \left (D a^2-b C a+b^2 B\right )}{b^3}}{\left (b x^2+a\right )^{11/4}}dx}{11 a}+\frac {2 x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{11 a \left (a+b x^2\right )^{11/4}}\)

\(\Big \downarrow \) 1471

\(\displaystyle \frac {\frac {2 x \left (\frac {24 a^2 D-13 a b C+2 b^2 B}{b^3}+\frac {9 A}{a}\right )}{7 \left (a+b x^2\right )^{7/4}}-\frac {2 \int -\frac {\left (45 A+\frac {2 a \left (-17 D a^2+6 b C a+5 b^2 B\right )}{b^3}\right ) b^2+77 a^2 D x^2}{2 b^2 \left (b x^2+a\right )^{7/4}}dx}{7 a}}{11 a}+\frac {2 x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{11 a \left (a+b x^2\right )^{11/4}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {45 A b^2+77 a^2 D x^2+2 a \left (-\frac {17 D a^2}{b}+6 C a+5 b B\right )}{\left (b x^2+a\right )^{7/4}}dx}{7 a b^2}+\frac {2 x \left (\frac {24 a^2 D-13 a b C+2 b^2 B}{b^3}+\frac {9 A}{a}\right )}{7 \left (a+b x^2\right )^{7/4}}}{11 a}+\frac {2 x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{11 a \left (a+b x^2\right )^{11/4}}\)

\(\Big \downarrow \) 298

\(\displaystyle \frac {\frac {\frac {\left (120 a^3 D+12 a^2 b C+10 a b^2 B+45 A b^3\right ) \int \frac {1}{\left (b x^2+a\right )^{3/4}}dx}{3 a b}+\frac {2 x \left (a \left (-111 a^2 D+12 a b C+10 b^2 B\right )+45 A b^3\right )}{3 a b \left (a+b x^2\right )^{3/4}}}{7 a b^2}+\frac {2 x \left (\frac {24 a^2 D-13 a b C+2 b^2 B}{b^3}+\frac {9 A}{a}\right )}{7 \left (a+b x^2\right )^{7/4}}}{11 a}+\frac {2 x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{11 a \left (a+b x^2\right )^{11/4}}\)

\(\Big \downarrow \) 231

\(\displaystyle \frac {\frac {\frac {\left (\frac {b x^2}{a}+1\right )^{3/4} \left (120 a^3 D+12 a^2 b C+10 a b^2 B+45 A b^3\right ) \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{3 a b \left (a+b x^2\right )^{3/4}}+\frac {2 x \left (a \left (-111 a^2 D+12 a b C+10 b^2 B\right )+45 A b^3\right )}{3 a b \left (a+b x^2\right )^{3/4}}}{7 a b^2}+\frac {2 x \left (\frac {24 a^2 D-13 a b C+2 b^2 B}{b^3}+\frac {9 A}{a}\right )}{7 \left (a+b x^2\right )^{7/4}}}{11 a}+\frac {2 x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{11 a \left (a+b x^2\right )^{11/4}}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {2 x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{11 a \left (a+b x^2\right )^{11/4}}+\frac {\frac {2 x \left (\frac {24 a^2 D-13 a b C+2 b^2 B}{b^3}+\frac {9 A}{a}\right )}{7 \left (a+b x^2\right )^{7/4}}+\frac {\frac {2 x \left (a \left (-111 a^2 D+12 a b C+10 b^2 B\right )+45 A b^3\right )}{3 a b \left (a+b x^2\right )^{3/4}}+\frac {2 \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right ) \left (120 a^3 D+12 a^2 b C+10 a b^2 B+45 A b^3\right )}{3 \sqrt {a} b^{3/2} \left (a+b x^2\right )^{3/4}}}{7 a b^2}}{11 a}\)

Input:

Int[(A + B*x^2 + C*x^4 + D*x^6)/(a + b*x^2)^(15/4),x]
 

Output:

(2*(A - (a*(b^2*B - a*b*C + a^2*D))/b^3)*x)/(11*a*(a + b*x^2)^(11/4)) + (( 
2*((9*A)/a + (2*b^2*B - 13*a*b*C + 24*a^2*D)/b^3)*x)/(7*(a + b*x^2)^(7/4)) 
 + ((2*(45*A*b^3 + a*(10*b^2*B + 12*a*b*C - 111*a^2*D))*x)/(3*a*b*(a + b*x 
^2)^(3/4)) + (2*(45*A*b^3 + 10*a*b^2*B + 12*a^2*b*C + 120*a^3*D)*(1 + (b*x 
^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(3*Sqrt[a]*b^(3/ 
2)*(a + b*x^2)^(3/4)))/(7*a*b^2))/(11*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 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 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]

\[\int \frac {D x^{6}+C \,x^{4}+x^{2} B +A}{\left (b \,x^{2}+a \right )^{\frac {15}{4}}}d x\]

Input:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(15/4),x)
 

Output:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(15/4),x)
 

Fricas [F]

\[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{15/4}} \, dx=\int { \frac {D x^{6} + C x^{4} + B x^{2} + A}{{\left (b x^{2} + a\right )}^{\frac {15}{4}}} \,d x } \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(15/4),x, algorithm="fricas")
 

Output:

integral((D*x^6 + C*x^4 + B*x^2 + A)*(b*x^2 + a)^(1/4)/(b^4*x^8 + 4*a*b^3* 
x^6 + 6*a^2*b^2*x^4 + 4*a^3*b*x^2 + a^4), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 37.91 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.52 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{15/4}} \, dx=\frac {A x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {15}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {15}{4}}} + \frac {B x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {15}{4} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 a^{\frac {15}{4}}} + \frac {C x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, \frac {15}{4} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5 a^{\frac {15}{4}}} + \frac {D x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{2}, \frac {15}{4} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7 a^{\frac {15}{4}}} \] Input:

integrate((D*x**6+C*x**4+B*x**2+A)/(b*x**2+a)**(15/4),x)
 

Output:

A*x*hyper((1/2, 15/4), (3/2,), b*x**2*exp_polar(I*pi)/a)/a**(15/4) + B*x** 
3*hyper((3/2, 15/4), (5/2,), b*x**2*exp_polar(I*pi)/a)/(3*a**(15/4)) + C*x 
**5*hyper((5/2, 15/4), (7/2,), b*x**2*exp_polar(I*pi)/a)/(5*a**(15/4)) + D 
*x**7*hyper((7/2, 15/4), (9/2,), b*x**2*exp_polar(I*pi)/a)/(7*a**(15/4))
 

Maxima [F]

\[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{15/4}} \, dx=\int { \frac {D x^{6} + C x^{4} + B x^{2} + A}{{\left (b x^{2} + a\right )}^{\frac {15}{4}}} \,d x } \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(15/4),x, algorithm="maxima")
 

Output:

integrate((D*x^6 + C*x^4 + B*x^2 + A)/(b*x^2 + a)^(15/4), x)
 

Giac [F]

\[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{15/4}} \, dx=\int { \frac {D x^{6} + C x^{4} + B x^{2} + A}{{\left (b x^{2} + a\right )}^{\frac {15}{4}}} \,d x } \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(15/4),x, algorithm="giac")
 

Output:

integrate((D*x^6 + C*x^4 + B*x^2 + A)/(b*x^2 + a)^(15/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{15/4}} \, dx=\int \frac {A+B\,x^2+C\,x^4+x^6\,D}{{\left (b\,x^2+a\right )}^{15/4}} \,d x \] Input:

int((A + B*x^2 + C*x^4 + x^6*D)/(a + b*x^2)^(15/4),x)
 

Output:

int((A + B*x^2 + C*x^4 + x^6*D)/(a + b*x^2)^(15/4), x)
 

Reduce [F]

\[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{15/4}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} a^{3}+3 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a^{2} b \,x^{2}+3 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a \,b^{2} x^{4}+\left (b \,x^{2}+a \right )^{\frac {3}{4}} b^{3} x^{6}}d x \right ) d +\left (\int \frac {x^{4}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} a^{3}+3 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a^{2} b \,x^{2}+3 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a \,b^{2} x^{4}+\left (b \,x^{2}+a \right )^{\frac {3}{4}} b^{3} x^{6}}d x \right ) c +\left (\int \frac {x^{2}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} a^{3}+3 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a^{2} b \,x^{2}+3 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a \,b^{2} x^{4}+\left (b \,x^{2}+a \right )^{\frac {3}{4}} b^{3} x^{6}}d x \right ) b +\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} a^{3}+3 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a^{2} b \,x^{2}+3 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a \,b^{2} x^{4}+\left (b \,x^{2}+a \right )^{\frac {3}{4}} b^{3} x^{6}}d x \right ) a \] Input:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(15/4),x)
 

Output:

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