\(\int \frac {A+B x^2+C x^4+D x^6}{(a+b x^2)^{5/4}} \, dx\) [145]

Optimal result

Integrand size = 29, antiderivative size = 173 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {2 \left (15 b^2 B-21 a b C+25 a^2 D\right ) x}{15 b^3 \sqrt [4]{a+b x^2}}+\frac {2 (3 b C-5 a D) x \left (a+b x^2\right )^{3/4}}{15 b^3}+\frac {2 D x^3 \left (a+b x^2\right )^{3/4}}{9 b^2}+\frac {2 \left (15 A b^3-30 a b^2 B+36 a^2 b C-40 a^3 D\right ) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 \sqrt {a} b^{7/2} \sqrt [4]{a+b x^2}} \] Output:

2/15*(15*B*b^2-21*C*a*b+25*D*a^2)*x/b^3/(b*x^2+a)^(1/4)+2/15*(3*C*b-5*D*a) 
*x*(b*x^2+a)^(3/4)/b^3+2/9*D*x^3*(b*x^2+a)^(3/4)/b^2+2/15*(15*A*b^3-30*B*a 
*b^2+36*C*a^2*b-40*D*a^3)*(1+b*x^2/a)^(1/4)*EllipticE(sin(1/2*arctan(b^(1/ 
2)*x/a^(1/2))),2^(1/2))/a^(1/2)/b^(7/2)/(b*x^2+a)^(1/4)
 

Mathematica [C] (verified)

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

Time = 10.11 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.78 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {x \left (90 A b^3+2 a \left (-60 a^2 D+2 a b \left (27 C-5 D x^2\right )+b^2 \left (-45 B+9 C x^2+5 D x^4\right )\right )+3 \left (-15 A b^3+30 a b^2 B-36 a^2 b C+40 a^3 D\right ) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{45 a b^3 \sqrt [4]{a+b x^2}} \] Input:

Integrate[(A + B*x^2 + C*x^4 + D*x^6)/(a + b*x^2)^(5/4),x]
 

Output:

(x*(90*A*b^3 + 2*a*(-60*a^2*D + 2*a*b*(27*C - 5*D*x^2) + b^2*(-45*B + 9*C* 
x^2 + 5*D*x^4)) + 3*(-15*A*b^3 + 30*a*b^2*B - 36*a^2*b*C + 40*a^3*D)*(1 + 
(b*x^2)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^2)/a)]))/(45*a*b^ 
3*(a + b*x^2)^(1/4))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.25, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2345, 27, 1473, 27, 299, 227, 225, 212}

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.

\(\displaystyle \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{5/4}} \, dx\)

\(\Big \downarrow \) 2345

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1473

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 299

\(\displaystyle \frac {2 x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{a \sqrt [4]{a+b x^2}}-\frac {\frac {\frac {\left (-40 a^3 D+36 a^2 b C-30 a b^2 B+15 A b^3\right ) \int \frac {1}{\sqrt [4]{b x^2+a}}dx}{5 b^2}-\frac {2 a x \left (a+b x^2\right )^{3/4} (3 b C-5 a D)}{5 b^2}}{3 b}-\frac {2 a D x^3 \left (a+b x^2\right )^{3/4}}{9 b^2}}{a}\)

\(\Big \downarrow \) 227

\(\displaystyle \frac {2 x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{a \sqrt [4]{a+b x^2}}-\frac {\frac {\frac {\sqrt [4]{\frac {b x^2}{a}+1} \left (-40 a^3 D+36 a^2 b C-30 a b^2 B+15 A b^3\right ) \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1}}dx}{5 b^2 \sqrt [4]{a+b x^2}}-\frac {2 a x \left (a+b x^2\right )^{3/4} (3 b C-5 a D)}{5 b^2}}{3 b}-\frac {2 a D x^3 \left (a+b x^2\right )^{3/4}}{9 b^2}}{a}\)

\(\Big \downarrow \) 225

\(\displaystyle \frac {2 x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{a \sqrt [4]{a+b x^2}}-\frac {\frac {\frac {\sqrt [4]{\frac {b x^2}{a}+1} \left (-40 a^3 D+36 a^2 b C-30 a b^2 B+15 A b^3\right ) \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx\right )}{5 b^2 \sqrt [4]{a+b x^2}}-\frac {2 a x \left (a+b x^2\right )^{3/4} (3 b C-5 a D)}{5 b^2}}{3 b}-\frac {2 a D x^3 \left (a+b x^2\right )^{3/4}}{9 b^2}}{a}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {2 x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{a \sqrt [4]{a+b x^2}}-\frac {\frac {\frac {\sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right ) \left (-40 a^3 D+36 a^2 b C-30 a b^2 B+15 A b^3\right )}{5 b^2 \sqrt [4]{a+b x^2}}-\frac {2 a x \left (a+b x^2\right )^{3/4} (3 b C-5 a D)}{5 b^2}}{3 b}-\frac {2 a D x^3 \left (a+b x^2\right )^{3/4}}{9 b^2}}{a}\)

Input:

Int[(A + B*x^2 + C*x^4 + D*x^6)/(a + b*x^2)^(5/4),x]
 

Output:

(2*(A - (a*(b^2*B - a*b*C + a^2*D))/b^3)*x)/(a*(a + b*x^2)^(1/4)) - ((-2*a 
*D*x^3*(a + b*x^2)^(3/4))/(9*b^2) + ((-2*a*(3*b*C - 5*a*D)*x*(a + b*x^2)^( 
3/4))/(5*b^2) + ((15*A*b^3 - 30*a*b^2*B + 36*a^2*b*C - 40*a^3*D)*(1 + (b*x 
^2)/a)^(1/4)*((2*x)/(1 + (b*x^2)/a)^(1/4) - (2*Sqrt[a]*EllipticE[ArcTan[(S 
qrt[b]*x)/Sqrt[a]]/2, 2])/Sqrt[b]))/(5*b^2*(a + b*x^2)^(1/4)))/(3*b))/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 212

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) 
)*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 225

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) 
, x] - Simp[a   Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ 
a, 0] && PosQ[b/a]
 

rule 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(1/4)/( 
a + b*x^2)^(1/4)   Int[1/(1 + b*(x^2/a))^(1/4), 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 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]

\[\int \frac {D x^{6}+C \,x^{4}+x^{2} B +A}{\left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]

Input:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(5/4),x)
 

Output:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(5/4),x)
 

Fricas [F]

\[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {D x^{6} + C x^{4} + B x^{2} + A}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(5/4),x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.32 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.68 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {A x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {5}{4}}} + \frac {B x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 a^{\frac {5}{4}}} + \frac {C x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5 a^{\frac {5}{4}}} + \frac {D x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7 a^{\frac {5}{4}}} \] Input:

integrate((D*x**6+C*x**4+B*x**2+A)/(b*x**2+a)**(5/4),x)
 

Output:

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

Maxima [F]

\[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {D x^{6} + C x^{4} + B x^{2} + A}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(5/4),x, algorithm="maxima")
 

Output:

integrate((D*x^6 + C*x^4 + B*x^2 + A)/(b*x^2 + a)^(5/4), x)
 

Giac [F]

\[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {D x^{6} + C x^{4} + B x^{2} + A}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(5/4),x, algorithm="giac")
 

Output:

integrate((D*x^6 + C*x^4 + B*x^2 + A)/(b*x^2 + a)^(5/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{5/4}} \, dx=\int \frac {A+B\,x^2+C\,x^4+x^6\,D}{{\left (b\,x^2+a\right )}^{5/4}} \,d x \] Input:

int((A + B*x^2 + C*x^4 + x^6*D)/(a + b*x^2)^(5/4),x)
 

Output:

int((A + B*x^2 + C*x^4 + x^6*D)/(a + b*x^2)^(5/4), x)
 

Reduce [F]

\[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{5/4}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) d +\left (\int \frac {x^{4}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) c +\left (\int \frac {x^{2}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) b +\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) a \] Input:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(5/4),x)
 

Output:

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