\(\int \sqrt [4]{a+b x^2} (A+B x^2+C x^4+D x^6) \, dx\) [142]

Optimal result

Integrand size = 29, antiderivative size = 220 \[ \int \sqrt [4]{a+b x^2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {2}{231} \left (77 A-\frac {2 a \left (11 b^2 B-6 a b C+4 a^2 D\right )}{b^3}\right ) x \sqrt [4]{a+b x^2}+\frac {2 \left (11 b^2 B-6 a b C+4 a^2 D\right ) x \left (a+b x^2\right )^{5/4}}{77 b^3}+\frac {2 (3 b C-2 a D) x^3 \left (a+b x^2\right )^{5/4}}{33 b^2}+\frac {2 D x^5 \left (a+b x^2\right )^{5/4}}{15 b}+\frac {2 a^{3/2} \left (77 A b^3-2 a \left (11 b^2 B-6 a b C+4 a^2 D\right )\right ) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{231 b^{7/2} \left (a+b x^2\right )^{3/4}} \] Output:

2/231*(77*A-2*a*(11*B*b^2-6*C*a*b+4*D*a^2)/b^3)*x*(b*x^2+a)^(1/4)+2/77*(11 
*B*b^2-6*C*a*b+4*D*a^2)*x*(b*x^2+a)^(5/4)/b^3+2/33*(3*C*b-2*D*a)*x^3*(b*x^ 
2+a)^(5/4)/b^2+2/15*D*x^5*(b*x^2+a)^(5/4)/b+2/231*a^(3/2)*(77*A*b^3-2*a*(1 
1*B*b^2-6*C*a*b+4*D*a^2))*(1+b*x^2/a)^(3/4)*InverseJacobiAM(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 = 7.48 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.60 \[ \int \sqrt [4]{a+b x^2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {x \sqrt [4]{a+b x^2} \left (2 \left (a+b x^2\right ) \left (60 a^2 D-10 a b \left (9 C+7 D x^2\right )+b^2 \left (165 B+105 C x^2+77 D x^4\right )\right )+\frac {15 \left (77 A b^3-2 a \left (11 b^2 B-6 a b C+4 a^2 D\right )\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )}{\sqrt [4]{1+\frac {b x^2}{a}}}\right )}{1155 b^3} \] Input:

Integrate[(a + b*x^2)^(1/4)*(A + B*x^2 + C*x^4 + D*x^6),x]
 

Output:

(x*(a + b*x^2)^(1/4)*(2*(a + b*x^2)*(60*a^2*D - 10*a*b*(9*C + 7*D*x^2) + b 
^2*(165*B + 105*C*x^2 + 77*D*x^4)) + (15*(77*A*b^3 - 2*a*(11*b^2*B - 6*a*b 
*C + 4*a^2*D))*Hypergeometric2F1[-1/4, 1/2, 3/2, -((b*x^2)/a)])/(1 + (b*x^ 
2)/a)^(1/4)))/(1155*b^3)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2346, 27, 1473, 27, 299, 211, 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.

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

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1473

\(\displaystyle \frac {\frac {2 \int \frac {3}{2} \sqrt [4]{b x^2+a} \left (11 A b^2+\left (4 D a^2-6 b C a+11 b^2 B\right ) x^2\right )dx}{11 b}+\frac {2 x^3 \left (a+b x^2\right )^{5/4} (3 b C-2 a D)}{11 b}}{3 b}+\frac {2 D x^5 \left (a+b x^2\right )^{5/4}}{15 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \int \sqrt [4]{b x^2+a} \left (11 A b^2+\left (4 D a^2-6 b C a+11 b^2 B\right ) x^2\right )dx}{11 b}+\frac {2 x^3 \left (a+b x^2\right )^{5/4} (3 b C-2 a D)}{11 b}}{3 b}+\frac {2 D x^5 \left (a+b x^2\right )^{5/4}}{15 b}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {3 \left (\frac {\left (77 A b^3-2 a \left (4 a^2 D-6 a b C+11 b^2 B\right )\right ) \int \sqrt [4]{b x^2+a}dx}{7 b}+\frac {2 x \left (a+b x^2\right )^{5/4} \left (4 a^2 D-6 a b C+11 b^2 B\right )}{7 b}\right )}{11 b}+\frac {2 x^3 \left (a+b x^2\right )^{5/4} (3 b C-2 a D)}{11 b}}{3 b}+\frac {2 D x^5 \left (a+b x^2\right )^{5/4}}{15 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {3 \left (\frac {\left (77 A b^3-2 a \left (4 a^2 D-6 a b C+11 b^2 B\right )\right ) \left (\frac {1}{3} a \int \frac {1}{\left (b x^2+a\right )^{3/4}}dx+\frac {2}{3} x \sqrt [4]{a+b x^2}\right )}{7 b}+\frac {2 x \left (a+b x^2\right )^{5/4} \left (4 a^2 D-6 a b C+11 b^2 B\right )}{7 b}\right )}{11 b}+\frac {2 x^3 \left (a+b x^2\right )^{5/4} (3 b C-2 a D)}{11 b}}{3 b}+\frac {2 D x^5 \left (a+b x^2\right )^{5/4}}{15 b}\)

\(\Big \downarrow \) 231

\(\displaystyle \frac {\frac {3 \left (\frac {\left (77 A b^3-2 a \left (4 a^2 D-6 a b C+11 b^2 B\right )\right ) \left (\frac {a \left (\frac {b x^2}{a}+1\right )^{3/4} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{3 \left (a+b x^2\right )^{3/4}}+\frac {2}{3} x \sqrt [4]{a+b x^2}\right )}{7 b}+\frac {2 x \left (a+b x^2\right )^{5/4} \left (4 a^2 D-6 a b C+11 b^2 B\right )}{7 b}\right )}{11 b}+\frac {2 x^3 \left (a+b x^2\right )^{5/4} (3 b C-2 a D)}{11 b}}{3 b}+\frac {2 D x^5 \left (a+b x^2\right )^{5/4}}{15 b}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {\frac {3 \left (\frac {2 x \left (a+b x^2\right )^{5/4} \left (4 a^2 D-6 a b C+11 b^2 B\right )}{7 b}+\frac {\left (\frac {2 a^{3/2} \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{3 \sqrt {b} \left (a+b x^2\right )^{3/4}}+\frac {2}{3} x \sqrt [4]{a+b x^2}\right ) \left (77 A b^3-2 a \left (4 a^2 D-6 a b C+11 b^2 B\right )\right )}{7 b}\right )}{11 b}+\frac {2 x^3 \left (a+b x^2\right )^{5/4} (3 b C-2 a D)}{11 b}}{3 b}+\frac {2 D x^5 \left (a+b x^2\right )^{5/4}}{15 b}\)

Input:

Int[(a + b*x^2)^(1/4)*(A + B*x^2 + C*x^4 + D*x^6),x]
 

Output:

(2*D*x^5*(a + b*x^2)^(5/4))/(15*b) + ((2*(3*b*C - 2*a*D)*x^3*(a + b*x^2)^( 
5/4))/(11*b) + (3*((2*(11*b^2*B - 6*a*b*C + 4*a^2*D)*x*(a + b*x^2)^(5/4))/ 
(7*b) + ((77*A*b^3 - 2*a*(11*b^2*B - 6*a*b*C + 4*a^2*D))*((2*x*(a + b*x^2) 
^(1/4))/3 + (2*a^(3/2)*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/ 
Sqrt[a]]/2, 2])/(3*Sqrt[b]*(a + b*x^2)^(3/4))))/(7*b)))/(11*b))/(3*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 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

integrate((b*x^2+a)^(1/4)*(D*x^6+C*x^4+B*x^2+A),x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.71 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.56 \[ \int \sqrt [4]{a+b x^2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=A \sqrt [4]{a} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {B \sqrt [4]{a} x^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + \frac {C \sqrt [4]{a} x^{5} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} + \frac {D \sqrt [4]{a} x^{7} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7} \] Input:

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

Output:

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

Maxima [F]

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

integrate((b*x^2+a)^(1/4)*(D*x^6+C*x^4+B*x^2+A),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((b*x^2+a)^(1/4)*(D*x^6+C*x^4+B*x^2+A),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt [4]{a+b x^2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {40 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{3} d x -60 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2} b c x -20 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2} b d \,x^{3}+880 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a \,b^{3} x +30 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a \,b^{2} c \,x^{3}+14 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a \,b^{2} d \,x^{5}+330 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{4} x^{3}+210 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{3} c \,x^{5}+154 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{3} d \,x^{7}-40 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a^{4} d +60 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a^{3} b c +275 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a^{2} b^{3}}{1155 b^{3}} \] Input:

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

Output:

(40*(a + b*x**2)**(1/4)*a**3*d*x - 60*(a + b*x**2)**(1/4)*a**2*b*c*x - 20* 
(a + b*x**2)**(1/4)*a**2*b*d*x**3 + 880*(a + b*x**2)**(1/4)*a*b**3*x + 30* 
(a + b*x**2)**(1/4)*a*b**2*c*x**3 + 14*(a + b*x**2)**(1/4)*a*b**2*d*x**5 + 
 330*(a + b*x**2)**(1/4)*b**4*x**3 + 210*(a + b*x**2)**(1/4)*b**3*c*x**5 + 
 154*(a + b*x**2)**(1/4)*b**3*d*x**7 - 40*int((a + b*x**2)**(1/4)/(a + b*x 
**2),x)*a**4*d + 60*int((a + b*x**2)**(1/4)/(a + b*x**2),x)*a**3*b*c + 275 
*int((a + b*x**2)**(1/4)/(a + b*x**2),x)*a**2*b**3)/(1155*b**3)