\(\int \sqrt [4]{a+b x^2} (c+d x^2)^2 \, dx\) [406]

Optimal result

Integrand size = 21, antiderivative size = 172 \[ \int \sqrt [4]{a+b x^2} \left (c+d x^2\right )^2 \, dx=\frac {2}{231} \left (77 c^2-\frac {4 a d (11 b c-3 a d)}{b^2}\right ) x \sqrt [4]{a+b x^2}+\frac {4 d (11 b c-3 a d) x \left (a+b x^2\right )^{5/4}}{77 b^2}+\frac {2 d^2 x^3 \left (a+b x^2\right )^{5/4}}{11 b}+\frac {2 a^{3/2} \left (77 b^2 c^2-4 a d (11 b c-3 a d)\right ) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{231 b^{5/2} \left (a+b x^2\right )^{3/4}} \] Output:

2/231*(77*c^2-4*a*d*(-3*a*d+11*b*c)/b^2)*x*(b*x^2+a)^(1/4)+4/77*d*(-3*a*d+ 
11*b*c)*x*(b*x^2+a)^(5/4)/b^2+2/11*d^2*x^3*(b*x^2+a)^(5/4)/b+2/231*a^(3/2) 
*(77*b^2*c^2-4*a*d*(-3*a*d+11*b*c))*(1+b*x^2/a)^(3/4)*InverseJacobiAM(1/2* 
arctan(b^(1/2)*x/a^(1/2)),2^(1/2))/b^(5/2)/(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 = 8.74 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.03 \[ \int \sqrt [4]{a+b x^2} \left (c+d x^2\right )^2 \, dx=\frac {x \sqrt [4]{a+b x^2} \left (7 a \left (15 c^2+10 c d x^2+3 d^2 x^4\right ) \operatorname {Gamma}\left (-\frac {1}{4}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {7}{2},-\frac {b x^2}{a}\right )-8 b x^2 \left (2 c^2+3 c d x^2+d^2 x^4\right ) \operatorname {Gamma}\left (\frac {3}{4}\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {9}{2},-\frac {b x^2}{a}\right )-4 b x^2 \left (c+d x^2\right )^2 \operatorname {Gamma}\left (\frac {3}{4}\right ) \, _3F_2\left (\frac {3}{4},\frac {3}{2},2;1,\frac {9}{2};-\frac {b x^2}{a}\right )\right )}{105 a \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Gamma}\left (-\frac {1}{4}\right )} \] Input:

Integrate[(a + b*x^2)^(1/4)*(c + d*x^2)^2,x]
 

Output:

(x*(a + b*x^2)^(1/4)*(7*a*(15*c^2 + 10*c*d*x^2 + 3*d^2*x^4)*Gamma[-1/4]*Hy 
pergeometric2F1[-1/4, 1/2, 7/2, -((b*x^2)/a)] - 8*b*x^2*(2*c^2 + 3*c*d*x^2 
 + d^2*x^4)*Gamma[3/4]*Hypergeometric2F1[3/4, 3/2, 9/2, -((b*x^2)/a)] - 4* 
b*x^2*(c + d*x^2)^2*Gamma[3/4]*HypergeometricPFQ[{3/4, 3/2, 2}, {1, 9/2}, 
-((b*x^2)/a)]))/(105*a*(1 + (b*x^2)/a)^(1/4)*Gamma[-1/4])
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {318, 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.

\(\displaystyle \int \sqrt [4]{a+b x^2} \left (c+d x^2\right )^2 \, dx\)

\(\Big \downarrow \) 318

\(\displaystyle \frac {2 \int \frac {1}{2} \sqrt [4]{b x^2+a} \left (3 d (5 b c-2 a d) x^2+c (11 b c-2 a d)\right )dx}{11 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}{11 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sqrt [4]{b x^2+a} \left (3 d (5 b c-2 a d) x^2+c (11 b c-2 a d)\right )dx}{11 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}{11 b}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\left (12 a^2 d^2-44 a b c d+77 b^2 c^2\right ) \int \sqrt [4]{b x^2+a}dx}{7 b}+\frac {6 d x \left (a+b x^2\right )^{5/4} (5 b c-2 a d)}{7 b}}{11 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}{11 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {\left (12 a^2 d^2-44 a b c d+77 b^2 c^2\right ) \left (\frac {1}{3} a \int \frac {1}{\left (b x^2+a\right )^{3/4}}dx+\frac {2}{3} x \sqrt [4]{a+b x^2}\right )}{7 b}+\frac {6 d x \left (a+b x^2\right )^{5/4} (5 b c-2 a d)}{7 b}}{11 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}{11 b}\)

\(\Big \downarrow \) 231

\(\displaystyle \frac {\frac {\left (12 a^2 d^2-44 a b c d+77 b^2 c^2\right ) \left (\frac {a \left (\frac {b x^2}{a}+1\right )^{3/4} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{3 \left (a+b x^2\right )^{3/4}}+\frac {2}{3} x \sqrt [4]{a+b x^2}\right )}{7 b}+\frac {6 d x \left (a+b x^2\right )^{5/4} (5 b c-2 a d)}{7 b}}{11 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}{11 b}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {\frac {\left (12 a^2 d^2-44 a b c d+77 b^2 c^2\right ) \left (\frac {2 a^{3/2} \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{3 \sqrt {b} \left (a+b x^2\right )^{3/4}}+\frac {2}{3} x \sqrt [4]{a+b x^2}\right )}{7 b}+\frac {6 d x \left (a+b x^2\right )^{5/4} (5 b c-2 a d)}{7 b}}{11 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )}{11 b}\)

Input:

Int[(a + b*x^2)^(1/4)*(c + d*x^2)^2,x]
 

Output:

(2*d*x*(a + b*x^2)^(5/4)*(c + d*x^2))/(11*b) + ((6*d*(5*b*c - 2*a*d)*x*(a 
+ b*x^2)^(5/4))/(7*b) + ((77*b^2*c^2 - 44*a*b*c*d + 12*a^2*d^2)*((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)
 

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 318

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S 
imp[1/(b*(2*(p + q) + 1))   Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b 
*c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 
 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G 
tQ[q, 1] && NeQ[2*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[a, b, c, 
d, 2, p, q, x]
 

Maple [F]

\[\int \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (x^{2} d +c \right )^{2}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate((b*x^2+a)^(1/4)*(d*x^2+c)^2,x, algorithm="fricas")
 

Output:

integral((d^2*x^4 + 2*c*d*x^2 + c^2)*(b*x^2 + a)^(1/4), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.58 \[ \int \sqrt [4]{a+b x^2} \left (c+d x^2\right )^2 \, dx=\sqrt [4]{a} c^{2} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {2 \sqrt [4]{a} c d x^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + \frac {\sqrt [4]{a} d^{2} x^{5} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} \] Input:

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

Output:

a**(1/4)*c**2*x*hyper((-1/4, 1/2), (3/2,), b*x**2*exp_polar(I*pi)/a) + 2*a 
**(1/4)*c*d*x**3*hyper((-1/4, 3/2), (5/2,), b*x**2*exp_polar(I*pi)/a)/3 + 
a**(1/4)*d**2*x**5*hyper((-1/4, 5/2), (7/2,), b*x**2*exp_polar(I*pi)/a)/5
 

Maxima [F]

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

integrate((b*x^2+a)^(1/4)*(d*x^2+c)^2,x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^(1/4)*(d*x^2 + c)^2, x)
 

Giac [F]

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

integrate((b*x^2+a)^(1/4)*(d*x^2+c)^2,x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^(1/4)*(d*x^2 + c)^2, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt [4]{a+b x^2} \left (c+d x^2\right )^2 \, dx=\frac {-12 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2} d^{2} x +44 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b c d x +6 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b \,d^{2} x^{3}+154 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} c^{2} x +132 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} c d \,x^{3}+42 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} d^{2} x^{5}+12 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a^{3} d^{2}-44 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a^{2} b c d +77 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a \,b^{2} c^{2}}{231 b^{2}} \] Input:

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

Output:

( - 12*(a + b*x**2)**(1/4)*a**2*d**2*x + 44*(a + b*x**2)**(1/4)*a*b*c*d*x 
+ 6*(a + b*x**2)**(1/4)*a*b*d**2*x**3 + 154*(a + b*x**2)**(1/4)*b**2*c**2* 
x + 132*(a + b*x**2)**(1/4)*b**2*c*d*x**3 + 42*(a + b*x**2)**(1/4)*b**2*d* 
*2*x**5 + 12*int((a + b*x**2)**(1/4)/(a + b*x**2),x)*a**3*d**2 - 44*int((a 
 + b*x**2)**(1/4)/(a + b*x**2),x)*a**2*b*c*d + 77*int((a + b*x**2)**(1/4)/ 
(a + b*x**2),x)*a*b**2*c**2)/(231*b**2)