\(\int \sqrt [4]{a+b x^2} (c+d x^2)^3 \, dx\) [405]

Optimal result

Integrand size = 21, antiderivative size = 247 \[ \int \sqrt [4]{a+b x^2} \left (c+d x^2\right )^3 \, dx=\frac {2}{231} \left (77 c^3-\frac {2 a d \left (33 b^2 c^2-18 a b c d+4 a^2 d^2\right )}{b^3}\right ) x \sqrt [4]{a+b x^2}+\frac {2 d \left (33 b^2 c^2-18 a b c d+4 a^2 d^2\right ) x \left (a+b x^2\right )^{5/4}}{77 b^3}+\frac {2 d^2 (9 b c-2 a d) x^3 \left (a+b x^2\right )^{5/4}}{33 b^2}+\frac {2 d^3 x^5 \left (a+b x^2\right )^{5/4}}{15 b}+\frac {2 a^{3/2} \left (77 b^3 c^3-2 a d \left (33 b^2 c^2-18 a b c d+4 a^2 d^2\right )\right ) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{231 b^{7/2} \left (a+b x^2\right )^{3/4}} \] Output:

2/231*(77*c^3-2*a*d*(4*a^2*d^2-18*a*b*c*d+33*b^2*c^2)/b^3)*x*(b*x^2+a)^(1/ 
4)+2/77*d*(4*a^2*d^2-18*a*b*c*d+33*b^2*c^2)*x*(b*x^2+a)^(5/4)/b^3+2/33*d^2 
*(-2*a*d+9*b*c)*x^3*(b*x^2+a)^(5/4)/b^2+2/15*d^3*x^5*(b*x^2+a)^(5/4)/b+2/2 
31*a^(3/2)*(77*b^3*c^3-2*a*d*(4*a^2*d^2-18*a*b*c*d+33*b^2*c^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 = 11.63 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.78 \[ \int \sqrt [4]{a+b x^2} \left (c+d x^2\right )^3 \, dx=\frac {x \left (2 \left (a+b x^2\right ) \left (20 a^3 d^3-10 a^2 b d^2 \left (9 c+d x^2\right )+a b^2 d \left (165 c^2+45 c d x^2+7 d^2 x^4\right )+b^3 \left (385 c^3+495 c^2 d x^2+315 c d^2 x^4+77 d^3 x^6\right )\right )+5 a \left (77 b^3 c^3-66 a b^2 c^2 d+36 a^2 b c d^2-8 a^3 d^3\right ) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{1155 b^3 \left (a+b x^2\right )^{3/4}} \] Input:

Integrate[(a + b*x^2)^(1/4)*(c + d*x^2)^3,x]
 

Output:

(x*(2*(a + b*x^2)*(20*a^3*d^3 - 10*a^2*b*d^2*(9*c + d*x^2) + a*b^2*d*(165* 
c^2 + 45*c*d*x^2 + 7*d^2*x^4) + b^3*(385*c^3 + 495*c^2*d*x^2 + 315*c*d^2*x 
^4 + 77*d^3*x^6)) + 5*a*(77*b^3*c^3 - 66*a*b^2*c^2*d + 36*a^2*b*c*d^2 - 8* 
a^3*d^3)*(1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, -((b*x^2)/ 
a)]))/(1155*b^3*(a + b*x^2)^(3/4))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {318, 27, 403, 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.

\(\displaystyle \int \sqrt [4]{a+b x^2} \left (c+d x^2\right )^3 \, dx\)

\(\Big \downarrow \) 318

\(\displaystyle \frac {2 \int \frac {1}{2} \sqrt [4]{b x^2+a} \left (d x^2+c\right ) \left (d (23 b c-10 a d) x^2+c (15 b c-2 a d)\right )dx}{15 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^2}{15 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sqrt [4]{b x^2+a} \left (d x^2+c\right ) \left (d (23 b c-10 a d) x^2+c (15 b c-2 a d)\right )dx}{15 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^2}{15 b}\)

\(\Big \downarrow \) 403

\(\displaystyle \frac {\frac {2 \int \frac {1}{2} \sqrt [4]{b x^2+a} \left (d \left (257 b^2 c^2-200 a b d c+60 a^2 d^2\right ) x^2+c \left (165 b^2 c^2-68 a b d c+20 a^2 d^2\right )\right )dx}{11 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right ) (23 b c-10 a d)}{11 b}}{15 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^2}{15 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \sqrt [4]{b x^2+a} \left (d \left (257 b^2 c^2-200 a b d c+60 a^2 d^2\right ) x^2+c \left (165 b^2 c^2-68 a b d c+20 a^2 d^2\right )\right )dx}{11 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right ) (23 b c-10 a d)}{11 b}}{15 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^2}{15 b}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\frac {15 \left (-8 a^3 d^3+36 a^2 b c d^2-66 a b^2 c^2 d+77 b^3 c^3\right ) \int \sqrt [4]{b x^2+a}dx}{7 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (60 a^2 d^2-200 a b c d+257 b^2 c^2\right )}{7 b}}{11 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right ) (23 b c-10 a d)}{11 b}}{15 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^2}{15 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {\frac {15 \left (-8 a^3 d^3+36 a^2 b c d^2-66 a b^2 c^2 d+77 b^3 c^3\right ) \left (\frac {1}{3} a \int \frac {1}{\left (b x^2+a\right )^{3/4}}dx+\frac {2}{3} x \sqrt [4]{a+b x^2}\right )}{7 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (60 a^2 d^2-200 a b c d+257 b^2 c^2\right )}{7 b}}{11 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right ) (23 b c-10 a d)}{11 b}}{15 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^2}{15 b}\)

\(\Big \downarrow \) 231

\(\displaystyle \frac {\frac {\frac {15 \left (-8 a^3 d^3+36 a^2 b c d^2-66 a b^2 c^2 d+77 b^3 c^3\right ) \left (\frac {a \left (\frac {b x^2}{a}+1\right )^{3/4} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{3 \left (a+b x^2\right )^{3/4}}+\frac {2}{3} x \sqrt [4]{a+b x^2}\right )}{7 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (60 a^2 d^2-200 a b c d+257 b^2 c^2\right )}{7 b}}{11 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right ) (23 b c-10 a d)}{11 b}}{15 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^2}{15 b}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {\frac {\frac {2 d x \left (a+b x^2\right )^{5/4} \left (60 a^2 d^2-200 a b c d+257 b^2 c^2\right )}{7 b}+\frac {15 \left (-8 a^3 d^3+36 a^2 b c d^2-66 a b^2 c^2 d+77 b^3 c^3\right ) \left (\frac {2 a^{3/2} \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{3 \sqrt {b} \left (a+b x^2\right )^{3/4}}+\frac {2}{3} x \sqrt [4]{a+b x^2}\right )}{7 b}}{11 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right ) (23 b c-10 a d)}{11 b}}{15 b}+\frac {2 d x \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^2}{15 b}\)

Input:

Int[(a + b*x^2)^(1/4)*(c + d*x^2)^3,x]
 

Output:

(2*d*x*(a + b*x^2)^(5/4)*(c + d*x^2)^2)/(15*b) + ((2*d*(23*b*c - 10*a*d)*x 
*(a + b*x^2)^(5/4)*(c + d*x^2))/(11*b) + ((2*d*(257*b^2*c^2 - 200*a*b*c*d 
+ 60*a^2*d^2)*x*(a + b*x^2)^(5/4))/(7*b) + (15*(77*b^3*c^3 - 66*a*b^2*c^2* 
d + 36*a^2*b*c*d^2 - 8*a^3*d^3)*((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))/(15*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]
 

rule 403

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( 
x_)^2), x_Symbol] :> Simp[f*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b*(2*(p + 
 q + 1) + 1))), x] + Simp[1/(b*(2*(p + q + 1) + 1))   Int[(a + b*x^2)^p*(c 
+ d*x^2)^(q - 1)*Simp[c*(b*e - a*f + b*e*2*(p + q + 1)) + (d*(b*e - a*f) + 
f*2*q*(b*c - a*d) + b*d*e*2*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, 
 d, e, f, p}, x] && GtQ[q, 0] && NeQ[2*(p + q + 1) + 1, 0]
 

Maple [F]

\[\int \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (x^{2} d +c \right )^{3}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \sqrt [4]{a+b x^2} \left (c+d x^2\right )^3 \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}^{3} \,d x } \] Input:

integrate((b*x^2+a)^(1/4)*(d*x^2+c)^3,x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.57 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.54 \[ \int \sqrt [4]{a+b x^2} \left (c+d x^2\right )^3 \, dx=\sqrt [4]{a} c^{3} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \sqrt [4]{a} c^{2} d x^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {3 \sqrt [4]{a} c d^{2} x^{5} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} + \frac {\sqrt [4]{a} d^{3} x^{7} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7} \] Input:

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

Output:

a**(1/4)*c**3*x*hyper((-1/4, 1/2), (3/2,), b*x**2*exp_polar(I*pi)/a) + a** 
(1/4)*c**2*d*x**3*hyper((-1/4, 3/2), (5/2,), b*x**2*exp_polar(I*pi)/a) + 3 
*a**(1/4)*c*d**2*x**5*hyper((-1/4, 5/2), (7/2,), b*x**2*exp_polar(I*pi)/a) 
/5 + a**(1/4)*d**3*x**7*hyper((-1/4, 7/2), (9/2,), b*x**2*exp_polar(I*pi)/ 
a)/7
 

Maxima [F]

\[ \int \sqrt [4]{a+b x^2} \left (c+d x^2\right )^3 \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}^{3} \,d x } \] Input:

integrate((b*x^2+a)^(1/4)*(d*x^2+c)^3,x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^(1/4)*(d*x^2 + c)^3, x)
 

Giac [F]

\[ \int \sqrt [4]{a+b x^2} \left (c+d x^2\right )^3 \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}^{3} \,d x } \] Input:

integrate((b*x^2+a)^(1/4)*(d*x^2+c)^3,x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^(1/4)*(d*x^2 + c)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt [4]{a+b x^2} \left (c+d x^2\right )^3 \, dx=\int {\left (b\,x^2+a\right )}^{1/4}\,{\left (d\,x^2+c\right )}^3 \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \sqrt [4]{a+b x^2} \left (c+d x^2\right )^3 \, dx=\frac {40 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{3} d^{3} x -180 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2} b c \,d^{2} x -20 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2} b \,d^{3} x^{3}+330 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a \,b^{2} c^{2} d x +90 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a \,b^{2} c \,d^{2} x^{3}+14 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a \,b^{2} d^{3} x^{5}+770 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{3} c^{3} x +990 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{3} c^{2} d \,x^{3}+630 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{3} c \,d^{2} x^{5}+154 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{3} d^{3} x^{7}-40 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a^{4} d^{3}+180 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a^{3} b c \,d^{2}-330 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a^{2} b^{2} c^{2} d +385 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a \,b^{3} c^{3}}{1155 b^{3}} \] Input:

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

Output:

(40*(a + b*x**2)**(1/4)*a**3*d**3*x - 180*(a + b*x**2)**(1/4)*a**2*b*c*d** 
2*x - 20*(a + b*x**2)**(1/4)*a**2*b*d**3*x**3 + 330*(a + b*x**2)**(1/4)*a* 
b**2*c**2*d*x + 90*(a + b*x**2)**(1/4)*a*b**2*c*d**2*x**3 + 14*(a + b*x**2 
)**(1/4)*a*b**2*d**3*x**5 + 770*(a + b*x**2)**(1/4)*b**3*c**3*x + 990*(a + 
 b*x**2)**(1/4)*b**3*c**2*d*x**3 + 630*(a + b*x**2)**(1/4)*b**3*c*d**2*x** 
5 + 154*(a + b*x**2)**(1/4)*b**3*d**3*x**7 - 40*int((a + b*x**2)**(1/4)/(a 
 + b*x**2),x)*a**4*d**3 + 180*int((a + b*x**2)**(1/4)/(a + b*x**2),x)*a**3 
*b*c*d**2 - 330*int((a + b*x**2)**(1/4)/(a + b*x**2),x)*a**2*b**2*c**2*d + 
 385*int((a + b*x**2)**(1/4)/(a + b*x**2),x)*a*b**3*c**3)/(1155*b**3)