\(\int \frac {x^8}{(a+b x^2)^{3/4} (c+d x^2)^2} \, dx\) [1542]

Optimal result

Integrand size = 24, antiderivative size = 435 \[ \int \frac {x^8}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx=-\frac {\left (77 b^2 c^2-32 a b c d-24 a^2 d^2\right ) x \sqrt [4]{a+b x^2}}{42 b^2 d^3 (b c-a d)}+\frac {(11 b c-4 a d) x^3 \sqrt [4]{a+b x^2}}{14 b d^2 (b c-a d)}-\frac {c x^5 \sqrt [4]{a+b x^2}}{2 d (b c-a d) \left (c+d x^2\right )}+\frac {\sqrt {a} \left (231 b^3 c^3-140 a b^2 c^2 d-64 a^2 b c d^2-48 a^3 d^3\right ) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{42 b^{5/2} d^4 (b c-a d) \left (a+b x^2\right )^{3/4}}-\frac {\sqrt [4]{a} c^3 (11 b c-14 a d) \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{4 d^4 (b c-a d)^2 x}-\frac {\sqrt [4]{a} c^3 (11 b c-14 a d) \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{4 d^4 (b c-a d)^2 x} \] Output:

-1/42*(-24*a^2*d^2-32*a*b*c*d+77*b^2*c^2)*x*(b*x^2+a)^(1/4)/b^2/d^3/(-a*d+ 
b*c)+1/14*(-4*a*d+11*b*c)*x^3*(b*x^2+a)^(1/4)/b/d^2/(-a*d+b*c)-1/2*c*x^5*( 
b*x^2+a)^(1/4)/d/(-a*d+b*c)/(d*x^2+c)+1/42*a^(1/2)*(-48*a^3*d^3-64*a^2*b*c 
*d^2-140*a*b^2*c^2*d+231*b^3*c^3)*(1+b*x^2/a)^(3/4)*InverseJacobiAM(1/2*ar 
ctan(b^(1/2)*x/a^(1/2)),2^(1/2))/b^(5/2)/d^4/(-a*d+b*c)/(b*x^2+a)^(3/4)-1/ 
4*a^(1/4)*c^3*(-14*a*d+11*b*c)*(-b*x^2/a)^(1/2)*EllipticPi((b*x^2+a)^(1/4) 
/a^(1/4),-a^(1/2)*d^(1/2)/(a*d-b*c)^(1/2),I)/d^4/(-a*d+b*c)^2/x-1/4*a^(1/4 
)*c^3*(-14*a*d+11*b*c)*(-b*x^2/a)^(1/2)*EllipticPi((b*x^2+a)^(1/4)/a^(1/4) 
,a^(1/2)*d^(1/2)/(a*d-b*c)^(1/2),I)/d^4/(-a*d+b*c)^2/x
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 10.92 (sec) , antiderivative size = 404, normalized size of antiderivative = 0.93 \[ \int \frac {x^8}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx=\frac {x \left (-\frac {\left (-231 b^3 c^3+140 a b^2 c^2 d+64 a^2 b c d^2+48 a^3 d^3\right ) x^2 \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{b^2 c (b c-a d)}+6 \left (-\frac {56 a c}{b}-\frac {24 a^2 d}{b^2}-56 c x^2-\frac {12 a d x^2}{b}+12 d x^4+\frac {21 a c^3}{(-b c+a d) \left (c+d x^2\right )}+\frac {21 b c^3 x^2}{(-b c+a d) \left (c+d x^2\right )}+\frac {6 a^2 c^2 \left (-77 b^2 c^2+32 a b c d+24 a^2 d^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{b^2 (b c-a d) \left (c+d x^2\right ) \left (-6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}\right )\right )}{252 d^3 \left (a+b x^2\right )^{3/4}} \] Input:

Integrate[x^8/((a + b*x^2)^(3/4)*(c + d*x^2)^2),x]
 

Output:

(x*(-(((-231*b^3*c^3 + 140*a*b^2*c^2*d + 64*a^2*b*c*d^2 + 48*a^3*d^3)*x^2* 
(1 + (b*x^2)/a)^(3/4)*AppellF1[3/2, 3/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c 
)])/(b^2*c*(b*c - a*d))) + 6*((-56*a*c)/b - (24*a^2*d)/b^2 - 56*c*x^2 - (1 
2*a*d*x^2)/b + 12*d*x^4 + (21*a*c^3)/((-(b*c) + a*d)*(c + d*x^2)) + (21*b* 
c^3*x^2)/((-(b*c) + a*d)*(c + d*x^2)) + (6*a^2*c^2*(-77*b^2*c^2 + 32*a*b*c 
*d + 24*a^2*d^2)*AppellF1[1/2, 3/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)])/( 
b^2*(b*c - a*d)*(c + d*x^2)*(-6*a*c*AppellF1[1/2, 3/4, 1, 3/2, -((b*x^2)/a 
), -((d*x^2)/c)] + x^2*(4*a*d*AppellF1[3/2, 3/4, 2, 5/2, -((b*x^2)/a), -(( 
d*x^2)/c)] + 3*b*c*AppellF1[3/2, 7/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)]) 
)))))/(252*d^3*(a + b*x^2)^(3/4))
 

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 0.20 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.15, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {395, 394}

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.

\(\displaystyle \int \frac {x^8}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx\)

\(\Big \downarrow \) 395

\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{3/4} \int \frac {x^8}{\left (\frac {b x^2}{a}+1\right )^{3/4} \left (d x^2+c\right )^2}dx}{\left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 394

\(\displaystyle \frac {x^9 \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {AppellF1}\left (\frac {9}{2},\frac {3}{4},2,\frac {11}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{9 c^2 \left (a+b x^2\right )^{3/4}}\)

Input:

Int[x^8/((a + b*x^2)^(3/4)*(c + d*x^2)^2),x]
 

Output:

(x^9*(1 + (b*x^2)/a)^(3/4)*AppellF1[9/2, 3/4, 2, 11/2, -((b*x^2)/a), -((d* 
x^2)/c)])/(9*c^2*(a + b*x^2)^(3/4))
 

Defintions of rubi rules used

rule 394

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 
, -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, 
 d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int 
egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 395

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

Maple [F]

\[\int \frac {x^{8}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (x^{2} d +c \right )^{2}}d x\]

Input:

int(x^8/(b*x^2+a)^(3/4)/(d*x^2+c)^2,x)
 

Output:

int(x^8/(b*x^2+a)^(3/4)/(d*x^2+c)^2,x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {x^8}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x^8/(b*x^2+a)^(3/4)/(d*x^2+c)^2,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {x^8}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx=\int \frac {x^{8}}{\left (a + b x^{2}\right )^{\frac {3}{4}} \left (c + d x^{2}\right )^{2}}\, dx \] Input:

integrate(x**8/(b*x**2+a)**(3/4)/(d*x**2+c)**2,x)
 

Output:

Integral(x**8/((a + b*x**2)**(3/4)*(c + d*x**2)**2), x)
 

Maxima [F]

\[ \int \frac {x^8}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx=\int { \frac {x^{8}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (d x^{2} + c\right )}^{2}} \,d x } \] Input:

integrate(x^8/(b*x^2+a)^(3/4)/(d*x^2+c)^2,x, algorithm="maxima")
 

Output:

integrate(x^8/((b*x^2 + a)^(3/4)*(d*x^2 + c)^2), x)
 

Giac [F]

\[ \int \frac {x^8}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx=\int { \frac {x^{8}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (d x^{2} + c\right )}^{2}} \,d x } \] Input:

integrate(x^8/(b*x^2+a)^(3/4)/(d*x^2+c)^2,x, algorithm="giac")
 

Output:

integrate(x^8/((b*x^2 + a)^(3/4)*(d*x^2 + c)^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^8}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx=\int \frac {x^8}{{\left (b\,x^2+a\right )}^{3/4}\,{\left (d\,x^2+c\right )}^2} \,d x \] Input:

int(x^8/((a + b*x^2)^(3/4)*(c + d*x^2)^2),x)
 

Output:

int(x^8/((a + b*x^2)^(3/4)*(c + d*x^2)^2), x)
 

Reduce [F]

\[ \int \frac {x^8}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx=\int \frac {x^{8}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} c^{2}+2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} c d \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {3}{4}} d^{2} x^{4}}d x \] Input:

int(x^8/(b*x^2+a)^(3/4)/(d*x^2+c)^2,x)
 

Output:

int(x**8/((a + b*x**2)**(3/4)*c**2 + 2*(a + b*x**2)**(3/4)*c*d*x**2 + (a + 
 b*x**2)**(3/4)*d**2*x**4),x)