\(\int \frac {1}{x^4 (a+b x^2)^{3/4} (c+d x^2)^2} \, dx\) [1548]

Optimal result

Integrand size = 24, antiderivative size = 423 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx=-\frac {(2 b c-5 a d) \sqrt [4]{a+b x^2}}{6 a c^2 (b c-a d) x^3}+\frac {\left (5 b^2 c^2+7 a b c d-15 a^2 d^2\right ) \sqrt [4]{a+b x^2}}{6 a^2 c^3 (b c-a d) x}-\frac {d \sqrt [4]{a+b x^2}}{2 c (b c-a d) x^3 \left (c+d x^2\right )}+\frac {\sqrt {b} \left (5 b^2 c^2+7 a b c d-15 a^2 d^2\right ) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{6 a^{3/2} c^3 (b c-a d) \left (a+b x^2\right )^{3/4}}+\frac {\sqrt [4]{a} d^2 (13 b c-10 a d) \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{4 c^3 (b c-a d)^2 x}+\frac {\sqrt [4]{a} d^2 (13 b c-10 a d) \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{4 c^3 (b c-a d)^2 x} \] Output:

-1/6*(-5*a*d+2*b*c)*(b*x^2+a)^(1/4)/a/c^2/(-a*d+b*c)/x^3+1/6*(-15*a^2*d^2+ 
7*a*b*c*d+5*b^2*c^2)*(b*x^2+a)^(1/4)/a^2/c^3/(-a*d+b*c)/x-1/2*d*(b*x^2+a)^ 
(1/4)/c/(-a*d+b*c)/x^3/(d*x^2+c)+1/6*b^(1/2)*(-15*a^2*d^2+7*a*b*c*d+5*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))/a^(3/2)/c^3/(-a*d+b*c)/(b*x^2+a)^(3/4)+1/4*a^(1/4)*d^2*(-10*a*d+13*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)/c^3/(-a*d+b*c)^2/x+1/4*a^(1/4)*d^2*(-10*a*d+13*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)/c^3/(-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 = 10.86 (sec) , antiderivative size = 379, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx=\frac {b d \left (-5 b^2 c^2-7 a b c d+15 a^2 d^2\right ) x^6 \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+\frac {6 c \left (\left (a+b x^2\right ) \left (-5 b^2 c^2 x^2 \left (c+d x^2\right )+a b c \left (2 c^2-5 c d x^2-7 d^2 x^4\right )+a^2 d \left (-2 c^2+10 c d x^2+15 d^2 x^4\right )\right )-\frac {3 a c \left (5 b^3 c^3+7 a b^2 c^2 d+24 a^2 b c d^2-30 a^3 d^3\right ) x^4 \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )-x^2 \left (4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}\right )}{c+d x^2}}{36 a^2 c^4 (-b c+a d) x^3 \left (a+b x^2\right )^{3/4}} \] Input:

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

Output:

(b*d*(-5*b^2*c^2 - 7*a*b*c*d + 15*a^2*d^2)*x^6*(1 + (b*x^2)/a)^(3/4)*Appel 
lF1[3/2, 3/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)] + (6*c*((a + b*x^2)*(-5* 
b^2*c^2*x^2*(c + d*x^2) + a*b*c*(2*c^2 - 5*c*d*x^2 - 7*d^2*x^4) + a^2*d*(- 
2*c^2 + 10*c*d*x^2 + 15*d^2*x^4)) - (3*a*c*(5*b^3*c^3 + 7*a*b^2*c^2*d + 24 
*a^2*b*c*d^2 - 30*a^3*d^3)*x^4*AppellF1[1/2, 3/4, 1, 3/2, -((b*x^2)/a), -( 
(d*x^2)/c)])/(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)]))))/(c + d*x^2)) 
/(36*a^2*c^4*(-(b*c) + a*d)*x^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.

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 394

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{x^4 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^2} \, dx=\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} c^{2} x^{4}+2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} c d \,x^{6}+\left (b \,x^{2}+a \right )^{\frac {3}{4}} d^{2} x^{8}}d x \] Input:

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

Output:

int(1/((a + b*x**2)**(3/4)*c**2*x**4 + 2*(a + b*x**2)**(3/4)*c*d*x**6 + (a 
 + b*x**2)**(3/4)*d**2*x**8),x)