\(\int \frac {\sqrt [4]{c+\frac {d}{x^2}}}{(a+\frac {b}{x^2})^{15/4}} \, dx\) [4]

Optimal result

Integrand size = 23, antiderivative size = 687 \[ \int \frac {\sqrt [4]{c+\frac {d}{x^2}}}{\left (a+\frac {b}{x^2}\right )^{15/4}} \, dx=\frac {2 b d (41 b c-35 a d) \sqrt [4]{c+\frac {d}{x^2}}}{33 a^3 c (b c-a d) \left (a+\frac {b}{x^2}\right )^{7/4} x^3}-\frac {d \sqrt [4]{c+\frac {d}{x^2}}}{a c \left (a+\frac {b}{x^2}\right )^{11/4} x}+\frac {b (9 b c-11 a d) (13 b c-a d) \sqrt [4]{c+\frac {d}{x^2}}}{77 a^3 (b c-a d)^2 \left (a+\frac {b}{x^2}\right )^{7/4} x}+\frac {b (5 b c-11 a d) (9 b c-11 a d) (13 b c-a d) \sqrt [4]{c+\frac {d}{x^2}}}{231 a^4 (b c-a d)^3 \left (a+\frac {b}{x^2}\right )^{3/4} x}+\frac {b \sqrt [4]{c+\frac {d}{x^2}} \left (13 b c-a d+\frac {12 b d}{x^2}\right )}{11 a^2 (b c-a d) \left (a+\frac {b}{x^2}\right )^{11/4} x}+\frac {\left (c+\frac {d}{x^2}\right )^{5/4} x}{a c \left (a+\frac {b}{x^2}\right )^{11/4}}+\frac {(9 b c-11 a d) (13 b c-a d) \left (5 b^2 c^2-14 a b c d+21 a^2 d^2\right ) \left (\frac {c \left (a+\frac {b}{x^2}\right )}{a \left (c+\frac {d}{x^2}\right )}\right )^{3/4} \sqrt [4]{c+\frac {d}{x^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\frac {b c-a d}{a \left (c+\frac {d}{x^2}\right ) x^2}\right )}{462 a^4 c (b c-a d)^3 \left (a+\frac {b}{x^2}\right )^{3/4} x}+\frac {24 b^2 d^2 \left (\frac {a \left (c+\frac {d}{x^2}\right )}{c \left (a+\frac {b}{x^2}\right )}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{2},\frac {b c-a d}{c \left (a+\frac {b}{x^2}\right ) x^2}\right )}{55 a^3 (b c-a d) \left (a+\frac {b}{x^2}\right )^{7/4} \left (c+\frac {d}{x^2}\right )^{3/4} x^5}+\frac {b d (41 b c-35 a d) (b c-7 a d) \left (\frac {a \left (c+\frac {d}{x^2}\right )}{c \left (a+\frac {b}{x^2}\right )}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{2},\frac {b c-a d}{c \left (a+\frac {b}{x^2}\right ) x^2}\right )}{165 a^4 c (b c-a d) \left (a+\frac {b}{x^2}\right )^{7/4} \left (c+\frac {d}{x^2}\right )^{3/4} x^5} \] Output:

2/33*b*d*(-35*a*d+41*b*c)*(c+d/x^2)^(1/4)/a^3/c/(-a*d+b*c)/(a+b/x^2)^(7/4) 
/x^3-d*(c+d/x^2)^(1/4)/a/c/(a+b/x^2)^(11/4)/x+1/77*b*(-11*a*d+9*b*c)*(-a*d 
+13*b*c)*(c+d/x^2)^(1/4)/a^3/(-a*d+b*c)^2/(a+b/x^2)^(7/4)/x+1/231*b*(-11*a 
*d+5*b*c)*(-11*a*d+9*b*c)*(-a*d+13*b*c)*(c+d/x^2)^(1/4)/a^4/(-a*d+b*c)^3/( 
a+b/x^2)^(3/4)/x+1/11*b*(c+d/x^2)^(1/4)*(13*b*c-a*d+12*b*d/x^2)/a^2/(-a*d+ 
b*c)/(a+b/x^2)^(11/4)/x+(c+d/x^2)^(5/4)*x/a/c/(a+b/x^2)^(11/4)+1/462*(-11* 
a*d+9*b*c)*(-a*d+13*b*c)*(21*a^2*d^2-14*a*b*c*d+5*b^2*c^2)*(c*(a+b/x^2)/a/ 
(c+d/x^2))^(3/4)*(c+d/x^2)^(1/4)*hypergeom([1/2, 3/4],[3/2],-(-a*d+b*c)/a/ 
(c+d/x^2)/x^2)/a^4/c/(-a*d+b*c)^3/(a+b/x^2)^(3/4)/x+24/55*b^2*d^2*(a*(c+d/ 
x^2)/c/(a+b/x^2))^(3/4)*hypergeom([3/4, 5/2],[7/2],(-a*d+b*c)/c/(a+b/x^2)/ 
x^2)/a^3/(-a*d+b*c)/(a+b/x^2)^(7/4)/(c+d/x^2)^(3/4)/x^5+1/165*b*d*(-35*a*d 
+41*b*c)*(-7*a*d+b*c)*(a*(c+d/x^2)/c/(a+b/x^2))^(3/4)*hypergeom([3/4, 5/2] 
,[7/2],(-a*d+b*c)/c/(a+b/x^2)/x^2)/a^4/c/(-a*d+b*c)/(a+b/x^2)^(7/4)/(c+d/x 
^2)^(3/4)/x^5
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 4.05 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.35 \[ \int \frac {\sqrt [4]{c+\frac {d}{x^2}}}{\left (a+\frac {b}{x^2}\right )^{15/4}} \, dx=\frac {\sqrt [4]{c+\frac {d}{x^2}} \left (b+a x^2\right ) \left (77 x^4 \left (d+c x^2\right )+\frac {b \left (d+c x^2\right ) \left (117 b^3 c^2+154 a^3 d^2 x^2+11 a^2 b d \left (7 d-34 c x^2\right )+2 a b^2 c \left (-103 d+104 c x^2\right )\right )}{a^2 (b c-a d)^2}+\frac {\left (195 b^3 c^3-495 a b^2 c^2 d+385 a^2 b c d^2-77 a^3 d^3\right ) \left (b+a x^2\right )^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{4},\frac {1}{4},\frac {c \left (b+a x^2\right )}{b c-a d}\right )}{a^3 (b c-a d)^2 \sqrt [4]{\frac {a \left (d+c x^2\right )}{-b c+a d}}}\right )}{77 a c \left (a+\frac {b}{x^2}\right )^{15/4} x^7} \] Input:

Integrate[(c + d/x^2)^(1/4)/(a + b/x^2)^(15/4),x]
 

Output:

((c + d/x^2)^(1/4)*(b + a*x^2)*(77*x^4*(d + c*x^2) + (b*(d + c*x^2)*(117*b 
^3*c^2 + 154*a^3*d^2*x^2 + 11*a^2*b*d*(7*d - 34*c*x^2) + 2*a*b^2*c*(-103*d 
 + 104*c*x^2)))/(a^2*(b*c - a*d)^2) + ((195*b^3*c^3 - 495*a*b^2*c^2*d + 38 
5*a^2*b*c*d^2 - 77*a^3*d^3)*(b + a*x^2)^2*Hypergeometric2F1[-3/4, -1/4, 1/ 
4, (c*(b + a*x^2))/(b*c - a*d)])/(a^3*(b*c - a*d)^2*((a*(d + c*x^2))/(-(b* 
c) + a*d))^(1/4))))/(77*a*c*(a + b/x^2)^(15/4)*x^7)
 

Rubi [F]

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.

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

\(\Big \downarrow \) 899

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

\(\Big \downarrow \) 395

\(\displaystyle -\frac {\left (\frac {b}{a x^2}+1\right )^{3/4} \int \frac {\sqrt [4]{c+\frac {d}{x^2}} x^2}{\left (\frac {b}{a x^2}+1\right )^{15/4}}d\frac {1}{x}}{a^3 \left (a+\frac {b}{x^2}\right )^{3/4}}\)

\(\Big \downarrow \) 395

\(\displaystyle -\frac {\left (\frac {b}{a x^2}+1\right )^{3/4} \sqrt [4]{c+\frac {d}{x^2}} \int \frac {\sqrt [4]{\frac {d}{c x^2}+1} x^2}{\left (\frac {b}{a x^2}+1\right )^{15/4}}d\frac {1}{x}}{a^3 \left (a+\frac {b}{x^2}\right )^{3/4} \sqrt [4]{\frac {d}{c x^2}+1}}\)

\(\Big \downarrow \) 7299

\(\displaystyle -\frac {\left (\frac {b}{a x^2}+1\right )^{3/4} \sqrt [4]{c+\frac {d}{x^2}} \int \frac {\sqrt [4]{\frac {d}{c x^2}+1} x^2}{\left (\frac {b}{a x^2}+1\right )^{15/4}}d\frac {1}{x}}{a^3 \left (a+\frac {b}{x^2}\right )^{3/4} \sqrt [4]{\frac {d}{c x^2}+1}}\)

Input:

Int[(c + d/x^2)^(1/4)/(a + b/x^2)^(15/4),x]
 

Output:

$Aborted
 

Defintions of rubi rules used

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])
 

rule 899

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol 
] :> -Subst[Int[(a + b/x^n)^p*((c + d/x^n)^q/x^2), x], x, 1/x] /; FreeQ[{a, 
 b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]
 

rule 7299

Int[u_, x_] :> CannotIntegrate[u, x]
 

Maple [F]

\[\int \frac {\left (c +\frac {d}{x^{2}}\right )^{\frac {1}{4}}}{\left (a +\frac {b}{x^{2}}\right )^{\frac {15}{4}}}d x\]

Input:

int((c+1/x^2*d)^(1/4)/(a+b/x^2)^(15/4),x)
 

Output:

int((c+1/x^2*d)^(1/4)/(a+b/x^2)^(15/4),x)
 

Fricas [F]

\[ \int \frac {\sqrt [4]{c+\frac {d}{x^2}}}{\left (a+\frac {b}{x^2}\right )^{15/4}} \, dx=\int { \frac {{\left (c + \frac {d}{x^{2}}\right )}^{\frac {1}{4}}}{{\left (a + \frac {b}{x^{2}}\right )}^{\frac {15}{4}}} \,d x } \] Input:

integrate((c+d/x^2)^(1/4)/(a+b/x^2)^(15/4),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{c+\frac {d}{x^2}}}{\left (a+\frac {b}{x^2}\right )^{15/4}} \, dx=\text {Timed out} \] Input:

integrate((c+d/x**2)**(1/4)/(a+b/x**2)**(15/4),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sqrt [4]{c+\frac {d}{x^2}}}{\left (a+\frac {b}{x^2}\right )^{15/4}} \, dx=\int { \frac {{\left (c + \frac {d}{x^{2}}\right )}^{\frac {1}{4}}}{{\left (a + \frac {b}{x^{2}}\right )}^{\frac {15}{4}}} \,d x } \] Input:

integrate((c+d/x^2)^(1/4)/(a+b/x^2)^(15/4),x, algorithm="maxima")
 

Output:

integrate((c + d/x^2)^(1/4)/(a + b/x^2)^(15/4), x)
 

Giac [F]

\[ \int \frac {\sqrt [4]{c+\frac {d}{x^2}}}{\left (a+\frac {b}{x^2}\right )^{15/4}} \, dx=\int { \frac {{\left (c + \frac {d}{x^{2}}\right )}^{\frac {1}{4}}}{{\left (a + \frac {b}{x^{2}}\right )}^{\frac {15}{4}}} \,d x } \] Input:

integrate((c+d/x^2)^(1/4)/(a+b/x^2)^(15/4),x, algorithm="giac")
 

Output:

integrate((c + d/x^2)^(1/4)/(a + b/x^2)^(15/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{c+\frac {d}{x^2}}}{\left (a+\frac {b}{x^2}\right )^{15/4}} \, dx=\int \frac {{\left (c+\frac {d}{x^2}\right )}^{1/4}}{{\left (a+\frac {b}{x^2}\right )}^{15/4}} \,d x \] Input:

int((c + d/x^2)^(1/4)/(a + b/x^2)^(15/4),x)
 

Output:

int((c + d/x^2)^(1/4)/(a + b/x^2)^(15/4), x)
 

Reduce [F]

\[ \int \frac {\sqrt [4]{c+\frac {d}{x^2}}}{\left (a+\frac {b}{x^2}\right )^{15/4}} \, dx=\int \frac {\left (c \,x^{2}+d \right )^{\frac {1}{4}} x^{7}}{\left (a \,x^{2}+b \right )^{\frac {3}{4}} a^{3} x^{6}+3 \left (a \,x^{2}+b \right )^{\frac {3}{4}} a^{2} b \,x^{4}+3 \left (a \,x^{2}+b \right )^{\frac {3}{4}} a \,b^{2} x^{2}+\left (a \,x^{2}+b \right )^{\frac {3}{4}} b^{3}}d x \] Input:

int((c+d/x^2)^(1/4)/(a+b/x^2)^(15/4),x)
 

Output:

int(((c*x**2 + d)**(1/4)*x**7)/((a*x**2 + b)**(3/4)*a**3*x**6 + 3*(a*x**2 
+ b)**(3/4)*a**2*b*x**4 + 3*(a*x**2 + b)**(3/4)*a*b**2*x**2 + (a*x**2 + b) 
**(3/4)*b**3),x)