\(\int \frac {1}{x^2 (a+b x^2)^{5/4} (c+d x^2)^{5/4}} \, dx\) [1586]

Optimal result

Integrand size = 26, antiderivative size = 211 \[ \int \frac {1}{x^2 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=-\frac {1}{a c x \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}-\frac {b (3 b c-a d) x}{a^2 c (b c-a d) \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}+\frac {\left (3 b^2 c^2-2 a b c d+3 a^2 d^2\right ) x \sqrt [4]{\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {(b c-a d) x^2}{a \left (c+d x^2\right )}\right )}{2 a^2 c^2 (b c-a d) \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 10.94 (sec) , antiderivative size = 532, normalized size of antiderivative = 2.52 \[ \int \frac {1}{x^2 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\frac {2 b d \left (3 b^2 c^2-2 a b c d+3 a^2 d^2\right ) x^3 \sqrt [4]{1+\frac {b x^2}{a}} \sqrt [4]{1+\frac {d x^2}{c}} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},\frac {1}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+\frac {9 a c \left (3 b^3 c^2 x^2 \left (c+2 d x^2\right )+a^3 d^2 \left (2 c+3 d x^2\right )+a b^2 c \left (2 c^2-3 c d x^2-4 d^2 x^4\right )+a^2 b d \left (-4 c^2-3 c d x^2+6 d^2 x^4\right )\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},\frac {1}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )-3 x^2 \left (3 b^3 c^2 x^2 \left (c+d x^2\right )+a^3 d^2 \left (c+3 d x^2\right )+a b^2 c \left (c^2-c d x^2-2 d^2 x^4\right )+a^2 b d \left (-2 c^2-c d x^2+3 d^2 x^4\right )\right ) \left (a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},\frac {5}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},\frac {1}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}{-6 a c x \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},\frac {1}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^3 \left (a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},\frac {5}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},\frac {1}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}}{3 a^2 c^2 (b c-a d)^2 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Input:

Integrate[1/(x^2*(a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)),x]
 

Output:

(2*b*d*(3*b^2*c^2 - 2*a*b*c*d + 3*a^2*d^2)*x^3*(1 + (b*x^2)/a)^(1/4)*(1 + 
(d*x^2)/c)^(1/4)*AppellF1[3/2, 1/4, 1/4, 5/2, -((b*x^2)/a), -((d*x^2)/c)] 
+ (9*a*c*(3*b^3*c^2*x^2*(c + 2*d*x^2) + a^3*d^2*(2*c + 3*d*x^2) + a*b^2*c* 
(2*c^2 - 3*c*d*x^2 - 4*d^2*x^4) + a^2*b*d*(-4*c^2 - 3*c*d*x^2 + 6*d^2*x^4) 
)*AppellF1[1/2, 1/4, 1/4, 3/2, -((b*x^2)/a), -((d*x^2)/c)] - 3*x^2*(3*b^3* 
c^2*x^2*(c + d*x^2) + a^3*d^2*(c + 3*d*x^2) + a*b^2*c*(c^2 - c*d*x^2 - 2*d 
^2*x^4) + a^2*b*d*(-2*c^2 - c*d*x^2 + 3*d^2*x^4))*(a*d*AppellF1[3/2, 1/4, 
5/4, 5/2, -((b*x^2)/a), -((d*x^2)/c)] + b*c*AppellF1[3/2, 5/4, 1/4, 5/2, - 
((b*x^2)/a), -((d*x^2)/c)]))/(-6*a*c*x*AppellF1[1/2, 1/4, 1/4, 3/2, -((b*x 
^2)/a), -((d*x^2)/c)] + x^3*(a*d*AppellF1[3/2, 1/4, 5/4, 5/2, -((b*x^2)/a) 
, -((d*x^2)/c)] + b*c*AppellF1[3/2, 5/4, 1/4, 5/2, -((b*x^2)/a), -((d*x^2) 
/c)])))/(3*a^2*c^2*(b*c - a*d)^2*(a + b*x^2)^(1/4)*(c + d*x^2)^(1/4))
 

Rubi [A] (warning: unable to verify)

Time = 4.41 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.11, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {395, 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.

\(\displaystyle \int \frac {1}{x^2 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx\)

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 394

\(\displaystyle -\frac {2 x^2 \left (c+d x^2\right )^2 (b c-a d) \, _3F_2\left (2,2,\frac {9}{4};1,\frac {7}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )+2 x^2 \left (2 c^2+5 c d x^2+3 d^2 x^4\right ) (b c-a d) \operatorname {Hypergeometric2F1}\left (2,\frac {9}{4},\frac {7}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )+c \left (a+b x^2\right ) \left (3 c^2+12 c d x^2+8 d^2 x^4\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {5}{4},\frac {5}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )}{3 a c^4 x \left (a+b x^2\right )^{5/4} \left (\frac {b x^2}{a}+1\right ) \sqrt [4]{c+d x^2}}\)

Input:

Int[1/(x^2*(a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)),x]
 

Output:

-1/3*(c*(a + b*x^2)*(3*c^2 + 12*c*d*x^2 + 8*d^2*x^4)*Hypergeometric2F1[1, 
5/4, 5/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 2*(b*c - a*d)*x^2*(2*c^2 + 
5*c*d*x^2 + 3*d^2*x^4)*Hypergeometric2F1[2, 9/4, 7/2, ((b*c - a*d)*x^2)/(c 
*(a + b*x^2))] + 2*(b*c - a*d)*x^2*(c + d*x^2)^2*HypergeometricPFQ[{2, 2, 
9/4}, {1, 7/2}, ((b*c - a*d)*x^2)/(c*(a + b*x^2))])/(a*c^4*x*(a + b*x^2)^( 
5/4)*(1 + (b*x^2)/a)*(c + d*x^2)^(1/4))
 

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]

\[\int \frac {1}{x^{2} \left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (x^{2} d +c \right )^{\frac {5}{4}}}d x\]

Input:

int(1/x^2/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x)
 

Output:

int(1/x^2/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x)
 

Fricas [F]

\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {5}{4}} x^{2}} \,d x } \] Input:

integrate(1/x^2/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(3/4)*(d*x^2 + c)^(3/4)/(b^2*d^2*x^10 + 2*(b^2*c*d + 
a*b*d^2)*x^8 + (b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^6 + a^2*c^2*x^2 + 2*(a*b* 
c^2 + a^2*c*d)*x^4), x)
 

Sympy [F]

\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int \frac {1}{x^{2} \left (a + b x^{2}\right )^{\frac {5}{4}} \left (c + d x^{2}\right )^{\frac {5}{4}}}\, dx \] Input:

integrate(1/x**2/(b*x**2+a)**(5/4)/(d*x**2+c)**(5/4),x)
 

Output:

Integral(1/(x**2*(a + b*x**2)**(5/4)*(c + d*x**2)**(5/4)), x)
 

Maxima [F]

\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {5}{4}} x^{2}} \,d x } \] Input:

integrate(1/x^2/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^2 + a)^(5/4)*(d*x^2 + c)^(5/4)*x^2), x)
 

Giac [F]

\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {5}{4}} x^{2}} \,d x } \] Input:

integrate(1/x^2/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x, algorithm="giac")
 

Output:

integrate(1/((b*x^2 + a)^(5/4)*(d*x^2 + c)^(5/4)*x^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int \frac {1}{x^2\,{\left (b\,x^2+a\right )}^{5/4}\,{\left (d\,x^2+c\right )}^{5/4}} \,d x \] Input:

int(1/(x^2*(a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)),x)
 

Output:

int(1/(x^2*(a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)), x)
 

Reduce [F]

\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int \frac {1}{\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} a c \,x^{2}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} a d \,x^{4}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b c \,x^{4}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b d \,x^{6}}d x \] Input:

int(1/x^2/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x)
 

Output:

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