\(\int \frac {1}{x^2 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \, dx\) [1560]

Optimal result

Integrand size = 26, antiderivative size = 89 \[ \int \frac {1}{x^2 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \, dx=-\frac {\left (a+b x^2\right )^{3/4} \sqrt [4]{\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )}{a x \sqrt [4]{c+d x^2}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 2.01 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \, dx=-\frac {\sqrt [4]{1+\frac {b x^2}{a}} \left (1+\frac {d x^2}{c}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},\frac {(-b c+a d) x^2}{a \left (c+d x^2\right )}\right )}{x \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Input:

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

Output:

-(((1 + (b*x^2)/a)^(1/4)*(1 + (d*x^2)/c)^(3/4)*Hypergeometric2F1[-1/2, 1/4 
, 1/2, ((-(b*c) + a*d)*x^2)/(a*(c + d*x^2))])/(x*(a + b*x^2)^(1/4)*(c + d* 
x^2)^(1/4)))
 

Rubi [A] (warning: unable to verify)

Time = 0.25 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.07, 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.

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 394

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

Input:

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

Output:

-(((1 + (b*x^2)/a)^(1/4)*(1 + (d*x^2)/c)^(3/4)*Hypergeometric2F1[-1/2, 1/4 
, 1/2, -((c*((b*x^2)/a - (d*x^2)/c))/(c + d*x^2))])/(x*(a + b*x^2)^(1/4)*( 
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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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