Integrand size = 23, antiderivative size = 148 \[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\frac {2 b x}{a (b c-a d) \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}-\frac {(b c+a d) x \sqrt [4]{\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {(b c-a d) x^2}{a \left (c+d x^2\right )}\right )}{a c (b c-a d) \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Output:
2*b*x/a/(-a*d+b*c)/(b*x^2+a)^(1/4)/(d*x^2+c)^(1/4)-(a*d+b*c)*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/c/(-a*d+b*c)/(b*x^2+a)^(1/4)/(d*x^2+c)^(1/4)
Time = 0.03 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\frac {x \left (c \left (3 c+2 d x^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {5}{4},\frac {5}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )+\frac {(b c-a d) x^2 \left (c+d x^2\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {9}{4},\frac {7}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )}{a+b x^2}\right )}{3 c^3 \left (a+b x^2\right )^{5/4} \sqrt [4]{c+d x^2}} \] Input:
Integrate[1/((a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)),x]
Output:
(x*(c*(3*c + 2*d*x^2)*Hypergeometric2F1[1, 5/4, 5/2, ((b*c - a*d)*x^2)/(c* (a + b*x^2))] + ((b*c - a*d)*x^2*(c + d*x^2)*Hypergeometric2F1[2, 9/4, 7/2 , ((b*c - a*d)*x^2)/(c*(a + b*x^2))])/(a + b*x^2)))/(3*c^3*(a + b*x^2)^(5/ 4)*(c + d*x^2)^(1/4))
Time = 0.24 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {296, 294}
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.
\(\displaystyle \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 296 |
\(\displaystyle \frac {2 b x}{a \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2} (b c-a d)}-\frac {(a d+b c) \int \frac {1}{\sqrt [4]{b x^2+a} \left (d x^2+c\right )^{5/4}}dx}{a (b c-a d)}\) |
\(\Big \downarrow \) 294 |
\(\displaystyle \frac {2 b x}{a \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2} (b c-a d)}-\frac {x (a d+b c) \sqrt [4]{\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {(b c-a d) x^2}{a \left (d x^2+c\right )}\right )}{a c \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2} (b c-a d)}\) |
Input:
Int[1/((a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)),x]
Output:
(2*b*x)/(a*(b*c - a*d)*(a + b*x^2)^(1/4)*(c + d*x^2)^(1/4)) - ((b*c + a*d) *x*((c*(a + b*x^2))/(a*(c + d*x^2)))^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2 , -(((b*c - a*d)*x^2)/(a*(c + d*x^2)))])/(a*c*(b*c - a*d)*(a + b*x^2)^(1/4 )*(c + d*x^2)^(1/4))
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[x*((a + b*x^2)^p/(c*(c*((a + b*x^2)/(a*(c + d*x^2))))^p*(c + d*x^2)^(1/2 + p)))*Hypergeometric2F1[1/2, -p, 3/2, (-(b*c - a*d))*(x^2/(a*(c + d*x^2))) ], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[2*(p + q + 1) + 1, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d)) Int[ (a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] || !LtQ[q, -1 ]) && NeQ[p, -1]
\[\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (x^{2} d +c \right )^{\frac {5}{4}}}d x\]
Input:
int(1/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x)
Output:
int(1/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x)
\[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate(1/(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^8 + 2*(b^2*c*d + a *b*d^2)*x^6 + (b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^4 + a^2*c^2 + 2*(a*b*c^2 + a^2*c*d)*x^2), x)
\[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int \frac {1}{\left (a + b x^{2}\right )^{\frac {5}{4}} \left (c + d x^{2}\right )^{\frac {5}{4}}}\, dx \] Input:
integrate(1/(b*x**2+a)**(5/4)/(d*x**2+c)**(5/4),x)
Output:
Integral(1/((a + b*x**2)**(5/4)*(c + d*x**2)**(5/4)), x)
\[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate(1/(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)
\[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate(1/(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)
Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int \frac {1}{{\left (b\,x^2+a\right )}^{5/4}\,{\left (d\,x^2+c\right )}^{5/4}} \,d x \] Input:
int(1/((a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)),x)
Output:
int(1/((a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)), x)
\[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int \frac {1}{\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} a c +\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} a d \,x^{2}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b c \,x^{2}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b d \,x^{4}}d x \] Input:
int(1/(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 + (c + d*x**2)**(1/4)*( a + b*x**2)**(1/4)*a*d*x**2 + (c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*b*c* x**2 + (c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*b*d*x**4),x)