Integrand size = 23, antiderivative size = 212 \[ \int \frac {1}{\left (a+b x^2\right )^{7/4} \left (c+d x^2\right )^{7/4}} \, dx=\frac {2 b x}{3 a (b c-a d) \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}}-\frac {2 b (b c-7 a d) x \sqrt [4]{a+b x^2}}{3 a^2 (b c-a d)^2 \left (c+d x^2\right )^{3/4}}+\frac {\left (b^2 c^2-6 a b c d+a^2 d^2\right ) x \sqrt [4]{a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {(b c-a d) x^2}{a \left (c+d x^2\right )}\right )}{a^2 c (b c-a d)^2 \sqrt [4]{\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \left (c+d x^2\right )^{3/4}} \] Output:
2/3*b*x/a/(-a*d+b*c)/(b*x^2+a)^(3/4)/(d*x^2+c)^(3/4)-2/3*b*(-7*a*d+b*c)*x* (b*x^2+a)^(1/4)/a^2/(-a*d+b*c)^2/(d*x^2+c)^(3/4)+(a^2*d^2-6*a*b*c*d+b^2*c^ 2)*x*(b*x^2+a)^(1/4)*hypergeom([-1/4, 1/2],[3/2],-(-a*d+b*c)*x^2/a/(d*x^2+ c))/a^2/c/(-a*d+b*c)^2/(c*(b*x^2+a)/a/(d*x^2+c))^(1/4)/(d*x^2+c)^(3/4)
Time = 10.08 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\left (a+b x^2\right )^{7/4} \left (c+d x^2\right )^{7/4}} \, dx=\frac {x \operatorname {Gamma}\left (\frac {3}{4}\right ) \left (c \left (a+b x^2\right ) \left (15 c^2+20 c d x^2+8 d^2 x^4\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {7}{4},\frac {7}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )+2 (b c-a d) x^2 \left (4 c^2+7 c d x^2+3 d^2 x^4\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {11}{4},\frac {9}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )+2 (b c-a d) x^2 \left (c+d x^2\right )^2 \, _3F_2\left (2,2,\frac {11}{4};1,\frac {9}{2};\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )\right )}{20 c^4 \left (a+b x^2\right )^{11/4} \left (c+d x^2\right )^{3/4} \operatorname {Gamma}\left (\frac {7}{4}\right )} \] Input:
Integrate[1/((a + b*x^2)^(7/4)*(c + d*x^2)^(7/4)),x]
Output:
(x*Gamma[3/4]*(c*(a + b*x^2)*(15*c^2 + 20*c*d*x^2 + 8*d^2*x^4)*Hypergeomet ric2F1[1, 7/4, 7/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 2*(b*c - a*d)*x^2 *(4*c^2 + 7*c*d*x^2 + 3*d^2*x^4)*Hypergeometric2F1[2, 11/4, 9/2, ((b*c - a *d)*x^2)/(c*(a + b*x^2))] + 2*(b*c - a*d)*x^2*(c + d*x^2)^2*Hypergeometric PFQ[{2, 2, 11/4}, {1, 9/2}, ((b*c - a*d)*x^2)/(c*(a + b*x^2))]))/(20*c^4*( a + b*x^2)^(11/4)*(c + d*x^2)^(3/4)*Gamma[7/4])
Time = 4.57 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.14, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {334, 334, 333}
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.
| \(\displaystyle \int \frac {1}{\left (a+b x^2\right )^{7/4} \left (c+d x^2\right )^{7/4}} \, dx\) |
| \(\Big \downarrow \) 334 |
| \(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{3/4} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{7/4} \left (d x^2+c\right )^{7/4}}dx}{a \left (a+b x^2\right )^{3/4}}\) |
| \(\Big \downarrow \) 334 |
| \(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{3/4} \left (\frac {d x^2}{c}+1\right )^{3/4} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{7/4} \left (\frac {d x^2}{c}+1\right )^{7/4}}dx}{a c \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}}\) |
| \(\Big \downarrow \) 333 |
| \(\displaystyle \frac {x \operatorname {Gamma}\left (\frac {3}{4}\right ) \left (2 x^2 \left (c+d x^2\right )^2 (b c-a d) \, _3F_2\left (2,2,\frac {11}{4};1,\frac {9}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )+2 x^2 \left (4 c^2+7 c d x^2+3 d^2 x^4\right ) (b c-a d) \operatorname {Hypergeometric2F1}\left (2,\frac {11}{4},\frac {9}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )+c \left (a+b x^2\right ) \left (15 c^2+20 c d x^2+8 d^2 x^4\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {7}{4},\frac {7}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )\right )}{20 a c^4 \operatorname {Gamma}\left (\frac {7}{4}\right ) \left (a+b x^2\right )^{7/4} \left (\frac {b x^2}{a}+1\right ) \left (c+d x^2\right )^{3/4}}\) |
Input:
Int[1/((a + b*x^2)^(7/4)*(c + d*x^2)^(7/4)),x]
Output:
(x*Gamma[3/4]*(c*(a + b*x^2)*(15*c^2 + 20*c*d*x^2 + 8*d^2*x^4)*Hypergeomet ric2F1[1, 7/4, 7/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 2*(b*c - a*d)*x^2 *(4*c^2 + 7*c*d*x^2 + 3*d^2*x^4)*Hypergeometric2F1[2, 11/4, 9/2, ((b*c - a *d)*x^2)/(c*(a + b*x^2))] + 2*(b*c - a*d)*x^2*(c + d*x^2)^2*Hypergeometric PFQ[{2, 2, 11/4}, {1, 9/2}, ((b*c - a*d)*x^2)/(c*(a + b*x^2))]))/(20*a*c^4 *(a + b*x^2)^(7/4)*(1 + (b*x^2)/a)*(c + d*x^2)^(3/4)*Gamma[7/4])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[ (1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && !(IntegerQ[p] || GtQ[a, 0])
\[\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {7}{4}} \left (x^{2} d +c \right )^{\frac {7}{4}}}d x\]
Input:
int(1/(b*x^2+a)^(7/4)/(d*x^2+c)^(7/4),x)
Output:
int(1/(b*x^2+a)^(7/4)/(d*x^2+c)^(7/4),x)
\[ \int \frac {1}{\left (a+b x^2\right )^{7/4} \left (c+d x^2\right )^{7/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {7}{4}} {\left (d x^{2} + c\right )}^{\frac {7}{4}}} \,d x } \] Input:
integrate(1/(b*x^2+a)^(7/4)/(d*x^2+c)^(7/4),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(1/4)*(d*x^2 + c)^(1/4)/(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)
\[ \int \frac {1}{\left (a+b x^2\right )^{7/4} \left (c+d x^2\right )^{7/4}} \, dx=\int \frac {1}{\left (a + b x^{2}\right )^{\frac {7}{4}} \left (c + d x^{2}\right )^{\frac {7}{4}}}\, dx \] Input:
integrate(1/(b*x**2+a)**(7/4)/(d*x**2+c)**(7/4),x)
Output:
Integral(1/((a + b*x**2)**(7/4)*(c + d*x**2)**(7/4)), x)
\[ \int \frac {1}{\left (a+b x^2\right )^{7/4} \left (c+d x^2\right )^{7/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {7}{4}} {\left (d x^{2} + c\right )}^{\frac {7}{4}}} \,d x } \] Input:
integrate(1/(b*x^2+a)^(7/4)/(d*x^2+c)^(7/4),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 + a)^(7/4)*(d*x^2 + c)^(7/4)), x)
\[ \int \frac {1}{\left (a+b x^2\right )^{7/4} \left (c+d x^2\right )^{7/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {7}{4}} {\left (d x^{2} + c\right )}^{\frac {7}{4}}} \,d x } \] Input:
integrate(1/(b*x^2+a)^(7/4)/(d*x^2+c)^(7/4),x, algorithm="giac")
Output:
integrate(1/((b*x^2 + a)^(7/4)*(d*x^2 + c)^(7/4)), x)
Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^{7/4} \left (c+d x^2\right )^{7/4}} \, dx=\int \frac {1}{{\left (b\,x^2+a\right )}^{7/4}\,{\left (d\,x^2+c\right )}^{7/4}} \,d x \] Input:
int(1/((a + b*x^2)^(7/4)*(c + d*x^2)^(7/4)),x)
Output:
int(1/((a + b*x^2)^(7/4)*(c + d*x^2)^(7/4)), x)
\[ \int \frac {1}{\left (a+b x^2\right )^{7/4} \left (c+d x^2\right )^{7/4}} \, dx=\int \frac {1}{\left (d \,x^{2}+c \right )^{\frac {3}{4}} \left (b \,x^{2}+a \right )^{\frac {3}{4}} a c +\left (d \,x^{2}+c \right )^{\frac {3}{4}} \left (b \,x^{2}+a \right )^{\frac {3}{4}} a d \,x^{2}+\left (d \,x^{2}+c \right )^{\frac {3}{4}} \left (b \,x^{2}+a \right )^{\frac {3}{4}} b c \,x^{2}+\left (d \,x^{2}+c \right )^{\frac {3}{4}} \left (b \,x^{2}+a \right )^{\frac {3}{4}} b d \,x^{4}}d x \] Input:
int(1/(b*x^2+a)^(7/4)/(d*x^2+c)^(7/4),x)
Output:
int(1/((c + d*x**2)**(3/4)*(a + b*x**2)**(3/4)*a*c + (c + d*x**2)**(3/4)*( a + b*x**2)**(3/4)*a*d*x**2 + (c + d*x**2)**(3/4)*(a + b*x**2)**(3/4)*b*c* x**2 + (c + d*x**2)**(3/4)*(a + b*x**2)**(3/4)*b*d*x**4),x)