Integrand size = 26, antiderivative size = 294 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=-\frac {1}{3 a c x^3 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}-\frac {b (7 b c-a d)}{3 a^2 c (b c-a d) x \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}-\frac {d \left (7 b^2 c^2-2 a b c d+7 a^2 d^2\right ) \left (a+b x^2\right )^{3/4}}{3 a^2 c^2 (b c-a d)^2 x \sqrt [4]{c+d x^2}}+\frac {(b c+a d) \left (7 b^2 c^2-10 a b c d+7 a^2 d^2\right ) \left (a+b x^2\right )^{3/4} \sqrt [4]{\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )}{2 a^3 c^2 (b c-a d)^2 x \sqrt [4]{c+d x^2}} \] Output:
-1/3/a/c/x^3/(b*x^2+a)^(1/4)/(d*x^2+c)^(1/4)-1/3*b*(-a*d+7*b*c)/a^2/c/(-a* d+b*c)/x/(b*x^2+a)^(1/4)/(d*x^2+c)^(1/4)-1/3*d*(7*a^2*d^2-2*a*b*c*d+7*b^2* c^2)*(b*x^2+a)^(3/4)/a^2/c^2/(-a*d+b*c)^2/x/(d*x^2+c)^(1/4)+1/2*(a*d+b*c)* (7*a^2*d^2-10*a*b*c*d+7*b^2*c^2)*(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^3/c^2/(-a *d+b*c)^2/x/(d*x^2+c)^(1/4)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 11.19 (sec) , antiderivative size = 689, normalized size of antiderivative = 2.34 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\frac {-2 b d \left (7 b^3 c^3-3 a b^2 c^2 d-3 a^2 b c d^2+7 a^3 d^3\right ) x^6 \sqrt [4]{1+\frac {b x^2}{a}} \sqrt [4]{1+\frac {d x^2}{c}} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},\frac {1}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+\frac {3 a c \left (-21 b^4 c^3 x^4 \left (c+2 d x^2\right )+a^4 d^2 \left (4 c^2-14 c d x^2-21 d^2 x^4\right )+2 a b^3 c^2 x^2 \left (-7 c^2+8 c d x^2+9 d^2 x^4\right )+2 a^2 b^2 c \left (2 c^3+7 c^2 d x^2+5 c d^2 x^4+9 d^3 x^6\right )-2 a^3 b d \left (4 c^3-7 c^2 d x^2-8 c d^2 x^4+21 d^3 x^6\right )\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},\frac {1}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (21 b^4 c^3 x^4 \left (c+d x^2\right )+a b^3 c^2 x^2 \left (7 c^2-2 c d x^2-9 d^2 x^4\right )+a^4 d^2 \left (-2 c^2+7 c d x^2+21 d^2 x^4\right )-a^2 b^2 c \left (2 c^3+7 c^2 d x^2+14 c d^2 x^4+9 d^3 x^6\right )+a^3 b d \left (4 c^3-7 c^2 d x^2-2 c d^2 x^4+21 d^3 x^6\right )\right ) \left (a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},\frac {5}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},\frac {1}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}{-6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},\frac {1}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},\frac {5}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},\frac {1}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}}{6 a^3 c^3 (b c-a d)^2 x^3 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Input:
Integrate[1/(x^4*(a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)),x]
Output:
(-2*b*d*(7*b^3*c^3 - 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + 7*a^3*d^3)*x^6*(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)] + (3*a*c*(-21*b^4*c^3*x^4*(c + 2*d*x^2) + a^4*d^2*(4* c^2 - 14*c*d*x^2 - 21*d^2*x^4) + 2*a*b^3*c^2*x^2*(-7*c^2 + 8*c*d*x^2 + 9*d ^2*x^4) + 2*a^2*b^2*c*(2*c^3 + 7*c^2*d*x^2 + 5*c*d^2*x^4 + 9*d^3*x^6) - 2* a^3*b*d*(4*c^3 - 7*c^2*d*x^2 - 8*c*d^2*x^4 + 21*d^3*x^6))*AppellF1[1/2, 1/ 4, 1/4, 3/2, -((b*x^2)/a), -((d*x^2)/c)] + x^2*(21*b^4*c^3*x^4*(c + d*x^2) + a*b^3*c^2*x^2*(7*c^2 - 2*c*d*x^2 - 9*d^2*x^4) + a^4*d^2*(-2*c^2 + 7*c*d *x^2 + 21*d^2*x^4) - a^2*b^2*c*(2*c^3 + 7*c^2*d*x^2 + 14*c*d^2*x^4 + 9*d^3 *x^6) + a^3*b*d*(4*c^3 - 7*c^2*d*x^2 - 2*c*d^2*x^4 + 21*d^3*x^6))*(a*d*App ellF1[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*AppellF1[1/2, 1/4, 1/ 4, 3/2, -((b*x^2)/a), -((d*x^2)/c)] + x^2*(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^3*c^3*(b*c - a*d)^2*x^3*(a + b*x^2)^(1/4)*(c + d*x^2)^(1/4))
Leaf count is larger than twice the leaf count of optimal. \(846\) vs. \(2(294)=588\).
Time = 5.91 (sec) , antiderivative size = 846, normalized size of antiderivative = 2.88, 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.
| \(\displaystyle \int \frac {1}{x^4 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx\) |
| \(\Big \downarrow \) 395 |
| \(\displaystyle \frac {\sqrt [4]{\frac {b x^2}{a}+1} \int \frac {1}{x^4 \left (\frac {b x^2}{a}+1\right )^{5/4} \left (d x^2+c\right )^{5/4}}dx}{a \sqrt [4]{a+b x^2}}\) |
| \(\Big \downarrow \) 395 |
| \(\displaystyle \frac {\sqrt [4]{\frac {b x^2}{a}+1} \sqrt [4]{\frac {d x^2}{c}+1} \int \frac {1}{x^4 \left (\frac {b x^2}{a}+1\right )^{5/4} \left (\frac {d x^2}{c}+1\right )^{5/4}}dx}{a c \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}\) |
| \(\Big \downarrow \) 394 |
| \(\displaystyle -\frac {-48 b c d^3 \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 ) x^8+44 a d^4 \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 ) x^8-44 b c d^3 \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 ) x^8-48 a c d^3 \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 ) x^6-72 b c^2 d^2 \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 ) x^6+78 a c d^3 \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 ) x^6-78 b c^2 d^2 \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 ) x^6-72 a c^2 d^2 \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 ) x^4-18 b c^3 d \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 ) x^4+33 a c^2 d^2 \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 ) x^4-33 b c^3 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 ) x^4+3 b c^4 \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 ) x^2-18 a c^3 d \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 ) x^2+b c^4 \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 ) x^2-a c^3 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 ) x^2-6 (b c-a d) \left (d x^2+c\right )^2 \left (4 d x^2+c\right ) \, _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 ) x^2-4 (b c-a d) \left (d x^2+c\right )^3 \, _4F_3\left (2,2,2,\frac {9}{4};1,1,\frac {7}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^2+3 a c^4 \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 )}{9 a c^5 x^3 \left (b x^2+a\right )^{5/4} \left (\frac {b x^2}{a}+1\right ) \sqrt [4]{d x^2+c}}\) |
Input:
Int[1/(x^4*(a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)),x]
Output:
-1/9*(3*a*c^4*Hypergeometric2F1[1, 5/4, 5/2, ((b*c - a*d)*x^2)/(c*(a + b*x ^2))] + 3*b*c^4*x^2*Hypergeometric2F1[1, 5/4, 5/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 18*a*c^3*d*x^2*Hypergeometric2F1[1, 5/4, 5/2, ((b*c - a*d)*x ^2)/(c*(a + b*x^2))] - 18*b*c^3*d*x^4*Hypergeometric2F1[1, 5/4, 5/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 72*a*c^2*d^2*x^4*Hypergeometric2F1[1, 5/4, 5/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 72*b*c^2*d^2*x^6*Hypergeometric 2F1[1, 5/4, 5/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 48*a*c*d^3*x^6*Hyper geometric2F1[1, 5/4, 5/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 48*b*c*d^3* x^8*Hypergeometric2F1[1, 5/4, 5/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + b* c^4*x^2*Hypergeometric2F1[2, 9/4, 7/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] - a*c^3*d*x^2*Hypergeometric2F1[2, 9/4, 7/2, ((b*c - a*d)*x^2)/(c*(a + b*x ^2))] - 33*b*c^3*d*x^4*Hypergeometric2F1[2, 9/4, 7/2, ((b*c - a*d)*x^2)/(c *(a + b*x^2))] + 33*a*c^2*d^2*x^4*Hypergeometric2F1[2, 9/4, 7/2, ((b*c - a *d)*x^2)/(c*(a + b*x^2))] - 78*b*c^2*d^2*x^6*Hypergeometric2F1[2, 9/4, 7/2 , ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 78*a*c*d^3*x^6*Hypergeometric2F1[2, 9/4, 7/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 44*b*c*d^3*x^8*Hypergeomet ric2F1[2, 9/4, 7/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 44*a*d^4*x^8*Hype rgeometric2F1[2, 9/4, 7/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 6*(b*c - a *d)*x^2*(c + d*x^2)^2*(c + 4*d*x^2)*HypergeometricPFQ[{2, 2, 9/4}, {1, 7/2 }, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 4*(b*c - a*d)*x^2*(c + d*x^2)^3...
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])
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])
\[\int \frac {1}{x^{4} \left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (x^{2} d +c \right )^{\frac {5}{4}}}d x\]
Input:
int(1/x^4/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x)
Output:
int(1/x^4/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x)
\[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {5}{4}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(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^12 + 2*(b^2*c*d + a*b*d^2)*x^10 + (b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^8 + a^2*c^2*x^4 + 2*(a*b *c^2 + a^2*c*d)*x^6), x)
\[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int \frac {1}{x^{4} \left (a + b x^{2}\right )^{\frac {5}{4}} \left (c + d x^{2}\right )^{\frac {5}{4}}}\, dx \] Input:
integrate(1/x**4/(b*x**2+a)**(5/4)/(d*x**2+c)**(5/4),x)
Output:
Integral(1/(x**4*(a + b*x**2)**(5/4)*(c + d*x**2)**(5/4)), x)
\[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {5}{4}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(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^4), x)
\[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {5}{4}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(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^4), x)
Timed out. \[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int \frac {1}{x^4\,{\left (b\,x^2+a\right )}^{5/4}\,{\left (d\,x^2+c\right )}^{5/4}} \,d x \] Input:
int(1/(x^4*(a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)),x)
Output:
int(1/(x^4*(a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)), x)
\[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int \frac {1}{\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} a c \,x^{4}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} a d \,x^{6}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b c \,x^{6}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b d \,x^{8}}d x \] Input:
int(1/x^4/(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**4 + (c + d*x**2)**(1 /4)*(a + b*x**2)**(1/4)*a*d*x**6 + (c + d*x**2)**(1/4)*(a + b*x**2)**(1/4) *b*c*x**6 + (c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*b*d*x**8),x)