\(\int \frac {1}{x^6 (a+b x^2)^{3/4} (c+d x^2)^{3/4}} \, dx\) [1575]

Optimal result

Integrand size = 26, antiderivative size = 259 \[ \int \frac {1}{x^6 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=-\frac {\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}{5 a c x^5}+\frac {3 (b c+a d) \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}{10 a^2 c^2 x^3}-\frac {\left (15 b^2 c^2+14 a b c d+15 a^2 d^2\right ) \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}{20 a^3 c^3 x}-\frac {(b c+a d) \left (3 b^2 c^2-2 a b c d+3 a^2 d^2\right ) x \left (\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}\right )^{3/4} \sqrt [4]{c+d x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\frac {(b c-a d) x^2}{a \left (c+d x^2\right )}\right )}{8 a^3 c^4 \left (a+b x^2\right )^{3/4}} \] Output:

-1/5*(b*x^2+a)^(1/4)*(d*x^2+c)^(1/4)/a/c/x^5+3/10*(a*d+b*c)*(b*x^2+a)^(1/4 
)*(d*x^2+c)^(1/4)/a^2/c^2/x^3-1/20*(15*a^2*d^2+14*a*b*c*d+15*b^2*c^2)*(b*x 
^2+a)^(1/4)*(d*x^2+c)^(1/4)/a^3/c^3/x-1/8*(a*d+b*c)*(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))^(3/4)*(d*x^2+c)^(1/4)*hypergeom([1/2 
, 3/4],[3/2],-(-a*d+b*c)*x^2/a/(d*x^2+c))/a^3/c^4/(b*x^2+a)^(3/4)
 

Mathematica [A] (verified)

Time = 11.12 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^6 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=-\frac {\sqrt [4]{c+d x^2} \left (2 c \left (a+b x^2\right ) \left (15 b^2 c^2 x^4+2 a b c x^2 \left (-3 c+7 d x^2\right )+a^2 \left (4 c^2-6 c d x^2+15 d^2 x^4\right )\right )+5 \left (3 b^3 c^3+a b^2 c^2 d+a^2 b c d^2+3 a^3 d^3\right ) x^6 \left (\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\frac {(-b c+a d) x^2}{a \left (c+d x^2\right )}\right )\right )}{40 a^3 c^4 x^5 \left (a+b x^2\right )^{3/4}} \] Input:

Integrate[1/(x^6*(a + b*x^2)^(3/4)*(c + d*x^2)^(3/4)),x]
 

Output:

-1/40*((c + d*x^2)^(1/4)*(2*c*(a + b*x^2)*(15*b^2*c^2*x^4 + 2*a*b*c*x^2*(- 
3*c + 7*d*x^2) + a^2*(4*c^2 - 6*c*d*x^2 + 15*d^2*x^4)) + 5*(3*b^3*c^3 + a* 
b^2*c^2*d + a^2*b*c*d^2 + 3*a^3*d^3)*x^6*((c*(a + b*x^2))/(a*(c + d*x^2))) 
^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, ((-(b*c) + a*d)*x^2)/(a*(c + d*x^2 
))]))/(a^3*c^4*x^5*(a + b*x^2)^(3/4))
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(841\) vs. \(2(259)=518\).

Time = 4.88 (sec) , antiderivative size = 841, normalized size of antiderivative = 3.25, 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.

\(\displaystyle \int \frac {1}{x^6 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx\)

\(\Big \downarrow \) 395

\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{3/4} \int \frac {1}{x^6 \left (\frac {b x^2}{a}+1\right )^{3/4} \left (d x^2+c\right )^{3/4}}dx}{\left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 395

\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{3/4} \left (\frac {d x^2}{c}+1\right )^{3/4} \int \frac {1}{x^6 \left (\frac {b x^2}{a}+1\right )^{3/4} \left (\frac {d x^2}{c}+1\right )^{3/4}}dx}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}}\)

\(\Big \downarrow \) 394

\(\displaystyle -\frac {\left (\frac {d x^2}{c}+1\right ) \left (48 b c d^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {3}{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 (\frac {7}{4},2,\frac {5}{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 (\frac {7}{4},2,\frac {5}{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 (\frac {3}{4},1,\frac {3}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^6+24 b c^2 d^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {3}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^6-42 a c d^3 \operatorname {Hypergeometric2F1}\left (\frac {7}{4},2,\frac {5}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^6+42 b c^2 d^2 \operatorname {Hypergeometric2F1}\left (\frac {7}{4},2,\frac {5}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^6+24 a c^2 d^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {3}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^4-6 b c^3 d \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {3}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^4+3 a c^2 d^2 \operatorname {Hypergeometric2F1}\left (\frac {7}{4},2,\frac {5}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^4-3 b c^3 d \operatorname {Hypergeometric2F1}\left (\frac {7}{4},2,\frac {5}{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 (\frac {3}{4},1,\frac {3}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^2-6 a c^3 d \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {3}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^2-b c^4 \operatorname {Hypergeometric2F1}\left (\frac {7}{4},2,\frac {5}{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 (\frac {7}{4},2,\frac {5}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^2-6 (b c-a d) \left (c-4 d x^2\right ) \left (d x^2+c\right )^2 \, _3F_2\left (\frac {7}{4},2,2;1,\frac {5}{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 (\frac {7}{4},2,2,2;1,1,\frac {5}{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 (\frac {3}{4},1,\frac {3}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )\right )}{15 c^4 x^5 \left (b x^2+a\right )^{7/4} \left (d x^2+c\right )^{3/4}}\)

Input:

Int[1/(x^6*(a + b*x^2)^(3/4)*(c + d*x^2)^(3/4)),x]
 

Output:

-1/15*((1 + (d*x^2)/c)*(3*a*c^4*Hypergeometric2F1[3/4, 1, 3/2, ((b*c - a*d 
)*x^2)/(c*(a + b*x^2))] + 3*b*c^4*x^2*Hypergeometric2F1[3/4, 1, 3/2, ((b*c 
 - a*d)*x^2)/(c*(a + b*x^2))] - 6*a*c^3*d*x^2*Hypergeometric2F1[3/4, 1, 3/ 
2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 6*b*c^3*d*x^4*Hypergeometric2F1[3/ 
4, 1, 3/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 24*a*c^2*d^2*x^4*Hypergeom 
etric2F1[3/4, 1, 3/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 24*b*c^2*d^2*x^ 
6*Hypergeometric2F1[3/4, 1, 3/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 48*a 
*c*d^3*x^6*Hypergeometric2F1[3/4, 1, 3/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2) 
)] + 48*b*c*d^3*x^8*Hypergeometric2F1[3/4, 1, 3/2, ((b*c - a*d)*x^2)/(c*(a 
 + b*x^2))] - b*c^4*x^2*Hypergeometric2F1[7/4, 2, 5/2, ((b*c - a*d)*x^2)/( 
c*(a + b*x^2))] + a*c^3*d*x^2*Hypergeometric2F1[7/4, 2, 5/2, ((b*c - a*d)* 
x^2)/(c*(a + b*x^2))] - 3*b*c^3*d*x^4*Hypergeometric2F1[7/4, 2, 5/2, ((b*c 
 - a*d)*x^2)/(c*(a + b*x^2))] + 3*a*c^2*d^2*x^4*Hypergeometric2F1[7/4, 2, 
5/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 42*b*c^2*d^2*x^6*Hypergeometric2 
F1[7/4, 2, 5/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 42*a*c*d^3*x^6*Hyperg 
eometric2F1[7/4, 2, 5/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 44*b*c*d^3*x 
^8*Hypergeometric2F1[7/4, 2, 5/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 44* 
a*d^4*x^8*Hypergeometric2F1[7/4, 2, 5/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2)) 
] - 6*(b*c - a*d)*x^2*(c - 4*d*x^2)*(c + d*x^2)^2*HypergeometricPFQ[{7/4, 
2, 2}, {1, 5/2}, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 4*(b*c - a*d)*x^2...
 

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]

\[\int \frac {1}{x^{6} \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (x^{2} d +c \right )^{\frac {3}{4}}}d x\]

Input:

int(1/x^6/(b*x^2+a)^(3/4)/(d*x^2+c)^(3/4),x)
 

Output:

int(1/x^6/(b*x^2+a)^(3/4)/(d*x^2+c)^(3/4),x)
 

Fricas [F]

\[ \int \frac {1}{x^6 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (d x^{2} + c\right )}^{\frac {3}{4}} x^{6}} \,d x } \] Input:

integrate(1/x^6/(b*x^2+a)^(3/4)/(d*x^2+c)^(3/4),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(1/4)*(d*x^2 + c)^(1/4)/(b*d*x^10 + (b*c + a*d)*x^8 + 
 a*c*x^6), x)
 

Sympy [F]

\[ \int \frac {1}{x^6 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=\int \frac {1}{x^{6} \left (a + b x^{2}\right )^{\frac {3}{4}} \left (c + d x^{2}\right )^{\frac {3}{4}}}\, dx \] Input:

integrate(1/x**6/(b*x**2+a)**(3/4)/(d*x**2+c)**(3/4),x)
 

Output:

Integral(1/(x**6*(a + b*x**2)**(3/4)*(c + d*x**2)**(3/4)), x)
 

Maxima [F]

\[ \int \frac {1}{x^6 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (d x^{2} + c\right )}^{\frac {3}{4}} x^{6}} \,d x } \] Input:

integrate(1/x^6/(b*x^2+a)^(3/4)/(d*x^2+c)^(3/4),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^2 + a)^(3/4)*(d*x^2 + c)^(3/4)*x^6), x)
 

Giac [F]

\[ \int \frac {1}{x^6 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (d x^{2} + c\right )}^{\frac {3}{4}} x^{6}} \,d x } \] Input:

integrate(1/x^6/(b*x^2+a)^(3/4)/(d*x^2+c)^(3/4),x, algorithm="giac")
 

Output:

integrate(1/((b*x^2 + a)^(3/4)*(d*x^2 + c)^(3/4)*x^6), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^6 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=\int \frac {1}{x^6\,{\left (b\,x^2+a\right )}^{3/4}\,{\left (d\,x^2+c\right )}^{3/4}} \,d x \] Input:

int(1/(x^6*(a + b*x^2)^(3/4)*(c + d*x^2)^(3/4)),x)
 

Output:

int(1/(x^6*(a + b*x^2)^(3/4)*(c + d*x^2)^(3/4)), x)
 

Reduce [F]

\[ \int \frac {1}{x^6 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=\int \frac {1}{\left (d \,x^{2}+c \right )^{\frac {3}{4}} \left (b \,x^{2}+a \right )^{\frac {3}{4}} x^{6}}d x \] Input:

int(1/x^6/(b*x^2+a)^(3/4)/(d*x^2+c)^(3/4),x)
 

Output:

int(1/((c + d*x**2)**(3/4)*(a + b*x**2)**(3/4)*x**6),x)