\(\int \frac {1}{x^4 (a+b x^2)^{3/4} (c+d x^2)^{3/4}} \, dx\) [1574]

Optimal result

Integrand size = 26, antiderivative size = 194 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=-\frac {\sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}{3 a c x^3}+\frac {5 (b c+a d) \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}{6 a^2 c^2 x}+\frac {\left (5 b^2 c^2+2 a b c d+5 a^2 d^2\right ) x \left (\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}\right )^{3/4} \sqrt [4]{c+d x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\frac {(b c-a d) x^2}{a \left (c+d x^2\right )}\right )}{12 a^2 c^3 \left (a+b x^2\right )^{3/4}} \] Output:

-1/3*(b*x^2+a)^(1/4)*(d*x^2+c)^(1/4)/a/c/x^3+5/6*(a*d+b*c)*(b*x^2+a)^(1/4) 
*(d*x^2+c)^(1/4)/a^2/c^2/x+1/12*(5*a^2*d^2+2*a*b*c*d+5*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^2/c^3/(b*x^2+a)^(3/4)
 

Mathematica [A] (verified)

Time = 10.32 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.16 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \, dx=-\frac {4 \left (1+\frac {d x^2}{c}\right ) \operatorname {Gamma}\left (\frac {3}{4}\right ) \left (c \left (a+b x^2\right ) \left (-c^2+4 c d x^2+8 d^2 x^4\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {3}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )+6 d (b c-a d) x^4 \left (c+d x^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {7}{4},2,\frac {5}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )+2 (b c-a d) x^2 \left (c+d x^2\right )^2 \, _3F_2\left (\frac {7}{4},2,2;1,\frac {5}{2};\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )\right )}{3 c^3 x^3 \left (a+b x^2\right )^{7/4} \left (c+d x^2\right )^{3/4} \operatorname {Gamma}\left (-\frac {1}{4}\right )} \] Input:

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

Output:

(-4*(1 + (d*x^2)/c)*Gamma[3/4]*(c*(a + b*x^2)*(-c^2 + 4*c*d*x^2 + 8*d^2*x^ 
4)*Hypergeometric2F1[3/4, 1, 3/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 6*d 
*(b*c - a*d)*x^4*(c + d*x^2)*Hypergeometric2F1[7/4, 2, 5/2, ((b*c - a*d)*x 
^2)/(c*(a + b*x^2))] + 2*(b*c - a*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))]))/(3*c^3*x^3*(a + 
 b*x^2)^(7/4)*(c + d*x^2)^(3/4)*Gamma[-1/4])
 

Rubi [A] (verified)

Time = 3.24 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.15, 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.

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 394

\(\displaystyle \frac {4 \operatorname {Gamma}\left (\frac {3}{4}\right ) \left (\frac {d x^2}{c}+1\right ) \left (-2 x^2 \left (c+d x^2\right )^2 (b c-a d) \, _3F_2\left (\frac {7}{4},2,2;1,\frac {5}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )+c \left (a+b x^2\right ) \left (c^2-4 c d x^2-8 d^2 x^4\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {3}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )-6 d x^4 \left (c+d x^2\right ) (b c-a d) \operatorname {Hypergeometric2F1}\left (\frac {7}{4},2,\frac {5}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )\right )}{3 c^3 x^3 \operatorname {Gamma}\left (-\frac {1}{4}\right ) \left (a+b x^2\right )^{7/4} \left (c+d x^2\right )^{3/4}}\)

Input:

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

Output:

(4*(1 + (d*x^2)/c)*Gamma[3/4]*(c*(a + b*x^2)*(c^2 - 4*c*d*x^2 - 8*d^2*x^4) 
*Hypergeometric2F1[3/4, 1, 3/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 6*d*( 
b*c - a*d)*x^4*(c + d*x^2)*Hypergeometric2F1[7/4, 2, 5/2, ((b*c - a*d)*x^2 
)/(c*(a + b*x^2))] - 2*(b*c - a*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))]))/(3*c^3*x^3*(a + b 
*x^2)^(7/4)*(c + d*x^2)^(3/4)*Gamma[-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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(1/x^4/(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^4), x)
 

Giac [F]

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

integrate(1/x^4/(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^4), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(1/x^4/(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**4),x)