\(\int \frac {1}{x^3 (a+b x^2)^{5/4} (c+d x^2)^{5/4}} \, dx\) [1580]

Optimal result

Integrand size = 26, antiderivative size = 92 \[ \int \frac {1}{x^3 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=-\frac {2 b \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{5/4} \operatorname {AppellF1}\left (-\frac {1}{4},2,\frac {5}{4},\frac {3}{4},1+\frac {b x^2}{a},-\frac {d \left (a+b x^2\right )}{b c-a d}\right )}{a^2 \sqrt [4]{a+b x^2} \left (c+d x^2\right )^{5/4}} \] Output:

-2*b*(b*(d*x^2+c)/(-a*d+b*c))^(5/4)*AppellF1(-1/4,5/4,2,3/4,-d*(b*x^2+a)/( 
-a*d+b*c),1+b*x^2/a)/a^2/(b*x^2+a)^(1/4)/(d*x^2+c)^(5/4)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(299\) vs. \(2(92)=184\).

Time = 5.40 (sec) , antiderivative size = 299, normalized size of antiderivative = 3.25 \[ \int \frac {1}{x^3 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\frac {-2 \left (5 b^3 c^2 x^2 \left (c+d x^2\right )+a^3 d^2 \left (c+5 d x^2\right )+a b^2 c \left (c^2-c d x^2-2 d^2 x^4\right )+a^2 b d \left (-2 c^2-c d x^2+5 d^2 x^4\right )\right )+5 (b c-a d)^2 (b c+a d) \sqrt [4]{1+\frac {a}{b x^2}} \sqrt [4]{1+\frac {c}{d x^2}} x^2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},\frac {1}{4},\frac {3}{2},-\frac {a}{b x^2},-\frac {c}{d x^2}\right )+b d \left (5 b^2 c^2-2 a b c d+5 a^2 d^2\right ) x^4 \sqrt [4]{1+\frac {b x^2}{a}} \sqrt [4]{1+\frac {d x^2}{c}} \operatorname {AppellF1}\left (1,\frac {1}{4},\frac {1}{4},2,-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{4 a^2 c^2 (b c-a d)^2 x^2 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Input:

Integrate[1/(x^3*(a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)),x]
 

Output:

(-2*(5*b^3*c^2*x^2*(c + d*x^2) + a^3*d^2*(c + 5*d*x^2) + a*b^2*c*(c^2 - c* 
d*x^2 - 2*d^2*x^4) + a^2*b*d*(-2*c^2 - c*d*x^2 + 5*d^2*x^4)) + 5*(b*c - a* 
d)^2*(b*c + a*d)*(1 + a/(b*x^2))^(1/4)*(1 + c/(d*x^2))^(1/4)*x^2*AppellF1[ 
1/2, 1/4, 1/4, 3/2, -(a/(b*x^2)), -(c/(d*x^2))] + b*d*(5*b^2*c^2 - 2*a*b*c 
*d + 5*a^2*d^2)*x^4*(1 + (b*x^2)/a)^(1/4)*(1 + (d*x^2)/c)^(1/4)*AppellF1[1 
, 1/4, 1/4, 2, -((b*x^2)/a), -((d*x^2)/c)])/(4*a^2*c^2*(b*c - a*d)^2*x^2*( 
a + b*x^2)^(1/4)*(c + d*x^2)^(1/4))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {354, 154, 153}

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.

\(\displaystyle \int \frac {1}{x^3 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (b x^2+a\right )^{5/4} \left (d x^2+c\right )^{5/4}}dx^2\)

\(\Big \downarrow \) 154

\(\displaystyle \frac {b \sqrt [4]{\frac {b \left (c+d x^2\right )}{b c-a d}} \int \frac {1}{x^4 \left (b x^2+a\right )^{5/4} \left (\frac {b d x^2}{b c-a d}+\frac {b c}{b c-a d}\right )^{5/4}}dx^2}{2 \sqrt [4]{c+d x^2} (b c-a d)}\)

\(\Big \downarrow \) 153

\(\displaystyle -\frac {2 b^2 \sqrt [4]{\frac {b \left (c+d x^2\right )}{b c-a d}} \operatorname {AppellF1}\left (-\frac {1}{4},\frac {5}{4},2,\frac {3}{4},-\frac {d \left (b x^2+a\right )}{b c-a d},\frac {b x^2+a}{a}\right )}{a^2 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2} (b c-a d)}\)

Input:

Int[1/(x^3*(a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)),x]
 

Output:

(-2*b^2*((b*(c + d*x^2))/(b*c - a*d))^(1/4)*AppellF1[-1/4, 5/4, 2, 3/4, -( 
(d*(a + b*x^2))/(b*c - a*d)), (a + b*x^2)/a])/(a^2*(b*c - a*d)*(a + b*x^2) 
^(1/4)*(c + d*x^2)^(1/4))
 

Defintions of rubi rules used

rule 153

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(b*e - a*f)^p*((a + b*x)^(m + 1)/(b^(p + 1)*(m + 1)*Simp 
lify[b/(b*c - a*d)]^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c 
 - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, 
n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && IntegerQ[p] && GtQ[Simplify[b/( 
b*c - a*d)], 0] &&  !(GtQ[Simplify[d/(d*a - c*b)], 0] && SimplerQ[c + d*x, 
a + b*x])
 

rule 154

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[b/(b*c - a*d)]^IntPart[n 
]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c 
 - a*d)) + b*d*(x/(b*c - a*d)), x]^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && IntegerQ[p] &&  !G 
tQ[Simplify[b/(b*c - a*d)], 0] &&  !SimplerQ[c + d*x, a + b*x]
 

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

Maple [F]

\[\int \frac {1}{x^{3} \left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (x^{2} d +c \right )^{\frac {5}{4}}}d x\]

Input:

int(1/x^3/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x)
 

Output:

int(1/x^3/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{x^3 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\text {Timed out} \] Input:

integrate(1/x^3/(b*x^2+a)^(5/4)/(d*x^2+c)^(5/4),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {1}{x^3 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int \frac {1}{x^{3} \left (a + b x^{2}\right )^{\frac {5}{4}} \left (c + d x^{2}\right )^{\frac {5}{4}}}\, dx \] Input:

integrate(1/x**3/(b*x**2+a)**(5/4)/(d*x**2+c)**(5/4),x)
 

Output:

Integral(1/(x**3*(a + b*x**2)**(5/4)*(c + d*x**2)**(5/4)), x)
 

Maxima [F]

\[ \int \frac {1}{x^3 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {5}{4}} x^{3}} \,d x } \] Input:

integrate(1/x^3/(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^3), x)
 

Giac [F]

\[ \int \frac {1}{x^3 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {5}{4}} x^{3}} \,d x } \] Input:

integrate(1/x^3/(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^3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^3 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int \frac {1}{x^3\,{\left (b\,x^2+a\right )}^{5/4}\,{\left (d\,x^2+c\right )}^{5/4}} \,d x \] Input:

int(1/(x^3*(a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)),x)
 

Output:

int(1/(x^3*(a + b*x^2)^(5/4)*(c + d*x^2)^(5/4)), x)
 

Reduce [F]

\[ \int \frac {1}{x^3 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\int \frac {1}{\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} a c \,x^{3}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} a d \,x^{5}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b c \,x^{5}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b d \,x^{7}}d x \] Input:

int(1/x^3/(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**3 + (c + d*x**2)**(1 
/4)*(a + b*x**2)**(1/4)*a*d*x**5 + (c + d*x**2)**(1/4)*(a + b*x**2)**(1/4) 
*b*c*x**5 + (c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*b*d*x**7),x)