\(\int \frac {x^4}{(a+b x^2)^{5/4} (c+d x^2)^{5/4}} \, dx\) [1583]

Optimal result

Integrand size = 26, antiderivative size = 94 \[ \int \frac {x^4}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\frac {x^5 \sqrt [4]{1+\frac {b x^2}{a}} \sqrt [4]{1+\frac {d x^2}{c}} \operatorname {AppellF1}\left (\frac {5}{2},\frac {5}{4},\frac {5}{4},\frac {7}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{5 a c \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Output:

1/5*x^5*(1+b*x^2/a)^(1/4)*(1+d*x^2/c)^(1/4)*AppellF1(5/2,5/4,5/4,7/2,-b*x^ 
2/a,-d*x^2/c)/a/c/(b*x^2+a)^(1/4)/(d*x^2+c)^(1/4)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(277\) vs. \(2(94)=188\).

Time = 5.66 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.95 \[ \int \frac {x^4}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\frac {2 x \left (6 a c+3 b c x^2+3 a d x^2-2 (b c+a d) x^2 \sqrt [4]{1+\frac {b x^2}{a}} \sqrt [4]{1+\frac {d x^2}{c}} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},\frac {1}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+\frac {36 a^2 c^2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},\frac {1}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{-6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},\frac {1}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},\frac {5}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},\frac {1}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}\right )}{3 (b c-a d)^2 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Input:

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

Output:

(2*x*(6*a*c + 3*b*c*x^2 + 3*a*d*x^2 - 2*(b*c + a*d)*x^2*(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)] + (36*a^2*c^2*AppellF1[1/2, 1/4, 1/4, 3/2, -((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*Ap 
pellF1[3/2, 5/4, 1/4, 5/2, -((b*x^2)/a), -((d*x^2)/c)]))))/(3*(b*c - a*d)^ 
2*(a + b*x^2)^(1/4)*(c + d*x^2)^(1/4))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, 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.

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 394

\(\displaystyle \frac {x^5 \sqrt [4]{\frac {b x^2}{a}+1} \sqrt [4]{\frac {d x^2}{c}+1} \operatorname {AppellF1}\left (\frac {5}{2},\frac {5}{4},\frac {5}{4},\frac {7}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{5 a c \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}\)

Input:

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

Output:

(x^5*(1 + (b*x^2)/a)^(1/4)*(1 + (d*x^2)/c)^(1/4)*AppellF1[5/2, 5/4, 5/4, 7 
/2, -((b*x^2)/a), -((d*x^2)/c)])/(5*a*c*(a + b*x^2)^(1/4)*(c + d*x^2)^(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]

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

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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