\(\int \frac {x^2}{(a+b x^2)^{5/4} (c+d x^2)^{5/4}} \, dx\) [1584]

Optimal result

Integrand size = 26, antiderivative size = 91 \[ \int \frac {x^2}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\frac {x^3 \left (\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}\right )^{5/4} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {3}{2},\frac {5}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )}{3 a \sqrt [4]{a+b x^2} \left (c+d x^2\right )^{5/4}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 5.12 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.04 \[ \int \frac {x^2}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\frac {c x^3 \sqrt [4]{1+\frac {b x^2}{a}} \left (1+\frac {d x^2}{c}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {3}{2},\frac {5}{2},\frac {(-b c+a d) x^2}{a \left (c+d x^2\right )}\right )}{3 a \sqrt [4]{a+b x^2} \left (c+d x^2\right )^{9/4}} \] Input:

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

Output:

(c*x^3*(1 + (b*x^2)/a)^(1/4)*(1 + (d*x^2)/c)^(3/4)*Hypergeometric2F1[5/4, 
3/2, 5/2, ((-(b*c) + a*d)*x^2)/(a*(c + d*x^2))])/(3*a*(a + b*x^2)^(1/4)*(c 
 + d*x^2)^(9/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.13, 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.

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 394

\(\displaystyle \frac {x^3 \sqrt [4]{\frac {b x^2}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {3}{2},\frac {5}{2},-\frac {c \left (\frac {b x^2}{a}-\frac {d x^2}{c}\right )}{d x^2+c}\right )}{3 a c \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2} \left (\frac {d x^2}{c}+1\right )^{5/4}}\)

Input:

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

Output:

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(x**2/((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)