\(\int \frac {e+f x^2}{(a+b x^2)^{5/4} (c+d x^2)^{5/4}} \, dx\) [565]

Optimal result

Integrand size = 30, antiderivative size = 162 \[ \int \frac {e+f x^2}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\frac {2 (b e-a f) x}{a (b c-a d) \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}-\frac {(b c e+a d e-2 a c f) x \sqrt [4]{\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {(b c-a d) x^2}{a \left (c+d x^2\right )}\right )}{a c (b c-a d) \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 10.33 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.34 \[ \int \frac {e+f x^2}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{5/4}} \, dx=\frac {x \left (c e \left (3 c+2 d x^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {5}{4},\frac {5}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )+\frac {c^2 f x^2 \left (a+b x^2\right ) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {3}{2},\frac {5}{2},\frac {(-b c+a d) x^2}{a \left (c+d x^2\right )}\right )}{a \left (1+\frac {d x^2}{c}\right )^{5/4}}+\frac {(b c-a d) e x^2 \left (c+d x^2\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {9}{4},\frac {7}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )}{a+b x^2}\right )}{3 c^3 \left (a+b x^2\right )^{5/4} \sqrt [4]{c+d x^2}} \] Input:

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

Output:

(x*(c*e*(3*c + 2*d*x^2)*Hypergeometric2F1[1, 5/4, 5/2, ((b*c - a*d)*x^2)/( 
c*(a + b*x^2))] + (c^2*f*x^2*(a + b*x^2)*(1 + (b*x^2)/a)^(1/4)*Hypergeomet 
ric2F1[5/4, 3/2, 5/2, ((-(b*c) + a*d)*x^2)/(a*(c + d*x^2))])/(a*(1 + (d*x^ 
2)/c)^(5/4)) + ((b*c - a*d)*e*x^2*(c + d*x^2)*Hypergeometric2F1[2, 9/4, 7/ 
2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))])/(a + b*x^2)))/(3*c^3*(a + b*x^2)^(5 
/4)*(c + d*x^2)^(1/4))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {402, 27, 294}

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.

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 294

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

Input:

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

Output:

(2*(b*e - a*f)*x)/(a*(b*c - a*d)*(a + b*x^2)^(1/4)*(c + d*x^2)^(1/4)) - (( 
b*c*e + a*d*e - 2*a*c*f)*x*((c*(a + b*x^2))/(a*(c + d*x^2)))^(1/4)*Hyperge 
ometric2F1[1/4, 1/2, 3/2, -(((b*c - a*d)*x^2)/(a*(c + d*x^2)))])/(a*c*(b*c 
 - a*d)*(a + b*x^2)^(1/4)*(c + d*x^2)^(1/4))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 294

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[x*((a + b*x^2)^p/(c*(c*((a + b*x^2)/(a*(c + d*x^2))))^p*(c + d*x^2)^(1/2 
+ p)))*Hypergeometric2F1[1/2, -p, 3/2, (-(b*c - a*d))*(x^2/(a*(c + d*x^2))) 
], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[2*(p + q 
+ 1) + 1, 0]
 

rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate((f*x^2+e)/(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)*(f*x^2 + e)/(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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((f*x^2+e)/(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)*f + int(1/( 
(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)*e