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

Optimal result

Integrand size = 30, antiderivative size = 269 \[ \int \frac {\sqrt {a+b x^2} \left (e+f x^2\right )}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {(d e-c f) x \sqrt {a+b x^2}}{3 c d \left (c+d x^2\right )^{3/2}}-\frac {(a d (2 d e+c f)-b c (d e+2 c f)) \sqrt {a+b x^2} E\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 c^{3/2} d^{3/2} (b c-a d) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}+\frac {b (d e-c f) \sqrt {a+b x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{3 \sqrt {c} d^{3/2} (b c-a d) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}} \] Output:

1/3*(-c*f+d*e)*x*(b*x^2+a)^(1/2)/c/d/(d*x^2+c)^(3/2)-1/3*(a*d*(c*f+2*d*e)- 
b*c*(2*c*f+d*e))*(b*x^2+a)^(1/2)*EllipticE(d^(1/2)*x/c^(1/2)/(1+d*x^2/c)^( 
1/2),(1-b*c/a/d)^(1/2))/c^(3/2)/d^(3/2)/(-a*d+b*c)/(c*(b*x^2+a)/a/(d*x^2+c 
))^(1/2)/(d*x^2+c)^(1/2)+1/3*b*(-c*f+d*e)*(b*x^2+a)^(1/2)*InverseJacobiAM( 
arctan(d^(1/2)*x/c^(1/2)),(1-b*c/a/d)^(1/2))/c^(1/2)/d^(3/2)/(-a*d+b*c)/(c 
*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 4.31 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.11 \[ \int \frac {\sqrt {a+b x^2} \left (e+f x^2\right )}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {\sqrt {\frac {b}{a}} d x \left (a+b x^2\right ) \left (a d^2 \left (3 c e+2 d e x^2+c f x^2\right )-b c \left (c^2 f+d^2 e x^2+2 c d \left (e+f x^2\right )\right )\right )+i b c (a d (2 d e+c f)-b c (d e+2 c f)) \sqrt {1+\frac {b x^2}{a}} \left (c+d x^2\right ) \sqrt {1+\frac {d x^2}{c}} E\left (i \text {arcsinh}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )-i b c (-b c+a d) (d e+2 c f) \sqrt {1+\frac {b x^2}{a}} \left (c+d x^2\right ) \sqrt {1+\frac {d x^2}{c}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {b}{a}} x\right ),\frac {a d}{b c}\right )}{3 \sqrt {\frac {b}{a}} c^2 d^2 (-b c+a d) \sqrt {a+b x^2} \left (c+d x^2\right )^{3/2}} \] Input:

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

Output:

(Sqrt[b/a]*d*x*(a + b*x^2)*(a*d^2*(3*c*e + 2*d*e*x^2 + c*f*x^2) - b*c*(c^2 
*f + d^2*e*x^2 + 2*c*d*(e + f*x^2))) + I*b*c*(a*d*(2*d*e + c*f) - b*c*(d*e 
 + 2*c*f))*Sqrt[1 + (b*x^2)/a]*(c + d*x^2)*Sqrt[1 + (d*x^2)/c]*EllipticE[I 
*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] - I*b*c*(-(b*c) + a*d)*(d*e + 2*c*f)*S 
qrt[1 + (b*x^2)/a]*(c + d*x^2)*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sqr 
t[b/a]*x], (a*d)/(b*c)])/(3*Sqrt[b/a]*c^2*d^2*(-(b*c) + a*d)*Sqrt[a + b*x^ 
2]*(c + d*x^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {401, 25, 400, 313, 320}

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.

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

\(\Big \downarrow \) 401

\(\displaystyle \frac {x \sqrt {a+b x^2} (d e-c f)}{3 c d \left (c+d x^2\right )^{3/2}}-\frac {\int -\frac {b (d e+2 c f) x^2+a (2 d e+c f)}{\sqrt {b x^2+a} \left (d x^2+c\right )^{3/2}}dx}{3 c d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {b (d e+2 c f) x^2+a (2 d e+c f)}{\sqrt {b x^2+a} \left (d x^2+c\right )^{3/2}}dx}{3 c d}+\frac {x \sqrt {a+b x^2} (d e-c f)}{3 c d \left (c+d x^2\right )^{3/2}}\)

\(\Big \downarrow \) 400

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

\(\Big \downarrow \) 313

\(\displaystyle \frac {\frac {a b (d e-c f) \int \frac {1}{\sqrt {b x^2+a} \sqrt {d x^2+c}}dx}{b c-a d}-\frac {\sqrt {a+b x^2} (a d (c f+2 d e)-b c (2 c f+d e)) E\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{\sqrt {c} \sqrt {d} \sqrt {c+d x^2} (b c-a d) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}}{3 c d}+\frac {x \sqrt {a+b x^2} (d e-c f)}{3 c d \left (c+d x^2\right )^{3/2}}\)

\(\Big \downarrow \) 320

\(\displaystyle \frac {\frac {b \sqrt {c} \sqrt {a+b x^2} (d e-c f) \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{\sqrt {d} \sqrt {c+d x^2} (b c-a d) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {\sqrt {a+b x^2} (a d (c f+2 d e)-b c (2 c f+d e)) E\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{\sqrt {c} \sqrt {d} \sqrt {c+d x^2} (b c-a d) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}}{3 c d}+\frac {x \sqrt {a+b x^2} (d e-c f)}{3 c d \left (c+d x^2\right )^{3/2}}\)

Input:

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

Output:

((d*e - c*f)*x*Sqrt[a + b*x^2])/(3*c*d*(c + d*x^2)^(3/2)) + (-(((a*d*(2*d* 
e + c*f) - b*c*(d*e + 2*c*f))*Sqrt[a + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x) 
/Sqrt[c]], 1 - (b*c)/(a*d)])/(Sqrt[c]*Sqrt[d]*(b*c - a*d)*Sqrt[(c*(a + b*x 
^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2])) + (b*Sqrt[c]*(d*e - c*f)*Sqrt[a + 
b*x^2]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(Sqrt[d]*( 
b*c - a*d)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]))/(3*c*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 313

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Sim 
p[(Sqrt[a + b*x^2]/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*(c 
+ d*x^2)))]))*EllipticE[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /; FreeQ 
[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]
 

rule 320

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(Sqrt[a + b*x^2]/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*( 
c + d*x^2)))]))*EllipticF[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /; Fre 
eQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]
 

rule 400

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)^ 
(3/2)), x_Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(Sqrt[a + b*x^2]* 
Sqrt[c + d*x^2]), x], x] - Simp[(d*e - c*f)/(b*c - a*d)   Int[Sqrt[a + b*x^ 
2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[b/a] & 
& PosQ[d/c]
 

rule 401

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/(a*b*2*(p + 1))), x] + Simp[1/(a*b*2*(p + 1))   Int[(a + b*x^2)^(p + 1)*( 
c + d*x^2)^(q - 1)*Simp[c*(b*e*2*(p + 1) + b*e - a*f) + d*(b*e*2*(p + 1) + 
(b*e - a*f)*(2*q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && L 
tQ[p, -1] && GtQ[q, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(517\) vs. \(2(246)=492\).

Time = 6.00 (sec) , antiderivative size = 518, normalized size of antiderivative = 1.93

method	result	size
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \left (-\frac {\left (c f -d e \right ) x \sqrt {b d \,x^{4}+a d \,x^{2}+x^{2} b c +a c}}{3 c \,d^{3} \left (x^{2}+\frac {c}{d}\right )^{2}}+\frac {\left (b d \,x^{2}+a d \right ) x \left (a c d f +2 a \,d^{2} e -2 b \,c^{2} f -b c d e \right )}{3 c^{2} d^{2} \left (a d -b c \right ) \sqrt {\left (x^{2}+\frac {c}{d}\right ) \left (b d \,x^{2}+a d \right )}}+\frac {\left (\frac {f b}{d^{2}}-\frac {\left (c f -d e \right ) b}{3 d^{2} c}+\frac {a c d f +2 a \,d^{2} e -2 b \,c^{2} f -b c d e}{3 d^{2} c^{2}}-\frac {a \left (a c d f +2 a \,d^{2} e -2 b \,c^{2} f -b c d e \right )}{3 d \,c^{2} \left (a d -b c \right )}\right ) \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \operatorname {EllipticF}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+x^{2} b c +a c}}+\frac {b \left (a c d f +2 a \,d^{2} e -2 b \,c^{2} f -b c d e \right ) \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \left (\operatorname {EllipticF}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )-\operatorname {EllipticE}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )\right )}{3 d^{2} \left (a d -b c \right ) c \sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+x^{2} b c +a c}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}}\)	\(518\)
default	\(\text {Expression too large to display}\)	\(1261\)

Input:

int((b*x^2+a)^(1/2)*(f*x^2+e)/(d*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

((b*x^2+a)*(d*x^2+c))^(1/2)/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2)*(-1/3*(c*f-d*e 
)/c/d^3*x*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)/(x^2+c/d)^2+1/3*(b*d*x^2+a*d 
)/c^2/d^2/(a*d-b*c)*x*(a*c*d*f+2*a*d^2*e-2*b*c^2*f-b*c*d*e)/((x^2+c/d)*(b* 
d*x^2+a*d))^(1/2)+(f*b/d^2-1/3*(c*f-d*e)/d^2*b/c+1/3/d^2*(a*c*d*f+2*a*d^2* 
e-2*b*c^2*f-b*c*d*e)/c^2-1/3*a/d/c^2/(a*d-b*c)*(a*c*d*f+2*a*d^2*e-2*b*c^2* 
f-b*c*d*e))/(-b/a)^(1/2)*(1+b*x^2/a)^(1/2)*(1+d*x^2/c)^(1/2)/(b*d*x^4+a*d* 
x^2+b*c*x^2+a*c)^(1/2)*EllipticF(x*(-b/a)^(1/2),(-1+(a*d+b*c)/c/b)^(1/2))+ 
1/3*b/d^2*(a*c*d*f+2*a*d^2*e-2*b*c^2*f-b*c*d*e)/(a*d-b*c)/c/(-b/a)^(1/2)*( 
1+b*x^2/a)^(1/2)*(1+d*x^2/c)^(1/2)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(El 
lipticF(x*(-b/a)^(1/2),(-1+(a*d+b*c)/c/b)^(1/2))-EllipticE(x*(-b/a)^(1/2), 
(-1+(a*d+b*c)/c/b)^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 550 vs. \(2 (246) = 492\).

Time = 0.09 (sec) , antiderivative size = 550, normalized size of antiderivative = 2.04 \[ \int \frac {\sqrt {a+b x^2} \left (e+f x^2\right )}{\left (c+d x^2\right )^{5/2}} \, dx=-\frac {{\left ({\left ({\left (b^{2} c d^{3} - 2 \, a b d^{4}\right )} e + {\left (2 \, b^{2} c^{2} d^{2} - a b c d^{3}\right )} f\right )} x^{4} + 2 \, {\left ({\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3}\right )} e + {\left (2 \, b^{2} c^{3} d - a b c^{2} d^{2}\right )} f\right )} x^{2} + {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2}\right )} e + {\left (2 \, b^{2} c^{4} - a b c^{3} d\right )} f\right )} \sqrt {a c} \sqrt {-\frac {b}{a}} E(\arcsin \left (x \sqrt {-\frac {b}{a}}\right )\,|\,\frac {a d}{b c}) - {\left ({\left ({\left (b^{2} c d^{3} - {\left (a^{2} + 2 \, a b\right )} d^{4}\right )} e + {\left (2 \, b^{2} c^{2} d^{2} + {\left (a^{2} - a b\right )} c d^{3}\right )} f\right )} x^{4} + 2 \, {\left ({\left (b^{2} c^{2} d^{2} - {\left (a^{2} + 2 \, a b\right )} c d^{3}\right )} e + {\left (2 \, b^{2} c^{3} d + {\left (a^{2} - a b\right )} c^{2} d^{2}\right )} f\right )} x^{2} + {\left (b^{2} c^{3} d - {\left (a^{2} + 2 \, a b\right )} c^{2} d^{2}\right )} e + {\left (2 \, b^{2} c^{4} + {\left (a^{2} - a b\right )} c^{3} d\right )} f\right )} \sqrt {a c} \sqrt {-\frac {b}{a}} F(\arcsin \left (x \sqrt {-\frac {b}{a}}\right )\,|\,\frac {a d}{b c}) - {\left ({\left ({\left (a b c d^{3} - 2 \, a^{2} d^{4}\right )} e + {\left (2 \, a b c^{2} d^{2} - a^{2} c d^{3}\right )} f\right )} x^{3} + {\left (a b c^{3} d f + {\left (2 \, a b c^{2} d^{2} - 3 \, a^{2} c d^{3}\right )} e\right )} x\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{3 \, {\left (a b c^{5} d^{2} - a^{2} c^{4} d^{3} + {\left (a b c^{3} d^{4} - a^{2} c^{2} d^{5}\right )} x^{4} + 2 \, {\left (a b c^{4} d^{3} - a^{2} c^{3} d^{4}\right )} x^{2}\right )}} \] Input:

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

Output:

-1/3*((((b^2*c*d^3 - 2*a*b*d^4)*e + (2*b^2*c^2*d^2 - a*b*c*d^3)*f)*x^4 + 2 
*((b^2*c^2*d^2 - 2*a*b*c*d^3)*e + (2*b^2*c^3*d - a*b*c^2*d^2)*f)*x^2 + (b^ 
2*c^3*d - 2*a*b*c^2*d^2)*e + (2*b^2*c^4 - a*b*c^3*d)*f)*sqrt(a*c)*sqrt(-b/ 
a)*elliptic_e(arcsin(x*sqrt(-b/a)), a*d/(b*c)) - (((b^2*c*d^3 - (a^2 + 2*a 
*b)*d^4)*e + (2*b^2*c^2*d^2 + (a^2 - a*b)*c*d^3)*f)*x^4 + 2*((b^2*c^2*d^2 
- (a^2 + 2*a*b)*c*d^3)*e + (2*b^2*c^3*d + (a^2 - a*b)*c^2*d^2)*f)*x^2 + (b 
^2*c^3*d - (a^2 + 2*a*b)*c^2*d^2)*e + (2*b^2*c^4 + (a^2 - a*b)*c^3*d)*f)*s 
qrt(a*c)*sqrt(-b/a)*elliptic_f(arcsin(x*sqrt(-b/a)), a*d/(b*c)) - (((a*b*c 
*d^3 - 2*a^2*d^4)*e + (2*a*b*c^2*d^2 - a^2*c*d^3)*f)*x^3 + (a*b*c^3*d*f + 
(2*a*b*c^2*d^2 - 3*a^2*c*d^3)*e)*x)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(a*b* 
c^5*d^2 - a^2*c^4*d^3 + (a*b*c^3*d^4 - a^2*c^2*d^5)*x^4 + 2*(a*b*c^4*d^3 - 
 a^2*c^3*d^4)*x^2)
 

Sympy [F]

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

integrate((b*x**2+a)**(1/2)*(f*x**2+e)/(d*x**2+c)**(5/2),x)
 

Output:

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

Maxima [F]

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

integrate((b*x^2+a)^(1/2)*(f*x^2+e)/(d*x^2+c)^(5/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((b*x^2+a)^(1/2)*(f*x^2+e)/(d*x^2+c)^(5/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - sqrt(c + d*x**2)*sqrt(a + b*x**2)*a*f*x - sqrt(c + d*x**2)*sqrt(a + b* 
x**2)*b*e*x + int((sqrt(c + d*x**2)*sqrt(a + b*x**2)*x**4)/(a**2*c**3*d + 
3*a**2*c**2*d**2*x**2 + 3*a**2*c*d**3*x**4 + a**2*d**4*x**6 - a*b*c**4 - 2 
*a*b*c**3*d*x**2 + 2*a*b*c*d**3*x**6 + a*b*d**4*x**8 - b**2*c**4*x**2 - 3* 
b**2*c**3*d*x**4 - 3*b**2*c**2*d**2*x**6 - b**2*c*d**3*x**8),x)*a**2*b*c** 
2*d**2*f + 2*int((sqrt(c + d*x**2)*sqrt(a + b*x**2)*x**4)/(a**2*c**3*d + 3 
*a**2*c**2*d**2*x**2 + 3*a**2*c*d**3*x**4 + a**2*d**4*x**6 - a*b*c**4 - 2* 
a*b*c**3*d*x**2 + 2*a*b*c*d**3*x**6 + a*b*d**4*x**8 - b**2*c**4*x**2 - 3*b 
**2*c**3*d*x**4 - 3*b**2*c**2*d**2*x**6 - b**2*c*d**3*x**8),x)*a**2*b*c*d* 
*3*f*x**2 + int((sqrt(c + d*x**2)*sqrt(a + b*x**2)*x**4)/(a**2*c**3*d + 3* 
a**2*c**2*d**2*x**2 + 3*a**2*c*d**3*x**4 + a**2*d**4*x**6 - a*b*c**4 - 2*a 
*b*c**3*d*x**2 + 2*a*b*c*d**3*x**6 + a*b*d**4*x**8 - b**2*c**4*x**2 - 3*b* 
*2*c**3*d*x**4 - 3*b**2*c**2*d**2*x**6 - b**2*c*d**3*x**8),x)*a**2*b*d**4* 
f*x**4 - 3*int((sqrt(c + d*x**2)*sqrt(a + b*x**2)*x**4)/(a**2*c**3*d + 3*a 
**2*c**2*d**2*x**2 + 3*a**2*c*d**3*x**4 + a**2*d**4*x**6 - a*b*c**4 - 2*a* 
b*c**3*d*x**2 + 2*a*b*c*d**3*x**6 + a*b*d**4*x**8 - b**2*c**4*x**2 - 3*b** 
2*c**3*d*x**4 - 3*b**2*c**2*d**2*x**6 - b**2*c*d**3*x**8),x)*a*b**2*c**3*d 
*f - int((sqrt(c + d*x**2)*sqrt(a + b*x**2)*x**4)/(a**2*c**3*d + 3*a**2*c* 
*2*d**2*x**2 + 3*a**2*c*d**3*x**4 + a**2*d**4*x**6 - a*b*c**4 - 2*a*b*c**3 
*d*x**2 + 2*a*b*c*d**3*x**6 + a*b*d**4*x**8 - b**2*c**4*x**2 - 3*b**2*c...