\(\int \frac {\sqrt {c+d x^2} (e+f x^2)}{(a+b x^2)^{5/2}} \, dx\) [35]

Optimal result

Integrand size = 30, antiderivative size = 272 \[ \int \frac {\sqrt {c+d x^2} \left (e+f x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {(b e-a f) x \sqrt {c+d x^2}}{3 a b \left (a+b x^2\right )^{3/2}}+\frac {\left (2 b^2 c e-2 a^2 d f-a b (d e-c f)\right ) \sqrt {c+d x^2} E\left (\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )|1-\frac {a d}{b c}\right )}{3 a^{3/2} b^{3/2} (b c-a d) \sqrt {a+b x^2} \sqrt {\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}}-\frac {d (b e-a f) \sqrt {c+d x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),1-\frac {a d}{b c}\right )}{3 \sqrt {a} b^{3/2} (b c-a d) \sqrt {a+b x^2} \sqrt {\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 4.25 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {c+d x^2} \left (e+f x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {\sqrt {\frac {b}{a}} x \left (c+d x^2\right ) \left (a^3 d f-2 b^3 c e x^2+2 a^2 b d \left (e+f x^2\right )+a b^2 \left (d e x^2-c \left (3 e+f x^2\right )\right )\right )+i c \left (-2 b^2 c e+2 a^2 d f+a b (d e-c f)\right ) \left (a+b x^2\right ) \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} E\left (i \text {arcsinh}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )-i c (-b c+a d) (2 b e+a f) \left (a+b x^2\right ) \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {b}{a}} x\right ),\frac {a d}{b c}\right )}{3 a^3 \left (\frac {b}{a}\right )^{3/2} (-b c+a d) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 277, 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.

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

\(\Big \downarrow \) 401

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 400

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

\(\Big \downarrow \) 313

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

\(\Big \downarrow \) 320

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

Input:

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

Output:

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

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. \(2(249)=498\).

Time = 6.02 (sec) , antiderivative size = 518, normalized size of antiderivative = 1.90

method	result	size
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \left (-\frac {\left (a f -b e \right ) x \sqrt {b d \,x^{4}+a d \,x^{2}+x^{2} b c +a c}}{3 a \,b^{3} \left (x^{2}+\frac {a}{b}\right )^{2}}+\frac {\left (b d \,x^{2}+b c \right ) x \left (2 a^{2} d f -a b c f +a b d e -2 c e \,b^{2}\right )}{3 b^{2} a^{2} \left (a d -b c \right ) \sqrt {\left (x^{2}+\frac {a}{b}\right ) \left (b d \,x^{2}+b c \right )}}+\frac {\left (\frac {d f}{b^{2}}-\frac {\left (a f -b e \right ) d}{3 b^{2} a}-\frac {2 a^{2} d f -a b c f +a b d e -2 c e \,b^{2}}{3 b^{2} a^{2}}-\frac {c \left (2 a^{2} d f -a b c f +a b d e -2 c e \,b^{2}\right )}{3 b \,a^{2} \left (a d -b c \right )}\right ) \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \operatorname {EllipticF}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+x^{2} b c +a c}}+\frac {\left (2 a^{2} d f -a b c f +a b d e -2 c e \,b^{2}\right ) c \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \left (\operatorname {EllipticF}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )-\operatorname {EllipticE}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )\right )}{3 b \left (a d -b c \right ) a^{2} \sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+x^{2} b c +a c}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}}\)	\(518\)
default	\(\text {Expression too large to display}\)	\(1236\)

Input:

int((d*x^2+c)^(1/2)*(f*x^2+e)/(b*x^2+a)^(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*(a*f-b*e 
)/a/b^3*x*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)/(x^2+a/b)^2+1/3*(b*d*x^2+b*c 
)/b^2/a^2/(a*d-b*c)*x*(2*a^2*d*f-a*b*c*f+a*b*d*e-2*b^2*c*e)/((x^2+a/b)*(b* 
d*x^2+b*c))^(1/2)+(d*f/b^2-1/3*(a*f-b*e)/b^2*d/a-1/3/b^2*(2*a^2*d*f-a*b*c* 
f+a*b*d*e-2*b^2*c*e)/a^2-1/3/b*c/a^2/(a*d-b*c)*(2*a^2*d*f-a*b*c*f+a*b*d*e- 
2*b^2*c*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*(2*a^2*d*f-a*b*c*f+a*b*d*e-2*b^2*c*e)/(a*d-b*c)/a^2*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. 534 vs. \(2 (249) = 498\).

Time = 0.11 (sec) , antiderivative size = 534, normalized size of antiderivative = 1.96 \[ \int \frac {\sqrt {c+d x^2} \left (e+f x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {{\left ({\left ({\left (2 \, b^{5} c - a b^{4} d\right )} e + {\left (a b^{4} c - 2 \, a^{2} b^{3} d\right )} f\right )} x^{4} + 2 \, {\left ({\left (2 \, a b^{4} c - a^{2} b^{3} d\right )} e + {\left (a^{2} b^{3} c - 2 \, a^{3} b^{2} d\right )} f\right )} x^{2} + {\left (2 \, a^{2} b^{3} c - a^{3} b^{2} d\right )} e + {\left (a^{3} b^{2} c - 2 \, a^{4} b d\right )} f\right )} \sqrt {a c} \sqrt {-\frac {b}{a}} E(\arcsin \left (x \sqrt {-\frac {b}{a}}\right )\,|\,\frac {a d}{b c}) - {\left ({\left ({\left (2 \, b^{5} c + {\left (a^{2} b^{3} - a b^{4}\right )} d\right )} e + {\left (a b^{4} c - {\left (a^{3} b^{2} + 2 \, a^{2} b^{3}\right )} d\right )} f\right )} x^{4} + 2 \, {\left ({\left (2 \, a b^{4} c + {\left (a^{3} b^{2} - a^{2} b^{3}\right )} d\right )} e + {\left (a^{2} b^{3} c - {\left (a^{4} b + 2 \, a^{3} b^{2}\right )} d\right )} f\right )} x^{2} + {\left (2 \, a^{2} b^{3} c + {\left (a^{4} b - a^{3} b^{2}\right )} d\right )} e + {\left (a^{3} b^{2} c - {\left (a^{5} + 2 \, a^{4} b\right )} d\right )} f\right )} \sqrt {a c} \sqrt {-\frac {b}{a}} F(\arcsin \left (x \sqrt {-\frac {b}{a}}\right )\,|\,\frac {a d}{b c}) - {\left ({\left ({\left (2 \, a b^{4} c - a^{2} b^{3} d\right )} e + {\left (a^{2} b^{3} c - 2 \, a^{3} b^{2} d\right )} f\right )} x^{3} - {\left (a^{4} b d f - {\left (3 \, a^{2} b^{3} c - 2 \, a^{3} b^{2} d\right )} e\right )} x\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{3 \, {\left (a^{5} b^{3} c - a^{6} b^{2} d + {\left (a^{3} b^{5} c - a^{4} b^{4} d\right )} x^{4} + 2 \, {\left (a^{4} b^{4} c - a^{5} b^{3} d\right )} x^{2}\right )}} \] Input:

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

Output:

-1/3*((((2*b^5*c - a*b^4*d)*e + (a*b^4*c - 2*a^2*b^3*d)*f)*x^4 + 2*((2*a*b 
^4*c - a^2*b^3*d)*e + (a^2*b^3*c - 2*a^3*b^2*d)*f)*x^2 + (2*a^2*b^3*c - a^ 
3*b^2*d)*e + (a^3*b^2*c - 2*a^4*b*d)*f)*sqrt(a*c)*sqrt(-b/a)*elliptic_e(ar 
csin(x*sqrt(-b/a)), a*d/(b*c)) - (((2*b^5*c + (a^2*b^3 - a*b^4)*d)*e + (a* 
b^4*c - (a^3*b^2 + 2*a^2*b^3)*d)*f)*x^4 + 2*((2*a*b^4*c + (a^3*b^2 - a^2*b 
^3)*d)*e + (a^2*b^3*c - (a^4*b + 2*a^3*b^2)*d)*f)*x^2 + (2*a^2*b^3*c + (a^ 
4*b - a^3*b^2)*d)*e + (a^3*b^2*c - (a^5 + 2*a^4*b)*d)*f)*sqrt(a*c)*sqrt(-b 
/a)*elliptic_f(arcsin(x*sqrt(-b/a)), a*d/(b*c)) - (((2*a*b^4*c - a^2*b^3*d 
)*e + (a^2*b^3*c - 2*a^3*b^2*d)*f)*x^3 - (a^4*b*d*f - (3*a^2*b^3*c - 2*a^3 
*b^2*d)*e)*x)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(a^5*b^3*c - a^6*b^2*d + (a 
^3*b^5*c - a^4*b^4*d)*x^4 + 2*(a^4*b^4*c - a^5*b^3*d)*x^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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