\(\int \frac {A+B x^2}{\sqrt {-d+e x^2} \sqrt {-d e+(d^2+e^2) x^2-d e x^4}} \, dx\) [140]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F(-1)]
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 48, antiderivative size = 150 \[ \int \frac {A+B x^2}{\sqrt {-d+e x^2} \sqrt {-d e+\left (d^2+e^2\right ) x^2-d e x^4}} \, dx=\frac {B \arctan \left (\frac {\sqrt {d} x \sqrt {-d+e x^2}}{\sqrt {-d e+\left (d^2+e^2\right ) x^2-d e x^4}}\right )}{\sqrt {d} e}-\frac {(B d+A e) \arctan \left (\frac {\sqrt {d^2-e^2} x \sqrt {-d+e x^2}}{\sqrt {d} \sqrt {-d e+\left (d^2+e^2\right ) x^2-d e x^4}}\right )}{\sqrt {d} e \sqrt {d^2-e^2}} \] Output:

B*arctan(d^(1/2)*x*(e*x^2-d)^(1/2)/(-d*e+(d^2+e^2)*x^2-d*e*x^4)^(1/2))/d^( 
1/2)/e-(A*e+B*d)*arctan((d^2-e^2)^(1/2)*x*(e*x^2-d)^(1/2)/d^(1/2)/(-d*e+(d 
^2+e^2)*x^2-d*e*x^4)^(1/2))/d^(1/2)/e/(d^2-e^2)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 11.78 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.19 \[ \int \frac {A+B x^2}{\sqrt {-d+e x^2} \sqrt {-d e+\left (d^2+e^2\right ) x^2-d e x^4}} \, dx=\frac {-\frac {(B d+A e) \sqrt {e-d x^2} \sqrt {-d+e x^2} \text {arctanh}\left (\frac {\sqrt {-d^2+e^2} x}{\sqrt {d} \sqrt {e-d x^2}}\right )}{\sqrt {-d^2+e^2} \sqrt {d^2 x^2+e^2 x^2-d e \left (1+x^4\right )}}-i B \log \left (-2 i \sqrt {d} x-\frac {2 \sqrt {d^2 x^2+e^2 x^2-d e \left (1+x^4\right )}}{\sqrt {-d+e x^2}}\right )}{\sqrt {d} e} \] Input:

Integrate[(A + B*x^2)/(Sqrt[-d + e*x^2]*Sqrt[-(d*e) + (d^2 + e^2)*x^2 - d* 
e*x^4]),x]
 

Output:

(-(((B*d + A*e)*Sqrt[e - d*x^2]*Sqrt[-d + e*x^2]*ArcTanh[(Sqrt[-d^2 + e^2] 
*x)/(Sqrt[d]*Sqrt[e - d*x^2])])/(Sqrt[-d^2 + e^2]*Sqrt[d^2*x^2 + e^2*x^2 - 
 d*e*(1 + x^4)])) - I*B*Log[(-2*I)*Sqrt[d]*x - (2*Sqrt[d^2*x^2 + e^2*x^2 - 
 d*e*(1 + x^4)])/Sqrt[-d + e*x^2]])/(Sqrt[d]*e)
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {1395, 25, 398, 224, 216, 291, 218}

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.

\(\displaystyle \int \frac {A+B x^2}{\sqrt {e x^2-d} \sqrt {x^2 \left (d^2+e^2\right )-d e x^4-d e}} \, dx\)

\(\Big \downarrow \) 1395

\(\displaystyle \frac {\sqrt {e-d x^2} \sqrt {e x^2-d} \int -\frac {B x^2+A}{\sqrt {e-d x^2} \left (d-e x^2\right )}dx}{\sqrt {x^2 \left (d^2+e^2\right )-d e x^4-d e}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\sqrt {e-d x^2} \sqrt {e x^2-d} \int \frac {B x^2+A}{\sqrt {e-d x^2} \left (d-e x^2\right )}dx}{\sqrt {x^2 \left (d^2+e^2\right )-d e x^4-d e}}\)

\(\Big \downarrow \) 398

\(\displaystyle -\frac {\sqrt {e-d x^2} \sqrt {e x^2-d} \left (\frac {(A e+B d) \int \frac {1}{\sqrt {e-d x^2} \left (d-e x^2\right )}dx}{e}-\frac {B \int \frac {1}{\sqrt {e-d x^2}}dx}{e}\right )}{\sqrt {x^2 \left (d^2+e^2\right )-d e x^4-d e}}\)

\(\Big \downarrow \) 224

\(\displaystyle -\frac {\sqrt {e-d x^2} \sqrt {e x^2-d} \left (\frac {(A e+B d) \int \frac {1}{\sqrt {e-d x^2} \left (d-e x^2\right )}dx}{e}-\frac {B \int \frac {1}{\frac {d x^2}{e-d x^2}+1}d\frac {x}{\sqrt {e-d x^2}}}{e}\right )}{\sqrt {x^2 \left (d^2+e^2\right )-d e x^4-d e}}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {\sqrt {e-d x^2} \sqrt {e x^2-d} \left (\frac {(A e+B d) \int \frac {1}{\sqrt {e-d x^2} \left (d-e x^2\right )}dx}{e}-\frac {B \arctan \left (\frac {\sqrt {d} x}{\sqrt {e-d x^2}}\right )}{\sqrt {d} e}\right )}{\sqrt {x^2 \left (d^2+e^2\right )-d e x^4-d e}}\)

\(\Big \downarrow \) 291

\(\displaystyle -\frac {\sqrt {e-d x^2} \sqrt {e x^2-d} \left (\frac {(A e+B d) \int \frac {1}{d-\frac {\left (e^2-d^2\right ) x^2}{e-d x^2}}d\frac {x}{\sqrt {e-d x^2}}}{e}-\frac {B \arctan \left (\frac {\sqrt {d} x}{\sqrt {e-d x^2}}\right )}{\sqrt {d} e}\right )}{\sqrt {x^2 \left (d^2+e^2\right )-d e x^4-d e}}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {\sqrt {e-d x^2} \sqrt {e x^2-d} \left (\frac {(A e+B d) \arctan \left (\frac {x \sqrt {d^2-e^2}}{\sqrt {d} \sqrt {e-d x^2}}\right )}{\sqrt {d} e \sqrt {d^2-e^2}}-\frac {B \arctan \left (\frac {\sqrt {d} x}{\sqrt {e-d x^2}}\right )}{\sqrt {d} e}\right )}{\sqrt {x^2 \left (d^2+e^2\right )-d e x^4-d e}}\)

Input:

Int[(A + B*x^2)/(Sqrt[-d + e*x^2]*Sqrt[-(d*e) + (d^2 + e^2)*x^2 - d*e*x^4] 
),x]
 

Output:

-((Sqrt[e - d*x^2]*Sqrt[-d + e*x^2]*(-((B*ArcTan[(Sqrt[d]*x)/Sqrt[e - d*x^ 
2]])/(Sqrt[d]*e)) + ((B*d + A*e)*ArcTan[(Sqrt[d^2 - e^2]*x)/(Sqrt[d]*Sqrt[ 
e - d*x^2])])/(Sqrt[d]*e*Sqrt[d^2 - e^2])))/Sqrt[-(d*e) + (d^2 + e^2)*x^2 
- d*e*x^4])
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 216
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 218
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 224
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 291
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 398
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) 
, x_Symbol] :> Simp[f/b   Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ 
b   Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} 
, x]
 

rule 1395
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e_.)*( 
x_)^(n_))^(q_.), x_Symbol] :> Simp[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d 
+ e*x^n)^FracPart[p]*(a/d + c*(x^n/e))^FracPart[p])   Int[u*(d + e*x^n)^(p 
+ q)*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && E 
qQ[n2, 2*n] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] &&  !(EqQ[q, 
1] && EqQ[n, 2])
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(804\) vs. \(2(130)=260\).

Time = 0.09 (sec) , antiderivative size = 805, normalized size of antiderivative = 5.37

method result size
default \(\frac {\sqrt {e \,x^{2}-d}\, \sqrt {-d e \,x^{4}+d^{2} x^{2}+e^{2} x^{2}-d e}\, \sqrt {d}\, \left (-2 A \sqrt {d e}\, \arctan \left (\frac {\sqrt {d}\, x}{\sqrt {-d \,x^{2}+e}}\right ) \sqrt {-\frac {d^{2}-e^{2}}{e}}\, d e -2 B \sqrt {d e}\, \arctan \left (\frac {\sqrt {d}\, x}{\sqrt {-d \,x^{2}+e}}\right ) \sqrt {-\frac {d^{2}-e^{2}}{e}}\, d^{2}+2 A \sqrt {d e}\, \arctan \left (\frac {\sqrt {d}\, x}{\sqrt {\frac {\left (d x +\sqrt {d e}\right ) \left (-d x +\sqrt {d e}\right )}{d}}}\right ) \sqrt {-\frac {d^{2}-e^{2}}{e}}\, d e +A \ln \left (\frac {2 \sqrt {d e}\, d x -2 \sqrt {-d \,x^{2}+e}\, \sqrt {-\frac {d^{2}-e^{2}}{e}}\, e -2 e^{2}}{-e x +\sqrt {d e}}\right ) d^{\frac {5}{2}} e -A \ln \left (\frac {2 \sqrt {d e}\, d x -2 \sqrt {-d \,x^{2}+e}\, \sqrt {-\frac {d^{2}-e^{2}}{e}}\, e -2 e^{2}}{-e x +\sqrt {d e}}\right ) \sqrt {d}\, e^{3}-A \ln \left (\frac {2 \sqrt {d e}\, d x +2 \sqrt {-d \,x^{2}+e}\, \sqrt {-\frac {d^{2}-e^{2}}{e}}\, e +2 e^{2}}{e x +\sqrt {d e}}\right ) d^{\frac {5}{2}} e +A \ln \left (\frac {2 \sqrt {d e}\, d x +2 \sqrt {-d \,x^{2}+e}\, \sqrt {-\frac {d^{2}-e^{2}}{e}}\, e +2 e^{2}}{e x +\sqrt {d e}}\right ) \sqrt {d}\, e^{3}+2 B \sqrt {d e}\, \arctan \left (\frac {\sqrt {d}\, x}{\sqrt {\frac {\left (d x +\sqrt {d e}\right ) \left (-d x +\sqrt {d e}\right )}{d}}}\right ) \sqrt {-\frac {d^{2}-e^{2}}{e}}\, e^{2}+B \ln \left (\frac {2 \sqrt {d e}\, d x -2 \sqrt {-d \,x^{2}+e}\, \sqrt {-\frac {d^{2}-e^{2}}{e}}\, e -2 e^{2}}{-e x +\sqrt {d e}}\right ) d^{\frac {7}{2}}-B \ln \left (\frac {2 \sqrt {d e}\, d x -2 \sqrt {-d \,x^{2}+e}\, \sqrt {-\frac {d^{2}-e^{2}}{e}}\, e -2 e^{2}}{-e x +\sqrt {d e}}\right ) d^{\frac {3}{2}} e^{2}-B \ln \left (\frac {2 \sqrt {d e}\, d x +2 \sqrt {-d \,x^{2}+e}\, \sqrt {-\frac {d^{2}-e^{2}}{e}}\, e +2 e^{2}}{e x +\sqrt {d e}}\right ) d^{\frac {7}{2}}+B \ln \left (\frac {2 \sqrt {d e}\, d x +2 \sqrt {-d \,x^{2}+e}\, \sqrt {-\frac {d^{2}-e^{2}}{e}}\, e +2 e^{2}}{e x +\sqrt {d e}}\right ) d^{\frac {3}{2}} e^{2}\right )}{2 \left (-e \,x^{2}+d \right ) \sqrt {-d \,x^{2}+e}\, \left (d e \right )^{\frac {3}{2}} \left (d -e \right ) \left (d +e \right ) \sqrt {-\frac {d^{2}-e^{2}}{e}}}\) \(805\)

Input:

int((B*x^2+A)/(e*x^2-d)^(1/2)/(-d*e+(d^2+e^2)*x^2-d*e*x^4)^(1/2),x,method= 
_RETURNVERBOSE)
 

Output:

1/2*(e*x^2-d)^(1/2)*(-d*e*x^4+d^2*x^2+e^2*x^2-d*e)^(1/2)*d^(1/2)*(-2*A*(d* 
e)^(1/2)*arctan(d^(1/2)*x/(-d*x^2+e)^(1/2))*(-(d^2-e^2)/e)^(1/2)*d*e-2*B*( 
d*e)^(1/2)*arctan(d^(1/2)*x/(-d*x^2+e)^(1/2))*(-(d^2-e^2)/e)^(1/2)*d^2+2*A 
*(d*e)^(1/2)*arctan(d^(1/2)*x/(1/d*(d*x+(d*e)^(1/2))*(-d*x+(d*e)^(1/2)))^( 
1/2))*(-(d^2-e^2)/e)^(1/2)*d*e+A*ln(2*((d*e)^(1/2)*d*x-(-d*x^2+e)^(1/2)*(- 
(d^2-e^2)/e)^(1/2)*e-e^2)/(-e*x+(d*e)^(1/2)))*d^(5/2)*e-A*ln(2*((d*e)^(1/2 
)*d*x-(-d*x^2+e)^(1/2)*(-(d^2-e^2)/e)^(1/2)*e-e^2)/(-e*x+(d*e)^(1/2)))*d^( 
1/2)*e^3-A*ln(2*((d*e)^(1/2)*d*x+(-d*x^2+e)^(1/2)*(-(d^2-e^2)/e)^(1/2)*e+e 
^2)/(e*x+(d*e)^(1/2)))*d^(5/2)*e+A*ln(2*((d*e)^(1/2)*d*x+(-d*x^2+e)^(1/2)* 
(-(d^2-e^2)/e)^(1/2)*e+e^2)/(e*x+(d*e)^(1/2)))*d^(1/2)*e^3+2*B*(d*e)^(1/2) 
*arctan(d^(1/2)*x/(1/d*(d*x+(d*e)^(1/2))*(-d*x+(d*e)^(1/2)))^(1/2))*(-(d^2 
-e^2)/e)^(1/2)*e^2+B*ln(2*((d*e)^(1/2)*d*x-(-d*x^2+e)^(1/2)*(-(d^2-e^2)/e) 
^(1/2)*e-e^2)/(-e*x+(d*e)^(1/2)))*d^(7/2)-B*ln(2*((d*e)^(1/2)*d*x-(-d*x^2+ 
e)^(1/2)*(-(d^2-e^2)/e)^(1/2)*e-e^2)/(-e*x+(d*e)^(1/2)))*d^(3/2)*e^2-B*ln( 
2*((d*e)^(1/2)*d*x+(-d*x^2+e)^(1/2)*(-(d^2-e^2)/e)^(1/2)*e+e^2)/(e*x+(d*e) 
^(1/2)))*d^(7/2)+B*ln(2*((d*e)^(1/2)*d*x+(-d*x^2+e)^(1/2)*(-(d^2-e^2)/e)^( 
1/2)*e+e^2)/(e*x+(d*e)^(1/2)))*d^(3/2)*e^2)/(-e*x^2+d)/(-d*x^2+e)^(1/2)/(d 
*e)^(3/2)/(d-e)/(d+e)/(-(d^2-e^2)/e)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 896, normalized size of antiderivative = 5.97 \[ \int \frac {A+B x^2}{\sqrt {-d+e x^2} \sqrt {-d e+\left (d^2+e^2\right ) x^2-d e x^4}} \, dx =\text {Too large to display} \] Input:

integrate((B*x^2+A)/(e*x^2-d)^(1/2)/(-d*e+(d^2+e^2)*x^2-d*e*x^4)^(1/2),x, 
algorithm="fricas")
 

Output:

[-1/2*(sqrt(-d^3 + d*e^2)*(B*d + A*e)*log(-(2*d^3*x^2 - (2*d^2*e - e^3)*x^ 
4 - d^2*e - 2*sqrt(-d*e*x^4 + (d^2 + e^2)*x^2 - d*e)*sqrt(-d^3 + d*e^2)*sq 
rt(e*x^2 - d)*x)/(e^2*x^4 - 2*d*e*x^2 + d^2)) + (B*d^2 - B*e^2)*sqrt(-d)*l 
og(-(2*d*e*x^4 - (2*d^2 + e^2)*x^2 - 2*sqrt(-d*e*x^4 + (d^2 + e^2)*x^2 - d 
*e)*sqrt(e*x^2 - d)*sqrt(-d)*x + d*e)/(e*x^2 - d)))/(d^3*e - d*e^3), -1/2* 
(2*(B*d^2 - B*e^2)*sqrt(d)*arctan(sqrt(-d*e*x^4 + (d^2 + e^2)*x^2 - d*e)*s 
qrt(e*x^2 - d)*sqrt(d)*x/(d*e*x^4 - (d^2 + e^2)*x^2 + d*e)) + sqrt(-d^3 + 
d*e^2)*(B*d + A*e)*log(-(2*d^3*x^2 - (2*d^2*e - e^3)*x^4 - d^2*e - 2*sqrt( 
-d*e*x^4 + (d^2 + e^2)*x^2 - d*e)*sqrt(-d^3 + d*e^2)*sqrt(e*x^2 - d)*x)/(e 
^2*x^4 - 2*d*e*x^2 + d^2)))/(d^3*e - d*e^3), 1/2*(2*sqrt(d^3 - d*e^2)*(B*d 
 + A*e)*arctan(sqrt(-d*e*x^4 + (d^2 + e^2)*x^2 - d*e)*sqrt(d^3 - d*e^2)*sq 
rt(e*x^2 - d)*x/(d^2*e*x^4 + d^2*e - (d^3 + d*e^2)*x^2)) - (B*d^2 - B*e^2) 
*sqrt(-d)*log(-(2*d*e*x^4 - (2*d^2 + e^2)*x^2 - 2*sqrt(-d*e*x^4 + (d^2 + e 
^2)*x^2 - d*e)*sqrt(e*x^2 - d)*sqrt(-d)*x + d*e)/(e*x^2 - d)))/(d^3*e - d* 
e^3), (sqrt(d^3 - d*e^2)*(B*d + A*e)*arctan(sqrt(-d*e*x^4 + (d^2 + e^2)*x^ 
2 - d*e)*sqrt(d^3 - d*e^2)*sqrt(e*x^2 - d)*x/(d^2*e*x^4 + d^2*e - (d^3 + d 
*e^2)*x^2)) - (B*d^2 - B*e^2)*sqrt(d)*arctan(sqrt(-d*e*x^4 + (d^2 + e^2)*x 
^2 - d*e)*sqrt(e*x^2 - d)*sqrt(d)*x/(d*e*x^4 - (d^2 + e^2)*x^2 + d*e)))/(d 
^3*e - d*e^3)]
 

Sympy [F]

\[ \int \frac {A+B x^2}{\sqrt {-d+e x^2} \sqrt {-d e+\left (d^2+e^2\right ) x^2-d e x^4}} \, dx=\int \frac {A + B x^{2}}{\sqrt {- \left (- d + e x^{2}\right ) \left (d x^{2} - e\right )} \sqrt {- d + e x^{2}}}\, dx \] Input:

integrate((B*x**2+A)/(e*x**2-d)**(1/2)/(-d*e+(d**2+e**2)*x**2-d*e*x**4)**( 
1/2),x)
 

Output:

Integral((A + B*x**2)/(sqrt(-(-d + e*x**2)*(d*x**2 - e))*sqrt(-d + e*x**2) 
), x)
 

Maxima [F]

\[ \int \frac {A+B x^2}{\sqrt {-d+e x^2} \sqrt {-d e+\left (d^2+e^2\right ) x^2-d e x^4}} \, dx=\int { \frac {B x^{2} + A}{\sqrt {-d e x^{4} + {\left (d^{2} + e^{2}\right )} x^{2} - d e} \sqrt {e x^{2} - d}} \,d x } \] Input:

integrate((B*x^2+A)/(e*x^2-d)^(1/2)/(-d*e+(d^2+e^2)*x^2-d*e*x^4)^(1/2),x, 
algorithm="maxima")
 

Output:

integrate((B*x^2 + A)/(sqrt(-d*e*x^4 + (d^2 + e^2)*x^2 - d*e)*sqrt(e*x^2 - 
 d)), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {A+B x^2}{\sqrt {-d+e x^2} \sqrt {-d e+\left (d^2+e^2\right ) x^2-d e x^4}} \, dx=\text {Timed out} \] Input:

integrate((B*x^2+A)/(e*x^2-d)^(1/2)/(-d*e+(d^2+e^2)*x^2-d*e*x^4)^(1/2),x, 
algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x^2}{\sqrt {-d+e x^2} \sqrt {-d e+\left (d^2+e^2\right ) x^2-d e x^4}} \, dx=\int \frac {B\,x^2+A}{\sqrt {e\,x^2-d}\,\sqrt {-d\,e\,x^4+\left (d^2+e^2\right )\,x^2-d\,e}} \,d x \] Input:

int((A + B*x^2)/((e*x^2 - d)^(1/2)*(x^2*(d^2 + e^2) - d*e - d*e*x^4)^(1/2) 
),x)
 

Output:

int((A + B*x^2)/((e*x^2 - d)^(1/2)*(x^2*(d^2 + e^2) - d*e - d*e*x^4)^(1/2) 
), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.53 \[ \int \frac {A+B x^2}{\sqrt {-d+e x^2} \sqrt {-d e+\left (d^2+e^2\right ) x^2-d e x^4}} \, dx=\frac {\sqrt {d}\, \left (\mathit {asin} \left (\frac {\sqrt {d}\, x}{\sqrt {e}}\right ) b \,d^{2}-\mathit {asin} \left (\frac {\sqrt {d}\, x}{\sqrt {e}}\right ) b \,e^{2}-\sqrt {d^{2}-e^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d}\, x}{\sqrt {e}}\right )}{2}\right ) d -e}{\sqrt {d^{2}-e^{2}}}\right ) a e -\sqrt {d^{2}-e^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d}\, x}{\sqrt {e}}\right )}{2}\right ) d -e}{\sqrt {d^{2}-e^{2}}}\right ) b d -\sqrt {d^{2}-e^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d}\, x}{\sqrt {e}}\right )}{2}\right ) d +e}{\sqrt {d^{2}-e^{2}}}\right ) a e -\sqrt {d^{2}-e^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d}\, x}{\sqrt {e}}\right )}{2}\right ) d +e}{\sqrt {d^{2}-e^{2}}}\right ) b d \right )}{d e \left (d^{2}-e^{2}\right )} \] Input:

int((B*x^2+A)/(e*x^2-d)^(1/2)/(-d*e+(d^2+e^2)*x^2-d*e*x^4)^(1/2),x)
 

Output:

(sqrt(d)*(asin((sqrt(d)*x)/sqrt(e))*b*d**2 - asin((sqrt(d)*x)/sqrt(e))*b*e 
**2 - sqrt(d**2 - e**2)*atan((tan(asin((sqrt(d)*x)/sqrt(e))/2)*d - e)/sqrt 
(d**2 - e**2))*a*e - sqrt(d**2 - e**2)*atan((tan(asin((sqrt(d)*x)/sqrt(e)) 
/2)*d - e)/sqrt(d**2 - e**2))*b*d - sqrt(d**2 - e**2)*atan((tan(asin((sqrt 
(d)*x)/sqrt(e))/2)*d + e)/sqrt(d**2 - e**2))*a*e - sqrt(d**2 - e**2)*atan( 
(tan(asin((sqrt(d)*x)/sqrt(e))/2)*d + e)/sqrt(d**2 - e**2))*b*d))/(d*e*(d* 
*2 - e**2))