\(\int \frac {A+B x^2}{\sqrt {-d-e x^2} \sqrt {-d^2+e^2 x^4}} \, dx\) [12]

Optimal result

Integrand size = 39, antiderivative size = 114 \[ \int \frac {A+B x^2}{\sqrt {-d-e x^2} \sqrt {-d^2+e^2 x^4}} \, dx=-\frac {B \arctan \left (\frac {\sqrt {e} x \sqrt {-d-e x^2}}{\sqrt {-d^2+e^2 x^4}}\right )}{e^{3/2}}+\frac {(B d-A e) \arctan \left (\frac {\sqrt {2} \sqrt {e} x \sqrt {-d-e x^2}}{\sqrt {-d^2+e^2 x^4}}\right )}{\sqrt {2} d e^{3/2}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 5.89 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.24 \[ \int \frac {A+B x^2}{\sqrt {-d-e x^2} \sqrt {-d^2+e^2 x^4}} \, dx=\frac {\frac {(B d-A e) \sqrt {-2 d^2+2 e^2 x^4} \arctan \left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {d-e x^2}}\right )}{d \sqrt {-d-e x^2} \sqrt {d-e x^2}}+2 i B \log \left (-2 i \sqrt {e} x-\frac {2 \sqrt {-d^2+e^2 x^4}}{\sqrt {-d-e x^2}}\right )}{2 e^{3/2}} \] Input:

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

Output:

(((B*d - A*e)*Sqrt[-2*d^2 + 2*e^2*x^4]*ArcTan[(Sqrt[2]*Sqrt[e]*x)/Sqrt[d - 
 e*x^2]])/(d*Sqrt[-d - e*x^2]*Sqrt[d - e*x^2]) + (2*I)*B*Log[(-2*I)*Sqrt[e 
]*x - (2*Sqrt[-d^2 + e^2*x^4])/Sqrt[-d - e*x^2]])/(2*e^(3/2))
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1396, 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.

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

\(\Big \downarrow \) 1396

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

\(\displaystyle -\frac {\sqrt {-d-e x^2} \sqrt {d-e x^2} \left (\frac {B \arctan \left (\frac {\sqrt {e} x}{\sqrt {d-e x^2}}\right )}{e^{3/2}}-\frac {(B d-A e) \int \frac {1}{\sqrt {d-e x^2} \left (e x^2+d\right )}dx}{e}\right )}{\sqrt {e^2 x^4-d^2}}\)

\(\Big \downarrow \) 291

\(\displaystyle -\frac {\sqrt {-d-e x^2} \sqrt {d-e x^2} \left (\frac {B \arctan \left (\frac {\sqrt {e} x}{\sqrt {d-e x^2}}\right )}{e^{3/2}}-\frac {(B d-A e) \int \frac {1}{\frac {2 d e x^2}{d-e x^2}+d}d\frac {x}{\sqrt {d-e x^2}}}{e}\right )}{\sqrt {e^2 x^4-d^2}}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {\sqrt {-d-e x^2} \sqrt {d-e x^2} \left (\frac {B \arctan \left (\frac {\sqrt {e} x}{\sqrt {d-e x^2}}\right )}{e^{3/2}}-\frac {(B d-A e) \arctan \left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {d-e x^2}}\right )}{\sqrt {2} d e^{3/2}}\right )}{\sqrt {e^2 x^4-d^2}}\)

Input:

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

Output:

-((Sqrt[-d - e*x^2]*Sqrt[d - e*x^2]*((B*ArcTan[(Sqrt[e]*x)/Sqrt[d - e*x^2] 
])/e^(3/2) - ((B*d - A*e)*ArcTan[(Sqrt[2]*Sqrt[e]*x)/Sqrt[d - e*x^2]])/(Sq 
rt[2]*d*e^(3/2))))/Sqrt[-d^2 + e^2*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 1396

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_)*((d_) + (e_.)*(x_)^(n_))^(q_.), x 
_Symbol] :> Simp[(a + 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, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + 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. \(437\) vs. \(2(95)=190\).

Time = 0.07 (sec) , antiderivative size = 438, normalized size of antiderivative = 3.84

method	result	size
default	\(\frac {\sqrt {e^{2} x^{4}-d^{2}}\, \left (A \sqrt {2}\, \sqrt {d}\, \ln \left (\frac {2 e \left (\sqrt {2}\, \sqrt {d}\, \sqrt {-e \,x^{2}+d}-\sqrt {-d e}\, x +d \right )}{e x -\sqrt {-d e}}\right ) e^{\frac {3}{2}}-A \sqrt {2}\, \sqrt {d}\, \ln \left (\frac {2 e \left (\sqrt {2}\, \sqrt {d}\, \sqrt {-e \,x^{2}+d}+\sqrt {-d e}\, x +d \right )}{e x +\sqrt {-d e}}\right ) e^{\frac {3}{2}}-B \sqrt {2}\, d^{\frac {3}{2}} \ln \left (\frac {2 e \left (\sqrt {2}\, \sqrt {d}\, \sqrt {-e \,x^{2}+d}-\sqrt {-d e}\, x +d \right )}{e x -\sqrt {-d e}}\right ) \sqrt {e}+B \sqrt {2}\, d^{\frac {3}{2}} \ln \left (\frac {2 e \left (\sqrt {2}\, \sqrt {d}\, \sqrt {-e \,x^{2}+d}+\sqrt {-d e}\, x +d \right )}{e x +\sqrt {-d e}}\right ) \sqrt {e}+2 A \sqrt {-d e}\, \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {-e \,x^{2}+d}}\right ) e -2 A \sqrt {-d e}\, \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {\frac {\left (-e x +\sqrt {d e}\right ) \left (e x +\sqrt {d e}\right )}{e}}}\right ) e -2 B \sqrt {-d e}\, \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {-e \,x^{2}+d}}\right ) d -2 B \sqrt {-d e}\, \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {\frac {\left (-e x +\sqrt {d e}\right ) \left (e x +\sqrt {d e}\right )}{e}}}\right ) d \right )}{2 \sqrt {-e \,x^{2}-d}\, \sqrt {-e \,x^{2}+d}\, \left (-\sqrt {-d e}+\sqrt {d e}\right ) \left (\sqrt {-d e}+\sqrt {d e}\right ) \sqrt {-d e}\, \sqrt {e}}\)	\(438\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.52 \[ \int \frac {A+B x^2}{\sqrt {-d-e x^2} \sqrt {-d^2+e^2 x^4}} \, dx=\left [-\frac {2 \, B d \sqrt {-e} \log \left (-\frac {2 \, e^{2} x^{4} + d e x^{2} - 2 \, \sqrt {e^{2} x^{4} - d^{2}} \sqrt {-e x^{2} - d} \sqrt {-e} x - d^{2}}{e x^{2} + d}\right ) - \sqrt {2} {\left (B d - A e\right )} \sqrt {-e} \log \left (-\frac {3 \, e^{2} x^{4} + 2 \, d e x^{2} - 2 \, \sqrt {2} \sqrt {e^{2} x^{4} - d^{2}} \sqrt {-e x^{2} - d} \sqrt {-e} x - d^{2}}{e^{2} x^{4} + 2 \, d e x^{2} + d^{2}}\right )}{4 \, d e^{2}}, -\frac {2 \, B d \sqrt {e} \arctan \left (\frac {\sqrt {-e x^{2} - d} \sqrt {e} x}{\sqrt {e^{2} x^{4} - d^{2}}}\right ) - \sqrt {2} {\left (B d - A e\right )} \sqrt {e} \arctan \left (\frac {\sqrt {2} \sqrt {-e x^{2} - d} \sqrt {e} x}{\sqrt {e^{2} x^{4} - d^{2}}}\right )}{2 \, d e^{2}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.17 \[ \int \frac {A+B x^2}{\sqrt {-d-e x^2} \sqrt {-d^2+e^2 x^4}} \, dx=\frac {B \log \left ({\left | -\sqrt {-e} x + \sqrt {-e x^{2} + d} \right |}\right )}{\sqrt {-e} e} - \frac {\sqrt {2} {\left (B d - A e\right )} \sqrt {-e} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-e} x - \sqrt {-e x^{2} + d}\right )}^{2} - 4 \, \sqrt {2} {\left | d \right |} - 6 \, d \right |}}{{\left | 2 \, {\left (\sqrt {-e} x - \sqrt {-e x^{2} + d}\right )}^{2} + 4 \, \sqrt {2} {\left | d \right |} - 6 \, d \right |}}\right )}{4 \, e^{2} {\left | d \right |}} \] Input:

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

Output:

B*log(abs(-sqrt(-e)*x + sqrt(-e*x^2 + d)))/(sqrt(-e)*e) - 1/4*sqrt(2)*(B*d 
 - A*e)*sqrt(-e)*log(abs(2*(sqrt(-e)*x - sqrt(-e*x^2 + d))^2 - 4*sqrt(2)*a 
bs(d) - 6*d)/abs(2*(sqrt(-e)*x - sqrt(-e*x^2 + d))^2 + 4*sqrt(2)*abs(d) - 
6*d))/(e^2*abs(d))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.35 \[ \int \frac {A+B x^2}{\sqrt {-d-e x^2} \sqrt {-d^2+e^2 x^4}} \, dx=\frac {\sqrt {e}\, i \left (-\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}-d}-\sqrt {d}\, \sqrt {2}\, i +\sqrt {d}\, i +\sqrt {e}\, x}{\sqrt {d}}\right ) a e +\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}-d}-\sqrt {d}\, \sqrt {2}\, i +\sqrt {d}\, i +\sqrt {e}\, x}{\sqrt {d}}\right ) b d -\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}-d}+\sqrt {d}\, \sqrt {2}\, i -\sqrt {d}\, i +\sqrt {e}\, x}{\sqrt {d}}\right ) a e +\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}-d}+\sqrt {d}\, \sqrt {2}\, i -\sqrt {d}\, i +\sqrt {e}\, x}{\sqrt {d}}\right ) b d +\sqrt {2}\, \mathrm {log}\left (\frac {2 \sqrt {e}\, \sqrt {e \,x^{2}-d}\, x +2 \sqrt {2}\, d +2 d +2 e \,x^{2}}{d}\right ) a e -\sqrt {2}\, \mathrm {log}\left (\frac {2 \sqrt {e}\, \sqrt {e \,x^{2}-d}\, x +2 \sqrt {2}\, d +2 d +2 e \,x^{2}}{d}\right ) b d -4 \,\mathrm {log}\left (\frac {\sqrt {e \,x^{2}-d}+\sqrt {e}\, x}{\sqrt {d}}\right ) b d \right )}{4 d \,e^{2}} \] Input:

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

Output:

(sqrt(e)*i*( - sqrt(2)*log((sqrt( - d + e*x**2) - sqrt(d)*sqrt(2)*i + sqrt 
(d)*i + sqrt(e)*x)/sqrt(d))*a*e + sqrt(2)*log((sqrt( - d + e*x**2) - sqrt( 
d)*sqrt(2)*i + sqrt(d)*i + sqrt(e)*x)/sqrt(d))*b*d - sqrt(2)*log((sqrt( - 
d + e*x**2) + sqrt(d)*sqrt(2)*i - sqrt(d)*i + sqrt(e)*x)/sqrt(d))*a*e + sq 
rt(2)*log((sqrt( - d + e*x**2) + sqrt(d)*sqrt(2)*i - sqrt(d)*i + sqrt(e)*x 
)/sqrt(d))*b*d + sqrt(2)*log((2*sqrt(e)*sqrt( - d + e*x**2)*x + 2*sqrt(2)* 
d + 2*d + 2*e*x**2)/d)*a*e - sqrt(2)*log((2*sqrt(e)*sqrt( - d + e*x**2)*x 
+ 2*sqrt(2)*d + 2*d + 2*e*x**2)/d)*b*d - 4*log((sqrt( - d + e*x**2) + sqrt 
(e)*x)/sqrt(d))*b*d))/(4*d*e**2)