\(\int \frac {A+B x^2}{\sqrt {-d-e x^2} \sqrt {d^2-e^2 x^4}} \, dx\) [8]

Optimal result

Integrand size = 38, antiderivative size = 112 \[ \int \frac {A+B x^2}{\sqrt {-d-e x^2} \sqrt {d^2-e^2 x^4}} \, dx=-\frac {B \text {arctanh}\left (\frac {\sqrt {e} x \sqrt {-d-e x^2}}{\sqrt {d^2-e^2 x^4}}\right )}{e^{3/2}}+\frac {(B d-A e) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} x \sqrt {-d-e x^2}}{\sqrt {d^2-e^2 x^4}}\right )}{\sqrt {2} d e^{3/2}} \] Output:

-B*arctanh(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)*arctanh(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 [A] (verified)

Time = 5.79 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.42 \[ \int \frac {A+B x^2}{\sqrt {-d-e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\frac {\frac {\sqrt {2} (B d-A e) \sqrt {d^2-e^2 x^4} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {-d+e x^2}}\right )}{d \sqrt {-d-e x^2} \sqrt {-d+e x^2}}+2 B \left (-\log \left (d+e x^2\right )+\log \left (d e x+e^2 x^3+\sqrt {e} \sqrt {-d-e x^2} \sqrt {d^2-e^2 x^4}\right )\right )}{2 e^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {1396, 25, 398, 224, 219, 291, 221}

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.

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

\(\Big \downarrow \) 1396

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 221

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

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*ArcTanh[(Sqrt[e]*x)/Sqrt[-d + e*x 
^2]])/e^(3/2) - ((B*d - A*e)*ArcTanh[(Sqrt[2]*Sqrt[e]*x)/Sqrt[-d + e*x^2]] 
)/(Sqrt[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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[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. \(512\) vs. \(2(93)=186\).

Time = 0.10 (sec) , antiderivative size = 513, normalized size of antiderivative = 4.58

method	result	size
default	\(\frac {\sqrt {-e^{2} x^{4}+d^{2}}\, \left (A \sqrt {-d e}\, \ln \left (\frac {\sqrt {e \,x^{2}-d}\, \sqrt {e}+e x}{\sqrt {e}}\right ) \sqrt {2}\, \sqrt {-d}\, e -A \sqrt {-d e}\, \ln \left (\frac {\sqrt {e}\, \sqrt {-\frac {\left (-e x +\sqrt {d e}\right ) \left (e x +\sqrt {d e}\right )}{e}}+e x}{\sqrt {e}}\right ) \sqrt {2}\, \sqrt {-d}\, e +A \ln \left (\frac {2 e \left (\sqrt {-d e}\, x +\sqrt {2}\, \sqrt {-d}\, \sqrt {e \,x^{2}-d}-d \right )}{e x -\sqrt {-d e}}\right ) e^{\frac {3}{2}} d -A \ln \left (\frac {2 e \left (\sqrt {2}\, \sqrt {-d}\, \sqrt {e \,x^{2}-d}-\sqrt {-d e}\, x -d \right )}{e x +\sqrt {-d e}}\right ) e^{\frac {3}{2}} d -B \sqrt {-d e}\, \ln \left (\frac {\sqrt {e \,x^{2}-d}\, \sqrt {e}+e x}{\sqrt {e}}\right ) \sqrt {2}\, \sqrt {-d}\, d -B \sqrt {-d e}\, \ln \left (\frac {\sqrt {e}\, \sqrt {-\frac {\left (-e x +\sqrt {d e}\right ) \left (e x +\sqrt {d e}\right )}{e}}+e x}{\sqrt {e}}\right ) \sqrt {2}\, \sqrt {-d}\, d -B \ln \left (\frac {2 e \left (\sqrt {-d e}\, x +\sqrt {2}\, \sqrt {-d}\, \sqrt {e \,x^{2}-d}-d \right )}{e x -\sqrt {-d e}}\right ) \sqrt {e}\, d^{2}+B \ln \left (\frac {2 e \left (\sqrt {2}\, \sqrt {-d}\, \sqrt {e \,x^{2}-d}-\sqrt {-d e}\, x -d \right )}{e x +\sqrt {-d e}}\right ) \sqrt {e}\, d^{2}\right ) \sqrt {2}}{2 \sqrt {-e \,x^{2}-d}\, \sqrt {e \,x^{2}-d}\, \left (-\sqrt {-d e}+\sqrt {d e}\right ) \left (\sqrt {-d e}+\sqrt {d e}\right ) \sqrt {-d e}\, \sqrt {e}\, \sqrt {-d}}\)	\(513\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.79 \[ \int \frac {A+B x^2}{\sqrt {-d-e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\left [\frac {2 \, B d \sqrt {e} \log \left (-\frac {2 \, e^{2} x^{4} + d e x^{2} + 2 \, \sqrt {-e^{2} x^{4} + d^{2}} \sqrt {-e x^{2} - d} \sqrt {e} x - d^{2}}{e x^{2} + d}\right ) - \sqrt {2} {\left (B d - A e\right )} \sqrt {e} \log \left (-\frac {3 \, e^{2} x^{4} + 2 \, d e x^{2} + 2 \, \sqrt {2} \sqrt {-e^{2} x^{4} + d^{2}} \sqrt {-e x^{2} - d} \sqrt {e} x - d^{2}}{e^{2} x^{4} + 2 \, d e x^{2} + d^{2}}\right )}{4 \, d e^{2}}, -\frac {2 \, B d \sqrt {-e} \arctan \left (\frac {\sqrt {-e^{2} x^{4} + d^{2}} \sqrt {-e x^{2} - d} \sqrt {-e} x}{e^{2} x^{4} - d^{2}}\right ) - \sqrt {2} {\left (B d - A e\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {2} \sqrt {-e^{2} x^{4} + d^{2}} \sqrt {-e x^{2} - d} \sqrt {-e} x}{e^{2} x^{4} - d^{2}}\right )}{2 \, d e^{2}}\right ] \] Input:

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

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)*sqr 
t(-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*sq 
rt(-e)*arctan(sqrt(-e^2*x^4 + d^2)*sqrt(-e*x^2 - d)*sqrt(-e)*x/(e^2*x^4 - 
d^2)) - sqrt(2)*(B*d - A*e)*sqrt(-e)*arctan(sqrt(2)*sqrt(-e^2*x^4 + d^2)*s 
qrt(-e*x^2 - d)*sqrt(-e)*x/(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="ma 
xima")
 

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 = 0.21 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.07 \[ \int \frac {A+B x^2}{\sqrt {-d-e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\frac {B \log \left ({\left | -\sqrt {e} x + \sqrt {e x^{2} - d} \right |}\right )}{e^{\frac {3}{2}}} - \frac {\sqrt {2} {\left (B d - A e\right )} \log \left (\frac {{\left | 2 \, {\left (\sqrt {e} x - \sqrt {e x^{2} - d}\right )}^{2} - 4 \, \sqrt {2} {\left | d \right |} + 6 \, d \right |}}{{\left | 2 \, {\left (\sqrt {e} x - \sqrt {e x^{2} - d}\right )}^{2} + 4 \, \sqrt {2} {\left | d \right |} + 6 \, d \right |}}\right )}{4 \, e^{\frac {3}{2}} {\left | d \right |}} \] Input:

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

Output:

B*log(abs(-sqrt(e)*x + sqrt(e*x^2 - d)))/e^(3/2) - 1/4*sqrt(2)*(B*d - A*e) 
*log(abs(2*(sqrt(e)*x - sqrt(e*x^2 - d))^2 - 4*sqrt(2)*abs(d) + 6*d)/abs(2 
*(sqrt(e)*x - sqrt(e*x^2 - d))^2 + 4*sqrt(2)*abs(d) + 6*d))/(e^(3/2)*abs(d 
))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.63 \[ \int \frac {A+B x^2}{\sqrt {-d-e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\frac {\sqrt {e}\, \left (-4 \mathit {asin} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right ) b d i -2 \sqrt {2}\, \mathit {atan} \left (\frac {\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right )}{2}\right )}{\sqrt {2}+1}\right ) a e i +2 \sqrt {2}\, \mathit {atan} \left (\frac {\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right )}{2}\right )}{\sqrt {2}+1}\right ) b d i -\sqrt {2}\, \mathrm {log}\left (-\sqrt {2}\, i +\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right )}{2}\right )+i \right ) a e +\sqrt {2}\, \mathrm {log}\left (-\sqrt {2}\, i +\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right )}{2}\right )+i \right ) b d +\sqrt {2}\, \mathrm {log}\left (\sqrt {2}\, i +\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right )}{2}\right )-i \right ) a e -\sqrt {2}\, \mathrm {log}\left (\sqrt {2}\, i +\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right )}{2}\right )-i \right ) b d \right )}{4 d \,e^{2}} \] Input:

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

Output:

(sqrt(e)*( - 4*asin((sqrt(e)*x)/sqrt(d))*b*d*i - 2*sqrt(2)*atan(tan(asin(( 
sqrt(e)*x)/sqrt(d))/2)/(sqrt(2) + 1))*a*e*i + 2*sqrt(2)*atan(tan(asin((sqr 
t(e)*x)/sqrt(d))/2)/(sqrt(2) + 1))*b*d*i - sqrt(2)*log( - sqrt(2)*i + tan( 
asin((sqrt(e)*x)/sqrt(d))/2) + i)*a*e + sqrt(2)*log( - sqrt(2)*i + tan(asi 
n((sqrt(e)*x)/sqrt(d))/2) + i)*b*d + sqrt(2)*log(sqrt(2)*i + tan(asin((sqr 
t(e)*x)/sqrt(d))/2) - i)*a*e - sqrt(2)*log(sqrt(2)*i + tan(asin((sqrt(e)*x 
)/sqrt(d))/2) - i)*b*d))/(4*d*e**2)