\(\int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx\) [5]

Optimal result

Integrand size = 35, antiderivative size = 106 \[ \int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\frac {B \arctan \left (\frac {\sqrt {e} x \sqrt {d+e x^2}}{\sqrt {d^2-e^2 x^4}}\right )}{e^{3/2}}-\frac {(B d-A e) \arctan \left (\frac {\sqrt {2} \sqrt {e} x \sqrt {d+e x^2}}{\sqrt {d^2-e^2 x^4}}\right )}{\sqrt {2} d e^{3/2}} \] 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 = 5.75 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.30 \[ \int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\frac {(-B d+A e) \sqrt {d^2-e^2 x^4} \arctan \left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {d-e x^2}}\right )}{\sqrt {2} d e^{3/2} \sqrt {d-e x^2} \sqrt {d+e x^2}}+\frac {i B \log \left (-2 i \sqrt {e} x+\frac {2 \sqrt {d^2-e^2 x^4}}{\sqrt {d+e x^2}}\right )}{e^{3/2}} \] 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[d^2 - e^2*x^4]*ArcTan[(Sqrt[2]*Sqrt[e]*x)/Sqrt[d - e* 
x^2]])/(Sqrt[2]*d*e^(3/2)*Sqrt[d - e*x^2]*Sqrt[d + e*x^2]) + (I*B*Log[(-2* 
I)*Sqrt[e]*x + (2*Sqrt[d^2 - e^2*x^4])/Sqrt[d + e*x^2]])/e^(3/2)
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {1396, 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.

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

\(\Big \downarrow \) 1396

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 218

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

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*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]])/(Sqrt[ 
2]*d*e^(3/2))))/Sqrt[d^2 - e^2*x^4]
 

Defintions of rubi rules used

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. \(433\) vs. \(2(86)=172\).

Time = 0.34 (sec) , antiderivative size = 434, normalized size of antiderivative = 4.09

method	result	size
default	\(-\frac {\sqrt {-e^{2} x^{4}+d^{2}}\, \left (-A \sqrt {2}\, \sqrt {d}\, \ln \left (\frac {2 e \left (\sqrt {2}\, \sqrt {d}\, \sqrt {-e \,x^{2}+d}-\sqrt {-d e}\, x +d \right )}{e x -\sqrt {-d e}}\right ) e^{\frac {3}{2}}+A \sqrt {2}\, \sqrt {d}\, \ln \left (\frac {2 e \left (\sqrt {2}\, \sqrt {d}\, \sqrt {-e \,x^{2}+d}+\sqrt {-d e}\, x +d \right )}{e x +\sqrt {-d e}}\right ) e^{\frac {3}{2}}+B \sqrt {2}\, d^{\frac {3}{2}} \ln \left (\frac {2 e \left (\sqrt {2}\, \sqrt {d}\, \sqrt {-e \,x^{2}+d}-\sqrt {-d e}\, x +d \right )}{e x -\sqrt {-d e}}\right ) \sqrt {e}-B \sqrt {2}\, d^{\frac {3}{2}} \ln \left (\frac {2 e \left (\sqrt {2}\, \sqrt {d}\, \sqrt {-e \,x^{2}+d}+\sqrt {-d e}\, x +d \right )}{e x +\sqrt {-d e}}\right ) \sqrt {e}-2 A \sqrt {-d e}\, \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {-e \,x^{2}+d}}\right ) e +2 A \sqrt {-d e}\, \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {\frac {\left (-e x +\sqrt {d e}\right ) \left (e x +\sqrt {d e}\right )}{e}}}\right ) e +2 B \sqrt {-d e}\, \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {-e \,x^{2}+d}}\right ) d +2 B \sqrt {-d e}\, \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {\frac {\left (-e x +\sqrt {d e}\right ) \left (e x +\sqrt {d e}\right )}{e}}}\right ) d \right )}{2 \sqrt {e \,x^{2}+d}\, \sqrt {-e \,x^{2}+d}\, \left (\sqrt {-d e}-\sqrt {d e}\right ) \left (\sqrt {-d e}+\sqrt {d e}\right ) \sqrt {-d e}\, \sqrt {e}}\)	\(434\)

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) 
*arctan(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 = 0.09 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.84 \[ \int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\left [-\frac {2 \, B d \sqrt {-e} \log \left (-\frac {2 \, e^{2} x^{4} + d e x^{2} - 2 \, \sqrt {-e^{2} x^{4} + d^{2}} \sqrt {e x^{2} + d} \sqrt {-e} x - d^{2}}{e x^{2} + d}\right ) - \sqrt {2} {\left (B d - A e\right )} \sqrt {-e} \log \left (-\frac {3 \, e^{2} x^{4} + 2 \, d e x^{2} - 2 \, \sqrt {2} \sqrt {-e^{2} x^{4} + d^{2}} \sqrt {e x^{2} + d} \sqrt {-e} x - d^{2}}{e^{2} x^{4} + 2 \, d e x^{2} + d^{2}}\right )}{4 \, d e^{2}}, -\frac {2 \, B d \sqrt {e} \arctan \left (\frac {\sqrt {-e^{2} x^{4} + d^{2}} \sqrt {e x^{2} + d} \sqrt {e} x}{e^{2} x^{4} - d^{2}}\right ) - \sqrt {2} {\left (B d - A e\right )} \sqrt {e} \arctan \left (\frac {\sqrt {2} \sqrt {-e^{2} x^{4} + d^{2}} \sqrt {e x^{2} + d} \sqrt {e} x}{e^{2} x^{4} - d^{2}}\right )}{2 \, d e^{2}}\right ] \] Input:

integrate((B*x^2+A)/(e*x^2+d)^(1/2)/(-e^2*x^4+d^2)^(1/2),x, algorithm="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)*s 
qrt(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^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)*sq 
rt(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="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 = 134, normalized size of antiderivative = 1.26 \[ \int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=-\frac {B \log \left ({\left | -\sqrt {-e} x + \sqrt {-e x^{2} + d} \right |}\right )}{\sqrt {-e} e} + \frac {\sqrt {2} {\left (B d - A e\right )} \sqrt {-e} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-e} x - \sqrt {-e x^{2} + d}\right )}^{2} - 4 \, \sqrt {2} {\left | d \right |} - 6 \, d \right |}}{{\left | 2 \, {\left (\sqrt {-e} x - \sqrt {-e x^{2} + d}\right )}^{2} + 4 \, \sqrt {2} {\left | d \right |} - 6 \, d \right |}}\right )}{4 \, e^{2} {\left | d \right |}} \] 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)* 
abs(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. \[ \int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\int \frac {B\,x^2+A}{\sqrt {d^2-e^2\,x^4}\,\sqrt {e\,x^2+d}} \,d x \] 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 = 184, normalized size of antiderivative = 1.74 \[ \int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\frac {\sqrt {e}\, \left (4 \mathit {asin} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right ) b d +2 \sqrt {2}\, \mathit {atan} \left (\frac {\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right )}{2}\right )}{\sqrt {2}+1}\right ) a e -2 \sqrt {2}\, \mathit {atan} \left (\frac {\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right )}{2}\right )}{\sqrt {2}+1}\right ) b d -\sqrt {2}\, \mathrm {log}\left (-\sqrt {2}\, i +\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right )}{2}\right )+i \right ) a e i +\sqrt {2}\, \mathrm {log}\left (-\sqrt {2}\, i +\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right )}{2}\right )+i \right ) b d i +\sqrt {2}\, \mathrm {log}\left (\sqrt {2}\, i +\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right )}{2}\right )-i \right ) a e i -\sqrt {2}\, \mathrm {log}\left (\sqrt {2}\, i +\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right )}{2}\right )-i \right ) b d i \right )}{4 d \,e^{2}} \] 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 + 2*sqrt(2)*atan(tan(asin((sqrt( 
e)*x)/sqrt(d))/2)/(sqrt(2) + 1))*a*e - 2*sqrt(2)*atan(tan(asin((sqrt(e)*x) 
/sqrt(d))/2)/(sqrt(2) + 1))*b*d - sqrt(2)*log( - sqrt(2)*i + tan(asin((sqr 
t(e)*x)/sqrt(d))/2) + i)*a*e*i + sqrt(2)*log( - sqrt(2)*i + tan(asin((sqrt 
(e)*x)/sqrt(d))/2) + i)*b*d*i + sqrt(2)*log(sqrt(2)*i + tan(asin((sqrt(e)* 
x)/sqrt(d))/2) - i)*a*e*i - sqrt(2)*log(sqrt(2)*i + tan(asin((sqrt(e)*x)/s 
qrt(d))/2) - i)*b*d*i))/(4*d*e**2)