\(\int \frac {\sqrt {1+b x^2} (A+B x^2)}{\sqrt {1-b x^2}} \, dx\) [2]

Optimal result

Integrand size = 31, antiderivative size = 65 \[ \int \frac {\sqrt {1+b x^2} \left (A+B x^2\right )}{\sqrt {1-b x^2}} \, dx=-\frac {B x \sqrt {1-b^2 x^4}}{3 b}+\frac {(A b+B) E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{b^{3/2}}-\frac {2 B \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{3 b^{3/2}} \] Output:

-1/3*B*x*(-b^2*x^4+1)^(1/2)/b+(A*b+B)*EllipticE(b^(1/2)*x,I)/b^(3/2)-2/3*B 
*EllipticF(b^(1/2)*x,I)/b^(3/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.42 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.26 \[ \int \frac {\sqrt {1+b x^2} \left (A+B x^2\right )}{\sqrt {1-b x^2}} \, dx=\frac {\sqrt {-b} B x \sqrt {1-b^2 x^4}+3 i (A b+B) E\left (\left .i \text {arcsinh}\left (\sqrt {-b} x\right )\right |-1\right )-2 i B \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-b} x\right ),-1\right )}{3 (-b)^{3/2}} \] Input:

Integrate[(Sqrt[1 + b*x^2]*(A + B*x^2))/Sqrt[1 - b*x^2],x]
 

Output:

(Sqrt[-b]*B*x*Sqrt[1 - b^2*x^4] + (3*I)*(A*b + B)*EllipticE[I*ArcSinh[Sqrt 
[-b]*x], -1] - (2*I)*B*EllipticF[I*ArcSinh[Sqrt[-b]*x], -1])/(3*(-b)^(3/2) 
)
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.25, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {403, 25, 399, 284, 327, 762}

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.

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

\(\Big \downarrow \) 403

\(\displaystyle -\frac {\int -\frac {3 b (A b+B) x^2+3 A b+B}{\sqrt {1-b x^2} \sqrt {b x^2+1}}dx}{3 b}-\frac {B x \sqrt {1-b x^2} \sqrt {b x^2+1}}{3 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {3 b (A b+B) x^2+3 A b+B}{\sqrt {1-b x^2} \sqrt {b x^2+1}}dx}{3 b}-\frac {B x \sqrt {1-b x^2} \sqrt {b x^2+1}}{3 b}\)

\(\Big \downarrow \) 399

\(\displaystyle \frac {3 (A b+B) \int \frac {\sqrt {b x^2+1}}{\sqrt {1-b x^2}}dx-2 B \int \frac {1}{\sqrt {1-b x^2} \sqrt {b x^2+1}}dx}{3 b}-\frac {B x \sqrt {1-b x^2} \sqrt {b x^2+1}}{3 b}\)

\(\Big \downarrow \) 284

\(\displaystyle \frac {3 (A b+B) \int \frac {\sqrt {b x^2+1}}{\sqrt {1-b x^2}}dx-2 B \int \frac {1}{\sqrt {1-b^2 x^4}}dx}{3 b}-\frac {B x \sqrt {1-b x^2} \sqrt {b x^2+1}}{3 b}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {\frac {3 (A b+B) E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b}}-2 B \int \frac {1}{\sqrt {1-b^2 x^4}}dx}{3 b}-\frac {B x \sqrt {1-b x^2} \sqrt {b x^2+1}}{3 b}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {\frac {3 (A b+B) E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b}}-\frac {2 B \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{\sqrt {b}}}{3 b}-\frac {B x \sqrt {1-b x^2} \sqrt {b x^2+1}}{3 b}\)

Input:

Int[(Sqrt[1 + b*x^2]*(A + B*x^2))/Sqrt[1 - b*x^2],x]
 

Output:

-1/3*(B*x*Sqrt[1 - b*x^2]*Sqrt[1 + b*x^2])/b + ((3*(A*b + B)*EllipticE[Arc 
Sin[Sqrt[b]*x], -1])/Sqrt[b] - (2*B*EllipticF[ArcSin[Sqrt[b]*x], -1])/Sqrt 
[b])/(3*b)
 

Defintions of rubi rules used

rule 25

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

rule 284

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> I 
nt[(a*c + b*d*x^4)^p, x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[b*c + a*d, 0] 
&& (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0]))
 

rule 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) 
)], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
 

rule 399

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_) 
^2]), x_Symbol] :> Simp[f/b   Int[Sqrt[a + b*x^2]/Sqrt[c + d*x^2], x], x] + 
 Simp[(b*e - a*f)/b   Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; Fr 
eeQ[{a, b, c, d, e, f}, x] &&  !((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && 
(PosQ[d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c])))))
 

rule 403

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( 
x_)^2), x_Symbol] :> Simp[f*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b*(2*(p + 
 q + 1) + 1))), x] + Simp[1/(b*(2*(p + q + 1) + 1))   Int[(a + b*x^2)^p*(c 
+ d*x^2)^(q - 1)*Simp[c*(b*e - a*f + b*e*2*(p + q + 1)) + (d*(b*e - a*f) + 
f*2*q*(b*c - a*d) + b*d*e*2*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, 
 d, e, f, p}, x] && GtQ[q, 0] && NeQ[2*(p + q + 1) + 1, 0]
 

rule 762

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) 
)*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] 
 && GtQ[a, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (51 ) = 102\).

Time = 1.80 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.20

method	result	size
default	\(-\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (B \,b^{\frac {5}{2}} x^{5}+3 A \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \operatorname {EllipticE}\left (\sqrt {b}\, x , i\right ) b -2 B \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )+3 B \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )-B \sqrt {b}\, x \right )}{3 \left (b^{2} x^{4}-1\right ) b^{\frac {3}{2}}}\)	\(143\)
elliptic	\(\frac {\sqrt {-b^{2} x^{4}+1}\, \left (-\frac {B x \sqrt {-b^{2} x^{4}+1}}{3 b}+\frac {\left (A +\frac {B}{3 b}\right ) \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )}{\sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}-\frac {\left (b A +B \right ) \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{b^{\frac {3}{2}} \sqrt {-b^{2} x^{4}+1}}\right )}{\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}}\)	\(164\)
risch	\(\frac {B x \sqrt {b \,x^{2}+1}\, \left (b \,x^{2}-1\right ) \sqrt {\left (b \,x^{2}+1\right ) \left (-b \,x^{2}+1\right )}}{3 b \sqrt {-\left (b \,x^{2}+1\right ) \left (b \,x^{2}-1\right )}\, \sqrt {-b \,x^{2}+1}}+\frac {\left (\frac {B \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )}{\sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}-\frac {3 \left (b A +B \right ) \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{\sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}+\frac {3 \sqrt {b}\, A \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )}{\sqrt {-b^{2} x^{4}+1}}\right ) \sqrt {\left (b \,x^{2}+1\right ) \left (-b \,x^{2}+1\right )}}{3 b \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}}\)	\(263\)

Input:

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

Output:

-1/3*(b*x^2+1)^(1/2)*(-b*x^2+1)^(1/2)*(B*b^(5/2)*x^5+3*A*(b*x^2+1)^(1/2)*( 
-b*x^2+1)^(1/2)*EllipticE(b^(1/2)*x,I)*b-2*B*(b*x^2+1)^(1/2)*(-b*x^2+1)^(1 
/2)*EllipticF(b^(1/2)*x,I)+3*B*(b*x^2+1)^(1/2)*(-b*x^2+1)^(1/2)*EllipticE( 
b^(1/2)*x,I)-B*b^(1/2)*x)/(b^2*x^4-1)/b^(3/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (49) = 98\).

Time = 0.08 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.78 \[ \int \frac {\sqrt {1+b x^2} \left (A+B x^2\right )}{\sqrt {1-b x^2}} \, dx=-\frac {\frac {3 \, {\left (A b + B\right )} \sqrt {-b^{2}} x E(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}} - \frac {{\left (3 \, A b^{2} + {\left (3 \, A + B\right )} b + 3 \, B\right )} \sqrt {-b^{2}} x F(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}} + {\left (B b^{2} x^{2} + 3 \, A b^{2} + 3 \, B b\right )} \sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1}}{3 \, b^{3} x} \] Input:

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

Output:

-1/3*(3*(A*b + B)*sqrt(-b^2)*x*elliptic_e(arcsin(1/(sqrt(b)*x)), -1)/sqrt( 
b) - (3*A*b^2 + (3*A + B)*b + 3*B)*sqrt(-b^2)*x*elliptic_f(arcsin(1/(sqrt( 
b)*x)), -1)/sqrt(b) + (B*b^2*x^2 + 3*A*b^2 + 3*B*b)*sqrt(b*x^2 + 1)*sqrt(- 
b*x^2 + 1))/(b^3*x)
 

Sympy [F]

\[ \int \frac {\sqrt {1+b x^2} \left (A+B x^2\right )}{\sqrt {1-b x^2}} \, dx=\int \frac {\left (A + B x^{2}\right ) \sqrt {b x^{2} + 1}}{\sqrt {- b x^{2} + 1}}\, dx \] Input:

integrate((b*x**2+1)**(1/2)*(B*x**2+A)/(-b*x**2+1)**(1/2),x)
 

Output:

Integral((A + B*x**2)*sqrt(b*x**2 + 1)/sqrt(-b*x**2 + 1), x)
 

Maxima [F]

\[ \int \frac {\sqrt {1+b x^2} \left (A+B x^2\right )}{\sqrt {1-b x^2}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \sqrt {b x^{2} + 1}}{\sqrt {-b x^{2} + 1}} \,d x } \] Input:

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

Output:

integrate((B*x^2 + A)*sqrt(b*x^2 + 1)/sqrt(-b*x^2 + 1), x)
 

Giac [F]

\[ \int \frac {\sqrt {1+b x^2} \left (A+B x^2\right )}{\sqrt {1-b x^2}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \sqrt {b x^{2} + 1}}{\sqrt {-b x^{2} + 1}} \,d x } \] Input:

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

Output:

integrate((B*x^2 + A)*sqrt(b*x^2 + 1)/sqrt(-b*x^2 + 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {1+b x^2} \left (A+B x^2\right )}{\sqrt {1-b x^2}} \, dx=\int \frac {\left (B\,x^2+A\right )\,\sqrt {b\,x^2+1}}{\sqrt {1-b\,x^2}} \,d x \] Input:

int(((A + B*x^2)*(b*x^2 + 1)^(1/2))/(1 - b*x^2)^(1/2),x)
 

Output:

int(((A + B*x^2)*(b*x^2 + 1)^(1/2))/(1 - b*x^2)^(1/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {1+b x^2} \left (A+B x^2\right )}{\sqrt {1-b x^2}} \, dx=-\left (\int \frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, x^{2}}{b \,x^{2}-1}d x \right ) b -\left (\int \frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}}{b \,x^{2}-1}d x \right ) a \] Input:

int((b*x^2+1)^(1/2)*(B*x^2+A)/(-b*x^2+1)^(1/2),x)
 

Output:

 - (int((sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1)*x**2)/(b*x**2 - 1),x)*b + in 
t((sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1))/(b*x**2 - 1),x)*a)