\(\int \sqrt {2+b x^2} \sqrt {3+d x^2} \, dx\) [168]

Optimal result

Integrand size = 23, antiderivative size = 233 \[ \int \sqrt {2+b x^2} \sqrt {3+d x^2} \, dx=\frac {(3 b+2 d) x \sqrt {3+d x^2}}{3 d \sqrt {2+b x^2}}+\frac {1}{3} x \sqrt {2+b x^2} \sqrt {3+d x^2}-\frac {(3 b+2 d) \sqrt {3+d x^2} E\left (\arctan \left (\frac {\sqrt {b} x}{\sqrt {2}}\right )|1-\frac {2 d}{3 b}\right )}{\sqrt {3} \sqrt {b} d \sqrt {2+b x^2} \sqrt {\frac {3+d x^2}{2+b x^2}}}+\frac {4 \sqrt {3+d x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {b} x}{\sqrt {2}}\right ),1-\frac {2 d}{3 b}\right )}{\sqrt {3} \sqrt {b} \sqrt {2+b x^2} \sqrt {\frac {3+d x^2}{2+b x^2}}} \] Output:

1/3*(3*b+2*d)*x*(d*x^2+3)^(1/2)/d/(b*x^2+2)^(1/2)+1/3*x*(b*x^2+2)^(1/2)*(d 
*x^2+3)^(1/2)-1/3*(3*b+2*d)*(d*x^2+3)^(1/2)*EllipticE(b^(1/2)*x*2^(1/2)/(2 
*b*x^2+4)^(1/2),1/3*(9-6*d/b)^(1/2))*3^(1/2)/b^(1/2)/d/(b*x^2+2)^(1/2)/((d 
*x^2+3)/(b*x^2+2))^(1/2)+4/3*(d*x^2+3)^(1/2)*InverseJacobiAM(arctan(1/2*b^ 
(1/2)*x*2^(1/2)),1/3*(9-6*d/b)^(1/2))*3^(1/2)/b^(1/2)/(b*x^2+2)^(1/2)/((d* 
x^2+3)/(b*x^2+2))^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.86 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.55 \[ \int \sqrt {2+b x^2} \sqrt {3+d x^2} \, dx=\frac {\sqrt {b} d x \sqrt {2+b x^2} \sqrt {3+d x^2}-i \sqrt {3} (3 b+2 d) E\left (i \text {arcsinh}\left (\frac {\sqrt {b} x}{\sqrt {2}}\right )|\frac {2 d}{3 b}\right )+i \sqrt {3} (3 b-2 d) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {b} x}{\sqrt {2}}\right ),\frac {2 d}{3 b}\right )}{3 \sqrt {b} d} \] Input:

Integrate[Sqrt[2 + b*x^2]*Sqrt[3 + d*x^2],x]
 

Output:

(Sqrt[b]*d*x*Sqrt[2 + b*x^2]*Sqrt[3 + d*x^2] - I*Sqrt[3]*(3*b + 2*d)*Ellip 
ticE[I*ArcSinh[(Sqrt[b]*x)/Sqrt[2]], (2*d)/(3*b)] + I*Sqrt[3]*(3*b - 2*d)* 
EllipticF[I*ArcSinh[(Sqrt[b]*x)/Sqrt[2]], (2*d)/(3*b)])/(3*Sqrt[b]*d)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {319, 27, 406, 320, 388, 313}

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.

\(\displaystyle \int \sqrt {b x^2+2} \sqrt {d x^2+3} \, dx\)

\(\Big \downarrow \) 319

\(\displaystyle \frac {2}{3} \int \frac {(3 b+2 d) x^2+12}{2 \sqrt {b x^2+2} \sqrt {d x^2+3}}dx+\frac {1}{3} x \sqrt {b x^2+2} \sqrt {d x^2+3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \frac {(3 b+2 d) x^2+12}{\sqrt {b x^2+2} \sqrt {d x^2+3}}dx+\frac {1}{3} x \sqrt {b x^2+2} \sqrt {d x^2+3}\)

\(\Big \downarrow \) 406

\(\displaystyle \frac {1}{3} \left (12 \int \frac {1}{\sqrt {b x^2+2} \sqrt {d x^2+3}}dx+(3 b+2 d) \int \frac {x^2}{\sqrt {b x^2+2} \sqrt {d x^2+3}}dx\right )+\frac {1}{3} x \sqrt {b x^2+2} \sqrt {d x^2+3}\)

\(\Big \downarrow \) 320

\(\displaystyle \frac {1}{3} \left ((3 b+2 d) \int \frac {x^2}{\sqrt {b x^2+2} \sqrt {d x^2+3}}dx+\frac {6 \sqrt {2} \sqrt {b x^2+2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {3}}\right ),1-\frac {3 b}{2 d}\right )}{\sqrt {d} \sqrt {d x^2+3} \sqrt {\frac {b x^2+2}{d x^2+3}}}\right )+\frac {1}{3} x \sqrt {b x^2+2} \sqrt {d x^2+3}\)

\(\Big \downarrow \) 388

\(\displaystyle \frac {1}{3} \left ((3 b+2 d) \left (\frac {x \sqrt {b x^2+2}}{b \sqrt {d x^2+3}}-\frac {3 \int \frac {\sqrt {b x^2+2}}{\left (d x^2+3\right )^{3/2}}dx}{b}\right )+\frac {6 \sqrt {2} \sqrt {b x^2+2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {3}}\right ),1-\frac {3 b}{2 d}\right )}{\sqrt {d} \sqrt {d x^2+3} \sqrt {\frac {b x^2+2}{d x^2+3}}}\right )+\frac {1}{3} x \sqrt {b x^2+2} \sqrt {d x^2+3}\)

\(\Big \downarrow \) 313

\(\displaystyle \frac {1}{3} \left (\frac {6 \sqrt {2} \sqrt {b x^2+2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {3}}\right ),1-\frac {3 b}{2 d}\right )}{\sqrt {d} \sqrt {d x^2+3} \sqrt {\frac {b x^2+2}{d x^2+3}}}+(3 b+2 d) \left (\frac {x \sqrt {b x^2+2}}{b \sqrt {d x^2+3}}-\frac {\sqrt {2} \sqrt {b x^2+2} E\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {3}}\right )|1-\frac {3 b}{2 d}\right )}{b \sqrt {d} \sqrt {d x^2+3} \sqrt {\frac {b x^2+2}{d x^2+3}}}\right )\right )+\frac {1}{3} x \sqrt {b x^2+2} \sqrt {d x^2+3}\)

Input:

Int[Sqrt[2 + b*x^2]*Sqrt[3 + d*x^2],x]
 

Output:

(x*Sqrt[2 + b*x^2]*Sqrt[3 + d*x^2])/3 + ((3*b + 2*d)*((x*Sqrt[2 + b*x^2])/ 
(b*Sqrt[3 + d*x^2]) - (Sqrt[2]*Sqrt[2 + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x 
)/Sqrt[3]], 1 - (3*b)/(2*d)])/(b*Sqrt[d]*Sqrt[(2 + b*x^2)/(3 + d*x^2)]*Sqr 
t[3 + d*x^2])) + (6*Sqrt[2]*Sqrt[2 + b*x^2]*EllipticF[ArcTan[(Sqrt[d]*x)/S 
qrt[3]], 1 - (3*b)/(2*d)])/(Sqrt[d]*Sqrt[(2 + b*x^2)/(3 + d*x^2)]*Sqrt[3 + 
 d*x^2]))/3
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 313

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Sim 
p[(Sqrt[a + b*x^2]/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*(c 
+ d*x^2)))]))*EllipticE[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /; FreeQ 
[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]
 

rule 319

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

rule 320

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(Sqrt[a + b*x^2]/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*( 
c + d*x^2)))]))*EllipticF[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /; Fre 
eQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]
 

rule 388

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] 
 :> Simp[x*(Sqrt[a + b*x^2]/(b*Sqrt[c + d*x^2])), x] - Simp[c/b   Int[Sqrt[ 
a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]
 

rule 406

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

Maple [A] (verified)

Time = 1.44 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.08

method	result	size
risch	\(\frac {x \sqrt {b \,x^{2}+2}\, \sqrt {x^{2} d +3}}{3}+\frac {\left (-\frac {\left (3 b +2 d \right ) \sqrt {3 x^{2} d +9}\, \sqrt {2 b \,x^{2}+4}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )\right )}{3 \sqrt {-3 d}\, \sqrt {b d \,x^{4}+3 b \,x^{2}+2 x^{2} d +6}\, b}+\frac {2 \sqrt {3 x^{2} d +9}\, \sqrt {2 b \,x^{2}+4}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )}{\sqrt {-3 d}\, \sqrt {b d \,x^{4}+3 b \,x^{2}+2 x^{2} d +6}}\right ) \sqrt {\left (b \,x^{2}+2\right ) \left (x^{2} d +3\right )}}{\sqrt {b \,x^{2}+2}\, \sqrt {x^{2} d +3}}\)	\(252\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+2\right ) \left (x^{2} d +3\right )}\, \left (\frac {x \sqrt {b d \,x^{4}+3 b \,x^{2}+2 x^{2} d +6}}{3}+\frac {2 \sqrt {3 x^{2} d +9}\, \sqrt {2 b \,x^{2}+4}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )}{\sqrt {-3 d}\, \sqrt {b d \,x^{4}+3 b \,x^{2}+2 x^{2} d +6}}-\frac {\left (b +\frac {2 d}{3}\right ) \sqrt {3 x^{2} d +9}\, \sqrt {2 b \,x^{2}+4}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-3 d}}{3}, \frac {\sqrt {-4+\frac {6 b +4 d}{d}}}{2}\right )\right )}{\sqrt {-3 d}\, \sqrt {b d \,x^{4}+3 b \,x^{2}+2 x^{2} d +6}\, b}\right )}{\sqrt {b \,x^{2}+2}\, \sqrt {x^{2} d +3}}\)	\(253\)
default	\(\frac {\sqrt {b \,x^{2}+2}\, \sqrt {x^{2} d +3}\, \left (b^{2} d \,x^{5} \sqrt {-d}+3 b^{2} x^{3} \sqrt {-d}+2 b d \,x^{3} \sqrt {-d}+3 \sqrt {2}\, \operatorname {EllipticF}\left (\frac {x \sqrt {3}\, \sqrt {-d}}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {b}{d}}}{2}\right ) b \sqrt {b \,x^{2}+2}\, \sqrt {x^{2} d +3}-2 \sqrt {2}\, \operatorname {EllipticF}\left (\frac {x \sqrt {3}\, \sqrt {-d}}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {b}{d}}}{2}\right ) d \sqrt {b \,x^{2}+2}\, \sqrt {x^{2} d +3}+3 \sqrt {2}\, \operatorname {EllipticE}\left (\frac {x \sqrt {3}\, \sqrt {-d}}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {b}{d}}}{2}\right ) b \sqrt {b \,x^{2}+2}\, \sqrt {x^{2} d +3}+2 \sqrt {2}\, \operatorname {EllipticE}\left (\frac {x \sqrt {3}\, \sqrt {-d}}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {b}{d}}}{2}\right ) d \sqrt {b \,x^{2}+2}\, \sqrt {x^{2} d +3}+6 b x \sqrt {-d}\right )}{3 \left (b d \,x^{4}+3 b \,x^{2}+2 x^{2} d +6\right ) \sqrt {-d}\, b}\)	\(303\)

Input:

int((b*x^2+2)^(1/2)*(d*x^2+3)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/3*x*(b*x^2+2)^(1/2)*(d*x^2+3)^(1/2)+(-1/3*(3*b+2*d)/(-3*d)^(1/2)*(3*d*x^ 
2+9)^(1/2)*(2*b*x^2+4)^(1/2)/(b*d*x^4+3*b*x^2+2*d*x^2+6)^(1/2)/b*(Elliptic 
F(1/3*x*(-3*d)^(1/2),1/2*(-4+2*(3*b+2*d)/d)^(1/2))-EllipticE(1/3*x*(-3*d)^ 
(1/2),1/2*(-4+2*(3*b+2*d)/d)^(1/2)))+2/(-3*d)^(1/2)*(3*d*x^2+9)^(1/2)*(2*b 
*x^2+4)^(1/2)/(b*d*x^4+3*b*x^2+2*d*x^2+6)^(1/2)*EllipticF(1/3*x*(-3*d)^(1/ 
2),1/2*(-4+2*(3*b+2*d)/d)^(1/2)))*((b*x^2+2)*(d*x^2+3))^(1/2)/(b*x^2+2)^(1 
/2)/(d*x^2+3)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.64 \[ \int \sqrt {2+b x^2} \sqrt {3+d x^2} \, dx=-\frac {3 \, \sqrt {3} \sqrt {b d} {\left (3 \, b + 2 \, d\right )} x \sqrt {-\frac {1}{d}} E(\arcsin \left (\frac {\sqrt {3} \sqrt {-\frac {1}{d}}}{x}\right )\,|\,\frac {2 \, d}{3 \, b}) - \sqrt {3} \sqrt {b d} {\left (4 \, d^{2} + 9 \, b + 6 \, d\right )} x \sqrt {-\frac {1}{d}} F(\arcsin \left (\frac {\sqrt {3} \sqrt {-\frac {1}{d}}}{x}\right )\,|\,\frac {2 \, d}{3 \, b}) - {\left (b d^{2} x^{2} + 3 \, b d + 2 \, d^{2}\right )} \sqrt {b x^{2} + 2} \sqrt {d x^{2} + 3}}{3 \, b d^{2} x} \] Input:

integrate((b*x^2+2)^(1/2)*(d*x^2+3)^(1/2),x, algorithm="fricas")
 

Output:

-1/3*(3*sqrt(3)*sqrt(b*d)*(3*b + 2*d)*x*sqrt(-1/d)*elliptic_e(arcsin(sqrt( 
3)*sqrt(-1/d)/x), 2/3*d/b) - sqrt(3)*sqrt(b*d)*(4*d^2 + 9*b + 6*d)*x*sqrt( 
-1/d)*elliptic_f(arcsin(sqrt(3)*sqrt(-1/d)/x), 2/3*d/b) - (b*d^2*x^2 + 3*b 
*d + 2*d^2)*sqrt(b*x^2 + 2)*sqrt(d*x^2 + 3))/(b*d^2*x)
 

Sympy [F]

\[ \int \sqrt {2+b x^2} \sqrt {3+d x^2} \, dx=\int \sqrt {b x^{2} + 2} \sqrt {d x^{2} + 3}\, dx \] Input:

integrate((b*x**2+2)**(1/2)*(d*x**2+3)**(1/2),x)
 

Output:

Integral(sqrt(b*x**2 + 2)*sqrt(d*x**2 + 3), x)
 

Maxima [F]

\[ \int \sqrt {2+b x^2} \sqrt {3+d x^2} \, dx=\int { \sqrt {b x^{2} + 2} \sqrt {d x^{2} + 3} \,d x } \] Input:

integrate((b*x^2+2)^(1/2)*(d*x^2+3)^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*x^2 + 2)*sqrt(d*x^2 + 3), x)
 

Giac [F]

\[ \int \sqrt {2+b x^2} \sqrt {3+d x^2} \, dx=\int { \sqrt {b x^{2} + 2} \sqrt {d x^{2} + 3} \,d x } \] Input:

integrate((b*x^2+2)^(1/2)*(d*x^2+3)^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(b*x^2 + 2)*sqrt(d*x^2 + 3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {2+b x^2} \sqrt {3+d x^2} \, dx=\int \sqrt {b\,x^2+2}\,\sqrt {d\,x^2+3} \,d x \] Input:

int((b*x^2 + 2)^(1/2)*(d*x^2 + 3)^(1/2),x)
 

Output:

int((b*x^2 + 2)^(1/2)*(d*x^2 + 3)^(1/2), x)
 

Reduce [F]

\[ \int \sqrt {2+b x^2} \sqrt {3+d x^2} \, dx=\frac {\sqrt {d \,x^{2}+3}\, \sqrt {b \,x^{2}+2}\, x}{3}+\left (\int \frac {\sqrt {d \,x^{2}+3}\, \sqrt {b \,x^{2}+2}\, x^{2}}{b d \,x^{4}+3 b \,x^{2}+2 d \,x^{2}+6}d x \right ) b +\frac {2 \left (\int \frac {\sqrt {d \,x^{2}+3}\, \sqrt {b \,x^{2}+2}\, x^{2}}{b d \,x^{4}+3 b \,x^{2}+2 d \,x^{2}+6}d x \right ) d}{3}+4 \left (\int \frac {\sqrt {d \,x^{2}+3}\, \sqrt {b \,x^{2}+2}}{b d \,x^{4}+3 b \,x^{2}+2 d \,x^{2}+6}d x \right ) \] Input:

int((b*x^2+2)^(1/2)*(d*x^2+3)^(1/2),x)
 

Output:

(sqrt(d*x**2 + 3)*sqrt(b*x**2 + 2)*x + 3*int((sqrt(d*x**2 + 3)*sqrt(b*x**2 
 + 2)*x**2)/(b*d*x**4 + 3*b*x**2 + 2*d*x**2 + 6),x)*b + 2*int((sqrt(d*x**2 
 + 3)*sqrt(b*x**2 + 2)*x**2)/(b*d*x**4 + 3*b*x**2 + 2*d*x**2 + 6),x)*d + 1 
2*int((sqrt(d*x**2 + 3)*sqrt(b*x**2 + 2))/(b*d*x**4 + 3*b*x**2 + 2*d*x**2 
+ 6),x))/3