3.1.90 \(\int \frac {\sqrt {1+x^2} \sqrt {2+x^2}}{a+b x^2} \, dx\) [90]

3.1.90.1 Optimal result

Integrand size = 28, antiderivative size = 192 \[ \int \frac {\sqrt {1+x^2} \sqrt {2+x^2}}{a+b x^2} \, dx=\frac {x \sqrt {2+x^2}}{b \sqrt {1+x^2}}-\frac {\sqrt {2} \sqrt {2+x^2} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{b \sqrt {1+x^2} \sqrt {\frac {2+x^2}{1+x^2}}}+\frac {\sqrt {2+x^2} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{\sqrt {2} b \sqrt {1+x^2} \sqrt {\frac {2+x^2}{1+x^2}}}-\frac {(a-2 b) \sqrt {2+x^2} \operatorname {EllipticPi}\left (1-\frac {b}{a},\arctan (x),\frac {1}{2}\right )}{\sqrt {2} a b \sqrt {1+x^2} \sqrt {\frac {2+x^2}{1+x^2}}} \]

output

x*(x^2+2)^(1/2)/b/(x^2+1)^(1/2)+1/2*(1/(x^2+1))^(1/2)*EllipticF(x/(x^2+1)^ 
(1/2),1/2*2^(1/2))*(x^2+2)^(1/2)/b*2^(1/2)/((x^2+2)/(x^2+1))^(1/2)-1/2*(a- 
2*b)*(1/(x^2+1))^(1/2)*EllipticPi(x/(x^2+1)^(1/2),1-b/a,1/2*2^(1/2))*(x^2+ 
2)^(1/2)/a/b*2^(1/2)/((x^2+2)/(x^2+1))^(1/2)-(1/(x^2+1))^(1/2)*EllipticE(x 
/(x^2+1)^(1/2),1/2*2^(1/2))*2^(1/2)*(x^2+2)^(1/2)/b/((x^2+2)/(x^2+1))^(1/2 
)
 

3.1.90.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.02 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.41 \[ \int \frac {\sqrt {1+x^2} \sqrt {2+x^2}}{a+b x^2} \, dx=\frac {i \left (-a b E\left (\left .i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )+(a-2 b) \left (a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )+(-a+b) \operatorname {EllipticPi}\left (\frac {2 b}{a},i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )\right )\right )}{a b^2} \]

input

Integrate[(Sqrt[1 + x^2]*Sqrt[2 + x^2])/(a + b*x^2),x]
 

output

(I*(-(a*b*EllipticE[I*ArcSinh[x/Sqrt[2]], 2]) + (a - 2*b)*(a*EllipticF[I*A 
rcSinh[x/Sqrt[2]], 2] + (-a + b)*EllipticPi[(2*b)/a, I*ArcSinh[x/Sqrt[2]], 
 2])))/(a*b^2)
 

3.1.90.3 Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {409, 324, 320, 388, 313, 414}

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.

\(\displaystyle \int \frac {\sqrt {x^2+1} \sqrt {x^2+2}}{a+b x^2} \, dx\)

\(\Big \downarrow \) 409

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

\(\Big \downarrow \) 324

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

\(\Big \downarrow \) 320

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

\(\Big \downarrow \) 388

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

\(\Big \downarrow \) 313

\(\displaystyle \frac {\frac {\sqrt {x^2+2} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^2+1} \sqrt {\frac {x^2+2}{x^2+1}}}-\frac {\sqrt {2} \sqrt {x^2+2} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {x^2+1} \sqrt {\frac {x^2+2}{x^2+1}}}+\frac {\sqrt {x^2+2} x}{\sqrt {x^2+1}}}{b}-\frac {(a-2 b) \int \frac {\sqrt {x^2+1}}{\sqrt {x^2+2} \left (b x^2+a\right )}dx}{b}\)

\(\Big \downarrow \) 414

\(\displaystyle \frac {\frac {\sqrt {x^2+2} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^2+1} \sqrt {\frac {x^2+2}{x^2+1}}}-\frac {\sqrt {2} \sqrt {x^2+2} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {x^2+1} \sqrt {\frac {x^2+2}{x^2+1}}}+\frac {\sqrt {x^2+2} x}{\sqrt {x^2+1}}}{b}-\frac {\sqrt {x^2+2} (a-2 b) \operatorname {EllipticPi}\left (1-\frac {b}{a},\arctan (x),\frac {1}{2}\right )}{\sqrt {2} a b \sqrt {x^2+1} \sqrt {\frac {x^2+2}{x^2+1}}}\)

input

Int[(Sqrt[1 + x^2]*Sqrt[2 + x^2])/(a + b*x^2),x]
 

output

((x*Sqrt[2 + x^2])/Sqrt[1 + x^2] - (Sqrt[2]*Sqrt[2 + x^2]*EllipticE[ArcTan 
[x], 1/2])/(Sqrt[1 + x^2]*Sqrt[(2 + x^2)/(1 + x^2)]) + (Sqrt[2 + x^2]*Elli 
pticF[ArcTan[x], 1/2])/(Sqrt[2]*Sqrt[1 + x^2]*Sqrt[(2 + x^2)/(1 + x^2)]))/ 
b - ((a - 2*b)*Sqrt[2 + x^2]*EllipticPi[1 - b/a, ArcTan[x], 1/2])/(Sqrt[2] 
*a*b*Sqrt[1 + x^2]*Sqrt[(2 + x^2)/(1 + x^2)])
 

3.1.90.3.1 Defintions of rubi rules used

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 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 324

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
a   Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] + Simp[b   Int[x^2/(Sqr 
t[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] && PosQ[d/c 
] && PosQ[b/a]
 

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 409

Int[(Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2])/((a_) + (b_.)*(x_ 
)^2), x_Symbol] :> Simp[d/b   Int[Sqrt[e + f*x^2]/Sqrt[c + d*x^2], x], x] + 
 Simp[(b*c - a*d)/b   Int[Sqrt[e + f*x^2]/((a + b*x^2)*Sqrt[c + d*x^2]), x] 
, x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[d/c, 0] && GtQ[f/e, 0] &&  !Sim 
plerSqrtQ[d/c, f/e]
 

rule 414

Int[Sqrt[(c_) + (d_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)*Sqrt[(e_) + (f_.)*(x_) 
^2]), x_Symbol] :> Simp[c*(Sqrt[e + f*x^2]/(a*e*Rt[d/c, 2]*Sqrt[c + d*x^2]* 
Sqrt[c*((e + f*x^2)/(e*(c + d*x^2)))]))*EllipticPi[1 - b*(c/(a*d)), ArcTan[ 
Rt[d/c, 2]*x], 1 - c*(f/(d*e))], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ 
[d/c]
 

3.1.90.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 3.33 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.63

method	result	size
default	\(\frac {i \left (F\left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) a^{2}-2 F\left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) b a -a^{2} \Pi \left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right )+3 \Pi \left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right ) b a -2 \Pi \left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right ) b^{2}-E\left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) b a \right )}{a \,b^{2}}\)	\(121\)
elliptic	\(\frac {\sqrt {\left (x^{2}+1\right ) \left (x^{2}+2\right )}\, \left (-\frac {i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right )}{b \sqrt {x^{4}+3 x^{2}+2}}+\frac {i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) a}{2 \sqrt {x^{4}+3 x^{2}+2}\, b^{2}}-\frac {i a \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \Pi \left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right )}{b^{2} \sqrt {x^{4}+3 x^{2}+2}}-\frac {2 i \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \Pi \left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right )}{a \sqrt {x^{4}+3 x^{2}+2}}+\frac {3 i \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \Pi \left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right )}{b \sqrt {x^{4}+3 x^{2}+2}}-\frac {i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, E\left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right )}{2 b \sqrt {x^{4}+3 x^{2}+2}}\right )}{\sqrt {x^{2}+1}\, \sqrt {x^{2}+2}}\)	\(338\)

input

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

output

I*(EllipticF(1/2*I*x*2^(1/2),2^(1/2))*a^2-2*EllipticF(1/2*I*x*2^(1/2),2^(1 
/2))*b*a-a^2*EllipticPi(1/2*I*x*2^(1/2),2*b/a,2^(1/2))+3*EllipticPi(1/2*I* 
x*2^(1/2),2*b/a,2^(1/2))*b*a-2*EllipticPi(1/2*I*x*2^(1/2),2*b/a,2^(1/2))*b 
^2-EllipticE(1/2*I*x*2^(1/2),2^(1/2))*b*a)/a/b^2
 

3.1.90.5 Fricas [F]

\[ \int \frac {\sqrt {1+x^2} \sqrt {2+x^2}}{a+b x^2} \, dx=\int { \frac {\sqrt {x^{2} + 2} \sqrt {x^{2} + 1}}{b x^{2} + a} \,d x } \]

input

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

output

integral(sqrt(x^2 + 2)*sqrt(x^2 + 1)/(b*x^2 + a), x)
 

3.1.90.6 Sympy [F]

\[ \int \frac {\sqrt {1+x^2} \sqrt {2+x^2}}{a+b x^2} \, dx=\int \frac {\sqrt {x^{2} + 1} \sqrt {x^{2} + 2}}{a + b x^{2}}\, dx \]

input

integrate((x**2+1)**(1/2)*(x**2+2)**(1/2)/(b*x**2+a),x)
 

output

Integral(sqrt(x**2 + 1)*sqrt(x**2 + 2)/(a + b*x**2), x)
 

3.1.90.7 Maxima [F]

\[ \int \frac {\sqrt {1+x^2} \sqrt {2+x^2}}{a+b x^2} \, dx=\int { \frac {\sqrt {x^{2} + 2} \sqrt {x^{2} + 1}}{b x^{2} + a} \,d x } \]

input

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

output

integrate(sqrt(x^2 + 2)*sqrt(x^2 + 1)/(b*x^2 + a), x)
 

3.1.90.8 Giac [F]

\[ \int \frac {\sqrt {1+x^2} \sqrt {2+x^2}}{a+b x^2} \, dx=\int { \frac {\sqrt {x^{2} + 2} \sqrt {x^{2} + 1}}{b x^{2} + a} \,d x } \]

input

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

output

integrate(sqrt(x^2 + 2)*sqrt(x^2 + 1)/(b*x^2 + a), x)
 

3.1.90.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {1+x^2} \sqrt {2+x^2}}{a+b x^2} \, dx=\int \frac {\sqrt {x^2+1}\,\sqrt {x^2+2}}{b\,x^2+a} \,d x \]

input

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

output

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