\(\int \frac {\sqrt {1-2 x^2}}{\sqrt {3-5 x^2} \sqrt {7+11 x^2}} \, dx\) [498]

Optimal result

Integrand size = 34, antiderivative size = 194 \[ \int \frac {\sqrt {1-2 x^2}}{\sqrt {3-5 x^2} \sqrt {7+11 x^2}} \, dx=-\frac {\sqrt {3-5 x^2} \sqrt {\frac {7+11 x^2}{1-2 x^2}} \operatorname {EllipticF}\left (\arctan \left (\frac {x}{\sqrt {3} \sqrt {1-2 x^2}}\right ),-\frac {68}{7}\right )}{5 \sqrt {7} \sqrt {\frac {3-5 x^2}{1-2 x^2}} \sqrt {7+11 x^2}}+\frac {6 \sqrt {3-5 x^2} \sqrt {\frac {7+11 x^2}{1-2 x^2}} \operatorname {EllipticPi}\left (-5,\arctan \left (\frac {x}{\sqrt {3} \sqrt {1-2 x^2}}\right ),-\frac {68}{7}\right )}{5 \sqrt {7} \sqrt {\frac {3-5 x^2}{1-2 x^2}} \sqrt {7+11 x^2}} \] Output:

-1/35*(-5*x^2+3)^(1/2)*((11*x^2+7)/(-2*x^2+1))^(1/2)*InverseJacobiAM(arcta 
n(1/3*x*3^(1/2)/(-2*x^2+1)^(1/2)),2/7*I*119^(1/2))*7^(1/2)/((-5*x^2+3)/(-2 
*x^2+1))^(1/2)/(11*x^2+7)^(1/2)+6/35*(-5*x^2+3)^(1/2)*((11*x^2+7)/(-2*x^2+ 
1))^(1/2)*EllipticPi(x*3^(1/2)/(-2*x^2+1)^(1/2)/(9+3*x^2/(-2*x^2+1))^(1/2) 
,-5,2/7*I*119^(1/2))*7^(1/2)/((-5*x^2+3)/(-2*x^2+1))^(1/2)/(11*x^2+7)^(1/2 
)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.32 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.49 \[ \int \frac {\sqrt {1-2 x^2}}{\sqrt {3-5 x^2} \sqrt {7+11 x^2}} \, dx=-\frac {i \sqrt {3-5 x^2} \sqrt {\frac {7+11 x^2}{1-2 x^2}} \operatorname {EllipticPi}\left (\frac {14}{25},i \text {arcsinh}\left (\frac {5 x}{\sqrt {7-14 x^2}}\right ),\frac {7}{75}\right )}{5 \sqrt {\frac {9-15 x^2}{1-2 x^2}} \sqrt {7+11 x^2}} \] Input:

Integrate[Sqrt[1 - 2*x^2]/(Sqrt[3 - 5*x^2]*Sqrt[7 + 11*x^2]),x]
 

Output:

((-1/5*I)*Sqrt[3 - 5*x^2]*Sqrt[(7 + 11*x^2)/(1 - 2*x^2)]*EllipticPi[14/25, 
 I*ArcSinh[(5*x)/Sqrt[7 - 14*x^2]], 7/75])/(Sqrt[(9 - 15*x^2)/(1 - 2*x^2)] 
*Sqrt[7 + 11*x^2])
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.47, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {428, 27, 411, 320, 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.

\(\displaystyle \int \frac {\sqrt {1-2 x^2}}{\sqrt {3-5 x^2} \sqrt {11 x^2+7}} \, dx\)

\(\Big \downarrow \) 428

\(\displaystyle \frac {\sqrt {3-5 x^2} \sqrt {\frac {11 x^2+7}{1-2 x^2}} \int \frac {\sqrt {21}}{\sqrt {\frac {x^2}{1-2 x^2}+3} \left (\frac {2 x^2}{1-2 x^2}+1\right ) \sqrt {\frac {25 x^2}{1-2 x^2}+7}}d\frac {x}{\sqrt {1-2 x^2}}}{\sqrt {21} \sqrt {\frac {3-5 x^2}{1-2 x^2}} \sqrt {11 x^2+7}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {3-5 x^2} \sqrt {\frac {11 x^2+7}{1-2 x^2}} \int \frac {1}{\sqrt {\frac {x^2}{1-2 x^2}+3} \left (\frac {2 x^2}{1-2 x^2}+1\right ) \sqrt {\frac {25 x^2}{1-2 x^2}+7}}d\frac {x}{\sqrt {1-2 x^2}}}{\sqrt {\frac {3-5 x^2}{1-2 x^2}} \sqrt {11 x^2+7}}\)

\(\Big \downarrow \) 411

\(\displaystyle \frac {\sqrt {3-5 x^2} \sqrt {\frac {11 x^2+7}{1-2 x^2}} \left (\frac {2}{5} \int \frac {\sqrt {\frac {x^2}{1-2 x^2}+3}}{\left (\frac {2 x^2}{1-2 x^2}+1\right ) \sqrt {\frac {25 x^2}{1-2 x^2}+7}}d\frac {x}{\sqrt {1-2 x^2}}-\frac {1}{5} \int \frac {1}{\sqrt {\frac {x^2}{1-2 x^2}+3} \sqrt {\frac {25 x^2}{1-2 x^2}+7}}d\frac {x}{\sqrt {1-2 x^2}}\right )}{\sqrt {\frac {3-5 x^2}{1-2 x^2}} \sqrt {11 x^2+7}}\)

\(\Big \downarrow \) 320

\(\displaystyle \frac {\sqrt {3-5 x^2} \sqrt {\frac {11 x^2+7}{1-2 x^2}} \left (\frac {2}{5} \int \frac {\sqrt {\frac {x^2}{1-2 x^2}+3}}{\left (\frac {2 x^2}{1-2 x^2}+1\right ) \sqrt {\frac {25 x^2}{1-2 x^2}+7}}d\frac {x}{\sqrt {1-2 x^2}}-\frac {\sqrt {\frac {25 x^2}{1-2 x^2}+7} \operatorname {EllipticF}\left (\arctan \left (\frac {x}{\sqrt {3} \sqrt {1-2 x^2}}\right ),-\frac {68}{7}\right )}{5 \sqrt {7} \sqrt {\frac {x^2}{1-2 x^2}+3} \sqrt {\frac {\frac {25 x^2}{1-2 x^2}+7}{\frac {x^2}{1-2 x^2}+3}}}\right )}{\sqrt {\frac {3-5 x^2}{1-2 x^2}} \sqrt {11 x^2+7}}\)

\(\Big \downarrow \) 414

\(\displaystyle \frac {\sqrt {3-5 x^2} \sqrt {\frac {11 x^2+7}{1-2 x^2}} \left (\frac {6 \sqrt {\frac {25 x^2}{1-2 x^2}+7} \operatorname {EllipticPi}\left (-5,\arctan \left (\frac {x}{\sqrt {3} \sqrt {1-2 x^2}}\right ),-\frac {68}{7}\right )}{5 \sqrt {7} \sqrt {\frac {x^2}{1-2 x^2}+3} \sqrt {\frac {\frac {25 x^2}{1-2 x^2}+7}{\frac {x^2}{1-2 x^2}+3}}}-\frac {\sqrt {\frac {25 x^2}{1-2 x^2}+7} \operatorname {EllipticF}\left (\arctan \left (\frac {x}{\sqrt {3} \sqrt {1-2 x^2}}\right ),-\frac {68}{7}\right )}{5 \sqrt {7} \sqrt {\frac {x^2}{1-2 x^2}+3} \sqrt {\frac {\frac {25 x^2}{1-2 x^2}+7}{\frac {x^2}{1-2 x^2}+3}}}\right )}{\sqrt {\frac {3-5 x^2}{1-2 x^2}} \sqrt {11 x^2+7}}\)

Input:

Int[Sqrt[1 - 2*x^2]/(Sqrt[3 - 5*x^2]*Sqrt[7 + 11*x^2]),x]
 

Output:

(Sqrt[3 - 5*x^2]*Sqrt[(7 + 11*x^2)/(1 - 2*x^2)]*(-1/5*(Sqrt[7 + (25*x^2)/( 
1 - 2*x^2)]*EllipticF[ArcTan[x/(Sqrt[3]*Sqrt[1 - 2*x^2])], -68/7])/(Sqrt[7 
]*Sqrt[3 + x^2/(1 - 2*x^2)]*Sqrt[(7 + (25*x^2)/(1 - 2*x^2))/(3 + x^2/(1 - 
2*x^2))]) + (6*Sqrt[7 + (25*x^2)/(1 - 2*x^2)]*EllipticPi[-5, ArcTan[x/(Sqr 
t[3]*Sqrt[1 - 2*x^2])], -68/7])/(5*Sqrt[7]*Sqrt[3 + x^2/(1 - 2*x^2)]*Sqrt[ 
(7 + (25*x^2)/(1 - 2*x^2))/(3 + x^2/(1 - 2*x^2))])))/(Sqrt[(3 - 5*x^2)/(1 
- 2*x^2)]*Sqrt[7 + 11*x^2])
 

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 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 411

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

rule 428

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

Maple [F]

\[\int \frac {\sqrt {-2 x^{2}+1}}{\sqrt {-5 x^{2}+3}\, \sqrt {11 x^{2}+7}}d x\]

Input:

int((-2*x^2+1)^(1/2)/(-5*x^2+3)^(1/2)/(11*x^2+7)^(1/2),x)
 

Output:

int((-2*x^2+1)^(1/2)/(-5*x^2+3)^(1/2)/(11*x^2+7)^(1/2),x)
 

Fricas [F]

\[ \int \frac {\sqrt {1-2 x^2}}{\sqrt {3-5 x^2} \sqrt {7+11 x^2}} \, dx=\int { \frac {\sqrt {-2 \, x^{2} + 1}}{\sqrt {11 \, x^{2} + 7} \sqrt {-5 \, x^{2} + 3}} \,d x } \] Input:

integrate((-2*x^2+1)^(1/2)/(-5*x^2+3)^(1/2)/(11*x^2+7)^(1/2),x, algorithm= 
"fricas")
 

Output:

integral(-sqrt(11*x^2 + 7)*sqrt(-2*x^2 + 1)*sqrt(-5*x^2 + 3)/(55*x^4 + 2*x 
^2 - 21), x)
 

Sympy [F]

\[ \int \frac {\sqrt {1-2 x^2}}{\sqrt {3-5 x^2} \sqrt {7+11 x^2}} \, dx=\int \frac {\sqrt {1 - 2 x^{2}}}{\sqrt {3 - 5 x^{2}} \sqrt {11 x^{2} + 7}}\, dx \] Input:

integrate((-2*x**2+1)**(1/2)/(-5*x**2+3)**(1/2)/(11*x**2+7)**(1/2),x)
 

Output:

Integral(sqrt(1 - 2*x**2)/(sqrt(3 - 5*x**2)*sqrt(11*x**2 + 7)), x)
 

Maxima [F]

\[ \int \frac {\sqrt {1-2 x^2}}{\sqrt {3-5 x^2} \sqrt {7+11 x^2}} \, dx=\int { \frac {\sqrt {-2 \, x^{2} + 1}}{\sqrt {11 \, x^{2} + 7} \sqrt {-5 \, x^{2} + 3}} \,d x } \] Input:

integrate((-2*x^2+1)^(1/2)/(-5*x^2+3)^(1/2)/(11*x^2+7)^(1/2),x, algorithm= 
"maxima")
 

Output:

integrate(sqrt(-2*x^2 + 1)/(sqrt(11*x^2 + 7)*sqrt(-5*x^2 + 3)), x)
 

Giac [F]

\[ \int \frac {\sqrt {1-2 x^2}}{\sqrt {3-5 x^2} \sqrt {7+11 x^2}} \, dx=\int { \frac {\sqrt {-2 \, x^{2} + 1}}{\sqrt {11 \, x^{2} + 7} \sqrt {-5 \, x^{2} + 3}} \,d x } \] Input:

integrate((-2*x^2+1)^(1/2)/(-5*x^2+3)^(1/2)/(11*x^2+7)^(1/2),x, algorithm= 
"giac")
 

Output:

integrate(sqrt(-2*x^2 + 1)/(sqrt(11*x^2 + 7)*sqrt(-5*x^2 + 3)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {1-2 x^2}}{\sqrt {3-5 x^2} \sqrt {7+11 x^2}} \, dx=\int \frac {\sqrt {1-2\,x^2}}{\sqrt {3-5\,x^2}\,\sqrt {11\,x^2+7}} \,d x \] Input:

int((1 - 2*x^2)^(1/2)/((3 - 5*x^2)^(1/2)*(11*x^2 + 7)^(1/2)),x)
 

Output:

int((1 - 2*x^2)^(1/2)/((3 - 5*x^2)^(1/2)*(11*x^2 + 7)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {\sqrt {1-2 x^2}}{\sqrt {3-5 x^2} \sqrt {7+11 x^2}} \, dx=-\left (\int \frac {\sqrt {11 x^{2}+7}\, \sqrt {-2 x^{2}+1}\, \sqrt {-5 x^{2}+3}}{55 x^{4}+2 x^{2}-21}d x \right ) \] Input:

int((-2*x^2+1)^(1/2)/(-5*x^2+3)^(1/2)/(11*x^2+7)^(1/2),x)
 

Output:

 - int((sqrt(11*x**2 + 7)*sqrt( - 2*x**2 + 1)*sqrt( - 5*x**2 + 3))/(55*x** 
4 + 2*x**2 - 21),x)