\(\int \frac {1}{\sqrt {1+2 x^2} \sqrt {3+5 x^2} \sqrt {7+11 x^2}} \, dx\) [485]

Optimal result

Integrand size = 34, antiderivative size = 93 \[ \int \frac {1}{\sqrt {1+2 x^2} \sqrt {3+5 x^2} \sqrt {7+11 x^2}} \, dx=\frac {\sqrt {3+5 x^2} \sqrt {\frac {7+11 x^2}{1+2 x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{\sqrt {3} \sqrt {1+2 x^2}}\right ),\frac {9}{7}\right )}{\sqrt {7} \sqrt {\frac {3+5 x^2}{1+2 x^2}} \sqrt {7+11 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.14 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\sqrt {1+2 x^2} \sqrt {3+5 x^2} \sqrt {7+11 x^2}} \, dx=\frac {\sqrt {3+5 x^2} \sqrt {\frac {7+11 x^2}{1+2 x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{\sqrt {3+6 x^2}}\right ),\frac {9}{7}\right )}{\sqrt {7} \sqrt {\frac {3+5 x^2}{1+2 x^2}} \sqrt {7+11 x^2}} \] Input:

Integrate[1/(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)]*EllipticF[ArcSin[x/Sqrt[3 
+ 6*x^2]], 9/7])/(Sqrt[7]*Sqrt[(3 + 5*x^2)/(1 + 2*x^2)]*Sqrt[7 + 11*x^2])
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {427, 27, 321}

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.

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

\(\Big \downarrow \) 427

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 321

\(\displaystyle \frac {\sqrt {5 x^2+3} \sqrt {\frac {11 x^2+7}{2 x^2+1}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{\sqrt {3} \sqrt {2 x^2+1}}\right ),\frac {9}{7}\right )}{\sqrt {7} \sqrt {\frac {5 x^2+3}{2 x^2+1}} \sqrt {11 x^2+7}}\)

Input:

Int[1/(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)]*EllipticF[ArcSin[x/(Sqrt[3 
]*Sqrt[1 + 2*x^2])], 9/7])/(Sqrt[7]*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 321

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]))*EllipticF[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] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])
 

rule 427

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_. 
)*(x_)^2]), x_Symbol] :> Simp[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/(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]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(1/(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(5*x^2 + 3)*sqrt(2*x^2 + 1)/(110*x^6 + 191*x 
^4 + 110*x^2 + 21), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sqrt {1+2 x^2} \sqrt {3+5 x^2} \sqrt {7+11 x^2}} \, dx=\int \frac {\sqrt {2 x^{2}+1}\, \sqrt {5 x^{2}+3}\, \sqrt {11 x^{2}+7}}{110 x^{6}+191 x^{4}+110 x^{2}+21}d x \] Input:

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

Output:

int((sqrt(2*x**2 + 1)*sqrt(5*x**2 + 3)*sqrt(11*x**2 + 7))/(110*x**6 + 191* 
x**4 + 110*x**2 + 21),x)