\(\int \frac {\sqrt {1+2 x^2}}{\sqrt {3+5 x^2} \sqrt {7+11 x^2}} \, dx\) [493]

Optimal result

Integrand size = 34, antiderivative size = 94 \[ \int \frac {\sqrt {1+2 x^2}}{\sqrt {3+5 x^2} \sqrt {7+11 x^2}} \, dx=\frac {\sqrt {3+5 x^2} \sqrt {\frac {7+11 x^2}{1+2 x^2}} \operatorname {EllipticPi}\left (6,\arcsin \left (\frac {x}{\sqrt {3} \sqrt {1+2 x^2}}\right ),\frac {9}{7}\right )}{\sqrt {7} \sqrt {\frac {3+5 x^2}{1+2 x^2}} \sqrt {7+11 x^2}} \] Output:

1/7*(5*x^2+3)^(1/2)*((11*x^2+7)/(2*x^2+1))^(1/2)*EllipticPi(1/3*x*3^(1/2)/ 
(2*x^2+1)^(1/2),6,3/7*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 = 1.33 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {1+2 x^2}}{\sqrt {3+5 x^2} \sqrt {7+11 x^2}} \, dx=\frac {\sqrt {3+5 x^2} \sqrt {\frac {7+11 x^2}{1+2 x^2}} \operatorname {EllipticPi}\left (6,\arcsin \left (\frac {x}{\sqrt {3+6 x^2}}\right ),\frac {9}{7}\right )}{\sqrt {7} \sqrt {\frac {3+5 x^2}{1+2 x^2}} \sqrt {7+11 x^2}} \] Input:

Integrate[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)]*EllipticPi[6, ArcSin[x/Sqr 
t[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 = 94, 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 = {428, 27, 412}

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.

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

\(\Big \downarrow \) 428

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 412

\(\displaystyle \frac {\sqrt {5 x^2+3} \sqrt {\frac {11 x^2+7}{2 x^2+1}} \operatorname {EllipticPi}\left (6,\arcsin \left (\frac {x}{\sqrt {3} \sqrt {2 x^2+1}}\right ),\frac {9}{7}\right )}{\sqrt {7} \sqrt {\frac {5 x^2+3}{2 x^2+1}} \sqrt {11 x^2+7}}\)

Input:

Int[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)]*EllipticPi[6, ArcSin[x/(Sq 
rt[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 412

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x 
_)^2]), x_Symbol] :> Simp[(1/(a*Sqrt[c]*Sqrt[e]*Rt[-d/c, 2]))*EllipticPi[b* 
(c/(a*d)), ArcSin[Rt[-d/c, 2]*x], c*(f/(d*e))], x] /; FreeQ[{a, b, c, d, e, 
 f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && S 
implerSqrtQ[-f/e, -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="f 
ricas")
 

Output:

integral(sqrt(11*x^2 + 7)*sqrt(5*x^2 + 3)*sqrt(2*x^2 + 1)/(55*x^4 + 68*x^2 
 + 21), x)
 

Sympy [F]

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

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

Output:

Integral(sqrt(2*x**2 + 1)/(sqrt(5*x**2 + 3)*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="m 
axima")
 

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="g 
iac")
 

Output:

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

Mupad [F(-1)]

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

Reduce [F]

\[ \int \frac {\sqrt {1+2 x^2}}{\sqrt {3+5 x^2} \sqrt {7+11 x^2}} \, dx=\int \frac {\sqrt {2 x^{2}+1}\, \sqrt {5 x^{2}+3}\, \sqrt {11 x^{2}+7}}{55 x^{4}+68 x^{2}+21}d x \] Input:

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

Output:

int((sqrt(2*x**2 + 1)*sqrt(5*x**2 + 3)*sqrt(11*x**2 + 7))/(55*x**4 + 68*x* 
*2 + 21),x)