\(\int \frac {7+10 x^2}{\sqrt {\frac {2+3 x^2}{5+7 x^2}} (5+7 x^2)^2} \, dx\) [53]

Optimal result

Integrand size = 38, antiderivative size = 46 \[ \int \frac {7+10 x^2}{\sqrt {\frac {2+3 x^2}{5+7 x^2}} \left (5+7 x^2\right )^2} \, dx=\frac {1}{5} \sqrt {\frac {2}{7}} E\left (\arctan \left (\sqrt {\frac {7}{5}} x\right )|-\frac {1}{14}\right )+\frac {\operatorname {EllipticF}\left (\arctan \left (\sqrt {\frac {7}{5}} x\right ),-\frac {1}{14}\right )}{\sqrt {14}} \] Output:

1/35*14^(1/2)*EllipticE(35^(1/2)*x/(35*x^2+25)^(1/2),1/14*I*14^(1/2))+1/14 
*InverseJacobiAM(arctan(1/5*35^(1/2)*x),1/14*I*14^(1/2))*14^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.14 (sec) , antiderivative size = 131, normalized size of antiderivative = 2.85 \[ \int \frac {7+10 x^2}{\sqrt {\frac {2+3 x^2}{5+7 x^2}} \left (5+7 x^2\right )^2} \, dx=\frac {\sqrt {\frac {2+3 x^2}{5+7 x^2}} \left (28 x+42 x^3+2 i \sqrt {5+7 x^2} \sqrt {28+42 x^2} E\left (i \text {arcsinh}\left (\sqrt {\frac {7}{5}} x\right )|\frac {15}{14}\right )-7 i \sqrt {5+7 x^2} \sqrt {28+42 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {7}{5}} x\right ),\frac {15}{14}\right )\right )}{70 \left (2+3 x^2\right )} \] Input:

Integrate[(7 + 10*x^2)/(Sqrt[(2 + 3*x^2)/(5 + 7*x^2)]*(5 + 7*x^2)^2),x]
 

Output:

(Sqrt[(2 + 3*x^2)/(5 + 7*x^2)]*(28*x + 42*x^3 + (2*I)*Sqrt[5 + 7*x^2]*Sqrt 
[28 + 42*x^2]*EllipticE[I*ArcSinh[Sqrt[7/5]*x], 15/14] - (7*I)*Sqrt[5 + 7* 
x^2]*Sqrt[28 + 42*x^2]*EllipticF[I*ArcSinh[Sqrt[7/5]*x], 15/14]))/(70*(2 + 
 3*x^2))
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(132\) vs. \(2(46)=92\).

Time = 0.26 (sec) , antiderivative size = 132, normalized size of antiderivative = 2.87, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2050, 400, 313, 320}

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.

\(\displaystyle \int \frac {10 x^2+7}{\sqrt {\frac {3 x^2+2}{7 x^2+5}} \left (7 x^2+5\right )^2} \, dx\)

\(\Big \downarrow \) 2050

\(\displaystyle \int \frac {10 x^2+7}{\sqrt {3 x^2+2} \left (7 x^2+5\right )^{3/2}}dx\)

\(\Big \downarrow \) 400

\(\displaystyle \int \frac {\sqrt {3 x^2+2}}{\left (7 x^2+5\right )^{3/2}}dx+\int \frac {1}{\sqrt {3 x^2+2} \sqrt {7 x^2+5}}dx\)

\(\Big \downarrow \) 313

\(\displaystyle \int \frac {1}{\sqrt {3 x^2+2} \sqrt {7 x^2+5}}dx+\frac {\sqrt {\frac {2}{7}} \sqrt {3 x^2+2} E\left (\arctan \left (\sqrt {\frac {7}{5}} x\right )|-\frac {1}{14}\right )}{5 \sqrt {\frac {3 x^2+2}{7 x^2+5}} \sqrt {7 x^2+5}}\)

\(\Big \downarrow \) 320

\(\displaystyle \frac {\sqrt {3 x^2+2} \operatorname {EllipticF}\left (\arctan \left (\sqrt {\frac {7}{5}} x\right ),-\frac {1}{14}\right )}{\sqrt {14} \sqrt {\frac {3 x^2+2}{7 x^2+5}} \sqrt {7 x^2+5}}+\frac {\sqrt {\frac {2}{7}} \sqrt {3 x^2+2} E\left (\arctan \left (\sqrt {\frac {7}{5}} x\right )|-\frac {1}{14}\right )}{5 \sqrt {\frac {3 x^2+2}{7 x^2+5}} \sqrt {7 x^2+5}}\)

Input:

Int[(7 + 10*x^2)/(Sqrt[(2 + 3*x^2)/(5 + 7*x^2)]*(5 + 7*x^2)^2),x]
 

Output:

(Sqrt[2/7]*Sqrt[2 + 3*x^2]*EllipticE[ArcTan[Sqrt[7/5]*x], -1/14])/(5*Sqrt[ 
(2 + 3*x^2)/(5 + 7*x^2)]*Sqrt[5 + 7*x^2]) + (Sqrt[2 + 3*x^2]*EllipticF[Arc 
Tan[Sqrt[7/5]*x], -1/14])/(Sqrt[14]*Sqrt[(2 + 3*x^2)/(5 + 7*x^2)]*Sqrt[5 + 
 7*x^2])
 

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 400

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)^ 
(3/2)), x_Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(Sqrt[a + b*x^2]* 
Sqrt[c + d*x^2]), x], x] - Simp[(d*e - c*f)/(b*c - a*d)   Int[Sqrt[a + b*x^ 
2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[b/a] & 
& PosQ[d/c]
 

rule 2050

Int[(u_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p 
_), x_Symbol] :> Int[u*((a*e + b*e*x^n)^p/(c + d*x^n)^p), x] /; FreeQ[{a, b 
, c, d, e, n, p}, x] && GtQ[b*d*e, 0] && GtQ[c - a*(d/b), 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (46 ) = 92\).

Time = 4.08 (sec) , antiderivative size = 190, normalized size of antiderivative = 4.13

method	result	size
default	\(-\frac {7 i \sqrt {35}\, \sqrt {35 x^{2}+25}\, \sqrt {6 x^{2}+4}\, \operatorname {EllipticF}\left (\frac {i x \sqrt {35}}{5}, \frac {\sqrt {210}}{14}\right ) \sqrt {\left (3 x^{2}+2\right ) \left (7 x^{2}+5\right )}-2 i \sqrt {35}\, \sqrt {35 x^{2}+25}\, \sqrt {6 x^{2}+4}\, \sqrt {\left (3 x^{2}+2\right ) \left (7 x^{2}+5\right )}\, \operatorname {EllipticE}\left (\frac {i x \sqrt {35}}{5}, \frac {\sqrt {210}}{14}\right )-210 \sqrt {21 x^{4}+29 x^{2}+10}\, x^{3}-140 x \sqrt {21 x^{4}+29 x^{2}+10}}{350 \sqrt {\frac {3 x^{2}+2}{7 x^{2}+5}}\, \left (7 x^{2}+5\right ) \sqrt {21 x^{4}+29 x^{2}+10}}\)	\(190\)

Input:

int((10*x^2+7)/((3*x^2+2)/(7*x^2+5))^(1/2)/(7*x^2+5)^2,x,method=_RETURNVER 
BOSE)
 

Output:

-1/350*(7*I*35^(1/2)*(35*x^2+25)^(1/2)*(6*x^2+4)^(1/2)*EllipticF(1/5*I*x*3 
5^(1/2),1/14*210^(1/2))*((3*x^2+2)*(7*x^2+5))^(1/2)-2*I*35^(1/2)*(35*x^2+2 
5)^(1/2)*(6*x^2+4)^(1/2)*((3*x^2+2)*(7*x^2+5))^(1/2)*EllipticE(1/5*I*x*35^ 
(1/2),1/14*210^(1/2))-210*(21*x^4+29*x^2+10)^(1/2)*x^3-140*x*(21*x^4+29*x^ 
2+10)^(1/2))/((3*x^2+2)/(7*x^2+5))^(1/2)/(7*x^2+5)/(21*x^4+29*x^2+10)^(1/2 
)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.20 \[ \int \frac {7+10 x^2}{\sqrt {\frac {2+3 x^2}{5+7 x^2}} \left (5+7 x^2\right )^2} \, dx=-\frac {1}{25} \, \sqrt {10} \sqrt {-\frac {7}{5}} E(\arcsin \left (\sqrt {-\frac {7}{5}} x\right )\,|\,\frac {15}{14}) - \frac {11}{350} \, \sqrt {10} \sqrt {-\frac {7}{5}} F(\arcsin \left (\sqrt {-\frac {7}{5}} x\right )\,|\,\frac {15}{14}) + \frac {1}{5} \, x \sqrt {\frac {3 \, x^{2} + 2}{7 \, x^{2} + 5}} \] Input:

integrate((10*x^2+7)/((3*x^2+2)/(7*x^2+5))^(1/2)/(7*x^2+5)^2,x, algorithm= 
"fricas")
 

Output:

-1/25*sqrt(10)*sqrt(-7/5)*elliptic_e(arcsin(sqrt(-7/5)*x), 15/14) - 11/350 
*sqrt(10)*sqrt(-7/5)*elliptic_f(arcsin(sqrt(-7/5)*x), 15/14) + 1/5*x*sqrt( 
(3*x^2 + 2)/(7*x^2 + 5))
 

Sympy [F]

\[ \int \frac {7+10 x^2}{\sqrt {\frac {2+3 x^2}{5+7 x^2}} \left (5+7 x^2\right )^2} \, dx=\int \frac {10 x^{2} + 7}{\sqrt {\frac {3 x^{2} + 2}{7 x^{2} + 5}} \left (7 x^{2} + 5\right )^{2}}\, dx \] Input:

integrate((10*x**2+7)/((3*x**2+2)/(7*x**2+5))**(1/2)/(7*x**2+5)**2,x)
 

Output:

Integral((10*x**2 + 7)/(sqrt((3*x**2 + 2)/(7*x**2 + 5))*(7*x**2 + 5)**2), 
x)
 

Maxima [F]

\[ \int \frac {7+10 x^2}{\sqrt {\frac {2+3 x^2}{5+7 x^2}} \left (5+7 x^2\right )^2} \, dx=\int { \frac {10 \, x^{2} + 7}{{\left (7 \, x^{2} + 5\right )}^{2} \sqrt {\frac {3 \, x^{2} + 2}{7 \, x^{2} + 5}}} \,d x } \] Input:

integrate((10*x^2+7)/((3*x^2+2)/(7*x^2+5))^(1/2)/(7*x^2+5)^2,x, algorithm= 
"maxima")
 

Output:

integrate((10*x^2 + 7)/((7*x^2 + 5)^2*sqrt((3*x^2 + 2)/(7*x^2 + 5))), x)
 

Giac [F]

\[ \int \frac {7+10 x^2}{\sqrt {\frac {2+3 x^2}{5+7 x^2}} \left (5+7 x^2\right )^2} \, dx=\int { \frac {10 \, x^{2} + 7}{{\left (7 \, x^{2} + 5\right )}^{2} \sqrt {\frac {3 \, x^{2} + 2}{7 \, x^{2} + 5}}} \,d x } \] Input:

integrate((10*x^2+7)/((3*x^2+2)/(7*x^2+5))^(1/2)/(7*x^2+5)^2,x, algorithm= 
"giac")
 

Output:

integrate((10*x^2 + 7)/((7*x^2 + 5)^2*sqrt((3*x^2 + 2)/(7*x^2 + 5))), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {7+10 x^2}{\sqrt {\frac {2+3 x^2}{5+7 x^2}} \left (5+7 x^2\right )^2} \, dx=\int \frac {10\,x^2+7}{{\left (7\,x^2+5\right )}^2\,\sqrt {\frac {3\,x^2+2}{7\,x^2+5}}} \,d x \] Input:

int((10*x^2 + 7)/((7*x^2 + 5)^2*((3*x^2 + 2)/(7*x^2 + 5))^(1/2)),x)
 

Output:

int((10*x^2 + 7)/((7*x^2 + 5)^2*((3*x^2 + 2)/(7*x^2 + 5))^(1/2)), x)
 

Reduce [F]

\[ \int \frac {7+10 x^2}{\sqrt {\frac {2+3 x^2}{5+7 x^2}} \left (5+7 x^2\right )^2} \, dx=10 \left (\int \frac {\sqrt {3 x^{2}+2}\, \sqrt {7 x^{2}+5}\, x^{2}}{147 x^{6}+308 x^{4}+215 x^{2}+50}d x \right )+7 \left (\int \frac {\sqrt {3 x^{2}+2}\, \sqrt {7 x^{2}+5}}{147 x^{6}+308 x^{4}+215 x^{2}+50}d x \right ) \] Input:

int((10*x^2+7)/((3*x^2+2)/(7*x^2+5))^(1/2)/(7*x^2+5)^2,x)
                                                                                    
                                                                                    
 

Output:

10*int((sqrt(3*x**2 + 2)*sqrt(7*x**2 + 5)*x**2)/(147*x**6 + 308*x**4 + 215 
*x**2 + 50),x) + 7*int((sqrt(3*x**2 + 2)*sqrt(7*x**2 + 5))/(147*x**6 + 308 
*x**4 + 215*x**2 + 50),x)