3.4.10 \(\int \frac {(7+5 x^2)^2}{(2+3 x^2+x^4)^{3/2}} \, dx\) [310]

3.4.10.1 Optimal result
3.4.10.2 Mathematica [C] (verified)
3.4.10.3 Rubi [A] (verified)
3.4.10.4 Maple [C] (verified)
3.4.10.5 Fricas [C] (verification not implemented)
3.4.10.6 Sympy [F]
3.4.10.7 Maxima [F]
3.4.10.8 Giac [F]
3.4.10.9 Mupad [F(-1)]

3.4.10.1 Optimal result

Integrand size = 24, antiderivative size = 149 \[ \int \frac {\left (7+5 x^2\right )^2}{\left (2+3 x^2+x^4\right )^{3/2}} \, dx=-\frac {17 x \left (2+x^2\right )}{2 \sqrt {2+3 x^2+x^4}}+\frac {x \left (25+17 x^2\right )}{2 \sqrt {2+3 x^2+x^4}}+\frac {17 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {2} \sqrt {2+3 x^2+x^4}}+\frac {6 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{\sqrt {2+3 x^2+x^4}} \]

output
-17/2*x*(x^2+2)/(x^4+3*x^2+2)^(1/2)+1/2*x*(17*x^2+25)/(x^4+3*x^2+2)^(1/2)+ 
17/2*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*EllipticE(x/(x^2+1)^(1/2),1/2*2^(1/2) 
)*2^(1/2)*((x^2+2)/(x^2+1))^(1/2)/(x^4+3*x^2+2)^(1/2)+6*(x^2+1)^(3/2)*(1/( 
x^2+1))^(1/2)*EllipticF(x/(x^2+1)^(1/2),1/2*2^(1/2))*2^(1/2)*((x^2+2)/(x^2 
+1))^(1/2)/(x^4+3*x^2+2)^(1/2)
 
3.4.10.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.07 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.66 \[ \int \frac {\left (7+5 x^2\right )^2}{\left (2+3 x^2+x^4\right )^{3/2}} \, dx=\frac {25 x+17 x^3+17 i \sqrt {1+x^2} \sqrt {2+x^2} E\left (\left .i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-41 i \sqrt {1+x^2} \sqrt {2+x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )}{2 \sqrt {2+3 x^2+x^4}} \]

input
Integrate[(7 + 5*x^2)^2/(2 + 3*x^2 + x^4)^(3/2),x]
 
output
(25*x + 17*x^3 + (17*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticE[I*ArcSinh[x/ 
Sqrt[2]], 2] - (41*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticF[I*ArcSinh[x/Sq 
rt[2]], 2])/(2*Sqrt[2 + 3*x^2 + x^4])
 
3.4.10.3 Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1517, 25, 1503, 1412, 1455}

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.

\(\displaystyle \int \frac {\left (5 x^2+7\right )^2}{\left (x^4+3 x^2+2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1517

\(\displaystyle \frac {x \left (17 x^2+25\right )}{2 \sqrt {x^4+3 x^2+2}}-\frac {1}{2} \int -\frac {24-17 x^2}{\sqrt {x^4+3 x^2+2}}dx\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \int \frac {24-17 x^2}{\sqrt {x^4+3 x^2+2}}dx+\frac {x \left (17 x^2+25\right )}{2 \sqrt {x^4+3 x^2+2}}\)

\(\Big \downarrow \) 1503

\(\displaystyle \frac {1}{2} \left (24 \int \frac {1}{\sqrt {x^4+3 x^2+2}}dx-17 \int \frac {x^2}{\sqrt {x^4+3 x^2+2}}dx\right )+\frac {x \left (17 x^2+25\right )}{2 \sqrt {x^4+3 x^2+2}}\)

\(\Big \downarrow \) 1412

\(\displaystyle \frac {1}{2} \left (\frac {12 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{\sqrt {x^4+3 x^2+2}}-17 \int \frac {x^2}{\sqrt {x^4+3 x^2+2}}dx\right )+\frac {x \left (17 x^2+25\right )}{2 \sqrt {x^4+3 x^2+2}}\)

\(\Big \downarrow \) 1455

\(\displaystyle \frac {1}{2} \left (\frac {12 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{\sqrt {x^4+3 x^2+2}}-17 \left (\frac {x \left (x^2+2\right )}{\sqrt {x^4+3 x^2+2}}-\frac {\sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {x^4+3 x^2+2}}\right )\right )+\frac {x \left (17 x^2+25\right )}{2 \sqrt {x^4+3 x^2+2}}\)

input
Int[(7 + 5*x^2)^2/(2 + 3*x^2 + x^4)^(3/2),x]
 
output
(x*(25 + 17*x^2))/(2*Sqrt[2 + 3*x^2 + x^4]) + (-17*((x*(2 + x^2))/Sqrt[2 + 
 3*x^2 + x^4] - (Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[Arc 
Tan[x], 1/2])/Sqrt[2 + 3*x^2 + x^4]) + (12*Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2 
)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/Sqrt[2 + 3*x^2 + x^4])/2
 

3.4.10.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 1412
Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b 
^2 - 4*a*c, 2]}, Simp[(2*a + (b + q)*x^2)*(Sqrt[(2*a + (b - q)*x^2)/(2*a + 
(b + q)*x^2)]/(2*a*Rt[(b + q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]))*EllipticF 
[ArcTan[Rt[(b + q)/(2*a), 2]*x], 2*(q/(b + q))], x] /; PosQ[(b + q)/a] && 
!(PosQ[(b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[ 
{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]
 

rule 1455
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[b^2 - 4*a*c, 2]}, Simp[x*((b + q + 2*c*x^2)/(2*c*Sqrt[a + b*x^2 + c*x^4 
])), x] - Simp[Rt[(b + q)/(2*a), 2]*(2*a + (b + q)*x^2)*(Sqrt[(2*a + (b - q 
)*x^2)/(2*a + (b + q)*x^2)]/(2*c*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[ArcTan 
[Rt[(b + q)/(2*a), 2]*x], 2*(q/(b + q))], x] /; PosQ[(b + q)/a] &&  !(PosQ[ 
(b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[{a, b, 
c}, x] && GtQ[b^2 - 4*a*c, 0]
 

rule 1503
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[d   Int[1/Sqrt[a + b*x^2 + c*x^4] 
, x], x] + Simp[e   Int[x^2/Sqrt[a + b*x^2 + c*x^4], x], x] /; PosQ[(b + q) 
/a] || PosQ[(b - q)/a]] /; FreeQ[{a, b, c, d, e}, x] && GtQ[b^2 - 4*a*c, 0]
 

rule 1517
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x 
_Symbol] :> With[{f = Coeff[PolynomialRemainder[(d + e*x^2)^q, a + b*x^2 + 
c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[(d + e*x^2)^q, a + b*x^2 + 
c*x^4, x], x, 2]}, Simp[x*(a + b*x^2 + c*x^4)^(p + 1)*((a*b*g - f*(b^2 - 2* 
a*c) - c*(b*f - 2*a*g)*x^2)/(2*a*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(2*a* 
(p + 1)*(b^2 - 4*a*c))   Int[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToSum[2*a*(p 
 + 1)*(b^2 - 4*a*c)*PolynomialQuotient[(d + e*x^2)^q, a + b*x^2 + c*x^4, x] 
 + b^2*f*(2*p + 3) - 2*a*c*f*(4*p + 5) - a*b*g + c*(4*p + 7)*(b*f - 2*a*g)* 
x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ 
[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[q, 1] && LtQ[p, -1]
 
3.4.10.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.56 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.86

method result size
risch \(\frac {x \left (17 x^{2}+25\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}-\frac {6 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{\sqrt {x^{4}+3 x^{2}+2}}-\frac {17 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )-E\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{4 \sqrt {x^{4}+3 x^{2}+2}}\) \(128\)
elliptic \(-\frac {2 \left (-\frac {17}{4} x^{3}-\frac {25}{4} x \right )}{\sqrt {x^{4}+3 x^{2}+2}}-\frac {6 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{\sqrt {x^{4}+3 x^{2}+2}}-\frac {17 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )-E\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{4 \sqrt {x^{4}+3 x^{2}+2}}\) \(129\)
default \(-\frac {98 \left (-\frac {3}{4} x^{3}-\frac {5}{4} x \right )}{\sqrt {x^{4}+3 x^{2}+2}}-\frac {6 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{\sqrt {x^{4}+3 x^{2}+2}}-\frac {17 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )-E\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{4 \sqrt {x^{4}+3 x^{2}+2}}-\frac {50 \left (-\frac {3}{2} x^{3}-2 x \right )}{\sqrt {x^{4}+3 x^{2}+2}}-\frac {140 \left (x^{3}+\frac {3}{2} x \right )}{\sqrt {x^{4}+3 x^{2}+2}}\) \(173\)

input
int((5*x^2+7)^2/(x^4+3*x^2+2)^(3/2),x,method=_RETURNVERBOSE)
 
output
1/2*x*(17*x^2+25)/(x^4+3*x^2+2)^(1/2)-6*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^ 
(1/2)/(x^4+3*x^2+2)^(1/2)*EllipticF(1/2*I*2^(1/2)*x,2^(1/2))-17/4*I*2^(1/2 
)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*(EllipticF(1/2*I*2^(1/ 
2)*x,2^(1/2))-EllipticE(1/2*I*2^(1/2)*x,2^(1/2)))
 
3.4.10.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.56 \[ \int \frac {\left (7+5 x^2\right )^2}{\left (2+3 x^2+x^4\right )^{3/2}} \, dx=-\frac {17 \, {\left (i \, x^{4} + 3 i \, x^{2} + 2 i\right )} E(\arcsin \left (\frac {1}{2} i \, \sqrt {2} x\right )\,|\,2) + 31 \, {\left (i \, x^{4} + 3 i \, x^{2} + 2 i\right )} F(\arcsin \left (\frac {1}{2} i \, \sqrt {2} x\right )\,|\,2) - 2 \, \sqrt {x^{4} + 3 \, x^{2} + 2} {\left (17 \, x^{3} + 25 \, x\right )}}{4 \, {\left (x^{4} + 3 \, x^{2} + 2\right )}} \]

input
integrate((5*x^2+7)^2/(x^4+3*x^2+2)^(3/2),x, algorithm="fricas")
 
output
-1/4*(17*(I*x^4 + 3*I*x^2 + 2*I)*elliptic_e(arcsin(1/2*I*sqrt(2)*x), 2) + 
31*(I*x^4 + 3*I*x^2 + 2*I)*elliptic_f(arcsin(1/2*I*sqrt(2)*x), 2) - 2*sqrt 
(x^4 + 3*x^2 + 2)*(17*x^3 + 25*x))/(x^4 + 3*x^2 + 2)
 
3.4.10.6 Sympy [F]

\[ \int \frac {\left (7+5 x^2\right )^2}{\left (2+3 x^2+x^4\right )^{3/2}} \, dx=\int \frac {\left (5 x^{2} + 7\right )^{2}}{\left (\left (x^{2} + 1\right ) \left (x^{2} + 2\right )\right )^{\frac {3}{2}}}\, dx \]

input
integrate((5*x**2+7)**2/(x**4+3*x**2+2)**(3/2),x)
 
output
Integral((5*x**2 + 7)**2/((x**2 + 1)*(x**2 + 2))**(3/2), x)
 
3.4.10.7 Maxima [F]

\[ \int \frac {\left (7+5 x^2\right )^2}{\left (2+3 x^2+x^4\right )^{3/2}} \, dx=\int { \frac {{\left (5 \, x^{2} + 7\right )}^{2}}{{\left (x^{4} + 3 \, x^{2} + 2\right )}^{\frac {3}{2}}} \,d x } \]

input
integrate((5*x^2+7)^2/(x^4+3*x^2+2)^(3/2),x, algorithm="maxima")
 
output
integrate((5*x^2 + 7)^2/(x^4 + 3*x^2 + 2)^(3/2), x)
 
3.4.10.8 Giac [F]

\[ \int \frac {\left (7+5 x^2\right )^2}{\left (2+3 x^2+x^4\right )^{3/2}} \, dx=\int { \frac {{\left (5 \, x^{2} + 7\right )}^{2}}{{\left (x^{4} + 3 \, x^{2} + 2\right )}^{\frac {3}{2}}} \,d x } \]

input
integrate((5*x^2+7)^2/(x^4+3*x^2+2)^(3/2),x, algorithm="giac")
 
output
integrate((5*x^2 + 7)^2/(x^4 + 3*x^2 + 2)^(3/2), x)
 
3.4.10.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\left (7+5 x^2\right )^2}{\left (2+3 x^2+x^4\right )^{3/2}} \, dx=\int \frac {{\left (5\,x^2+7\right )}^2}{{\left (x^4+3\,x^2+2\right )}^{3/2}} \,d x \]

input
int((5*x^2 + 7)^2/(3*x^2 + x^4 + 2)^(3/2),x)
 
output
int((5*x^2 + 7)^2/(3*x^2 + x^4 + 2)^(3/2), x)