3.3.87 \(\int (7+5 x^2)^2 \sqrt {2+3 x^2+x^4} \, dx\) [287]

3.3.87.1 Optimal result

Integrand size = 24, antiderivative size = 168 \[ \int \left (7+5 x^2\right )^2 \sqrt {2+3 x^2+x^4} \, dx=\frac {31 x \left (2+x^2\right )}{\sqrt {2+3 x^2+x^4}}+\frac {1}{21} x \left (407+114 x^2\right ) \sqrt {2+3 x^2+x^4}+\frac {25}{7} x \left (2+3 x^2+x^4\right )^{3/2}-\frac {31 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {2+3 x^2+x^4}}+\frac {472 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{21 \sqrt {2+3 x^2+x^4}} \]

output

25/7*x*(x^4+3*x^2+2)^(3/2)+31*x*(x^2+2)/(x^4+3*x^2+2)^(1/2)-31*(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)+472/21*(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)+1/21*x*(114*x^2+407)*(x^4+3*x^2+2)^(1/2)
 

3.3.87.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 5.69 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.68 \[ \int \left (7+5 x^2\right )^2 \sqrt {2+3 x^2+x^4} \, dx=\frac {1114 x+2349 x^3+1724 x^5+564 x^7+75 x^9-651 i \sqrt {1+x^2} \sqrt {2+x^2} E\left (\left .i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-293 i \sqrt {1+x^2} \sqrt {2+x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )}{21 \sqrt {2+3 x^2+x^4}} \]

input

Integrate[(7 + 5*x^2)^2*Sqrt[2 + 3*x^2 + x^4],x]
 

output

(1114*x + 2349*x^3 + 1724*x^5 + 564*x^7 + 75*x^9 - (651*I)*Sqrt[1 + x^2]*S 
qrt[2 + x^2]*EllipticE[I*ArcSinh[x/Sqrt[2]], 2] - (293*I)*Sqrt[1 + x^2]*Sq 
rt[2 + x^2]*EllipticF[I*ArcSinh[x/Sqrt[2]], 2])/(21*Sqrt[2 + 3*x^2 + x^4])
 

3.3.87.3 Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1518, 1490, 27, 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.

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

\(\Big \downarrow \) 1518

\(\displaystyle \frac {1}{7} \int \left (190 x^2+293\right ) \sqrt {x^4+3 x^2+2}dx+\frac {25}{7} x \left (x^4+3 x^2+2\right )^{3/2}\)

\(\Big \downarrow \) 1490

\(\displaystyle \frac {1}{7} \left (\frac {1}{15} \int \frac {5 \left (651 x^2+944\right )}{\sqrt {x^4+3 x^2+2}}dx+\frac {1}{3} x \sqrt {x^4+3 x^2+2} \left (114 x^2+407\right )\right )+\frac {25}{7} x \left (x^4+3 x^2+2\right )^{3/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \int \frac {651 x^2+944}{\sqrt {x^4+3 x^2+2}}dx+\frac {1}{3} x \sqrt {x^4+3 x^2+2} \left (114 x^2+407\right )\right )+\frac {25}{7} x \left (x^4+3 x^2+2\right )^{3/2}\)

\(\Big \downarrow \) 1503

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (944 \int \frac {1}{\sqrt {x^4+3 x^2+2}}dx+651 \int \frac {x^2}{\sqrt {x^4+3 x^2+2}}dx\right )+\frac {1}{3} x \sqrt {x^4+3 x^2+2} \left (114 x^2+407\right )\right )+\frac {25}{7} x \left (x^4+3 x^2+2\right )^{3/2}\)

\(\Big \downarrow \) 1412

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (651 \int \frac {x^2}{\sqrt {x^4+3 x^2+2}}dx+\frac {472 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{\sqrt {x^4+3 x^2+2}}\right )+\frac {1}{3} x \sqrt {x^4+3 x^2+2} \left (114 x^2+407\right )\right )+\frac {25}{7} x \left (x^4+3 x^2+2\right )^{3/2}\)

\(\Big \downarrow \) 1455

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {472 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{\sqrt {x^4+3 x^2+2}}+651 \left (\frac {x \left (x^2+2\right )}{\sqrt {x^4+3 x^2+2}}-\frac {\sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {x^4+3 x^2+2}}\right )\right )+\frac {1}{3} x \sqrt {x^4+3 x^2+2} \left (114 x^2+407\right )\right )+\frac {25}{7} x \left (x^4+3 x^2+2\right )^{3/2}\)

input

Int[(7 + 5*x^2)^2*Sqrt[2 + 3*x^2 + x^4],x]
 

output

(25*x*(2 + 3*x^2 + x^4)^(3/2))/7 + ((x*(407 + 114*x^2)*Sqrt[2 + 3*x^2 + x^ 
4])/3 + (651*((x*(2 + x^2))/Sqrt[2 + 3*x^2 + x^4] - (Sqrt[2]*(1 + x^2)*Sqr 
t[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/Sqrt[2 + 3*x^2 + x^4]) + 
 (472*Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2 
])/Sqrt[2 + 3*x^2 + x^4])/3)/7
 

3.3.87.3.1 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 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 1490

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symb 
ol] :> Simp[x*(2*b*e*p + c*d*(4*p + 3) + c*e*(4*p + 1)*x^2)*((a + b*x^2 + c 
*x^4)^p/(c*(4*p + 1)*(4*p + 3))), x] + Simp[2*(p/(c*(4*p + 1)*(4*p + 3))) 
 Int[Simp[2*a*c*d*(4*p + 3) - a*b*e + (2*a*c*e*(4*p + 1) + b*c*d*(4*p + 3) 
- b^2*e*(2*p + 1))*x^2, x]*(a + b*x^2 + c*x^4)^(p - 1), 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] && 
 GtQ[p, 0] && FractionQ[p] && IntegerQ[2*p]
 

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 1518

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x 
_Symbol] :> Simp[e^q*x^(2*q - 3)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(4*p + 2*q 
 + 1))), x] + Simp[1/(c*(4*p + 2*q + 1))   Int[(a + b*x^2 + c*x^4)^p*Expand 
ToSum[c*(4*p + 2*q + 1)*(d + e*x^2)^q - a*(2*q - 3)*e^q*x^(2*q - 4) - b*(2* 
p + 2*q - 1)*e^q*x^(2*q - 2) - c*(4*p + 2*q + 1)*e^q*x^(2*q), x], x], x] /; 
 FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + 
 a*e^2, 0] && IGtQ[q, 1]
 

3.3.87.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.93 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.79

method	result	size
risch	\(\frac {x \left (75 x^{4}+339 x^{2}+557\right ) \sqrt {x^{4}+3 x^{2}+2}}{21}-\frac {472 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{21 \sqrt {x^{4}+3 x^{2}+2}}+\frac {31 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )-E\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}\)	\(133\)
default	\(\frac {557 x \sqrt {x^{4}+3 x^{2}+2}}{21}-\frac {472 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{21 \sqrt {x^{4}+3 x^{2}+2}}+\frac {31 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )-E\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}+\frac {25 x^{5} \sqrt {x^{4}+3 x^{2}+2}}{7}+\frac {113 x^{3} \sqrt {x^{4}+3 x^{2}+2}}{7}\)	\(155\)
elliptic	\(\frac {557 x \sqrt {x^{4}+3 x^{2}+2}}{21}-\frac {472 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{21 \sqrt {x^{4}+3 x^{2}+2}}+\frac {31 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )-E\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}+\frac {25 x^{5} \sqrt {x^{4}+3 x^{2}+2}}{7}+\frac {113 x^{3} \sqrt {x^{4}+3 x^{2}+2}}{7}\)	\(155\)

input

int((5*x^2+7)^2*(x^4+3*x^2+2)^(1/2),x,method=_RETURNVERBOSE)
 

output

1/21*x*(75*x^4+339*x^2+557)*(x^4+3*x^2+2)^(1/2)-472/21*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) 
)+31/2*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*(Ellipt 
icF(1/2*I*2^(1/2)*x,2^(1/2))-EllipticE(1/2*I*2^(1/2)*x,2^(1/2)))
 

3.3.87.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.35 \[ \int \left (7+5 x^2\right )^2 \sqrt {2+3 x^2+x^4} \, dx=\frac {-651 i \, x E(\arcsin \left (\frac {i}{x}\right )\,|\,2) + 1595 i \, x F(\arcsin \left (\frac {i}{x}\right )\,|\,2) + {\left (75 \, x^{6} + 339 \, x^{4} + 557 \, x^{2} + 651\right )} \sqrt {x^{4} + 3 \, x^{2} + 2}}{21 \, x} \]

input

integrate((5*x^2+7)^2*(x^4+3*x^2+2)^(1/2),x, algorithm="fricas")
 

output

1/21*(-651*I*x*elliptic_e(arcsin(I/x), 2) + 1595*I*x*elliptic_f(arcsin(I/x 
), 2) + (75*x^6 + 339*x^4 + 557*x^2 + 651)*sqrt(x^4 + 3*x^2 + 2))/x
 

3.3.87.6 Sympy [F]

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

input

integrate((5*x**2+7)**2*(x**4+3*x**2+2)**(1/2),x)
 

output

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

3.3.87.7 Maxima [F]

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

input

integrate((5*x^2+7)^2*(x^4+3*x^2+2)^(1/2),x, algorithm="maxima")
 

output

integrate(sqrt(x^4 + 3*x^2 + 2)*(5*x^2 + 7)^2, x)
 

3.3.87.8 Giac [F]

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

input

integrate((5*x^2+7)^2*(x^4+3*x^2+2)^(1/2),x, algorithm="giac")
 

output

integrate(sqrt(x^4 + 3*x^2 + 2)*(5*x^2 + 7)^2, x)
 

3.3.87.9 Mupad [F(-1)]

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

input

int((5*x^2 + 7)^2*(3*x^2 + x^4 + 2)^(1/2),x)
 

output

int((5*x^2 + 7)^2*(3*x^2 + x^4 + 2)^(1/2), x)