\(\int \frac {2+3 x^2+5 x^4}{x \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx\) [19]

Optimal result

Integrand size = 43, antiderivative size = 146 \[ \int \frac {2+3 x^2+5 x^4}{x \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\frac {5}{18} \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}-\frac {31 \sqrt {-1+9 x^2} \text {arcsinh}\left (\frac {\sqrt {-1+9 x^2}}{\sqrt {19}}\right )}{54 \sqrt {-1+3 x} \sqrt {1+3 x}}-\frac {\sqrt {2} \sqrt {-1+9 x^2} \arctan \left (\frac {\sqrt {2+x^2}}{\sqrt {2} \sqrt {-1+9 x^2}}\right )}{\sqrt {-1+3 x} \sqrt {1+3 x}} \] Output:

5/18*(-1+3*x)^(1/2)*(1+3*x)^(1/2)*(x^2+2)^(1/2)-31/54*(9*x^2-1)^(1/2)*arcs 
inh(1/19*(9*x^2-1)^(1/2)*19^(1/2))/(-1+3*x)^(1/2)/(1+3*x)^(1/2)-2^(1/2)*(9 
*x^2-1)^(1/2)*arctan(1/2*(x^2+2)^(1/2)*2^(1/2)/(9*x^2-1)^(1/2))/(-1+3*x)^( 
1/2)/(1+3*x)^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 10.13 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.94 \[ \int \frac {2+3 x^2+5 x^4}{x \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\frac {\sqrt {-1+9 x^2} \left (15 \sqrt {2+x^2} \sqrt {-\left (-1+9 x^2\right )^2}-31 \sqrt {-1+9 x^2} \arcsin \left (\frac {\sqrt {1-9 x^2}}{\sqrt {19}}\right )-54 \sqrt {2-18 x^2} \arctan \left (\frac {\sqrt {2+x^2}}{\sqrt {-2+18 x^2}}\right )\right )}{54 \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {1-9 x^2}} \] Input:

Integrate[(2 + 3*x^2 + 5*x^4)/(x*Sqrt[-1 + 3*x]*Sqrt[1 + 3*x]*Sqrt[2 + x^2 
]),x]
 

Output:

(Sqrt[-1 + 9*x^2]*(15*Sqrt[2 + x^2]*Sqrt[-(-1 + 9*x^2)^2] - 31*Sqrt[-1 + 9 
*x^2]*ArcSin[Sqrt[1 - 9*x^2]/Sqrt[19]] - 54*Sqrt[2 - 18*x^2]*ArcTan[Sqrt[2 
 + x^2]/Sqrt[-2 + 18*x^2]]))/(54*Sqrt[-1 + 3*x]*Sqrt[1 + 3*x]*Sqrt[1 - 9*x 
^2])
 

Rubi [A] (verified)

Time = 1.17 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.82, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.209, Rules used = {2038, 7282, 2118, 27, 175, 64, 104, 217, 222}

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.

\(\displaystyle \int \frac {5 x^4+3 x^2+2}{x \sqrt {3 x-1} \sqrt {3 x+1} \sqrt {x^2+2}} \, dx\)

\(\Big \downarrow \) 2038

\(\displaystyle \frac {\sqrt {9 x^2-1} \int \frac {5 x^4+3 x^2+2}{x \sqrt {x^2+2} \sqrt {9 x^2-1}}dx}{\sqrt {3 x-1} \sqrt {3 x+1}}\)

\(\Big \downarrow \) 7282

\(\displaystyle \frac {\sqrt {9 x^2-1} \int \frac {5 x^4+3 x^2+2}{x^2 \sqrt {x^2+2} \sqrt {9 x^2-1}}dx^2}{2 \sqrt {3 x-1} \sqrt {3 x+1}}\)

\(\Big \downarrow \) 2118

\(\displaystyle \frac {\sqrt {9 x^2-1} \left (\frac {1}{9} \int \frac {36-31 x^2}{2 x^2 \sqrt {x^2+2} \sqrt {9 x^2-1}}dx^2+\frac {5}{9} \sqrt {x^2+2} \sqrt {9 x^2-1}\right )}{2 \sqrt {3 x-1} \sqrt {3 x+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {9 x^2-1} \left (\frac {1}{18} \int \frac {36-31 x^2}{x^2 \sqrt {x^2+2} \sqrt {9 x^2-1}}dx^2+\frac {5}{9} \sqrt {x^2+2} \sqrt {9 x^2-1}\right )}{2 \sqrt {3 x-1} \sqrt {3 x+1}}\)

\(\Big \downarrow \) 175

\(\displaystyle \frac {\sqrt {9 x^2-1} \left (\frac {1}{18} \left (36 \int \frac {1}{x^2 \sqrt {x^2+2} \sqrt {9 x^2-1}}dx^2-31 \int \frac {1}{\sqrt {x^2+2} \sqrt {9 x^2-1}}dx^2\right )+\frac {5}{9} \sqrt {x^2+2} \sqrt {9 x^2-1}\right )}{2 \sqrt {3 x-1} \sqrt {3 x+1}}\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {\sqrt {9 x^2-1} \left (\frac {1}{18} \left (36 \int \frac {1}{x^2 \sqrt {x^2+2} \sqrt {9 x^2-1}}dx^2-\frac {62}{9} \int \frac {1}{\sqrt {\frac {x^4}{9}+\frac {19}{9}}}d\sqrt {9 x^2-1}\right )+\frac {5}{9} \sqrt {x^2+2} \sqrt {9 x^2-1}\right )}{2 \sqrt {3 x-1} \sqrt {3 x+1}}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {\sqrt {9 x^2-1} \left (\frac {1}{18} \left (72 \int \frac {1}{-x^4-2}d\frac {\sqrt {x^2+2}}{\sqrt {9 x^2-1}}-\frac {62}{9} \int \frac {1}{\sqrt {\frac {x^4}{9}+\frac {19}{9}}}d\sqrt {9 x^2-1}\right )+\frac {5}{9} \sqrt {x^2+2} \sqrt {9 x^2-1}\right )}{2 \sqrt {3 x-1} \sqrt {3 x+1}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\sqrt {9 x^2-1} \left (\frac {1}{18} \left (-\frac {62}{9} \int \frac {1}{\sqrt {\frac {x^4}{9}+\frac {19}{9}}}d\sqrt {9 x^2-1}-36 \sqrt {2} \arctan \left (\frac {\sqrt {x^2+2}}{\sqrt {2} \sqrt {9 x^2-1}}\right )\right )+\frac {5}{9} \sqrt {x^2+2} \sqrt {9 x^2-1}\right )}{2 \sqrt {3 x-1} \sqrt {3 x+1}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {\sqrt {9 x^2-1} \left (\frac {1}{18} \left (-\frac {62}{3} \text {arcsinh}\left (\frac {\sqrt {9 x^2-1}}{\sqrt {19}}\right )-36 \sqrt {2} \arctan \left (\frac {\sqrt {x^2+2}}{\sqrt {2} \sqrt {9 x^2-1}}\right )\right )+\frac {5}{9} \sqrt {x^2+2} \sqrt {9 x^2-1}\right )}{2 \sqrt {3 x-1} \sqrt {3 x+1}}\)

Input:

Int[(2 + 3*x^2 + 5*x^4)/(x*Sqrt[-1 + 3*x]*Sqrt[1 + 3*x]*Sqrt[2 + x^2]),x]
 

Output:

(Sqrt[-1 + 9*x^2]*((5*Sqrt[2 + x^2]*Sqrt[-1 + 9*x^2])/9 + ((-62*ArcSinh[Sq 
rt[-1 + 9*x^2]/Sqrt[19]])/3 - 36*Sqrt[2]*ArcTan[Sqrt[2 + x^2]/(Sqrt[2]*Sqr 
t[-1 + 9*x^2])])/18))/(2*Sqrt[-1 + 3*x]*Sqrt[1 + 3*x])
 

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 64

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp 
[2/b   Subst[Int[1/Sqrt[c - a*(d/b) + d*(x^2/b)], x], x, Sqrt[a + b*x]], x] 
 /; FreeQ[{a, b, c, d}, x] && GtQ[c - a*(d/b), 0] && ( !GtQ[a - c*(b/d), 0] 
 || PosQ[b])
 

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 175

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ 
)))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b   Int[(c + d*x)^n*(e + f*x)^p, x] 
, x] + Simp[(b*g - a*h)/b   Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] 
 /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

rule 2038

Int[(u_.)*((c_) + (d_.)*(x_)^(n_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p 
_)*((a2_) + (b2_.)*(x_)^(non2_.))^(p_), x_Symbol] :> Simp[(a1 + b1*x^(n/2)) 
^FracPart[p]*((a2 + b2*x^(n/2))^FracPart[p]/(a1*a2 + b1*b2*x^n)^FracPart[p] 
)   Int[u*(a1*a2 + b1*b2*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a1, b1, a2, 
 b2, c, d, n, p, q}, x] && EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] &&  !(Eq 
Q[n, 2] && IGtQ[q, 0])
 

rule 2118

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f 
_.)*(x_))^(p_.), x_Symbol] :> With[{q = Expon[Px, x], k = Coeff[Px, x, Expo 
n[Px, x]]}, Simp[k*(a + b*x)^(m + q - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 
1)/(d*f*b^(q - 1)*(m + n + p + q + 1))), x] + Simp[1/(d*f*b^q*(m + n + p + 
q + 1))   Int[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*ExpandToSum[d*f*b^q*(m + 
n + p + q + 1)*Px - d*f*k*(m + n + p + q + 1)*(a + b*x)^q + k*(a + b*x)^(q 
- 2)*(a^2*d*f*(m + n + p + q + 1) - b*(b*c*e*(m + q - 1) + a*(d*e*(n + 1) + 
 c*f*(p + 1))) + b*(a*d*f*(2*(m + q) + n + p) - b*(d*e*(m + q + n) + c*f*(m 
 + q + p)))*x), x], x], x] /; NeQ[m + n + p + q + 1, 0]] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] && PolyQ[Px, x]
 

rule 7282

Int[(u_)/(x_), x_Symbol] :> With[{lst = PowerVariableExpn[u, 0, x]}, Simp[1 
/lst[[2]]   Subst[Int[NormalizeIntegrand[Simplify[lst[[1]]/x], x], x], x, ( 
lst[[3]]*x)^lst[[2]]], x] /;  !FalseQ[lst] && NeQ[lst[[2]], 0]] /; NonsumQ[ 
u] &&  !RationalFunctionQ[u, x]
 

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.82

method	result	size
elliptic	\(\frac {\sqrt {\left (x^{2}+2\right ) \left (9 x^{2}-1\right )}\, \left (-\frac {31 \ln \left (\frac {\left (\frac {17}{2}+9 x^{2}\right ) \sqrt {9}}{9}+\sqrt {9 x^{4}+17 x^{2}-2}\right ) \sqrt {9}}{324}+\frac {\sqrt {2}\, \arctan \left (\frac {\left (17 x^{2}-4\right ) \sqrt {2}}{4 \sqrt {9 x^{4}+17 x^{2}-2}}\right )}{2}+\frac {5 \sqrt {9 x^{4}+17 x^{2}-2}}{18}\right )}{\sqrt {-1+3 x}\, \sqrt {1+3 x}\, \sqrt {x^{2}+2}}\)	\(120\)
default	\(\frac {\sqrt {-1+3 x}\, \sqrt {1+3 x}\, \sqrt {x^{2}+2}\, \left (54 \sqrt {2}\, \arctan \left (\frac {\left (17 x^{2}-4\right ) \sqrt {2}}{4 \sqrt {9 x^{4}+17 x^{2}-2}}\right )-225 \ln \left (\frac {17}{6}+3 x^{2}+\sqrt {9 x^{4}+17 x^{2}-2}\right )+194 \ln \left (6 x^{2}+\frac {17}{3}+2 \sqrt {9 x^{4}+17 x^{2}-2}\right )+30 \sqrt {9 x^{4}+17 x^{2}-2}\right )}{108 \sqrt {9 x^{4}+17 x^{2}-2}}\)	\(137\)

Input:

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

Output:

((x^2+2)*(9*x^2-1))^(1/2)/(-1+3*x)^(1/2)/(1+3*x)^(1/2)/(x^2+2)^(1/2)*(-31/ 
324*ln(1/9*(17/2+9*x^2)*9^(1/2)+(9*x^4+17*x^2-2)^(1/2))*9^(1/2)+1/2*2^(1/2 
)*arctan(1/4*(17*x^2-4)*2^(1/2)/(9*x^4+17*x^2-2)^(1/2))+5/18*(9*x^4+17*x^2 
-2)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.66 \[ \int \frac {2+3 x^2+5 x^4}{x \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\frac {5}{18} \, \sqrt {x^{2} + 2} \sqrt {3 \, x + 1} \sqrt {3 \, x - 1} + \sqrt {2} \arctan \left (-\frac {3}{2} \, \sqrt {2} x^{2} + \frac {1}{2} \, \sqrt {2} \sqrt {x^{2} + 2} \sqrt {3 \, x + 1} \sqrt {3 \, x - 1}\right ) + \frac {31}{108} \, \log \left (-18 \, x^{2} + 6 \, \sqrt {x^{2} + 2} \sqrt {3 \, x + 1} \sqrt {3 \, x - 1} - 17\right ) \] Input:

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

Output:

5/18*sqrt(x^2 + 2)*sqrt(3*x + 1)*sqrt(3*x - 1) + sqrt(2)*arctan(-3/2*sqrt( 
2)*x^2 + 1/2*sqrt(2)*sqrt(x^2 + 2)*sqrt(3*x + 1)*sqrt(3*x - 1)) + 31/108*l 
og(-18*x^2 + 6*sqrt(x^2 + 2)*sqrt(3*x + 1)*sqrt(3*x - 1) - 17)
 

Sympy [F]

\[ \int \frac {2+3 x^2+5 x^4}{x \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\int \frac {5 x^{4} + 3 x^{2} + 2}{x \sqrt {3 x - 1} \sqrt {3 x + 1} \sqrt {x^{2} + 2}}\, dx \] Input:

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

Output:

Integral((5*x**4 + 3*x**2 + 2)/(x*sqrt(3*x - 1)*sqrt(3*x + 1)*sqrt(x**2 + 
2)), x)
 

Maxima [F]

\[ \int \frac {2+3 x^2+5 x^4}{x \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\int { \frac {5 \, x^{4} + 3 \, x^{2} + 2}{\sqrt {x^{2} + 2} \sqrt {3 \, x + 1} \sqrt {3 \, x - 1} x} \,d x } \] Input:

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

Output:

integrate((5*x^4 + 3*x^2 + 2)/(sqrt(x^2 + 2)*sqrt(3*x + 1)*sqrt(3*x - 1)*x 
), x)
 

Giac [F]

\[ \int \frac {2+3 x^2+5 x^4}{x \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\int { \frac {5 \, x^{4} + 3 \, x^{2} + 2}{\sqrt {x^{2} + 2} \sqrt {3 \, x + 1} \sqrt {3 \, x - 1} x} \,d x } \] Input:

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

Output:

integrate((5*x^4 + 3*x^2 + 2)/(sqrt(x^2 + 2)*sqrt(3*x + 1)*sqrt(3*x - 1)*x 
), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {2+3 x^2+5 x^4}{x \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\int \frac {5\,x^4+3\,x^2+2}{x\,\sqrt {3\,x-1}\,\sqrt {3\,x+1}\,\sqrt {x^2+2}} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.86 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.55 \[ \int \frac {2+3 x^2+5 x^4}{x \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=-\sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {3 x +1}\, \sqrt {3 x -1}\, \sqrt {x^{2}+2}\, \sqrt {2}}{18 x^{2}-2}\right )+\frac {5 \sqrt {3 x +1}\, \sqrt {3 x -1}\, \sqrt {x^{2}+2}}{18}+\frac {31 \,\mathrm {log}\left (-3 \sqrt {x^{2}+2}+\sqrt {3 x +1}\, \sqrt {3 x -1}\right )}{54} \] Input:

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

Output:

( - 54*sqrt(2)*atan((sqrt(3*x + 1)*sqrt(3*x - 1)*sqrt(x**2 + 2)*sqrt(2))/( 
18*x**2 - 2)) + 15*sqrt(3*x + 1)*sqrt(3*x - 1)*sqrt(x**2 + 2) + 31*log( - 
3*sqrt(x**2 + 2) + sqrt(3*x + 1)*sqrt(3*x - 1)))/54