\(\int \frac {2+3 x^2+5 x^4}{x^3 \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx\) [20]

Optimal result

Integrand size = 43, antiderivative size = 151 \[ \int \frac {2+3 x^2+5 x^4}{x^3 \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\frac {\sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}}{2 x^2}+\frac {5 \sqrt {-1+9 x^2} \text {arcsinh}\left (\frac {\sqrt {-1+9 x^2}}{\sqrt {19}}\right )}{3 \sqrt {-1+3 x} \sqrt {1+3 x}}-\frac {23 \sqrt {-1+9 x^2} \arctan \left (\frac {\sqrt {2+x^2}}{\sqrt {2} \sqrt {-1+9 x^2}}\right )}{2 \sqrt {2} \sqrt {-1+3 x} \sqrt {1+3 x}} \] Output:

1/2*(-1+3*x)^(1/2)*(1+3*x)^(1/2)*(x^2+2)^(1/2)/x^2+5/3*(9*x^2-1)^(1/2)*arc 
sinh(1/19*(9*x^2-1)^(1/2)*19^(1/2))/(-1+3*x)^(1/2)/(1+3*x)^(1/2)-23/4*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] (verified)

Time = 10.15 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.86 \[ \int \frac {2+3 x^2+5 x^4}{x^3 \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\frac {\sqrt {-1+9 x^2} \left (\frac {6 \sqrt {2+x^2} \sqrt {-1+9 x^2}}{x^2}+\frac {20 \sqrt {-1+9 x^2} \arcsin \left (\frac {\sqrt {1-9 x^2}}{\sqrt {19}}\right )}{\sqrt {1-9 x^2}}-69 \sqrt {2} \arctan \left (\frac {\sqrt {2+x^2}}{\sqrt {-2+18 x^2}}\right )\right )}{12 \sqrt {-1+3 x} \sqrt {1+3 x}} \] Input:

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

Output:

(Sqrt[-1 + 9*x^2]*((6*Sqrt[2 + x^2]*Sqrt[-1 + 9*x^2])/x^2 + (20*Sqrt[-1 + 
9*x^2]*ArcSin[Sqrt[1 - 9*x^2]/Sqrt[19]])/Sqrt[1 - 9*x^2] - 69*Sqrt[2]*ArcT 
an[Sqrt[2 + x^2]/Sqrt[-2 + 18*x^2]]))/(12*Sqrt[-1 + 3*x]*Sqrt[1 + 3*x])
 

Rubi [A] (verified)

Time = 1.19 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.77, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {2038, 7293, 2009}

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.

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

\(\Big \downarrow \) 2038

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {9 x^2-1} \left (\frac {5}{3} \text {arcsinh}\left (\frac {\sqrt {9 x^2-1}}{\sqrt {19}}\right )-\frac {23 \arctan \left (\frac {\sqrt {x^2+2}}{\sqrt {2} \sqrt {9 x^2-1}}\right )}{2 \sqrt {2}}+\frac {\sqrt {x^2+2} \sqrt {9 x^2-1}}{2 x^2}\right )}{\sqrt {3 x-1} \sqrt {3 x+1}}\)

Input:

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

Output:

(Sqrt[-1 + 9*x^2]*((Sqrt[2 + x^2]*Sqrt[-1 + 9*x^2])/(2*x^2) + (5*ArcSinh[S 
qrt[-1 + 9*x^2]/Sqrt[19]])/3 - (23*ArcTan[Sqrt[2 + x^2]/(Sqrt[2]*Sqrt[-1 + 
 9*x^2])])/(2*Sqrt[2])))/(Sqrt[-1 + 3*x]*Sqrt[1 + 3*x])
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.81

method	result	size
elliptic	\(\frac {\sqrt {\left (x^{2}+2\right ) \left (9 x^{2}-1\right )}\, \left (\frac {5 \ln \left (\frac {\left (\frac {17}{2}+9 x^{2}\right ) \sqrt {9}}{9}+\sqrt {9 x^{4}+17 x^{2}-2}\right ) \sqrt {9}}{18}+\frac {\sqrt {9 x^{4}+17 x^{2}-2}}{2 x^{2}}+\frac {23 \sqrt {2}\, \arctan \left (\frac {\left (17 x^{2}-4\right ) \sqrt {2}}{4 \sqrt {9 x^{4}+17 x^{2}-2}}\right )}{8}\right )}{\sqrt {-1+3 x}\, \sqrt {1+3 x}\, \sqrt {x^{2}+2}}\)	\(123\)
risch	\(\frac {\sqrt {-1+3 x}\, \sqrt {1+3 x}\, \sqrt {x^{2}+2}}{2 x^{2}}+\frac {\left (\frac {23 \sqrt {2}\, \arctan \left (\frac {\left (17 x^{2}-4\right ) \sqrt {2}}{4 \sqrt {9 x^{4}+17 x^{2}-2}}\right )}{8}+\frac {5 \ln \left (\frac {\left (\frac {17}{2}+9 x^{2}\right ) \sqrt {9}}{9}+\sqrt {9 x^{4}+17 x^{2}-2}\right ) \sqrt {9}}{18}\right ) \sqrt {\left (-1+3 x \right ) \left (1+3 x \right ) \left (x^{2}+2\right )}}{\sqrt {-1+3 x}\, \sqrt {1+3 x}\, \sqrt {x^{2}+2}}\)	\(134\)
default	\(\frac {\sqrt {-1+3 x}\, \sqrt {1+3 x}\, \sqrt {x^{2}+2}\, \left (69 \sqrt {2}\, \arctan \left (\frac {\left (17 x^{2}-4\right ) \sqrt {2}}{4 \sqrt {9 x^{4}+17 x^{2}-2}}\right ) x^{2}-368 \ln \left (\frac {17}{6}+3 x^{2}+\sqrt {9 x^{4}+17 x^{2}-2}\right ) x^{2}+388 \ln \left (6 x^{2}+\frac {17}{3}+2 \sqrt {9 x^{4}+17 x^{2}-2}\right ) x^{2}+12 \sqrt {9 x^{4}+17 x^{2}-2}\right )}{24 \sqrt {9 x^{4}+17 x^{2}-2}\, x^{2}}\)	\(149\)

Input:

int((5*x^4+3*x^2+2)/x^3/(-1+3*x)^(1/2)/(1+3*x)^(1/2)/(x^2+2)^(1/2),x,metho 
d=_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)*(5/18 
*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*(9*x^4+17 
*x^2-2)^(1/2)/x^2+23/8*2^(1/2)*arctan(1/4*(17*x^2-4)*2^(1/2)/(9*x^4+17*x^2 
-2)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.75 \[ \int \frac {2+3 x^2+5 x^4}{x^3 \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\frac {69 \, \sqrt {2} x^{2} \arctan \left (-\frac {3}{2} \, \sqrt {2} x^{2} + \frac {1}{2} \, \sqrt {2} \sqrt {x^{2} + 2} \sqrt {3 \, x + 1} \sqrt {3 \, x - 1}\right ) - 10 \, x^{2} \log \left (-18 \, x^{2} + 6 \, \sqrt {x^{2} + 2} \sqrt {3 \, x + 1} \sqrt {3 \, x - 1} - 17\right ) + 18 \, x^{2} + 6 \, \sqrt {x^{2} + 2} \sqrt {3 \, x + 1} \sqrt {3 \, x - 1}}{12 \, x^{2}} \] Input:

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

Output:

1/12*(69*sqrt(2)*x^2*arctan(-3/2*sqrt(2)*x^2 + 1/2*sqrt(2)*sqrt(x^2 + 2)*s 
qrt(3*x + 1)*sqrt(3*x - 1)) - 10*x^2*log(-18*x^2 + 6*sqrt(x^2 + 2)*sqrt(3* 
x + 1)*sqrt(3*x - 1) - 17) + 18*x^2 + 6*sqrt(x^2 + 2)*sqrt(3*x + 1)*sqrt(3 
*x - 1))/x^2
 

Sympy [F]

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

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

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

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

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

Reduce [B] (verification not implemented)

Time = 1.24 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.62 \[ \int \frac {2+3 x^2+5 x^4}{x^3 \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\frac {-69 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {3 x +1}\, \sqrt {3 x -1}\, \sqrt {x^{2}+2}\, \sqrt {2}}{18 x^{2}-2}\right ) x^{2}+6 \sqrt {3 x +1}\, \sqrt {3 x -1}\, \sqrt {x^{2}+2}+20 \,\mathrm {log}\left (-3 \sqrt {x^{2}+2}-\sqrt {3 x +1}\, \sqrt {3 x -1}\right ) x^{2}}{12 x^{2}} \] Input:

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

Output:

( - 69*sqrt(2)*atan((sqrt(3*x + 1)*sqrt(3*x - 1)*sqrt(x**2 + 2)*sqrt(2))/( 
18*x**2 - 2))*x**2 + 6*sqrt(3*x + 1)*sqrt(3*x - 1)*sqrt(x**2 + 2) + 20*log 
( - 3*sqrt(x**2 + 2) - sqrt(3*x + 1)*sqrt(3*x - 1))*x**2)/(12*x**2)