\(\int \frac {x (2+3 x^2+5 x^4)}{\sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx\) [18]

Optimal result

Integrand size = 41, antiderivative size = 114 \[ \int \frac {x \left (2+3 x^2+5 x^4\right )}{\sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=-\frac {109}{216} \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}+\frac {5}{36} \sqrt {-1+3 x} \sqrt {1+3 x} \left (2+x^2\right )^{3/2}+\frac {1385 \sqrt {-1+9 x^2} \text {arcsinh}\left (\frac {\sqrt {-1+9 x^2}}{\sqrt {19}}\right )}{648 \sqrt {-1+3 x} \sqrt {1+3 x}} \] Output:

-109/216*(-1+3*x)^(1/2)*(1+3*x)^(1/2)*(x^2+2)^(1/2)+5/36*(-1+3*x)^(1/2)*(1 
+3*x)^(1/2)*(x^2+2)^(3/2)+1385/648*(9*x^2-1)^(1/2)*arcsinh(1/19*(9*x^2-1)^ 
(1/2)*19^(1/2))/(-1+3*x)^(1/2)/(1+3*x)^(1/2)
 

Mathematica [A] (verified)

Time = 10.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.68 \[ \int \frac {x \left (2+3 x^2+5 x^4\right )}{\sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\frac {3 \sqrt {2+x^2} \left (49-471 x^2+270 x^4\right )-1385 \sqrt {1-9 x^2} \arcsin \left (\frac {\sqrt {1-9 x^2}}{\sqrt {19}}\right )}{648 \sqrt {-1+3 x} \sqrt {1+3 x}} \] Input:

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

Output:

(3*Sqrt[2 + x^2]*(49 - 471*x^2 + 270*x^4) - 1385*Sqrt[1 - 9*x^2]*ArcSin[Sq 
rt[1 - 9*x^2]/Sqrt[19]])/(648*Sqrt[-1 + 3*x]*Sqrt[1 + 3*x])
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2038, 7266, 1194, 27, 90, 64, 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.

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

\(\Big \downarrow \) 2038

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

\(\Big \downarrow \) 7266

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

\(\Big \downarrow \) 1194

\(\displaystyle \frac {\sqrt {9 x^2-1} \left (\frac {1}{18} \int -\frac {3 \left (109 x^2+26\right )}{2 \sqrt {x^2+2} \sqrt {9 x^2-1}}dx^2+\frac {5}{18} \sqrt {9 x^2-1} \left (x^2+2\right )^{3/2}\right )}{2 \sqrt {3 x-1} \sqrt {3 x+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {9 x^2-1} \left (\frac {5}{18} \left (x^2+2\right )^{3/2} \sqrt {9 x^2-1}-\frac {1}{12} \int \frac {109 x^2+26}{\sqrt {x^2+2} \sqrt {9 x^2-1}}dx^2\right )}{2 \sqrt {3 x-1} \sqrt {3 x+1}}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {\sqrt {9 x^2-1} \left (\frac {1}{12} \left (\frac {1385}{18} \int \frac {1}{\sqrt {x^2+2} \sqrt {9 x^2-1}}dx^2-\frac {109}{9} \sqrt {x^2+2} \sqrt {9 x^2-1}\right )+\frac {5}{18} \sqrt {9 x^2-1} \left (x^2+2\right )^{3/2}\right )}{2 \sqrt {3 x-1} \sqrt {3 x+1}}\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {\sqrt {9 x^2-1} \left (\frac {1}{12} \left (\frac {1385}{81} \int \frac {1}{\sqrt {\frac {x^4}{9}+\frac {19}{9}}}d\sqrt {9 x^2-1}-\frac {109}{9} \sqrt {x^2+2} \sqrt {9 x^2-1}\right )+\frac {5}{18} \sqrt {9 x^2-1} \left (x^2+2\right )^{3/2}\right )}{2 \sqrt {3 x-1} \sqrt {3 x+1}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {\sqrt {9 x^2-1} \left (\frac {1}{12} \left (\frac {1385}{27} \text {arcsinh}\left (\frac {\sqrt {9 x^2-1}}{\sqrt {19}}\right )-\frac {109}{9} \sqrt {x^2+2} \sqrt {9 x^2-1}\right )+\frac {5}{18} \sqrt {9 x^2-1} \left (x^2+2\right )^{3/2}\right )}{2 \sqrt {3 x-1} \sqrt {3 x+1}}\)

Input:

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

Output:

(Sqrt[-1 + 9*x^2]*((5*(2 + x^2)^(3/2)*Sqrt[-1 + 9*x^2])/18 + ((-109*Sqrt[2 
 + x^2]*Sqrt[-1 + 9*x^2])/9 + (1385*ArcSinh[Sqrt[-1 + 9*x^2]/Sqrt[19]])/27 
)/12))/(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 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 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 1194

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[c^p*(d + e*x)^(m + 2*p)*((f + g*x 
)^(n + 1)/(g*e^(2*p)*(m + n + 2*p + 1))), x] + Simp[1/(g*e^(2*p)*(m + n + 2 
*p + 1))   Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^( 
2*p)*(a + b*x + c*x^2)^p - c^p*(d + e*x)^(2*p)) - c^p*(e*f - d*g)*(m + 2*p) 
*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ 
[p, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && NeQ[m + n + 2*p + 1, 0]
 

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 7266

Int[(u_)*(x_)^(m_.), x_Symbol] :> Simp[1/(m + 1)   Subst[Int[SubstFor[x^(m 
+ 1), u, x], x], x, x^(m + 1)], x] /; FreeQ[m, x] && NeQ[m, -1] && Function 
OfQ[x^(m + 1), u, x]
 

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.91

method	result	size
risch	\(\frac {\left (30 x^{2}-49\right ) \sqrt {-1+3 x}\, \sqrt {1+3 x}\, \sqrt {x^{2}+2}}{216}+\frac {1385 \ln \left (\frac {\left (\frac {17}{2}+9 x^{2}\right ) \sqrt {9}}{9}+\sqrt {9 x^{4}+17 x^{2}-2}\right ) \sqrt {9}\, \sqrt {\left (-1+3 x \right ) \left (1+3 x \right ) \left (x^{2}+2\right )}}{3888 \sqrt {-1+3 x}\, \sqrt {1+3 x}\, \sqrt {x^{2}+2}}\)	\(104\)
elliptic	\(\frac {\sqrt {\left (x^{2}+2\right ) \left (9 x^{2}-1\right )}\, \left (\frac {1385 \ln \left (\frac {\left (\frac {17}{2}+9 x^{2}\right ) \sqrt {9}}{9}+\sqrt {9 x^{4}+17 x^{2}-2}\right ) \sqrt {9}}{3888}-\frac {49 \sqrt {9 x^{4}+17 x^{2}-2}}{216}+\frac {5 \sqrt {9 x^{4}+17 x^{2}-2}\, x^{2}}{36}\right )}{\sqrt {-1+3 x}\, \sqrt {1+3 x}\, \sqrt {x^{2}+2}}\)	\(107\)
default	\(\frac {\sqrt {-1+3 x}\, \sqrt {1+3 x}\, \sqrt {x^{2}+2}\, \left (540 \sqrt {9 x^{4}+17 x^{2}-2}\, x^{2}-882 \sqrt {9 x^{4}+17 x^{2}-2}+3379 \ln \left (\frac {17}{6}+3 x^{2}+\sqrt {9 x^{4}+17 x^{2}-2}\right )+776 \ln \left (6 x^{2}+\frac {17}{3}+2 \sqrt {9 x^{4}+17 x^{2}-2}\right )\right )}{3888 \sqrt {9 x^{4}+17 x^{2}-2}}\)	\(124\)

Input:

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

Output:

1/216*(30*x^2-49)*(-1+3*x)^(1/2)*(1+3*x)^(1/2)*(x^2+2)^(1/2)+1385/3888*ln( 
1/9*(17/2+9*x^2)*9^(1/2)+(9*x^4+17*x^2-2)^(1/2))*9^(1/2)*((-1+3*x)*(1+3*x) 
*(x^2+2))^(1/2)/(-1+3*x)^(1/2)/(1+3*x)^(1/2)/(x^2+2)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.56 \[ \int \frac {x \left (2+3 x^2+5 x^4\right )}{\sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\frac {1}{216} \, {\left (30 \, x^{2} - 49\right )} \sqrt {x^{2} + 2} \sqrt {3 \, x + 1} \sqrt {3 \, x - 1} - \frac {1385}{1296} \, \log \left (-18 \, x^{2} + 6 \, \sqrt {x^{2} + 2} \sqrt {3 \, x + 1} \sqrt {3 \, x - 1} - 17\right ) \] Input:

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

Output:

1/216*(30*x^2 - 49)*sqrt(x^2 + 2)*sqrt(3*x + 1)*sqrt(3*x - 1) - 1385/1296* 
log(-18*x^2 + 6*sqrt(x^2 + 2)*sqrt(3*x + 1)*sqrt(3*x - 1) - 17)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.61 \[ \int \frac {x \left (2+3 x^2+5 x^4\right )}{\sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\frac {5 \sqrt {3 x +1}\, \sqrt {3 x -1}\, \sqrt {x^{2}+2}\, x^{2}}{36}-\frac {49 \sqrt {3 x +1}\, \sqrt {3 x -1}\, \sqrt {x^{2}+2}}{216}+\frac {1385 \,\mathrm {log}\left (-3 \sqrt {x^{2}+2}-\sqrt {3 x +1}\, \sqrt {3 x -1}\right )}{648} \] Input:

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

Output:

(90*sqrt(3*x + 1)*sqrt(3*x - 1)*sqrt(x**2 + 2)*x**2 - 147*sqrt(3*x + 1)*sq 
rt(3*x - 1)*sqrt(x**2 + 2) + 1385*log( - 3*sqrt(x**2 + 2) - sqrt(3*x + 1)* 
sqrt(3*x - 1)))/648