\(\int \frac {x^3 (2+3 x^2+5 x^4)}{\sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx\) [17]

Optimal result

Integrand size = 43, antiderivative size = 145 \[ \int \frac {x^3 \left (2+3 x^2+5 x^4\right )}{\sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\frac {13723 \sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}}{11664}-\frac {983 \sqrt {-1+3 x} \sqrt {1+3 x} \left (2+x^2\right )^{3/2}}{1944}+\frac {5}{54} \sqrt {-1+3 x} \sqrt {1+3 x} \left (2+x^2\right )^{5/2}-\frac {112511 \sqrt {-1+9 x^2} \text {arcsinh}\left (\frac {\sqrt {-1+9 x^2}}{\sqrt {19}}\right )}{34992 \sqrt {-1+3 x} \sqrt {1+3 x}} \] Output:

13723/11664*(-1+3*x)^(1/2)*(1+3*x)^(1/2)*(x^2+2)^(1/2)-983/1944*(-1+3*x)^( 
1/2)*(1+3*x)^(1/2)*(x^2+2)^(3/2)+5/54*(-1+3*x)^(1/2)*(1+3*x)^(1/2)*(x^2+2) 
^(5/2)-112511/34992*(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 = 10.14 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.57 \[ \int \frac {x^3 \left (2+3 x^2+5 x^4\right )}{\sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\frac {3 \sqrt {2+x^2} \left (-6247+57801 x^2-15282 x^4+9720 x^6\right )+112511 \sqrt {1-9 x^2} \arcsin \left (\frac {\sqrt {1-9 x^2}}{\sqrt {19}}\right )}{34992 \sqrt {-1+3 x} \sqrt {1+3 x}} \] Input:

Integrate[(x^3*(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]*(-6247 + 57801*x^2 - 15282*x^4 + 9720*x^6) + 112511*Sqrt[ 
1 - 9*x^2]*ArcSin[Sqrt[1 - 9*x^2]/Sqrt[19]])/(34992*Sqrt[-1 + 3*x]*Sqrt[1 
+ 3*x])
 

Rubi [A] (verified)

Time = 1.23 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.82, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {2038, 7283, 2118, 27, 164, 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.

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

\(\Big \downarrow \) 2038

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

\(\Big \downarrow \) 7283

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

\(\Big \downarrow \) 2118

\(\displaystyle \frac {\sqrt {9 x^2-1} \left (\frac {1}{27} \int \frac {x^2 \left (148-263 x^2\right )}{2 \sqrt {x^2+2} \sqrt {9 x^2-1}}dx^2+\frac {5}{27} \sqrt {x^2+2} \sqrt {9 x^2-1} x^4\right )}{2 \sqrt {3 x-1} \sqrt {3 x+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {9 x^2-1} \left (\frac {1}{54} \int \frac {x^2 \left (148-263 x^2\right )}{\sqrt {x^2+2} \sqrt {9 x^2-1}}dx^2+\frac {5}{27} \sqrt {x^2+2} \sqrt {9 x^2-1} x^4\right )}{2 \sqrt {3 x-1} \sqrt {3 x+1}}\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {\sqrt {9 x^2-1} \left (\frac {1}{54} \left (\frac {1}{108} \left (6247-1578 x^2\right ) \sqrt {x^2+2} \sqrt {9 x^2-1}-\frac {112511}{216} \int \frac {1}{\sqrt {x^2+2} \sqrt {9 x^2-1}}dx^2\right )+\frac {5}{27} \sqrt {x^2+2} \sqrt {9 x^2-1} x^4\right )}{2 \sqrt {3 x-1} \sqrt {3 x+1}}\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {\sqrt {9 x^2-1} \left (\frac {1}{54} \left (\frac {1}{108} \left (6247-1578 x^2\right ) \sqrt {x^2+2} \sqrt {9 x^2-1}-\frac {112511}{972} \int \frac {1}{\sqrt {\frac {x^4}{9}+\frac {19}{9}}}d\sqrt {9 x^2-1}\right )+\frac {5}{27} \sqrt {x^2+2} \sqrt {9 x^2-1} x^4\right )}{2 \sqrt {3 x-1} \sqrt {3 x+1}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {\sqrt {9 x^2-1} \left (\frac {1}{54} \left (\frac {1}{108} \left (6247-1578 x^2\right ) \sqrt {x^2+2} \sqrt {9 x^2-1}-\frac {112511}{324} \text {arcsinh}\left (\frac {\sqrt {9 x^2-1}}{\sqrt {19}}\right )\right )+\frac {5}{27} \sqrt {x^2+2} \sqrt {9 x^2-1} x^4\right )}{2 \sqrt {3 x-1} \sqrt {3 x+1}}\)

Input:

Int[(x^3*(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*x^4*Sqrt[2 + x^2]*Sqrt[-1 + 9*x^2])/27 + (((6247 - 1 
578*x^2)*Sqrt[2 + x^2]*Sqrt[-1 + 9*x^2])/108 - (112511*ArcSinh[Sqrt[-1 + 9 
*x^2]/Sqrt[19]])/324)/54))/(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 164

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ 
))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - 
 b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( 
c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h 
*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 
3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + 
d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3))   Int[( 
a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] 
&& NeQ[m + n + 2, 0] && NeQ[m + n + 3, 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 7283

Int[(u_)*(x_)^(m_.), x_Symbol] :> With[{lst = PowerVariableExpn[u, m + 1, 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]], m + 1 
]] /; IntegerQ[m] && NeQ[m, -1] && NonsumQ[u] && (GtQ[m, 0] ||  !AlgebraicF 
unctionQ[u, x])
 

Maple [A] (verified)

Time = 1.89 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.75

method	result	size
risch	\(\frac {\left (1080 x^{4}-1578 x^{2}+6247\right ) \sqrt {-1+3 x}\, \sqrt {1+3 x}\, \sqrt {x^{2}+2}}{11664}-\frac {112511 \ln \left (\frac {\left (\frac {17}{2}+9 x^{2}\right ) \sqrt {9}}{9}+\sqrt {9 x^{4}+17 x^{2}-2}\right ) \sqrt {9}\, \sqrt {\left (-1+3 x \right ) \left (1+3 x \right ) \left (x^{2}+2\right )}}{209952 \sqrt {-1+3 x}\, \sqrt {1+3 x}\, \sqrt {x^{2}+2}}\)	\(109\)
elliptic	\(\frac {\sqrt {\left (x^{2}+2\right ) \left (9 x^{2}-1\right )}\, \left (\frac {6247 \sqrt {9 x^{4}+17 x^{2}-2}}{11664}-\frac {112511 \ln \left (\frac {\left (\frac {17}{2}+9 x^{2}\right ) \sqrt {9}}{9}+\sqrt {9 x^{4}+17 x^{2}-2}\right ) \sqrt {9}}{209952}-\frac {263 \sqrt {9 x^{4}+17 x^{2}-2}\, x^{2}}{1944}+\frac {5 x^{4} \sqrt {9 x^{4}+17 x^{2}-2}}{54}\right )}{\sqrt {-1+3 x}\, \sqrt {1+3 x}\, \sqrt {x^{2}+2}}\)	\(126\)
default	\(-\frac {\sqrt {-1+3 x}\, \sqrt {1+3 x}\, \sqrt {x^{2}+2}\, \left (-6480 x^{4} \sqrt {9 x^{4}+17 x^{2}-2}+9468 \sqrt {9 x^{4}+17 x^{2}-2}\, x^{2}+114063 \ln \left (\frac {17}{6}+3 x^{2}+\sqrt {9 x^{4}+17 x^{2}-2}\right )-1552 \ln \left (6 x^{2}+\frac {17}{3}+2 \sqrt {9 x^{4}+17 x^{2}-2}\right )-37482 \sqrt {9 x^{4}+17 x^{2}-2}\right )}{69984 \sqrt {9 x^{4}+17 x^{2}-2}}\)	\(143\)

Input:

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

Output:

1/11664*(1080*x^4-1578*x^2+6247)*(-1+3*x)^(1/2)*(1+3*x)^(1/2)*(x^2+2)^(1/2 
)-112511/209952*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 = 0.06 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.48 \[ \int \frac {x^3 \left (2+3 x^2+5 x^4\right )}{\sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\frac {1}{11664} \, {\left (1080 \, x^{4} - 1578 \, x^{2} + 6247\right )} \sqrt {x^{2} + 2} \sqrt {3 \, x + 1} \sqrt {3 \, x - 1} + \frac {112511}{69984} \, \log \left (-18 \, x^{2} + 6 \, \sqrt {x^{2} + 2} \sqrt {3 \, x + 1} \sqrt {3 \, x - 1} - 17\right ) \] Input:

integrate(x^3*(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:

1/11664*(1080*x^4 - 1578*x^2 + 6247)*sqrt(x^2 + 2)*sqrt(3*x + 1)*sqrt(3*x 
- 1) + 112511/69984*log(-18*x^2 + 6*sqrt(x^2 + 2)*sqrt(3*x + 1)*sqrt(3*x - 
 1) - 17)
 

Sympy [F]

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

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

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

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

int((x^3*(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*(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 = 0.82 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.63 \[ \int \frac {x^3 \left (2+3 x^2+5 x^4\right )}{\sqrt {-1+3 x} \sqrt {1+3 x} \sqrt {2+x^2}} \, dx=\frac {5 \sqrt {3 x +1}\, \sqrt {3 x -1}\, \sqrt {x^{2}+2}\, x^{4}}{54}-\frac {263 \sqrt {3 x +1}\, \sqrt {3 x -1}\, \sqrt {x^{2}+2}\, x^{2}}{1944}+\frac {6247 \sqrt {3 x +1}\, \sqrt {3 x -1}\, \sqrt {x^{2}+2}}{11664}+\frac {112511 \,\mathrm {log}\left (-3 \sqrt {x^{2}+2}+\sqrt {3 x +1}\, \sqrt {3 x -1}\right )}{34992} \] Input:

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

Output:

(3240*sqrt(3*x + 1)*sqrt(3*x - 1)*sqrt(x**2 + 2)*x**4 - 4734*sqrt(3*x + 1) 
*sqrt(3*x - 1)*sqrt(x**2 + 2)*x**2 + 18741*sqrt(3*x + 1)*sqrt(3*x - 1)*sqr 
t(x**2 + 2) + 112511*log( - 3*sqrt(x**2 + 2) + sqrt(3*x + 1)*sqrt(3*x - 1) 
))/34992