\(\int \frac {x^5 (2+3 x^2)}{(1+x^2) \sqrt {1+x^2+x^4}} \, dx\) [93]

Optimal result

Integrand size = 30, antiderivative size = 82 \[ \int \frac {x^5 \left (2+3 x^2\right )}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=-\frac {13}{8} \sqrt {1+x^2+x^4}+\frac {3}{4} x^2 \sqrt {1+x^2+x^4}+\frac {9}{16} \text {arcsinh}\left (\frac {1+2 x^2}{\sqrt {3}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {1-x^2}{2 \sqrt {1+x^2+x^4}}\right ) \] Output:

-13/8*(x^4+x^2+1)^(1/2)+3/4*x^2*(x^4+x^2+1)^(1/2)+9/16*arcsinh(1/3*(2*x^2+ 
1)*3^(1/2))+1/2*arctanh(1/2*(-x^2+1)/(x^4+x^2+1)^(1/2))
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.88 \[ \int \frac {x^5 \left (2+3 x^2\right )}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=\frac {1}{8} \left (-13+6 x^2\right ) \sqrt {1+x^2+x^4}-\text {arctanh}\left (1+x^2-\sqrt {1+x^2+x^4}\right )-\frac {9}{16} \log \left (-1-2 x^2+2 \sqrt {1+x^2+x^4}\right ) \] Input:

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

Output:

((-13 + 6*x^2)*Sqrt[1 + x^2 + x^4])/8 - ArcTanh[1 + x^2 - Sqrt[1 + x^2 + x 
^4]] - (9*Log[-1 - 2*x^2 + 2*Sqrt[1 + x^2 + x^4]])/16
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2236, 2236, 2252, 2238, 1269, 1090, 222, 1154, 219}

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.

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

\(\Big \downarrow \) 2236

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

\(\Big \downarrow \) 2236

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \frac {\frac {13 \left (x^2+1\right ) \left (2 x^4+x^2\right )}{x}+2 \left (4 x^5 \left (3 x^2+2\right )-3 x \left (x^2+1\right ) \left (4 x^4+3 x^2+2\right )\right )}{\left (x^2+1\right ) \sqrt {x^4+x^2+1}}dx-\frac {13}{2} \sqrt {x^4+x^2+1}\right )+\frac {3}{4} \sqrt {x^4+x^2+1} x^2\)

\(\Big \downarrow \) 2252

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \frac {x \left (9 x^2+1\right )}{\left (x^2+1\right ) \sqrt {x^4+x^2+1}}dx-\frac {13}{2} \sqrt {x^4+x^2+1}\right )+\frac {3}{4} \sqrt {x^4+x^2+1} x^2\)

\(\Big \downarrow \) 2238

\(\displaystyle \frac {1}{4} \left (\frac {1}{4} \int \frac {9 x^2+1}{\left (x^2+1\right ) \sqrt {x^4+x^2+1}}dx^2-\frac {13}{2} \sqrt {x^4+x^2+1}\right )+\frac {3}{4} \sqrt {x^4+x^2+1} x^2\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {1}{4} \left (\frac {1}{4} \left (9 \int \frac {1}{\sqrt {x^4+x^2+1}}dx^2-8 \int \frac {1}{\left (x^2+1\right ) \sqrt {x^4+x^2+1}}dx^2\right )-\frac {13}{2} \sqrt {x^4+x^2+1}\right )+\frac {3}{4} \sqrt {x^4+x^2+1} x^2\)

\(\Big \downarrow \) 1090

\(\displaystyle \frac {1}{4} \left (\frac {1}{4} \left (3 \sqrt {3} \int \frac {1}{\sqrt {\frac {x^4}{3}+1}}d\left (2 x^2+1\right )-8 \int \frac {1}{\left (x^2+1\right ) \sqrt {x^4+x^2+1}}dx^2\right )-\frac {13}{2} \sqrt {x^4+x^2+1}\right )+\frac {3}{4} \sqrt {x^4+x^2+1} x^2\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {1}{4} \left (\frac {1}{4} \left (9 \text {arcsinh}\left (\frac {2 x^2+1}{\sqrt {3}}\right )-8 \int \frac {1}{\left (x^2+1\right ) \sqrt {x^4+x^2+1}}dx^2\right )-\frac {13}{2} \sqrt {x^4+x^2+1}\right )+\frac {3}{4} \sqrt {x^4+x^2+1} x^2\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {1}{4} \left (\frac {1}{4} \left (16 \int \frac {1}{4-x^4}d\frac {1-x^2}{\sqrt {x^4+x^2+1}}+9 \text {arcsinh}\left (\frac {2 x^2+1}{\sqrt {3}}\right )\right )-\frac {13}{2} \sqrt {x^4+x^2+1}\right )+\frac {3}{4} \sqrt {x^4+x^2+1} x^2\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{4} \left (\frac {1}{4} \left (9 \text {arcsinh}\left (\frac {2 x^2+1}{\sqrt {3}}\right )+8 \text {arctanh}\left (\frac {1-x^2}{2 \sqrt {x^4+x^2+1}}\right )\right )-\frac {13}{2} \sqrt {x^4+x^2+1}\right )+\frac {3}{4} \sqrt {x^4+x^2+1} x^2\)

Input:

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

Output:

(3*x^2*Sqrt[1 + x^2 + x^4])/4 + ((-13*Sqrt[1 + x^2 + x^4])/2 + (9*ArcSinh[ 
(1 + 2*x^2)/Sqrt[3]] + 8*ArcTanh[(1 - x^2)/(2*Sqrt[1 + x^2 + x^4])])/4)/4
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[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 1090

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* 
(c/(b^2 - 4*a*c)))^p)   Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, 
b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
 

rule 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-2   Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 
2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c 
, d, e}, x]
 

rule 1269

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

rule 2236

Int[(Px_)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), 
 x_Symbol] :> With[{q = Expon[Px, x]}, Simp[Coeff[Px, x, q]*x^(q - 5)*(Sqrt 
[a + b*x^2 + c*x^4]/(c*e*(q - 3))), x] + Simp[1/(c*e*(q - 3))   Int[(c*e*(q 
 - 3)*Px - Coeff[Px, x, q]*x^(q - 6)*(d + e*x^2)*(a*(q - 5) + b*(q - 4)*x^2 
 + c*(q - 3)*x^4))/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x] /; GtQ[q, 
4]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Px, x]
 

rule 2238

Int[(Px_)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ 
)^4)^(p_.), x_Symbol] :> Simp[1/2   Subst[Int[(Px /. x -> Sqrt[x])*(d + e*x 
)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] 
 && PolyQ[Px, x^2]
 

rule 2252

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^ 
(p_), x_Symbol] :> With[{m = Expon[Px, x, Min]}, Int[x^m*ExpandToSum[Px/x^m 
, x]*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x] /; GtQ[m, 0] &&  !MatchQ[Px, x 
^m*(u_.)]] /; FreeQ[{a, b, c, d, e, p, q}, x] && PolyQ[Px, x]
 

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84

\[\frac {9 \,\operatorname {arcsinh}\left (\frac {2 \sqrt {3}\, \left (x^{2}+\frac {1}{2}\right )}{3}\right )}{16}-\frac {13 \sqrt {x^{4}+x^{2}+1}}{8}+\frac {3 x^{2} \sqrt {x^{4}+x^{2}+1}}{4}+\frac {\operatorname {arctanh}\left (\frac {-x^{2}+1}{2 \sqrt {\left (x^{2}+1\right )^{2}-x^{2}}}\right )}{2}\]

Input:

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

Output:

9/16*arcsinh(2/3*3^(1/2)*(x^2+1/2))-13/8*(x^4+x^2+1)^(1/2)+3/4*x^2*(x^4+x^ 
2+1)^(1/2)+1/2*arctanh(1/2*(-x^2+1)/((x^2+1)^2-x^2)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.99 \[ \int \frac {x^5 \left (2+3 x^2\right )}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=\frac {1}{8} \, \sqrt {x^{4} + x^{2} + 1} {\left (6 \, x^{2} - 13\right )} + \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + x^{2} + 1}\right ) - \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + x^{2} + 1} - 2\right ) - \frac {9}{16} \, \log \left (-2 \, x^{2} + 2 \, \sqrt {x^{4} + x^{2} + 1} - 1\right ) \] Input:

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

Output:

1/8*sqrt(x^4 + x^2 + 1)*(6*x^2 - 13) + 1/2*log(-x^2 + sqrt(x^4 + x^2 + 1)) 
 - 1/2*log(-x^2 + sqrt(x^4 + x^2 + 1) - 2) - 9/16*log(-2*x^2 + 2*sqrt(x^4 
+ x^2 + 1) - 1)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.99 \[ \int \frac {x^5 \left (2+3 x^2\right )}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=\frac {1}{8} \, \sqrt {x^{4} + x^{2} + 1} {\left (6 \, x^{2} - 13\right )} - \frac {1}{2} \, \log \left (x^{2} - \sqrt {x^{4} + x^{2} + 1} + 2\right ) + \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + x^{2} + 1}\right ) - \frac {9}{16} \, \log \left (-2 \, x^{2} + 2 \, \sqrt {x^{4} + x^{2} + 1} - 1\right ) \] Input:

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

Output:

1/8*sqrt(x^4 + x^2 + 1)*(6*x^2 - 13) - 1/2*log(x^2 - sqrt(x^4 + x^2 + 1) + 
 2) + 1/2*log(-x^2 + sqrt(x^4 + x^2 + 1)) - 9/16*log(-2*x^2 + 2*sqrt(x^4 + 
 x^2 + 1) - 1)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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