\(\int \frac {\sqrt {1+6 x^2+x^4}}{x (1+x^2)} \, dx\) [1113]

Optimal result

Integrand size = 25, antiderivative size = 83 \[ \int \frac {\sqrt {1+6 x^2+x^4}}{x \left (1+x^2\right )} \, dx=-2 \arctan \left (\frac {1}{2}+\frac {x^2}{2}-\frac {1}{2} \sqrt {1+6 x^2+x^4}\right )-\text {arctanh}\left (2+x^2-\sqrt {1+6 x^2+x^4}\right )-\frac {1}{2} \log \left (1-x^2+\sqrt {1+6 x^2+x^4}\right ) \] Output:

-2*arctan(1/2+1/2*x^2-1/2*(x^4+6*x^2+1)^(1/2))-arctanh(2+x^2-(x^4+6*x^2+1) 
^(1/2))-1/2*ln(1-x^2+(x^4+6*x^2+1)^(1/2))
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {1+6 x^2+x^4}}{x \left (1+x^2\right )} \, dx=-2 \arctan \left (\frac {1}{2}+\frac {x^2}{2}-\frac {1}{2} \sqrt {1+6 x^2+x^4}\right )-\text {arctanh}\left (2+x^2-\sqrt {1+6 x^2+x^4}\right )-\frac {1}{2} \log \left (1-x^2+\sqrt {1+6 x^2+x^4}\right ) \] Input:

Integrate[Sqrt[1 + 6*x^2 + x^4]/(x*(1 + x^2)),x]
 

Output:

-2*ArcTan[1/2 + x^2/2 - Sqrt[1 + 6*x^2 + x^4]/2] - ArcTanh[2 + x^2 - Sqrt[ 
1 + 6*x^2 + x^4]] - Log[1 - x^2 + Sqrt[1 + 6*x^2 + x^4]]/2
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.92, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1578, 1270, 25, 1154, 219, 1269, 1092, 219, 1154, 217}

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.

\(\displaystyle \int \frac {\sqrt {x^4+6 x^2+1}}{x \left (x^2+1\right )} \, dx\)

\(\Big \downarrow \) 1578

\(\displaystyle \frac {1}{2} \int \frac {\sqrt {x^4+6 x^2+1}}{x^2 \left (x^2+1\right )}dx^2\)

\(\Big \downarrow \) 1270

\(\displaystyle \frac {1}{2} \left (\int \frac {1}{x^2 \sqrt {x^4+6 x^2+1}}dx^2-\int -\frac {x^2+5}{\left (x^2+1\right ) \sqrt {x^4+6 x^2+1}}dx^2\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (\int \frac {1}{x^2 \sqrt {x^4+6 x^2+1}}dx^2+\int \frac {x^2+5}{\left (x^2+1\right ) \sqrt {x^4+6 x^2+1}}dx^2\right )\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {1}{2} \left (\int \frac {x^2+5}{\left (x^2+1\right ) \sqrt {x^4+6 x^2+1}}dx^2-2 \int \frac {1}{4-x^4}d\frac {2 \left (3 x^2+1\right )}{\sqrt {x^4+6 x^2+1}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\int \frac {x^2+5}{\left (x^2+1\right ) \sqrt {x^4+6 x^2+1}}dx^2-\text {arctanh}\left (\frac {3 x^2+1}{\sqrt {x^4+6 x^2+1}}\right )\right )\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {1}{2} \left (\int \frac {1}{\sqrt {x^4+6 x^2+1}}dx^2+4 \int \frac {1}{\left (x^2+1\right ) \sqrt {x^4+6 x^2+1}}dx^2-\text {arctanh}\left (\frac {3 x^2+1}{\sqrt {x^4+6 x^2+1}}\right )\right )\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {1}{2} \left (2 \int \frac {1}{4-x^4}d\frac {2 \left (x^2+3\right )}{\sqrt {x^4+6 x^2+1}}+4 \int \frac {1}{\left (x^2+1\right ) \sqrt {x^4+6 x^2+1}}dx^2-\text {arctanh}\left (\frac {3 x^2+1}{\sqrt {x^4+6 x^2+1}}\right )\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (4 \int \frac {1}{\left (x^2+1\right ) \sqrt {x^4+6 x^2+1}}dx^2+\text {arctanh}\left (\frac {x^2+3}{\sqrt {x^4+6 x^2+1}}\right )-\text {arctanh}\left (\frac {3 x^2+1}{\sqrt {x^4+6 x^2+1}}\right )\right )\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {1}{2} \left (-8 \int \frac {1}{-x^4-16}d\left (-\frac {4 \left (1-x^2\right )}{\sqrt {x^4+6 x^2+1}}\right )+\text {arctanh}\left (\frac {x^2+3}{\sqrt {x^4+6 x^2+1}}\right )-\text {arctanh}\left (\frac {3 x^2+1}{\sqrt {x^4+6 x^2+1}}\right )\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{2} \left (-2 \arctan \left (\frac {1-x^2}{\sqrt {x^4+6 x^2+1}}\right )+\text {arctanh}\left (\frac {x^2+3}{\sqrt {x^4+6 x^2+1}}\right )-\text {arctanh}\left (\frac {3 x^2+1}{\sqrt {x^4+6 x^2+1}}\right )\right )\)

Input:

Int[Sqrt[1 + 6*x^2 + x^4]/(x*(1 + x^2)),x]
 

Output:

(-2*ArcTan[(1 - x^2)/Sqrt[1 + 6*x^2 + x^4]] + ArcTanh[(3 + x^2)/Sqrt[1 + 6 
*x^2 + x^4]] - ArcTanh[(1 + 3*x^2)/Sqrt[1 + 6*x^2 + x^4]])/2
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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 1092

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

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 1270

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

rule 1578

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ 
)^4)^(p_.), x_Symbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(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] && Int 
egerQ[(m - 1)/2]
 

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.77

method	result	size
pseudoelliptic	\(\frac {\ln \left (x^{2}+3+\sqrt {x^{4}+6 x^{2}+1}\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {3 x^{2}+1}{\sqrt {x^{4}+6 x^{2}+1}}\right )}{2}+\arctan \left (\frac {x^{2}-1}{\sqrt {x^{4}+6 x^{2}+1}}\right )\)	\(64\)
trager	\(\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}+\sqrt {x^{4}+6 x^{2}+1}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{x^{2}+1}\right )-\ln \left (\frac {1-x^{2}+\sqrt {x^{4}+6 x^{2}+1}}{x}\right )\)	\(74\)
default	\(\frac {\sqrt {x^{4}+6 x^{2}+1}}{2}+\frac {3 \ln \left (x^{2}+3+\sqrt {x^{4}+6 x^{2}+1}\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {6 x^{2}+2}{2 \sqrt {x^{4}+6 x^{2}+1}}\right )}{2}-\frac {\sqrt {\left (x^{2}+1\right )^{2}+4 x^{2}}}{2}-\ln \left (3+x^{2}+\sqrt {\left (x^{2}+1\right )^{2}+4 x^{2}}\right )+\arctan \left (\frac {4 x^{2}-4}{4 \sqrt {\left (x^{2}+1\right )^{2}+4 x^{2}}}\right )\)	\(125\)
elliptic	\(\frac {\sqrt {x^{4}+6 x^{2}+1}}{2}+\frac {3 \ln \left (x^{2}+3+\sqrt {x^{4}+6 x^{2}+1}\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {6 x^{2}+2}{2 \sqrt {x^{4}+6 x^{2}+1}}\right )}{2}-\frac {\sqrt {\left (x^{2}+1\right )^{2}+4 x^{2}}}{2}-\ln \left (3+x^{2}+\sqrt {\left (x^{2}+1\right )^{2}+4 x^{2}}\right )+\arctan \left (\frac {4 x^{2}-4}{4 \sqrt {\left (x^{2}+1\right )^{2}+4 x^{2}}}\right )\)	\(125\)

Input:

int((x^4+6*x^2+1)^(1/2)/x/(x^2+1),x,method=_RETURNVERBOSE)
 

Output:

1/2*ln(x^2+3+(x^4+6*x^2+1)^(1/2))-1/2*arctanh((3*x^2+1)/(x^4+6*x^2+1)^(1/2 
))+arctan((x^2-1)/(x^4+6*x^2+1)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {1+6 x^2+x^4}}{x \left (1+x^2\right )} \, dx=2 \, \arctan \left (-\frac {1}{2} \, x^{2} + \frac {1}{2} \, \sqrt {x^{4} + 6 \, x^{2} + 1} - \frac {1}{2}\right ) - \frac {1}{2} \, \log \left (x^{4} + 4 \, x^{2} - \sqrt {x^{4} + 6 \, x^{2} + 1} {\left (x^{2} + 1\right )} - 1\right ) + \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 6 \, x^{2} + 1} - 1\right ) \] Input:

integrate((x^4+6*x^2+1)^(1/2)/x/(x^2+1),x, algorithm="fricas")
 

Output:

2*arctan(-1/2*x^2 + 1/2*sqrt(x^4 + 6*x^2 + 1) - 1/2) - 1/2*log(x^4 + 4*x^2 
 - sqrt(x^4 + 6*x^2 + 1)*(x^2 + 1) - 1) + 1/2*log(-x^2 + sqrt(x^4 + 6*x^2 
+ 1) - 1)
 

Sympy [F]

\[ \int \frac {\sqrt {1+6 x^2+x^4}}{x \left (1+x^2\right )} \, dx=\int \frac {\sqrt {x^{4} + 6 x^{2} + 1}}{x \left (x^{2} + 1\right )}\, dx \] Input:

integrate((x**4+6*x**2+1)**(1/2)/x/(x**2+1),x)
 

Output:

Integral(sqrt(x**4 + 6*x**2 + 1)/(x*(x**2 + 1)), x)
 

Maxima [F]

\[ \int \frac {\sqrt {1+6 x^2+x^4}}{x \left (1+x^2\right )} \, dx=\int { \frac {\sqrt {x^{4} + 6 \, x^{2} + 1}}{{\left (x^{2} + 1\right )} x} \,d x } \] Input:

integrate((x^4+6*x^2+1)^(1/2)/x/(x^2+1),x, algorithm="maxima")
 

Output:

integrate(sqrt(x^4 + 6*x^2 + 1)/((x^2 + 1)*x), x)
 

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.10 \[ \int \frac {\sqrt {1+6 x^2+x^4}}{x \left (1+x^2\right )} \, dx=2 \, \arctan \left (-\frac {1}{2} \, x^{2} + \frac {1}{2} \, \sqrt {x^{4} + 6 \, x^{2} + 1} - \frac {1}{2}\right ) - \frac {1}{2} \, \log \left (x^{2} - \sqrt {x^{4} + 6 \, x^{2} + 1} + 3\right ) - \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 6 \, x^{2} + 1} + 1\right ) + \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 6 \, x^{2} + 1} - 1\right ) \] Input:

integrate((x^4+6*x^2+1)^(1/2)/x/(x^2+1),x, algorithm="giac")
 

Output:

2*arctan(-1/2*x^2 + 1/2*sqrt(x^4 + 6*x^2 + 1) - 1/2) - 1/2*log(x^2 - sqrt( 
x^4 + 6*x^2 + 1) + 3) - 1/2*log(-x^2 + sqrt(x^4 + 6*x^2 + 1) + 1) + 1/2*lo 
g(-x^2 + sqrt(x^4 + 6*x^2 + 1) - 1)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {1+6 x^2+x^4}}{x \left (1+x^2\right )} \, dx=\int \frac {\sqrt {x^4+6\,x^2+1}}{x\,\left (x^2+1\right )} \,d x \] Input:

int((6*x^2 + x^4 + 1)^(1/2)/(x*(x^2 + 1)),x)
 

Output:

int((6*x^2 + x^4 + 1)^(1/2)/(x*(x^2 + 1)), x)
 

Reduce [F]

\[ \int \frac {\sqrt {1+6 x^2+x^4}}{x \left (1+x^2\right )} \, dx=\int \frac {\sqrt {x^{4}+6 x^{2}+1}}{x^{3}+x}d x \] Input:

int((x^4+6*x^2+1)^(1/2)/x/(x^2+1),x)
 

Output:

int(sqrt(x**4 + 6*x**2 + 1)/(x**3 + x),x)