\(\int \frac {\sqrt {x+x^2+x^3}}{-1+x^4} \, dx\) [1218]

Optimal result

Integrand size = 20, antiderivative size = 89 \[ \int \frac {\sqrt {x+x^2+x^3}}{-1+x^4} \, dx=\frac {1}{4} \arctan \left (\frac {\sqrt {x+x^2+x^3}}{1+x+x^2}\right )+\frac {1}{2} \text {arctanh}\left (\frac {\sqrt {x+x^2+x^3}}{1+x+x^2}\right )-\frac {1}{4} \sqrt {3} \text {arctanh}\left (\frac {\sqrt {3} \sqrt {x+x^2+x^3}}{1+x+x^2}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {x+x^2+x^3}}{-1+x^4} \, dx=\frac {\sqrt {x} \sqrt {1+x+x^2} \left (\arctan \left (\frac {\sqrt {x}}{\sqrt {1+x+x^2}}\right )+2 \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt {1+x+x^2}}\right )-\sqrt {3} \text {arctanh}\left (\frac {\sqrt {3} \sqrt {x}}{\sqrt {1+x+x^2}}\right )\right )}{4 \sqrt {x \left (1+x+x^2\right )}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.17 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.16, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2467, 25, 2035, 7276, 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.

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 7276

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

rule 2009

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

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

rule 7276

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE 
xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ 
[n, 0]
 

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.93

method	result	size
default	\(-\frac {\ln \left (\frac {\sqrt {x \left (x^{2}+x +1\right )}-x}{x}\right )}{4}-\frac {\arctan \left (\frac {\sqrt {x \left (x^{2}+x +1\right )}}{x}\right )}{4}-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {x \left (x^{2}+x +1\right )}\, \sqrt {3}}{3 x}\right )}{4}+\frac {\ln \left (\frac {\sqrt {x \left (x^{2}+x +1\right )}+x}{x}\right )}{4}\)	\(83\)
pseudoelliptic	\(-\frac {\ln \left (\frac {\sqrt {x \left (x^{2}+x +1\right )}-x}{x}\right )}{4}-\frac {\arctan \left (\frac {\sqrt {x \left (x^{2}+x +1\right )}}{x}\right )}{4}-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {x \left (x^{2}+x +1\right )}\, \sqrt {3}}{3 x}\right )}{4}+\frac {\ln \left (\frac {\sqrt {x \left (x^{2}+x +1\right )}+x}{x}\right )}{4}\)	\(83\)
trager	\(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x^{2}+4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )+6 \sqrt {x^{3}+x^{2}+x}}{\left (-1+x \right )^{2}}\right )}{8}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}+2 \sqrt {x^{3}+x^{2}+x}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{\left (1+x \right )^{2}}\right )}{8}+\frac {\ln \left (\frac {x^{2}+2 \sqrt {x^{3}+x^{2}+x}+2 x +1}{x^{2}+1}\right )}{4}\)	\(130\)
elliptic	\(\text {Expression too large to display}\)	\(1500\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.37 \[ \int \frac {\sqrt {x+x^2+x^3}}{-1+x^4} \, dx=\frac {1}{16} \, \sqrt {3} \log \left (\frac {x^{4} + 20 \, x^{3} - 4 \, \sqrt {3} \sqrt {x^{3} + x^{2} + x} {\left (x^{2} + 4 \, x + 1\right )} + 30 \, x^{2} + 20 \, x + 1}{x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1}\right ) + \frac {1}{8} \, \arctan \left (\frac {2 \, \sqrt {x^{3} + x^{2} + x}}{x^{2} + 1}\right ) + \frac {1}{4} \, \log \left (\frac {x^{2} + 2 \, x + 2 \, \sqrt {x^{3} + x^{2} + x} + 1}{x^{2} + 1}\right ) \] Input:

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

Output:

1/16*sqrt(3)*log((x^4 + 20*x^3 - 4*sqrt(3)*sqrt(x^3 + x^2 + x)*(x^2 + 4*x 
+ 1) + 30*x^2 + 20*x + 1)/(x^4 - 4*x^3 + 6*x^2 - 4*x + 1)) + 1/8*arctan(2* 
sqrt(x^3 + x^2 + x)/(x^2 + 1)) + 1/4*log((x^2 + 2*x + 2*sqrt(x^3 + x^2 + x 
) + 1)/(x^2 + 1))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 565, normalized size of antiderivative = 6.35 \[ \int \frac {\sqrt {x+x^2+x^3}}{-1+x^4} \, dx=\text {Too large to display} \] Input:

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

Output:

(((3^(1/2)*1i)/2 - 1/2)*(x/((3^(1/2)*1i)/2 - 1/2))^(1/2)*(-(x - (3^(1/2)*1 
i)/2 + 1/2)/((3^(1/2)*1i)/2 - 1/2))^(1/2)*((x + (3^(1/2)*1i)/2 + 1/2)/((3^ 
(1/2)*1i)/2 + 1/2))^(1/2)*ellipticPi(- 3^(1/2)/2 - 1i/2, asin((x/((3^(1/2) 
*1i)/2 - 1/2))^(1/2)), -((3^(1/2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 + 1/2)))/(2 
*(x^2 + x^3 - x*((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2))^(1/2)) + (( 
(3^(1/2)*1i)/2 - 1/2)*(x/((3^(1/2)*1i)/2 - 1/2))^(1/2)*(-(x - (3^(1/2)*1i) 
/2 + 1/2)/((3^(1/2)*1i)/2 - 1/2))^(1/2)*((x + (3^(1/2)*1i)/2 + 1/2)/((3^(1 
/2)*1i)/2 + 1/2))^(1/2)*ellipticPi(3^(1/2)/2 + 1i/2, asin((x/((3^(1/2)*1i) 
/2 - 1/2))^(1/2)), -((3^(1/2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 + 1/2)))/(2*(x^ 
2 + x^3 - x*((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2))^(1/2)) + (((3^( 
1/2)*1i)/2 - 1/2)*(x/((3^(1/2)*1i)/2 - 1/2))^(1/2)*(-(x - (3^(1/2)*1i)/2 + 
 1/2)/((3^(1/2)*1i)/2 - 1/2))^(1/2)*((x + (3^(1/2)*1i)/2 + 1/2)/((3^(1/2)* 
1i)/2 + 1/2))^(1/2)*ellipticPi(1/2 - (3^(1/2)*1i)/2, asin((x/((3^(1/2)*1i) 
/2 - 1/2))^(1/2)), -((3^(1/2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 + 1/2)))/(2*(x^ 
2 + x^3 - x*((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2))^(1/2)) - (3*((3 
^(1/2)*1i)/2 - 1/2)*(x/((3^(1/2)*1i)/2 - 1/2))^(1/2)*(-(x - (3^(1/2)*1i)/2 
 + 1/2)/((3^(1/2)*1i)/2 - 1/2))^(1/2)*((x + (3^(1/2)*1i)/2 + 1/2)/((3^(1/2 
)*1i)/2 + 1/2))^(1/2)*ellipticPi((3^(1/2)*1i)/2 - 1/2, asin((x/((3^(1/2)*1 
i)/2 - 1/2))^(1/2)), -((3^(1/2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 + 1/2)))/(2*( 
x^2 + x^3 - x*((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2))^(1/2))
 

Reduce [F]

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

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

Output:

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