\(\int \frac {(1+x^2) \sqrt [4]{x^3+x^4}}{x^2 (-1+x^2)} \, dx\) [1418]

Optimal result

Integrand size = 27, antiderivative size = 101 \[ \int \frac {\left (1+x^2\right ) \sqrt [4]{x^3+x^4}}{x^2 \left (-1+x^2\right )} \, dx=\frac {4 \sqrt [4]{x^3+x^4}}{x}-2 \arctan \left (\frac {x}{\sqrt [4]{x^3+x^4}}\right )+2 \sqrt [4]{2} \arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^3+x^4}}\right )+2 \text {arctanh}\left (\frac {x}{\sqrt [4]{x^3+x^4}}\right )-2 \sqrt [4]{2} \text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^3+x^4}}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.28 \[ \int \frac {\left (1+x^2\right ) \sqrt [4]{x^3+x^4}}{x^2 \left (-1+x^2\right )} \, dx=\frac {2 x^2 (1+x)^{3/4} \left (2 \sqrt [4]{1+x}-\sqrt [4]{x} \arctan \left (\sqrt [4]{\frac {x}{1+x}}\right )+\sqrt [4]{2} \sqrt [4]{x} \arctan \left (\sqrt [4]{2} \sqrt [4]{\frac {x}{1+x}}\right )+\sqrt [4]{x} \text {arctanh}\left (\sqrt [4]{\frac {x}{1+x}}\right )-\sqrt [4]{2} \sqrt [4]{x} \text {arctanh}\left (\sqrt [4]{2} \sqrt [4]{\frac {x}{1+x}}\right )\right )}{\left (x^3 (1+x)\right )^{3/4}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.26, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2467, 25, 2003, 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.

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2003

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 7276

\(\displaystyle -\frac {4 \sqrt [4]{x^4+x^3} \int \left (-\frac {\sqrt {x}}{(x+1)^{3/4}}+\frac {1}{\left (-\sqrt {x}-1\right ) (x+1)^{3/4}}+\frac {1}{\left (1-\sqrt {x}\right ) (x+1)^{3/4}}+\frac {1}{(x+1)^{3/4} \sqrt {x}}\right )d\sqrt [4]{x}}{x^{3/4} \sqrt [4]{x+1}}\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

rule 2003

Int[(u_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] : 
> Int[u*(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, n, p} 
, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && 
  !IntegerQ[n]))
 

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 = 3.29 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.41

method	result	size
pseudoelliptic	\(\frac {-\ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}\right ) 2^{\frac {1}{4}} x -2 \arctan \left (\frac {2^{\frac {3}{4}} \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{2 x}\right ) 2^{\frac {1}{4}} x +2 \arctan \left (\frac {\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{x}\right ) x +\ln \left (\frac {x +\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{x}\right ) x -\ln \left (\frac {\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}-x}{x}\right ) x +4 \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{x}\)	\(142\)
trager	\(\frac {4 \left (x^{4}+x^{3}\right )^{\frac {1}{4}}}{x}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-2 \sqrt {x^{4}+x^{3}}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}+x^{3}\right )^{\frac {3}{4}}-2 \left (x^{4}+x^{3}\right )^{\frac {1}{4}} x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}}{x^{2}}\right )+\ln \left (\frac {2 \left (x^{4}+x^{3}\right )^{\frac {3}{4}}+2 \sqrt {x^{4}+x^{3}}\, x +2 \left (x^{4}+x^{3}\right )^{\frac {1}{4}} x^{2}+2 x^{3}+x^{2}}{x^{2}}\right )+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (-\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) x^{3}-\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) x^{2}+4 \left (x^{4}+x^{3}\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{2}+4 \sqrt {x^{4}+x^{3}}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) x -4 \left (x^{4}+x^{3}\right )^{\frac {3}{4}}}{\left (-1+x \right ) x^{2}}\right )-\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} x^{3}+4 \left (x^{4}+x^{3}\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} x^{2}+4 \sqrt {x^{4}+x^{3}}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) x +4 \left (x^{4}+x^{3}\right )^{\frac {3}{4}}}{\left (-1+x \right ) x^{2}}\right )\)	\(376\)
risch	\(\text {Expression too large to display}\)	\(975\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.60 \[ \int \frac {\left (1+x^2\right ) \sqrt [4]{x^3+x^4}}{x^2 \left (-1+x^2\right )} \, dx=\frac {2 \cdot 2^{\frac {1}{4}} x \arctan \left (\frac {2^{\frac {1}{4}} {\left (x^{4} + x^{3}\right )}^{\frac {3}{4}}}{x^{3} + x^{2}}\right ) - 2^{\frac {1}{4}} x \log \left (\frac {2^{\frac {1}{4}} x + {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + 2^{\frac {1}{4}} x \log \left (-\frac {2^{\frac {1}{4}} x - {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - 2 \, x \arctan \left (\frac {{\left (x^{4} + x^{3}\right )}^{\frac {3}{4}}}{x^{3} + x^{2}}\right ) + x \log \left (\frac {x + {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - x \log \left (-\frac {x - {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + 4 \, {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.96 \[ \int \frac {\left (1+x^2\right ) \sqrt [4]{x^3+x^4}}{x^2 \left (-1+x^2\right )} \, dx=-2 \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) - 2^{\frac {1}{4}} \log \left (2^{\frac {1}{4}} + {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) + 2^{\frac {1}{4}} \log \left ({\left | -2^{\frac {1}{4}} + {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} \right |}\right ) + 4 \, {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 2 \, \arctan \left ({\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) + \log \left ({\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 1\right ) - \log \left ({\left | {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (1+x^2\right ) \sqrt [4]{x^3+x^4}}{x^2 \left (-1+x^2\right )} \, dx=\int \frac {\left (x +1\right )^{\frac {1}{4}}}{x^{\frac {13}{4}}-x^{\frac {5}{4}}}d x +\int \frac {\left (x +1\right )^{\frac {1}{4}} x}{x^{\frac {9}{4}}-x^{\frac {1}{4}}}d x \] Input:

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

Output:

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