\(\int \frac {\sqrt [4]{x^3+x^4}}{-2+x+2 x^2} \, dx\) [1812]

Optimal result

Integrand size = 22, antiderivative size = 123 \[ \int \frac {\sqrt [4]{x^3+x^4}}{-2+x+2 x^2} \, dx=-\arctan \left (\frac {x}{\sqrt [4]{x^3+x^4}}\right )+\text {arctanh}\left (\frac {x}{\sqrt [4]{x^3+x^4}}\right )-\frac {1}{2} \text {RootSum}\left [1-5 \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{x^3+x^4}-x \text {$\#$1}\right )+2 \log (x) \text {$\#$1}^4-2 \log \left (\sqrt [4]{x^3+x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-5 \text {$\#$1}^3+4 \text {$\#$1}^7}\&\right ] \] Output:

Unintegrable
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.19 \[ \int \frac {\sqrt [4]{x^3+x^4}}{-2+x+2 x^2} \, dx=-\frac {x^{9/4} (1+x)^{3/4} \left (8 \arctan \left (\sqrt [4]{\frac {x}{1+x}}\right )-8 \text {arctanh}\left (\sqrt [4]{\frac {x}{1+x}}\right )+\text {RootSum}\left [1-5 \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {-\log (x)+4 \log \left (\sqrt [4]{1+x}-\sqrt [4]{x} \text {$\#$1}\right )+2 \log (x) \text {$\#$1}^4-8 \log \left (\sqrt [4]{1+x}-\sqrt [4]{x} \text {$\#$1}\right ) \text {$\#$1}^4}{-5 \text {$\#$1}^3+4 \text {$\#$1}^7}\&\right ]\right )}{8 \left (x^3 (1+x)\right )^{3/4}} \] Input:

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

Output:

-1/8*(x^(9/4)*(1 + x)^(3/4)*(8*ArcTan[(x/(1 + x))^(1/4)] - 8*ArcTanh[(x/(1 
 + x))^(1/4)] + RootSum[1 - 5*#1^4 + 2*#1^8 & , (-Log[x] + 4*Log[(1 + x)^( 
1/4) - x^(1/4)*#1] + 2*Log[x]*#1^4 - 8*Log[(1 + x)^(1/4) - x^(1/4)*#1]*#1^ 
4)/(-5*#1^3 + 4*#1^7) & ]))/(x^3*(1 + x))^(3/4)
 

Rubi [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(505\) vs. \(2(123)=246\).

Time = 1.40 (sec) , antiderivative size = 505, normalized size of antiderivative = 4.11, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {2467, 25, 1202, 25, 73, 854, 827, 216, 219, 2035, 7279, 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.

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1202

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 854

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 7279

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\sqrt [4]{x^4+x^3} \left (2 \left (\frac {\sqrt [4]{\frac {1}{2} \left (95+23 \sqrt {17}\right )} \arctan \left (\frac {\sqrt [4]{\frac {2}{5+\sqrt {17}}} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{2 \sqrt {17}}-\frac {\sqrt [4]{71+17 \sqrt {17}} \arctan \left (\frac {\sqrt [4]{\frac {2}{5+\sqrt {17}}} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{\sqrt {34}}-\frac {2 \sqrt {\frac {2}{17}} \sqrt [4]{5+\sqrt {17}} \arctan \left (\frac {\sqrt [4]{5+\sqrt {17}} \sqrt [4]{x}}{\sqrt {2} \sqrt [4]{x+1}}\right )}{3+\sqrt {17}}-\frac {\sqrt [4]{\frac {1}{2} \left (95-23 \sqrt {17}\right )} \arctan \left (\frac {\sqrt [4]{5+\sqrt {17}} \sqrt [4]{x}}{\sqrt {2} \sqrt [4]{x+1}}\right )}{2 \sqrt {17}}-\frac {\sqrt [4]{\frac {1}{2} \left (95+23 \sqrt {17}\right )} \text {arctanh}\left (\frac {\sqrt [4]{\frac {2}{5+\sqrt {17}}} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{2 \sqrt {17}}+\frac {\sqrt [4]{71+17 \sqrt {17}} \text {arctanh}\left (\frac {\sqrt [4]{\frac {2}{5+\sqrt {17}}} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{\sqrt {34}}+\frac {2 \sqrt {\frac {2}{17}} \sqrt [4]{5+\sqrt {17}} \text {arctanh}\left (\frac {\sqrt [4]{5+\sqrt {17}} \sqrt [4]{x}}{\sqrt {2} \sqrt [4]{x+1}}\right )}{3+\sqrt {17}}+\frac {\sqrt [4]{\frac {1}{2} \left (95-23 \sqrt {17}\right )} \text {arctanh}\left (\frac {\sqrt [4]{5+\sqrt {17}} \sqrt [4]{x}}{\sqrt {2} \sqrt [4]{x+1}}\right )}{2 \sqrt {17}}\right )-2 \left (\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )-\frac {1}{2} \arctan \left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )\right )\right )}{x^{3/4} \sqrt [4]{x+1}}\)

Input:

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

Output:

-(((x^3 + x^4)^(1/4)*(-2*(-1/2*ArcTan[x^(1/4)/(1 + x)^(1/4)] + ArcTanh[x^( 
1/4)/(1 + x)^(1/4)]/2) + 2*(-(((71 + 17*Sqrt[17])^(1/4)*ArcTan[((2/(5 + Sq 
rt[17]))^(1/4)*x^(1/4))/(1 + x)^(1/4)])/Sqrt[34]) + (((95 + 23*Sqrt[17])/2 
)^(1/4)*ArcTan[((2/(5 + Sqrt[17]))^(1/4)*x^(1/4))/(1 + x)^(1/4)])/(2*Sqrt[ 
17]) - (((95 - 23*Sqrt[17])/2)^(1/4)*ArcTan[((5 + Sqrt[17])^(1/4)*x^(1/4)) 
/(Sqrt[2]*(1 + x)^(1/4))])/(2*Sqrt[17]) - (2*Sqrt[2/17]*(5 + Sqrt[17])^(1/ 
4)*ArcTan[((5 + Sqrt[17])^(1/4)*x^(1/4))/(Sqrt[2]*(1 + x)^(1/4))])/(3 + Sq 
rt[17]) + ((71 + 17*Sqrt[17])^(1/4)*ArcTanh[((2/(5 + Sqrt[17]))^(1/4)*x^(1 
/4))/(1 + x)^(1/4)])/Sqrt[34] - (((95 + 23*Sqrt[17])/2)^(1/4)*ArcTanh[((2/ 
(5 + Sqrt[17]))^(1/4)*x^(1/4))/(1 + x)^(1/4)])/(2*Sqrt[17]) + (((95 - 23*S 
qrt[17])/2)^(1/4)*ArcTanh[((5 + Sqrt[17])^(1/4)*x^(1/4))/(Sqrt[2]*(1 + x)^ 
(1/4))])/(2*Sqrt[17]) + (2*Sqrt[2/17]*(5 + Sqrt[17])^(1/4)*ArcTanh[((5 + S 
qrt[17])^(1/4)*x^(1/4))/(Sqrt[2]*(1 + x)^(1/4))])/(3 + Sqrt[17]))))/(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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[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 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 854

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 
 1)/n)   Subst[Int[x^m/(1 - b*x^n)^(p + (m + 1)/n + 1), x], x, x/(a + b*x^n 
)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, - 
2^(-1)] && IntegersQ[m, p + (m + 1)/n]
 

rule 1202

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

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 7279

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ 
{v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su 
mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
 

Maple [N/A] (verified)

Time = 11.53 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.90

method	result	size
pseudoelliptic	\(\arctan \left (\frac {\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{x}\right )+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (2 \textit {\_Z}^{8}-5 \textit {\_Z}^{4}+1\right )}{\sum }\frac {\left (2 \textit {\_R}^{4}-1\right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{3} \left (4 \textit {\_R}^{4}-5\right )}\right )}{2}+\frac {\ln \left (\frac {x +\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{x}\right )}{2}-\frac {\ln \left (\frac {\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}-x}{x}\right )}{2}\)	\(111\)
trager	\(\text {Expression too large to display}\)	\(3013\)

Input:

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

Output:

arctan((x^3*(1+x))^(1/4)/x)+1/2*sum((2*_R^4-1)*ln((-_R*x+(x^3*(1+x))^(1/4) 
)/x)/_R^3/(4*_R^4-5),_R=RootOf(2*_Z^8-5*_Z^4+1))+1/2*ln((x+(x^3*(1+x))^(1/ 
4))/x)-1/2*ln(((x^3*(1+x))^(1/4)-x)/x)
 

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 0.09 (sec) , antiderivative size = 522, normalized size of antiderivative = 4.24 \[ \int \frac {\sqrt [4]{x^3+x^4}}{-2+x+2 x^2} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*sqrt(1/17)*sqrt(-sqrt(1/2*sqrt(17) + 23/2))*log((sqrt(1/17)*(sqrt(17) 
*x + 17*x)*sqrt(-sqrt(1/2*sqrt(17) + 23/2)) + 8*(x^4 + x^3)^(1/4))/x) + 1/ 
2*sqrt(1/17)*sqrt(-sqrt(1/2*sqrt(17) + 23/2))*log(-(sqrt(1/17)*(sqrt(17)*x 
 + 17*x)*sqrt(-sqrt(1/2*sqrt(17) + 23/2)) - 8*(x^4 + x^3)^(1/4))/x) + 1/2* 
sqrt(1/17)*sqrt(-sqrt(-1/2*sqrt(17) + 23/2))*log((sqrt(1/17)*(sqrt(17)*x - 
 17*x)*sqrt(-sqrt(-1/2*sqrt(17) + 23/2)) + 8*(x^4 + x^3)^(1/4))/x) - 1/2*s 
qrt(1/17)*sqrt(-sqrt(-1/2*sqrt(17) + 23/2))*log(-(sqrt(1/17)*(sqrt(17)*x - 
 17*x)*sqrt(-sqrt(-1/2*sqrt(17) + 23/2)) - 8*(x^4 + x^3)^(1/4))/x) - 1/2*s 
qrt(1/17)*(1/2*sqrt(17) + 23/2)^(1/4)*log((sqrt(1/17)*(sqrt(17)*x + 17*x)* 
(1/2*sqrt(17) + 23/2)^(1/4) + 8*(x^4 + x^3)^(1/4))/x) + 1/2*sqrt(1/17)*(1/ 
2*sqrt(17) + 23/2)^(1/4)*log(-(sqrt(1/17)*(sqrt(17)*x + 17*x)*(1/2*sqrt(17 
) + 23/2)^(1/4) - 8*(x^4 + x^3)^(1/4))/x) + 1/2*sqrt(1/17)*(-1/2*sqrt(17) 
+ 23/2)^(1/4)*log((sqrt(1/17)*(sqrt(17)*x - 17*x)*(-1/2*sqrt(17) + 23/2)^( 
1/4) + 8*(x^4 + x^3)^(1/4))/x) - 1/2*sqrt(1/17)*(-1/2*sqrt(17) + 23/2)^(1/ 
4)*log(-(sqrt(1/17)*(sqrt(17)*x - 17*x)*(-1/2*sqrt(17) + 23/2)^(1/4) - 8*( 
x^4 + x^3)^(1/4))/x) - arctan((x^4 + x^3)^(3/4)/(x^3 + x^2)) + 1/2*log((x 
+ (x^4 + x^3)^(1/4))/x) - 1/2*log(-(x - (x^4 + x^3)^(1/4))/x)
 

Sympy [N/A]

Not integrable

Time = 0.46 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.15 \[ \int \frac {\sqrt [4]{x^3+x^4}}{-2+x+2 x^2} \, dx=\int \frac {\sqrt [4]{x^{3} \left (x + 1\right )}}{2 x^{2} + x - 2}\, dx \] Input:

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

Output:

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

Maxima [N/A]

Not integrable

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.18 \[ \int \frac {\sqrt [4]{x^3+x^4}}{-2+x+2 x^2} \, dx=\int { \frac {{\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{2 \, x^{2} + x - 2} \,d x } \] Input:

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

Output:

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

Giac [N/A]

Not integrable

Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.18 \[ \int \frac {\sqrt [4]{x^3+x^4}}{-2+x+2 x^2} \, dx=\int { \frac {{\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{2 \, x^{2} + x - 2} \,d x } \] Input:

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

Output:

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

Mupad [N/A]

Not integrable

Time = 8.56 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.18 \[ \int \frac {\sqrt [4]{x^3+x^4}}{-2+x+2 x^2} \, dx=\int \frac {{\left (x^4+x^3\right )}^{1/4}}{2\,x^2+x-2} \,d x \] Input:

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

Output:

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

Reduce [N/A]

Not integrable

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.17 \[ \int \frac {\sqrt [4]{x^3+x^4}}{-2+x+2 x^2} \, dx=\int \frac {x^{\frac {3}{4}} \left (x +1\right )^{\frac {1}{4}}}{2 x^{2}+x -2}d x \] Input:

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

Output:

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