\(\int \frac {(x+x^2) \sqrt [4]{x^3+x^4}}{-1+x+x^2} \, dx\) [2743]

Optimal result

Integrand size = 25, antiderivative size = 255 \[ \int \frac {\left (x+x^2\right ) \sqrt [4]{x^3+x^4}}{-1+x+x^2} \, dx=\frac {1}{8} (1+4 x) \sqrt [4]{x^3+x^4}-\frac {29}{16} \arctan \left (\frac {x}{\sqrt [4]{x^3+x^4}}\right )+\sqrt {\frac {1}{5} \left (2+2 \sqrt {5}\right )} \arctan \left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{x^3+x^4}}\right )+\sqrt {\frac {1}{5} \left (-2+2 \sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{x^3+x^4}}\right )+\frac {29}{16} \text {arctanh}\left (\frac {x}{\sqrt [4]{x^3+x^4}}\right )-\sqrt {\frac {1}{5} \left (2+2 \sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{x^3+x^4}}\right )-\sqrt {\frac {1}{5} \left (-2+2 \sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{x^3+x^4}}\right ) \]

[Out]

1/8*(1+4*x)*(x^4+x^3)^(1/4)-29/16*arctan(x/(x^4+x^3)^(1/4))+1/5*(10+10*5^(1/2))^(1/2)*arctan(1/2*(-2+2*5^(1/2)
)^(1/2)*x/(x^4+x^3)^(1/4))+1/5*(-10+10*5^(1/2))^(1/2)*arctan(1/2*(2+2*5^(1/2))^(1/2)*x/(x^4+x^3)^(1/4))+29/16*
arctanh(x/(x^4+x^3)^(1/4))-1/5*(10+10*5^(1/2))^(1/2)*arctanh(1/2*(-2+2*5^(1/2))^(1/2)*x/(x^4+x^3)^(1/4))-1/5*(
-10+10*5^(1/2))^(1/2)*arctanh(1/2*(2+2*5^(1/2))^(1/2)*x/(x^4+x^3)^(1/4))

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.63, number of steps used = 26, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {1607, 2081, 917, 52, 65, 338, 304, 209, 212, 919, 925, 95} \[ \int \frac {\left (x+x^2\right ) \sqrt [4]{x^3+x^4}}{-1+x+x^2} \, dx=-\frac {29 \sqrt [4]{x^4+x^3} \arctan \left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{16 \sqrt [4]{x+1} x^{3/4}}+\frac {2^{3/4} \sqrt [4]{3+\sqrt {5}} \sqrt [4]{x^4+x^3} \arctan \left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{\sqrt {5} \sqrt [4]{x+1} x^{3/4}}+\frac {2^{3/4} \sqrt [4]{3-\sqrt {5}} \sqrt [4]{x^4+x^3} \arctan \left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{\sqrt {5} \sqrt [4]{x+1} x^{3/4}}+\frac {29 \sqrt [4]{x^4+x^3} \text {arctanh}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{16 \sqrt [4]{x+1} x^{3/4}}-\frac {2^{3/4} \sqrt [4]{3+\sqrt {5}} \sqrt [4]{x^4+x^3} \text {arctanh}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{\sqrt {5} \sqrt [4]{x+1} x^{3/4}}-\frac {2^{3/4} \sqrt [4]{3-\sqrt {5}} \sqrt [4]{x^4+x^3} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{\sqrt {5} \sqrt [4]{x+1} x^{3/4}}+\frac {1}{2} \sqrt [4]{x^4+x^3} x+\frac {1}{8} \sqrt [4]{x^4+x^3} \]

[In]

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

[Out]

(x^3 + x^4)^(1/4)/8 + (x*(x^3 + x^4)^(1/4))/2 - (29*(x^3 + x^4)^(1/4)*ArcTan[x^(1/4)/(1 + x)^(1/4)])/(16*x^(3/
4)*(1 + x)^(1/4)) + (2^(3/4)*(3 + Sqrt[5])^(1/4)*(x^3 + x^4)^(1/4)*ArcTan[((2/(3 + Sqrt[5]))^(1/4)*x^(1/4))/(1
+ x)^(1/4)])/(Sqrt[5]*x^(3/4)*(1 + x)^(1/4)) + (2^(3/4)*(3 - Sqrt[5])^(1/4)*(x^3 + x^4)^(1/4)*ArcTan[(((3 + S
qrt[5])/2)^(1/4)*x^(1/4))/(1 + x)^(1/4)])/(Sqrt[5]*x^(3/4)*(1 + x)^(1/4)) + (29*(x^3 + x^4)^(1/4)*ArcTanh[x^(1
/4)/(1 + x)^(1/4)])/(16*x^(3/4)*(1 + x)^(1/4)) - (2^(3/4)*(3 + Sqrt[5])^(1/4)*(x^3 + x^4)^(1/4)*ArcTanh[((2/(3
+ Sqrt[5]))^(1/4)*x^(1/4))/(1 + x)^(1/4)])/(Sqrt[5]*x^(3/4)*(1 + x)^(1/4)) - (2^(3/4)*(3 - Sqrt[5])^(1/4)*(x^
3 + x^4)^(1/4)*ArcTanh[(((3 + Sqrt[5])/2)^(1/4)*x^(1/4))/(1 + x)^(1/4)])/(Sqrt[5]*x^(3/4)*(1 + x)^(1/4))

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 209

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 304

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

Rule 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[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 917

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Di
st[g/c^2, Int[Simp[2*c*e*f + c*d*g - b*e*g + c*e*g*x, x]*(d + e*x)^(m - 1)*(f + g*x)^(n - 2), x], x] + Dist[1/
c^2, Int[Simp[c^2*d*f^2 - 2*a*c*e*f*g - a*c*d*g^2 + a*b*e*g^2 + (c^2*e*f^2 + 2*c^2*d*f*g - 2*b*c*e*f*g - b*c*d
*g^2 + b^2*e*g^2 - a*c*e*g^2)*x, x]*(d + e*x)^(m - 1)*((f + g*x)^(n - 2)/(a + b*x + c*x^2)), x], x] /; FreeQ[{
a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[m] &&  !Integer
Q[n] && GtQ[m, 0] && GtQ[n, 1]

Rule 919

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Di
st[e*(g/c), Int[(d + e*x)^(m - 1)*(f + g*x)^(n - 1), x], x] + Dist[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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[m, 0] &
& GtQ[n, 0]

Rule 925

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n, 1/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x
] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[m] &&  !IntegerQ[n]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x (1+x) \sqrt [4]{x^3+x^4}}{-1+x+x^2} \, dx \\ & = \frac {\sqrt [4]{x^3+x^4} \int \frac {x^{7/4} (1+x)^{5/4}}{-1+x+x^2} \, dx}{x^{3/4} \sqrt [4]{1+x}} \\ & = \frac {\sqrt [4]{x^3+x^4} \int x^{3/4} \sqrt [4]{1+x} \, dx}{x^{3/4} \sqrt [4]{1+x}}+\frac {\sqrt [4]{x^3+x^4} \int \frac {x^{3/4} \sqrt [4]{1+x}}{-1+x+x^2} \, dx}{x^{3/4} \sqrt [4]{1+x}} \\ & = \frac {1}{2} x \sqrt [4]{x^3+x^4}+\frac {\sqrt [4]{x^3+x^4} \int \frac {x^{3/4}}{(1+x)^{3/4}} \, dx}{8 x^{3/4} \sqrt [4]{1+x}}+\frac {\sqrt [4]{x^3+x^4} \int \frac {1}{\sqrt [4]{x} (1+x)^{3/4}} \, dx}{x^{3/4} \sqrt [4]{1+x}}+\frac {\sqrt [4]{x^3+x^4} \int \frac {1}{\sqrt [4]{x} (1+x)^{3/4} \left (-1+x+x^2\right )} \, dx}{x^{3/4} \sqrt [4]{1+x}} \\ & = \frac {1}{8} \sqrt [4]{x^3+x^4}+\frac {1}{2} x \sqrt [4]{x^3+x^4}-\frac {\left (3 \sqrt [4]{x^3+x^4}\right ) \int \frac {1}{\sqrt [4]{x} (1+x)^{3/4}} \, dx}{32 x^{3/4} \sqrt [4]{1+x}}+\frac {\sqrt [4]{x^3+x^4} \int \left (-\frac {2}{\sqrt {5} \left (-1+\sqrt {5}-2 x\right ) \sqrt [4]{x} (1+x)^{3/4}}-\frac {2}{\sqrt {5} \sqrt [4]{x} (1+x)^{3/4} \left (1+\sqrt {5}+2 x\right )}\right ) \, dx}{x^{3/4} \sqrt [4]{1+x}}+\frac {\left (4 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {x^2}{\left (1+x^4\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{x^{3/4} \sqrt [4]{1+x}} \\ & = \frac {1}{8} \sqrt [4]{x^3+x^4}+\frac {1}{2} x \sqrt [4]{x^3+x^4}-\frac {\left (3 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {x^2}{\left (1+x^4\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{8 x^{3/4} \sqrt [4]{1+x}}+\frac {\left (4 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{x^{3/4} \sqrt [4]{1+x}}-\frac {\left (2 \sqrt [4]{x^3+x^4}\right ) \int \frac {1}{\left (-1+\sqrt {5}-2 x\right ) \sqrt [4]{x} (1+x)^{3/4}} \, dx}{\sqrt {5} x^{3/4} \sqrt [4]{1+x}}-\frac {\left (2 \sqrt [4]{x^3+x^4}\right ) \int \frac {1}{\sqrt [4]{x} (1+x)^{3/4} \left (1+\sqrt {5}+2 x\right )} \, dx}{\sqrt {5} x^{3/4} \sqrt [4]{1+x}} \\ & = \frac {1}{8} \sqrt [4]{x^3+x^4}+\frac {1}{2} x \sqrt [4]{x^3+x^4}-\frac {\left (3 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{8 x^{3/4} \sqrt [4]{1+x}}+\frac {\left (2 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{x^{3/4} \sqrt [4]{1+x}}-\frac {\left (2 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{x^{3/4} \sqrt [4]{1+x}}-\frac {\left (8 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {x^2}{1+\sqrt {5}-\left (-1+\sqrt {5}\right ) x^4} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{\sqrt {5} x^{3/4} \sqrt [4]{1+x}}-\frac {\left (8 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {x^2}{-1+\sqrt {5}-\left (1+\sqrt {5}\right ) x^4} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{\sqrt {5} x^{3/4} \sqrt [4]{1+x}} \\ & = \frac {1}{8} \sqrt [4]{x^3+x^4}+\frac {1}{2} x \sqrt [4]{x^3+x^4}-\frac {2 \sqrt [4]{x^3+x^4} \arctan \left (\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{x^{3/4} \sqrt [4]{1+x}}+\frac {2 \sqrt [4]{x^3+x^4} \text {arctanh}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{x^{3/4} \sqrt [4]{1+x}}-\frac {\left (3 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{16 x^{3/4} \sqrt [4]{1+x}}+\frac {\left (3 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{16 x^{3/4} \sqrt [4]{1+x}}+\frac {\left (4 \sqrt {\frac {2}{5}} \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {3-\sqrt {5}}-\sqrt {2} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{\left (-1-\sqrt {5}\right ) x^{3/4} \sqrt [4]{1+x}}-\frac {\left (4 \sqrt {\frac {2}{5}} \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {3-\sqrt {5}}+\sqrt {2} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{\left (-1-\sqrt {5}\right ) x^{3/4} \sqrt [4]{1+x}}+\frac {\left (4 \sqrt {\frac {2}{5}} \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {3+\sqrt {5}}-\sqrt {2} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{\left (1-\sqrt {5}\right ) x^{3/4} \sqrt [4]{1+x}}-\frac {\left (4 \sqrt {\frac {2}{5}} \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {3+\sqrt {5}}+\sqrt {2} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{\left (1-\sqrt {5}\right ) x^{3/4} \sqrt [4]{1+x}} \\ & = \frac {1}{8} \sqrt [4]{x^3+x^4}+\frac {1}{2} x \sqrt [4]{x^3+x^4}-\frac {29 \sqrt [4]{x^3+x^4} \arctan \left (\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{16 x^{3/4} \sqrt [4]{1+x}}+\frac {2^{3/4} \sqrt [4]{3+\sqrt {5}} \sqrt [4]{x^3+x^4} \arctan \left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} \sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{\sqrt {5} x^{3/4} \sqrt [4]{1+x}}+\frac {2^{3/4} \sqrt [4]{3-\sqrt {5}} \sqrt [4]{x^3+x^4} \arctan \left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{\sqrt {5} x^{3/4} \sqrt [4]{1+x}}+\frac {29 \sqrt [4]{x^3+x^4} \text {arctanh}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{16 x^{3/4} \sqrt [4]{1+x}}-\frac {2^{3/4} \sqrt [4]{3+\sqrt {5}} \sqrt [4]{x^3+x^4} \text {arctanh}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} \sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{\sqrt {5} x^{3/4} \sqrt [4]{1+x}}-\frac {2^{3/4} \sqrt [4]{3-\sqrt {5}} \sqrt [4]{x^3+x^4} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{\sqrt {5} x^{3/4} \sqrt [4]{1+x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.00 \[ \int \frac {\left (x+x^2\right ) \sqrt [4]{x^3+x^4}}{-1+x+x^2} \, dx=\frac {x^{9/4} (1+x)^{3/4} \left (10 x^{3/4} \sqrt [4]{1+x}+40 x^{7/4} \sqrt [4]{1+x}-145 \arctan \left (\sqrt [4]{\frac {x}{1+x}}\right )+16 \sqrt {10 \left (1+\sqrt {5}\right )} \arctan \left (\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} \sqrt [4]{\frac {x}{1+x}}\right )+16 \sqrt {10 \left (-1+\sqrt {5}\right )} \arctan \left (\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \sqrt [4]{\frac {x}{1+x}}\right )+145 \text {arctanh}\left (\sqrt [4]{\frac {x}{1+x}}\right )-16 \sqrt {10 \left (1+\sqrt {5}\right )} \text {arctanh}\left (\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} \sqrt [4]{\frac {x}{1+x}}\right )-16 \sqrt {10 \left (-1+\sqrt {5}\right )} \text {arctanh}\left (\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \sqrt [4]{\frac {x}{1+x}}\right )\right )}{80 \left (x^3 (1+x)\right )^{3/4}} \]

[In]

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

[Out]

(x^(9/4)*(1 + x)^(3/4)*(10*x^(3/4)*(1 + x)^(1/4) + 40*x^(7/4)*(1 + x)^(1/4) - 145*ArcTan[(x/(1 + x))^(1/4)] +
16*Sqrt[10*(1 + Sqrt[5])]*ArcTan[Sqrt[(-1 + Sqrt[5])/2]*(x/(1 + x))^(1/4)] + 16*Sqrt[10*(-1 + Sqrt[5])]*ArcTan
[Sqrt[(1 + Sqrt[5])/2]*(x/(1 + x))^(1/4)] + 145*ArcTanh[(x/(1 + x))^(1/4)] - 16*Sqrt[10*(1 + Sqrt[5])]*ArcTanh
[Sqrt[(-1 + Sqrt[5])/2]*(x/(1 + x))^(1/4)] - 16*Sqrt[10*(-1 + Sqrt[5])]*ArcTanh[Sqrt[(1 + Sqrt[5])/2]*(x/(1 +
x))^(1/4)]))/(80*(x^3*(1 + x))^(3/4))

Maple [A] (verified)

Time = 15.51 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.96

method	result	size
pseudoelliptic	\(-\frac {x^{6} \left (\frac {5 \left (-\frac {1}{4}-x \right ) \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{2}+\left (\operatorname {arctanh}\left (\frac {2 \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{x \sqrt {2+2 \sqrt {5}}}\right )+\arctan \left (\frac {2 \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{x \sqrt {2+2 \sqrt {5}}}\right )\right ) \sqrt {5}\, \sqrt {-2+2 \sqrt {5}}+\sqrt {2+2 \sqrt {5}}\, \left (\operatorname {arctanh}\left (\frac {2 \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{x \sqrt {-2+2 \sqrt {5}}}\right )+\arctan \left (\frac {2 \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{x \sqrt {-2+2 \sqrt {5}}}\right )\right ) \sqrt {5}+\frac {145 \ln \left (\frac {\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}-x}{x}\right )}{32}-\frac {145 \ln \left (\frac {x +\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{x}\right )}{32}-\frac {145 \arctan \left (\frac {\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{x}\right )}{16}\right )}{5 {\left (x +\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}\right )}^{2} {\left (-\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}+x \right )}^{2} \left (x^{2}+\sqrt {x^{3} \left (1+x \right )}\right )^{2}}\)	\(244\)
trager	\(\text {Expression too large to display}\)	\(2153\)
risch	\(\text {Expression too large to display}\)	\(4381\)

[In]

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

[Out]

-1/5*x^6*(5/2*(-1/4-x)*(x^3*(1+x))^(1/4)+(arctanh(2*(x^3*(1+x))^(1/4)/x/(2+2*5^(1/2))^(1/2))+arctan(2*(x^3*(1+
x))^(1/4)/x/(2+2*5^(1/2))^(1/2)))*5^(1/2)*(-2+2*5^(1/2))^(1/2)+(2+2*5^(1/2))^(1/2)*(arctanh(2*(x^3*(1+x))^(1/4
)/x/(-2+2*5^(1/2))^(1/2))+arctan(2*(x^3*(1+x))^(1/4)/x/(-2+2*5^(1/2))^(1/2)))*5^(1/2)+145/32*ln(((x^3*(1+x))^(
1/4)-x)/x)-145/32*ln((x+(x^3*(1+x))^(1/4))/x)-145/16*arctan((x^3*(1+x))^(1/4)/x))/(x+(x^3*(1+x))^(1/4))^2/(-(x
^3*(1+x))^(1/4)+x)^2/(x^2+(x^3*(1+x))^(1/2))^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 468 vs. \(2 (177) = 354\).

Time = 0.28 (sec) , antiderivative size = 468, normalized size of antiderivative = 1.84 \[ \int \frac {\left (x+x^2\right ) \sqrt [4]{x^3+x^4}}{-1+x+x^2} \, dx=-\frac {1}{10} \, \sqrt {5} \sqrt {2 \, \sqrt {5} + 2} \log \left (\frac {{\left (\sqrt {5} x - x\right )} \sqrt {2 \, \sqrt {5} + 2} + 4 \, {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{10} \, \sqrt {5} \sqrt {2 \, \sqrt {5} + 2} \log \left (-\frac {{\left (\sqrt {5} x - x\right )} \sqrt {2 \, \sqrt {5} + 2} - 4 \, {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{10} \, \sqrt {5} \sqrt {2 \, \sqrt {5} - 2} \log \left (\frac {{\left (\sqrt {5} x + x\right )} \sqrt {2 \, \sqrt {5} - 2} + 4 \, {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{10} \, \sqrt {5} \sqrt {2 \, \sqrt {5} - 2} \log \left (-\frac {{\left (\sqrt {5} x + x\right )} \sqrt {2 \, \sqrt {5} - 2} - 4 \, {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{10} \, \sqrt {5} \sqrt {-2 \, \sqrt {5} + 2} \log \left (\frac {{\left (\sqrt {5} x + x\right )} \sqrt {-2 \, \sqrt {5} + 2} + 4 \, {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{10} \, \sqrt {5} \sqrt {-2 \, \sqrt {5} + 2} \log \left (-\frac {{\left (\sqrt {5} x + x\right )} \sqrt {-2 \, \sqrt {5} + 2} - 4 \, {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{10} \, \sqrt {5} \sqrt {-2 \, \sqrt {5} - 2} \log \left (\frac {{\left (\sqrt {5} x - x\right )} \sqrt {-2 \, \sqrt {5} - 2} + 4 \, {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{10} \, \sqrt {5} \sqrt {-2 \, \sqrt {5} - 2} \log \left (-\frac {{\left (\sqrt {5} x - x\right )} \sqrt {-2 \, \sqrt {5} - 2} - 4 \, {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{8} \, {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}} {\left (4 \, x + 1\right )} + \frac {29}{16} \, \arctan \left (\frac {{\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {29}{32} \, \log \left (\frac {x + {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {29}{32} \, \log \left (-\frac {x - {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) \]

[In]

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

[Out]

-1/10*sqrt(5)*sqrt(2*sqrt(5) + 2)*log(((sqrt(5)*x - x)*sqrt(2*sqrt(5) + 2) + 4*(x^4 + x^3)^(1/4))/x) + 1/10*sq
rt(5)*sqrt(2*sqrt(5) + 2)*log(-((sqrt(5)*x - x)*sqrt(2*sqrt(5) + 2) - 4*(x^4 + x^3)^(1/4))/x) - 1/10*sqrt(5)*s
qrt(2*sqrt(5) - 2)*log(((sqrt(5)*x + x)*sqrt(2*sqrt(5) - 2) + 4*(x^4 + x^3)^(1/4))/x) + 1/10*sqrt(5)*sqrt(2*sq
rt(5) - 2)*log(-((sqrt(5)*x + x)*sqrt(2*sqrt(5) - 2) - 4*(x^4 + x^3)^(1/4))/x) - 1/10*sqrt(5)*sqrt(-2*sqrt(5)
+ 2)*log(((sqrt(5)*x + x)*sqrt(-2*sqrt(5) + 2) + 4*(x^4 + x^3)^(1/4))/x) + 1/10*sqrt(5)*sqrt(-2*sqrt(5) + 2)*l
og(-((sqrt(5)*x + x)*sqrt(-2*sqrt(5) + 2) - 4*(x^4 + x^3)^(1/4))/x) - 1/10*sqrt(5)*sqrt(-2*sqrt(5) - 2)*log(((
sqrt(5)*x - x)*sqrt(-2*sqrt(5) - 2) + 4*(x^4 + x^3)^(1/4))/x) + 1/10*sqrt(5)*sqrt(-2*sqrt(5) - 2)*log(-((sqrt(
5)*x - x)*sqrt(-2*sqrt(5) - 2) - 4*(x^4 + x^3)^(1/4))/x) + 1/8*(x^4 + x^3)^(1/4)*(4*x + 1) + 29/16*arctan((x^4
+ x^3)^(1/4)/x) + 29/32*log((x + (x^4 + x^3)^(1/4))/x) - 29/32*log(-(x - (x^4 + x^3)^(1/4))/x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.93 \[ \int \frac {\left (x+x^2\right ) \sqrt [4]{x^3+x^4}}{-1+x+x^2} \, dx=\frac {1}{8} \, {\left ({\left (\frac {1}{x} + 1\right )}^{\frac {5}{4}} + 3 \, {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right )} x^{2} - \frac {1}{5} \, \sqrt {10 \, \sqrt {5} - 10} \arctan \left (\frac {{\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}}{\sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}}}\right ) - \frac {1}{5} \, \sqrt {10 \, \sqrt {5} + 10} \arctan \left (\frac {{\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}}{\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}}}\right ) - \frac {1}{10} \, \sqrt {10 \, \sqrt {5} - 10} \log \left (\sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} + {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{10} \, \sqrt {10 \, \sqrt {5} + 10} \log \left (\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} + {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{10} \, \sqrt {10 \, \sqrt {5} - 10} \log \left ({\left | -\sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} + {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} \right |}\right ) + \frac {1}{10} \, \sqrt {10 \, \sqrt {5} + 10} \log \left ({\left | -\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} + {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} \right |}\right ) + \frac {29}{16} \, \arctan \left ({\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) + \frac {29}{32} \, \log \left ({\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {29}{32} \, \log \left ({\left | {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \]

[In]

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

[Out]

1/8*((1/x + 1)^(5/4) + 3*(1/x + 1)^(1/4))*x^2 - 1/5*sqrt(10*sqrt(5) - 10)*arctan((1/x + 1)^(1/4)/sqrt(1/2*sqrt
(5) + 1/2)) - 1/5*sqrt(10*sqrt(5) + 10)*arctan((1/x + 1)^(1/4)/sqrt(1/2*sqrt(5) - 1/2)) - 1/10*sqrt(10*sqrt(5)
- 10)*log(sqrt(1/2*sqrt(5) + 1/2) + (1/x + 1)^(1/4)) - 1/10*sqrt(10*sqrt(5) + 10)*log(sqrt(1/2*sqrt(5) - 1/2)
+ (1/x + 1)^(1/4)) + 1/10*sqrt(10*sqrt(5) - 10)*log(abs(-sqrt(1/2*sqrt(5) + 1/2) + (1/x + 1)^(1/4))) + 1/10*s
qrt(10*sqrt(5) + 10)*log(abs(-sqrt(1/2*sqrt(5) - 1/2) + (1/x + 1)^(1/4))) + 29/16*arctan((1/x + 1)^(1/4)) + 29
/32*log((1/x + 1)^(1/4) + 1) - 29/32*log(abs((1/x + 1)^(1/4) - 1))

Mupad [F(-1)]

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

[In]

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

[Out]

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