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

Optimal. Leaf size=255 \[ \frac {1}{8} \sqrt [4]{x^4+x^3} (4 x+1)-\frac {29}{16} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+x^3}}\right )+\sqrt {\frac {1}{5} \left (2+2 \sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt {\frac {\sqrt {5}}{2}-\frac {1}{2}} x}{\sqrt [4]{x^4+x^3}}\right )+\sqrt {\frac {1}{5} \left (2 \sqrt {5}-2\right )} \tan ^{-1}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{x^4+x^3}}\right )+\frac {29}{16} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+x^3}}\right )-\sqrt {\frac {1}{5} \left (2+2 \sqrt {5}\right )} \tanh ^{-1}\left (\frac {\sqrt {\frac {\sqrt {5}}{2}-\frac {1}{2}} x}{\sqrt [4]{x^4+x^3}}\right )-\sqrt {\frac {1}{5} \left (2 \sqrt {5}-2\right )} \tanh ^{-1}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{x^4+x^3}}\right ) \]

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Rubi [A]  time = 0.45, antiderivative size = 416, normalized size of antiderivative = 1.63, number of steps used = 26, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {1593, 2056, 903, 50, 63, 331, 298, 203, 206, 905, 911, 93} \begin {gather*} \frac {1}{2} \sqrt [4]{x^4+x^3} x+\frac {1}{8} \sqrt [4]{x^4+x^3}-\frac {29 \sqrt [4]{x^4+x^3} \tan ^{-1}\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} \tan ^{-1}\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} \tan ^{-1}\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} \tanh ^{-1}\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} \tanh ^{-1}\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} \tanh ^{-1}\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}} \end {gather*}

Warning: Unable to verify antiderivative.

[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 50

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 63

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 93

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 203

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

Rule 206

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

Rule 298

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 331

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 903

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 905

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 911

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 1593

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 2056

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

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Mathematica [C]  time = 0.43, size = 186, normalized size = 0.73 \begin {gather*} \frac {4}{15} \sqrt [4]{x^3 (x+1)} \left (\frac {5 \, _2F_1\left (-\frac {5}{4},\frac {3}{4};\frac {7}{4};-x\right )}{\sqrt [4]{x+1}}-\frac {5 \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};-x\right )}{\sqrt [4]{x+1}}+\frac {5 \, _2F_1\left (\frac {3}{4},\frac {3}{4};\frac {7}{4};-x\right )}{\sqrt [4]{x+1}}-\frac {2 \sqrt {5} \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};\frac {\left (-1+\sqrt {5}\right ) x}{\left (1+\sqrt {5}\right ) (x+1)}\right )}{\left (1+\sqrt {5}\right ) (x+1)}-\frac {2 \sqrt {5} \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};\frac {\left (1+\sqrt {5}\right ) x}{\left (-1+\sqrt {5}\right ) (x+1)}\right )}{\left (\sqrt {5}-1\right ) (x+1)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(x^3*(1 + x))^(1/4)*((5*Hypergeometric2F1[-5/4, 3/4, 7/4, -x])/(1 + x)^(1/4) - (5*Hypergeometric2F1[-1/4, 3
/4, 7/4, -x])/(1 + x)^(1/4) + (5*Hypergeometric2F1[3/4, 3/4, 7/4, -x])/(1 + x)^(1/4) - (2*Sqrt[5]*Hypergeometr
ic2F1[3/4, 1, 7/4, ((-1 + Sqrt[5])*x)/((1 + Sqrt[5])*(1 + x))])/((1 + Sqrt[5])*(1 + x)) - (2*Sqrt[5]*Hypergeom
etric2F1[3/4, 1, 7/4, ((1 + Sqrt[5])*x)/((-1 + Sqrt[5])*(1 + x))])/((-1 + Sqrt[5])*(1 + x))))/15

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IntegrateAlgebraic [A]  time = 0.95, size = 255, normalized size = 1.00 \begin {gather*} \frac {1}{8} (1+4 x) \sqrt [4]{x^3+x^4}-\frac {29}{16} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^3+x^4}}\right )+\sqrt {\frac {1}{5} \left (2+2 \sqrt {5}\right )} \tan ^{-1}\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 )} \tan ^{-1}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{x^3+x^4}}\right )+\frac {29}{16} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^3+x^4}}\right )-\sqrt {\frac {1}{5} \left (2+2 \sqrt {5}\right )} \tanh ^{-1}\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 )} \tanh ^{-1}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{x^3+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.52, size = 438, normalized size = 1.72 \begin {gather*} \frac {2}{5} \, \sqrt {5} \sqrt {2 \, \sqrt {5} - 2} \arctan \left (\frac {\sqrt {2} x \sqrt {2 \, \sqrt {5} - 2} \sqrt {\frac {\sqrt {5} x^{2} + x^{2} + 2 \, \sqrt {x^{4} + x^{3}}}{x^{2}}} - 2 \, {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}} \sqrt {2 \, \sqrt {5} - 2}}{4 \, x}\right ) + \frac {2}{5} \, \sqrt {5} \sqrt {2 \, \sqrt {5} + 2} \arctan \left (\frac {\sqrt {2} x \sqrt {2 \, \sqrt {5} + 2} \sqrt {\frac {\sqrt {5} x^{2} - x^{2} + 2 \, \sqrt {x^{4} + x^{3}}}{x^{2}}} - 2 \, {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}} \sqrt {2 \, \sqrt {5} + 2}}{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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*sqrt(5)*sqrt(2*sqrt(5) - 2)*arctan(1/4*(sqrt(2)*x*sqrt(2*sqrt(5) - 2)*sqrt((sqrt(5)*x^2 + x^2 + 2*sqrt(x^4
+ x^3))/x^2) - 2*(x^4 + x^3)^(1/4)*sqrt(2*sqrt(5) - 2))/x) + 2/5*sqrt(5)*sqrt(2*sqrt(5) + 2)*arctan(1/4*(sqrt
(2)*x*sqrt(2*sqrt(5) + 2)*sqrt((sqrt(5)*x^2 - x^2 + 2*sqrt(x^4 + x^3))/x^2) - 2*(x^4 + x^3)^(1/4)*sqrt(2*sqrt(
5) + 2))/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
/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)

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giac [A]  time = 1.12, size = 238, normalized size = 0.93 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [C]  time = 26.69, size = 2150, normalized size = 8.43

method	result	size
trager	\(\text {Expression too large to display}\)	\(2150\)
risch	\(\text {Expression too large to display}\)	\(4380\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/8+1/2*x)*(x^4+x^3)^(1/4)-1/40*RootOf(_Z^2+25*RootOf(25*_Z^4+320*_Z^2-4096)^2+320)*ln((1075*RootOf(25*_Z^4+3
20*_Z^2-4096)^4*RootOf(_Z^2+25*RootOf(25*_Z^4+320*_Z^2-4096)^2+320)*x^3-1075*RootOf(_Z^2+25*RootOf(25*_Z^4+320
*_Z^2-4096)^2+320)*RootOf(25*_Z^4+320*_Z^2-4096)^4*x^2-104960*(x^4+x^3)^(1/2)*RootOf(25*_Z^4+320*_Z^2-4096)^2*
RootOf(_Z^2+25*RootOf(25*_Z^4+320*_Z^2-4096)^2+320)*x-104640*RootOf(25*_Z^4+320*_Z^2-4096)^2*RootOf(_Z^2+25*Ro
otOf(25*_Z^4+320*_Z^2-4096)^2+320)*x^3-1152000*(x^4+x^3)^(3/4)*RootOf(25*_Z^4+320*_Z^2-4096)^2-1894400*RootOf(
25*_Z^4+320*_Z^2-4096)^2*(x^4+x^3)^(1/4)*x^2-46720*RootOf(25*_Z^4+320*_Z^2-4096)^2*RootOf(_Z^2+25*RootOf(25*_Z
^4+320*_Z^2-4096)^2+320)*x^2+843776*(x^4+x^3)^(1/2)*RootOf(_Z^2+25*RootOf(25*_Z^4+320*_Z^2-4096)^2+320)*x+7454
72*RootOf(_Z^2+25*RootOf(25*_Z^4+320*_Z^2-4096)^2+320)*x^3+9502720*(x^4+x^3)^(3/4)[phone]*x^2*(x^4+x^3)^(1/4
)+425984*RootOf(_Z^2+25*RootOf(25*_Z^4+320*_Z^2-4096)^2+320)*x^2)/(5*x*RootOf(25*_Z^4+320*_Z^2-4096)^2-5*RootO
f(25*_Z^4+320*_Z^2-4096)^2-64*x)/x^2)-1/8*RootOf(25*_Z^4+320*_Z^2-4096)*ln((1075*x^3*RootOf(25*_Z^4+320*_Z^2-4
096)^5-1075*x^2*RootOf(25*_Z^4+320*_Z^2-4096)^5+104960*RootOf(25*_Z^4+320*_Z^2-4096)^3*(x^4+x^3)^(1/2)*x+13216
0*x^3*RootOf(25*_Z^4+320*_Z^2-4096)^3+19200*x^2*RootOf(25*_Z^4+320*_Z^2-4096)^3+230400*(x^4+x^3)^(3/4)*RootOf(
25*_Z^4+320*_Z^2-4096)^2+378880*RootOf(25*_Z^4+320*_Z^2-4096)^2*(x^4+x^3)^(1/4)*x^2+2187264*RootOf(25*_Z^4+320
*_Z^2-4096)*(x^4+x^3)^(1/2)*x+2260992*RootOf(25*_Z^4+320*_Z^2-4096)*x^3+847872*x^2*RootOf(25*_Z^4+320*_Z^2-409
6)+4849664*(x^4+x^3)^(3/4)+7798784*x^2*(x^4+x^3)^(1/4))/(5*x*RootOf(25*_Z^4+320*_Z^2-4096)^2-5*RootOf(25*_Z^4+
320*_Z^2-4096)^2+128*x-64)/x^2)+29/2048*RootOf(25*_Z^4+320*_Z^2-4096)*RootOf(_Z^2+25*RootOf(25*_Z^4+320*_Z^2-4
096)^2+320)*ln((2*(x^4+x^3)^(1/2)*RootOf(25*_Z^4+320*_Z^2-4096)*RootOf(_Z^2+25*RootOf(25*_Z^4+320*_Z^2-4096)^2
+320)*x-2*RootOf(_Z^2+25*RootOf(25*_Z^4+320*_Z^2-4096)^2+320)*RootOf(25*_Z^4+320*_Z^2-4096)*x^3-RootOf(_Z^2+25
*RootOf(25*_Z^4+320*_Z^2-4096)^2+320)*RootOf(25*_Z^4+320*_Z^2-4096)*x^2+128*(x^4+x^3)^(3/4)-128*x^2*(x^4+x^3)^
(1/4))/x^2)+29/32*ln((2*(x^4+x^3)^(3/4)+2*(x^4+x^3)^(1/2)*x+2*x^2*(x^4+x^3)^(1/4)+2*x^3+x^2)/x^2)+1/512*RootOf
(25*_Z^4+320*_Z^2-4096)^2*RootOf(_Z^2+25*RootOf(25*_Z^4+320*_Z^2-4096)^2+320)*ln((325*RootOf(25*_Z^4+320*_Z^2-
4096)^4*RootOf(_Z^2+25*RootOf(25*_Z^4+320*_Z^2-4096)^2+320)*x^3-325*RootOf(_Z^2+25*RootOf(25*_Z^4+320*_Z^2-409
6)^2+320)*RootOf(25*_Z^4+320*_Z^2-4096)^4*x^2-32960*(x^4+x^3)^(1/2)*RootOf(25*_Z^4+320*_Z^2-4096)^2*RootOf(_Z^
2+25*RootOf(25*_Z^4+320*_Z^2-4096)^2+320)*x+40160*RootOf(25*_Z^4+320*_Z^2-4096)^2*RootOf(_Z^2+25*RootOf(25*_Z^
4+320*_Z^2-4096)^2+320)*x^3+576000*(x^4+x^3)^(3/4)*RootOf(25*_Z^4+320*_Z^2-4096)^2-947200*RootOf(25*_Z^4+320*_
Z^2-4096)^2*(x^4+x^3)^(1/4)*x^2+5600*RootOf(25*_Z^4+320*_Z^2-4096)^2*RootOf(_Z^2+25*RootOf(25*_Z^4+320*_Z^2-40
96)^2+320)*x^2-671744*(x^4+x^3)^(1/2)*RootOf(_Z^2+25*RootOf(25*_Z^4+320*_Z^2-4096)^2+320)*x+704512*RootOf(_Z^2
+25*RootOf(25*_Z^4+320*_Z^2-4096)^2+320)*x^3+12124160*(x^4+x^3)^(3/4)-19496960*x^2*(x^4+x^3)^(1/4)+264192*Root
Of(_Z^2+25*RootOf(25*_Z^4+320*_Z^2-4096)^2+320)*x^2)/(5*x*RootOf(25*_Z^4+320*_Z^2-4096)^2-5*RootOf(25*_Z^4+320
*_Z^2-4096)^2+128*x-64)/x^2)+5/512*ln(-(325*x^3*RootOf(25*_Z^4+320*_Z^2-4096)^5-325*x^2*RootOf(25*_Z^4+320*_Z^
2-4096)^5+32960*RootOf(25*_Z^4+320*_Z^2-4096)^3*(x^4+x^3)^(1/2)*x-31840*x^3*RootOf(25*_Z^4+320*_Z^2-4096)^3-13
920*x^2*RootOf(25*_Z^4+320*_Z^2-4096)^3+115200*(x^4+x^3)^(3/4)*RootOf(25*_Z^4+320*_Z^2-4096)^2-189440*RootOf(2
5*_Z^4+320*_Z^2-4096)^2*(x^4+x^3)^(1/4)*x^2-249856*RootOf(25*_Z^4+320*_Z^2-4096)*(x^4+x^3)^(1/2)*x+243712*Root
Of(25*_Z^4+320*_Z^2-4096)*x^3+139264*x^2*RootOf(25*_Z^4+320*_Z^2-4096)-950272*(x^4+x^3)^(3/4)+1474560*x^2*(x^4
+x^3)^(1/4))/(5*x*RootOf(25*_Z^4+320*_Z^2-4096)^2-5*RootOf(25*_Z^4+320*_Z^2-4096)^2-64*x)/x^2)*RootOf(25*_Z^4+
320*_Z^2-4096)^3+1/8*ln(-(325*x^3*RootOf(25*_Z^4+320*_Z^2-4096)^5-325*x^2*RootOf(25*_Z^4+320*_Z^2-4096)^5+3296
0*RootOf(25*_Z^4+320*_Z^2-4096)^3*(x^4+x^3)^(1/2)*x-31840*x^3*RootOf(25*_Z^4+320*_Z^2-4096)^3-13920*x^2*RootOf
(25*_Z^4+320*_Z^2-4096)^3+115200*(x^4+x^3)^(3/4)*RootOf(25*_Z^4+320*_Z^2-4096)^2-189440*RootOf(25*_Z^4+320*_Z^
2-4096)^2*(x^4+x^3)^(1/4)*x^2-249856*RootOf(25*_Z^4+320*_Z^2-4096)*(x^4+x^3)^(1/2)*x+243712*RootOf(25*_Z^4+320
*_Z^2-4096)*x^3+139264*x^2*RootOf(25*_Z^4+320*_Z^2-4096)-950272*(x^4+x^3)^(3/4)+1474560*x^2*(x^4+x^3)^(1/4))/(
5*x*RootOf(25*_Z^4+320*_Z^2-4096)^2-5*RootOf(25*_Z^4+320*_Z^2-4096)^2-64*x)/x^2)*RootOf(25*_Z^4+320*_Z^2-4096)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} + x^{3}\right )}^{\frac {1}{4}} {\left (x^{2} + x\right )}}{x^{2} + x - 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (x^4+x^3\right )}^{1/4}\,\left (x^2+x\right )}{x^2+x-1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \sqrt [4]{x^{3} \left (x + 1\right )} \left (x + 1\right )}{x^{2} + x - 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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