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

Optimal. Leaf size=226 \[ 2 \sqrt [4]{x^4+x^3}+\tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+x^3}}\right )-\sqrt {2 \left (1+\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt {\frac {\sqrt {5}}{2}-\frac {1}{2}} x}{\sqrt [4]{x^4+x^3}}\right )+\sqrt {2 \left (\sqrt {5}-1\right )} \tan ^{-1}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{x^4+x^3}}\right )-\tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+x^3}}\right )+\sqrt {2 \left (1+\sqrt {5}\right )} \tanh ^{-1}\left (\frac {\sqrt {\frac {\sqrt {5}}{2}-\frac {1}{2}} x}{\sqrt [4]{x^4+x^3}}\right )-\sqrt {2 \left (\sqrt {5}-1\right )} \tanh ^{-1}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{x^4+x^3}}\right ) \]

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Rubi [B]  time = 0.74, antiderivative size = 496, normalized size of antiderivative = 2.19, number of steps used = 25, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2056, 6728, 101, 157, 63, 331, 298, 203, 206, 93} \begin {gather*} 2 \sqrt [4]{x^4+x^3}+\frac {\left (1+2 \sqrt {5}\right ) \sqrt [4]{x^4+x^3} \tan ^{-1}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{2 x^{3/4} \sqrt [4]{x+1}}+\frac {\left (1-2 \sqrt {5}\right ) \sqrt [4]{x^4+x^3} \tan ^{-1}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{2 x^{3/4} \sqrt [4]{x+1}}-\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 )}{x^{3/4} \sqrt [4]{x+1}}+\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 )}{x^{3/4} \sqrt [4]{x+1}}-\frac {\left (1+2 \sqrt {5}\right ) \sqrt [4]{x^4+x^3} \tanh ^{-1}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{2 x^{3/4} \sqrt [4]{x+1}}-\frac {\left (1-2 \sqrt {5}\right ) \sqrt [4]{x^4+x^3} \tanh ^{-1}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{2 x^{3/4} \sqrt [4]{x+1}}+\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 )}{x^{3/4} \sqrt [4]{x+1}}-\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 )}{x^{3/4} \sqrt [4]{x+1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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 101

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a +
b*x)^m*(c + d*x)^n*(e + f*x)^(p + 1))/(f*(m + n + p + 1)), x] - Dist[1/(f*(m + n + p + 1)), Int[(a + b*x)^(m -
1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a*f) + b*n*(d*e - c*f))
*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (Integ
ersQ[2*m, 2*n, 2*p] || (IntegersQ[m, n + p] || IntegersQ[p, m + n]))

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, 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 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]

Rule 6728

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] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(1+2 x) \sqrt [4]{x^3+x^4}}{-1+x+x^2} \, dx &=\frac {\sqrt [4]{x^3+x^4} \int \frac {x^{3/4} \sqrt [4]{1+x} (1+2 x)}{-1+x+x^2} \, dx}{x^{3/4} \sqrt [4]{1+x}}\\ &=\frac {\sqrt [4]{x^3+x^4} \int \left (\frac {2 x^{3/4} \sqrt [4]{1+x}}{1-\sqrt {5}+2 x}+\frac {2 x^{3/4} \sqrt [4]{1+x}}{1+\sqrt {5}+2 x}\right ) \, dx}{x^{3/4} \sqrt [4]{1+x}}\\ &=\frac {\left (2 \sqrt [4]{x^3+x^4}\right ) \int \frac {x^{3/4} \sqrt [4]{1+x}}{1-\sqrt {5}+2 x} \, dx}{x^{3/4} \sqrt [4]{1+x}}+\frac {\left (2 \sqrt [4]{x^3+x^4}\right ) \int \frac {x^{3/4} \sqrt [4]{1+x}}{1+\sqrt {5}+2 x} \, dx}{x^{3/4} \sqrt [4]{1+x}}\\ &=2 \sqrt [4]{x^3+x^4}-\frac {\sqrt [4]{x^3+x^4} \int \frac {\frac {3}{4} \left (1-\sqrt {5}\right )+\frac {1}{2} \left (1-2 \sqrt {5}\right ) x}{\sqrt [4]{x} (1+x)^{3/4} \left (1-\sqrt {5}+2 x\right )} \, dx}{x^{3/4} \sqrt [4]{1+x}}-\frac {\sqrt [4]{x^3+x^4} \int \frac {\frac {3}{4} \left (1+\sqrt {5}\right )+\frac {1}{2} \left (1+2 \sqrt {5}\right ) x}{\sqrt [4]{x} (1+x)^{3/4} \left (1+\sqrt {5}+2 x\right )} \, dx}{x^{3/4} \sqrt [4]{1+x}}\\ &=2 \sqrt [4]{x^3+x^4}+\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}{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}{x^{3/4} \sqrt [4]{1+x}}-\frac {\left (\left (1-2 \sqrt {5}\right ) \sqrt [4]{x^3+x^4}\right ) \int \frac {1}{\sqrt [4]{x} (1+x)^{3/4}} \, dx}{4 x^{3/4} \sqrt [4]{1+x}}-\frac {\left (\left (1+2 \sqrt {5}\right ) \sqrt [4]{x^3+x^4}\right ) \int \frac {1}{\sqrt [4]{x} (1+x)^{3/4}} \, dx}{4 x^{3/4} \sqrt [4]{1+x}}\\ &=2 \sqrt [4]{x^3+x^4}+\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 )}{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 )}{x^{3/4} \sqrt [4]{1+x}}-\frac {\left (\left (1-2 \sqrt {5}\right ) \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 {\left (\left (1+2 \sqrt {5}\right ) \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}}\\ &=2 \sqrt [4]{x^3+x^4}-\frac {\left (\left (1-2 \sqrt {5}\right ) \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 (4 \sqrt {2} \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 {2} \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 {2} \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 {2} \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 (\left (1+2 \sqrt {5}\right ) \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}}\\ &=2 \sqrt [4]{x^3+x^4}-\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 )}{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 )}{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 )}{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 )}{x^{3/4} \sqrt [4]{1+x}}-\frac {\left (\left (1-2 \sqrt {5}\right ) \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 )}{2 x^{3/4} \sqrt [4]{1+x}}+\frac {\left (\left (1-2 \sqrt {5}\right ) \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 )}{2 x^{3/4} \sqrt [4]{1+x}}-\frac {\left (\left (1+2 \sqrt {5}\right ) \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 )}{2 x^{3/4} \sqrt [4]{1+x}}+\frac {\left (\left (1+2 \sqrt {5}\right ) \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 )}{2 x^{3/4} \sqrt [4]{1+x}}\\ &=2 \sqrt [4]{x^3+x^4}+\frac {\left (1-2 \sqrt {5}\right ) \sqrt [4]{x^3+x^4} \tan ^{-1}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{2 x^{3/4} \sqrt [4]{1+x}}+\frac {\left (1+2 \sqrt {5}\right ) \sqrt [4]{x^3+x^4} \tan ^{-1}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{2 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 )}{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 )}{x^{3/4} \sqrt [4]{1+x}}-\frac {\left (1-2 \sqrt {5}\right ) \sqrt [4]{x^3+x^4} \tanh ^{-1}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{2 x^{3/4} \sqrt [4]{1+x}}-\frac {\left (1+2 \sqrt {5}\right ) \sqrt [4]{x^3+x^4} \tanh ^{-1}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{2 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 )}{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 )}{x^{3/4} \sqrt [4]{1+x}}\\ \end {align*}

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Mathematica [C]  time = 0.25, size = 147, normalized size = 0.65 \begin {gather*} \frac {2 \sqrt [4]{x^3 (x+1)} \left (4 (x+1) \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};-x\right )-2 (x+1) \, _2F_1\left (\frac {3}{4},\frac {3}{4};\frac {7}{4};-x\right )+\sqrt [4]{x+1} \left (\left (\sqrt {5}-1\right ) \, _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 ) \, _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 )\right )\right )}{3 (x+1)^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(x^3*(1 + x))^(1/4)*(4*(1 + x)*Hypergeometric2F1[-1/4, 3/4, 7/4, -x] - 2*(1 + x)*Hypergeometric2F1[3/4, 3/4
, 7/4, -x] + (1 + x)^(1/4)*((-1 + Sqrt[5])*Hypergeometric2F1[3/4, 1, 7/4, ((-1 + Sqrt[5])*x)/((1 + Sqrt[5])*(1
+ x))] - (1 + Sqrt[5])*Hypergeometric2F1[3/4, 1, 7/4, ((1 + Sqrt[5])*x)/((-1 + Sqrt[5])*(1 + x))])))/(3*(1 +
x)^(5/4))

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IntegrateAlgebraic [A]  time = 0.93, size = 226, normalized size = 1.00 \begin {gather*} 2 \sqrt [4]{x^3+x^4}+\tan ^{-1}\left (\frac {x}{\sqrt [4]{x^3+x^4}}\right )-\sqrt {2 \left (1+\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{x^3+x^4}}\right )+\sqrt {2 \left (-1+\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{x^3+x^4}}\right )-\tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^3+x^4}}\right )+\sqrt {2 \left (1+\sqrt {5}\right )} \tanh ^{-1}\left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{x^3+x^4}}\right )-\sqrt {2 \left (-1+\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[((1 + 2*x)*(x^3 + x^4)^(1/4))/(-1 + x + x^2),x]

[Out]

2*(x^3 + x^4)^(1/4) + ArcTan[x/(x^3 + x^4)^(1/4)] - Sqrt[2*(1 + Sqrt[5])]*ArcTan[(Sqrt[-1/2 + Sqrt[5]/2]*x)/(x
^3 + x^4)^(1/4)] + Sqrt[2*(-1 + Sqrt[5])]*ArcTan[(Sqrt[1/2 + Sqrt[5]/2]*x)/(x^3 + x^4)^(1/4)] - ArcTanh[x/(x^3
+ x^4)^(1/4)] + Sqrt[2*(1 + Sqrt[5])]*ArcTanh[(Sqrt[-1/2 + Sqrt[5]/2]*x)/(x^3 + x^4)^(1/4)] - Sqrt[2*(-1 + Sq
rt[5])]*ArcTanh[(Sqrt[1/2 + Sqrt[5]/2]*x)/(x^3 + x^4)^(1/4)]

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fricas [B]  time = 0.61, size = 415, normalized size = 1.84 \begin {gather*} 2 \, \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 ) - 2 \, \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}{2} \, \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}{2} \, \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}{2} \, \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}{2} \, \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 ) + 2 \, {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}} - \arctan \left (\frac {{\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{2} \, \log \left (\frac {x + {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{2} \, \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((1+2*x)*(x^4+x^3)^(1/4)/(x^2+x-1),x, algorithm="fricas")

[Out]

2*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*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) + 1/2*sq
rt(2*sqrt(5) + 2)*log(((sqrt(5)*x - x)*sqrt(2*sqrt(5) + 2) + 4*(x^4 + x^3)^(1/4))/x) - 1/2*sqrt(2*sqrt(5) + 2)
*log(-((sqrt(5)*x - x)*sqrt(2*sqrt(5) + 2) - 4*(x^4 + x^3)^(1/4))/x) - 1/2*sqrt(2*sqrt(5) - 2)*log(((sqrt(5)*x
+ x)*sqrt(2*sqrt(5) - 2) + 4*(x^4 + x^3)^(1/4))/x) + 1/2*sqrt(2*sqrt(5) - 2)*log(-((sqrt(5)*x + x)*sqrt(2*sqr
t(5) - 2) - 4*(x^4 + x^3)^(1/4))/x) + 2*(x^4 + x^3)^(1/4) - arctan((x^4 + x^3)^(1/4)/x) - 1/2*log((x + (x^4 +
x^3)^(1/4))/x) + 1/2*log(-(x - (x^4 + x^3)^(1/4))/x)

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giac [A]  time = 0.40, size = 225, normalized size = 1.00 \begin {gather*} -\sqrt {2 \, \sqrt {5} - 2} \arctan \left (\frac {{\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}}{\sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}}}\right ) + \sqrt {2 \, \sqrt {5} + 2} \arctan \left (\frac {{\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}}{\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}}}\right ) - \frac {1}{2} \, \sqrt {2 \, \sqrt {5} - 2} \log \left (\sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} + {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{2} \, \sqrt {2 \, \sqrt {5} + 2} \log \left (\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} + {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{2} \, \sqrt {2 \, \sqrt {5} - 2} \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}{2} \, \sqrt {2 \, \sqrt {5} + 2} \log \left ({\left | -\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} + {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} \right |}\right ) + 2 \, x {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} - \arctan \left ({\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \log \left ({\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \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((1+2*x)*(x^4+x^3)^(1/4)/(x^2+x-1),x, algorithm="giac")

[Out]

-sqrt(2*sqrt(5) - 2)*arctan((1/x + 1)^(1/4)/sqrt(1/2*sqrt(5) + 1/2)) + sqrt(2*sqrt(5) + 2)*arctan((1/x + 1)^(1
/4)/sqrt(1/2*sqrt(5) - 1/2)) - 1/2*sqrt(2*sqrt(5) - 2)*log(sqrt(1/2*sqrt(5) + 1/2) + (1/x + 1)^(1/4)) + 1/2*sq
rt(2*sqrt(5) + 2)*log(sqrt(1/2*sqrt(5) - 1/2) + (1/x + 1)^(1/4)) + 1/2*sqrt(2*sqrt(5) - 2)*log(abs(-sqrt(1/2*s
qrt(5) + 1/2) + (1/x + 1)^(1/4))) - 1/2*sqrt(2*sqrt(5) + 2)*log(abs(-sqrt(1/2*sqrt(5) - 1/2) + (1/x + 1)^(1/4)
)) + 2*x*(1/x + 1)^(1/4) - arctan((1/x + 1)^(1/4)) - 1/2*log((1/x + 1)^(1/4) + 1) + 1/2*log(abs((1/x + 1)^(1/4
) - 1))

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maple [C]  time = 23.74, size = 2021, normalized size = 8.94

method	result	size
trager	\(\text {Expression too large to display}\)	\(2021\)
risch	\(\text {Expression too large to display}\)	\(3988\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(x^4+x^3)^(1/4)-1/2*RootOf(_Z^2+1)*ln((2*(x^4+x^3)^(1/2)*RootOf(_Z^2+1)*x-2*RootOf(_Z^2+1)*x^3-RootOf(_Z^2+1
)*x^2+2*(x^4+x^3)^(3/4)-2*x^2*(x^4+x^3)^(1/4))/x^2)-RootOf(_Z^2+4*RootOf(16*_Z^4-4*_Z^2-1)^2-1)*ln(-(152*RootO
f(_Z^2+4*RootOf(16*_Z^4-4*_Z^2-1)^2-1)*RootOf(16*_Z^4-4*_Z^2-1)^4*x^3-152*RootOf(_Z^2+4*RootOf(16*_Z^4-4*_Z^2-
1)^2-1)*RootOf(16*_Z^4-4*_Z^2-1)^4*x^2-296*RootOf(_Z^2+4*RootOf(16*_Z^4-4*_Z^2-1)^2-1)*RootOf(16*_Z^4-4*_Z^2-1
)^2*(x^4+x^3)^(1/2)*x+290*RootOf(_Z^2+4*RootOf(16*_Z^4-4*_Z^2-1)^2-1)*RootOf(16*_Z^4-4*_Z^2-1)^2*x^3+128*RootO
f(_Z^2+4*RootOf(16*_Z^4-4*_Z^2-1)^2-1)*RootOf(16*_Z^4-4*_Z^2-1)^2*x^2-180*RootOf(16*_Z^4-4*_Z^2-1)^2*(x^4+x^3)
^(3/4)+296*RootOf(16*_Z^4-4*_Z^2-1)^2*(x^4+x^3)^(1/4)*x^2-45*RootOf(_Z^2+4*RootOf(16*_Z^4-4*_Z^2-1)^2-1)*(x^4+
x^3)^(1/2)*x+42*RootOf(_Z^2+4*RootOf(16*_Z^4-4*_Z^2-1)^2-1)*x^3+24*RootOf(_Z^2+4*RootOf(16*_Z^4-4*_Z^2-1)^2-1)
*x^2-29*(x^4+x^3)^(3/4)+45*x^2*(x^4+x^3)^(1/4))/(4*x*RootOf(16*_Z^4-4*_Z^2-1)^2-4*RootOf(16*_Z^4-4*_Z^2-1)^2+x
)/x^2)+2*RootOf(16*_Z^4-4*_Z^2-1)*ln((304*x^3*RootOf(16*_Z^4-4*_Z^2-1)^5-304*RootOf(16*_Z^4-4*_Z^2-1)^5*x^2+59
2*RootOf(16*_Z^4-4*_Z^2-1)^3*(x^4+x^3)^(1/2)*x-732*RootOf(16*_Z^4-4*_Z^2-1)^3*x^3-104*RootOf(16*_Z^4-4*_Z^2-1)
^3*x^2-180*RootOf(16*_Z^4-4*_Z^2-1)^2*(x^4+x^3)^(3/4)+296*RootOf(16*_Z^4-4*_Z^2-1)^2*(x^4+x^3)^(1/4)*x^2-238*R
ootOf(16*_Z^4-4*_Z^2-1)*(x^4+x^3)^(1/2)*x+248*RootOf(16*_Z^4-4*_Z^2-1)*x^3+93*RootOf(16*_Z^4-4*_Z^2-1)*x^2+74*
(x^4+x^3)^(3/4)-119*x^2*(x^4+x^3)^(1/4))/(4*x*RootOf(16*_Z^4-4*_Z^2-1)^2-4*RootOf(16*_Z^4-4*_Z^2-1)^2-2*x+1)/x
^2)+1/2*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)-8*RootOf(16*_Z^4-4*_Z^
2-1)^3*ln((192*x^3*RootOf(16*_Z^4-4*_Z^2-1)^5-192*RootOf(16*_Z^4-4*_Z^2-1)^5*x^2+360*RootOf(16*_Z^4-4*_Z^2-1)^
3*(x^4+x^3)^(1/2)*x+364*RootOf(16*_Z^4-4*_Z^2-1)^3*x^3+164*RootOf(16*_Z^4-4*_Z^2-1)^3*x^2+180*RootOf(16*_Z^4-4
*_Z^2-1)^2*(x^4+x^3)^(3/4)+296*RootOf(16*_Z^4-4*_Z^2-1)^2*(x^4+x^3)^(1/4)*x^2+58*RootOf(16*_Z^4-4*_Z^2-1)*(x^4
+x^3)^(1/2)*x+49*RootOf(16*_Z^4-4*_Z^2-1)*x^3+28*RootOf(16*_Z^4-4*_Z^2-1)*x^2+29*(x^4+x^3)^(3/4)+45*x^2*(x^4+x
^3)^(1/4))/(4*x*RootOf(16*_Z^4-4*_Z^2-1)^2-4*RootOf(16*_Z^4-4*_Z^2-1)^2+x)/x^2)+2*RootOf(16*_Z^4-4*_Z^2-1)*ln(
(192*x^3*RootOf(16*_Z^4-4*_Z^2-1)^5-192*RootOf(16*_Z^4-4*_Z^2-1)^5*x^2+360*RootOf(16*_Z^4-4*_Z^2-1)^3*(x^4+x^3
)^(1/2)*x+364*RootOf(16*_Z^4-4*_Z^2-1)^3*x^3+164*RootOf(16*_Z^4-4*_Z^2-1)^3*x^2+180*RootOf(16*_Z^4-4*_Z^2-1)^2
*(x^4+x^3)^(3/4)+296*RootOf(16*_Z^4-4*_Z^2-1)^2*(x^4+x^3)^(1/4)*x^2+58*RootOf(16*_Z^4-4*_Z^2-1)*(x^4+x^3)^(1/2
)*x+49*RootOf(16*_Z^4-4*_Z^2-1)*x^3+28*RootOf(16*_Z^4-4*_Z^2-1)*x^2+29*(x^4+x^3)^(3/4)+45*x^2*(x^4+x^3)^(1/4))
/(4*x*RootOf(16*_Z^4-4*_Z^2-1)^2-4*RootOf(16*_Z^4-4*_Z^2-1)^2+x)/x^2)+4*RootOf(16*_Z^4-4*_Z^2-1)^2*RootOf(_Z^2
+4*RootOf(16*_Z^4-4*_Z^2-1)^2-1)*ln((192*RootOf(_Z^2+4*RootOf(16*_Z^4-4*_Z^2-1)^2-1)*RootOf(16*_Z^4-4*_Z^2-1)^
4*x^3-192*RootOf(_Z^2+4*RootOf(16*_Z^4-4*_Z^2-1)^2-1)*RootOf(16*_Z^4-4*_Z^2-1)^4*x^2-360*RootOf(_Z^2+4*RootOf(
16*_Z^4-4*_Z^2-1)^2-1)*RootOf(16*_Z^4-4*_Z^2-1)^2*(x^4+x^3)^(1/2)*x-460*RootOf(_Z^2+4*RootOf(16*_Z^4-4*_Z^2-1)
^2-1)*RootOf(16*_Z^4-4*_Z^2-1)^2*x^3-68*RootOf(_Z^2+4*RootOf(16*_Z^4-4*_Z^2-1)^2-1)*RootOf(16*_Z^4-4*_Z^2-1)^2
*x^2-360*RootOf(16*_Z^4-4*_Z^2-1)^2*(x^4+x^3)^(3/4)-592*RootOf(16*_Z^4-4*_Z^2-1)^2*(x^4+x^3)^(1/4)*x^2+148*Roo
tOf(_Z^2+4*RootOf(16*_Z^4-4*_Z^2-1)^2-1)*(x^4+x^3)^(1/2)*x+152*RootOf(_Z^2+4*RootOf(16*_Z^4-4*_Z^2-1)^2-1)*x^3
+57*RootOf(_Z^2+4*RootOf(16*_Z^4-4*_Z^2-1)^2-1)*x^2+148*(x^4+x^3)^(3/4)+238*x^2*(x^4+x^3)^(1/4))/(4*x*RootOf(1
6*_Z^4-4*_Z^2-1)^2-4*RootOf(16*_Z^4-4*_Z^2-1)^2-2*x+1)/x^2)

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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 (2 \, x + 1\right )}}{x^{2} + x - 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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