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

Optimal. Leaf size=101 \[ \frac {4 \sqrt [4]{x^3+x^4}}{x}-2 \text {ArcTan}\left (\frac {x}{\sqrt [4]{x^3+x^4}}\right )+2 \sqrt [4]{2} \text {ArcTan}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^3+x^4}}\right )+2 \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^3+x^4}}\right )-2 \sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^3+x^4}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.72, antiderivative size = 193, normalized size of antiderivative = 1.91, number of steps used = 44, number of rules used = 20, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.741, Rules used = {2081, 1600, 6865, 6857, 270, 338, 304, 209, 212, 1254, 419, 243, 342, 281, 237, 416, 418, 1227, 551, 508} \begin {gather*} -\frac {2 \sqrt [4]{x^4+x^3} \text {ArcTan}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{x^{3/4} \sqrt [4]{x+1}}+\frac {2 \sqrt [4]{2} \sqrt [4]{x^4+x^3} \text {ArcTan}\left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{x^{3/4} \sqrt [4]{x+1}}+\frac {4 \sqrt [4]{x^4+x^3}}{x}+\frac {2 \sqrt [4]{x^4+x^3} \tanh ^{-1}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{x^{3/4} \sqrt [4]{x+1}}-\frac {2 \sqrt [4]{2} \sqrt [4]{x^4+x^3} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{x^{3/4} \sqrt [4]{x+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 237

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]))*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 243

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Dist[x^3*((1 + a/(b*x^4))^(3/4)/(a + b*x^4)^(3/4)), Int[1/(x^3*
(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ[{a, b}, x]

Rule 270

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

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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 342

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

Rule 416

Int[((a_) + (b_.)*(x_)^4)^(1/4)/((c_) + (d_.)*(x_)^4), x_Symbol] :> Dist[Sqrt[a + b*x^4]*Sqrt[a/(a + b*x^4)],
Subst[Int[1/(Sqrt[1 - b*x^4]*(c - (b*c - a*d)*x^4)), x], x, x/(a + b*x^4)^(1/4)], x] /; FreeQ[{a, b, c, d}, x]
&& NeQ[b*c - a*d, 0]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^4]*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1
- Rt[-d/c, 2]*x^2)), x], x] + Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1 + Rt[-d/c, 2]*x^2)), x], x] /; FreeQ[{a,
b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 419

Int[1/(((a_) + (b_.)*(x_)^4)^(3/4)*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x^4)^
(3/4), x], x] - Dist[d/(b*c - a*d), Int[(a + b*x^4)^(1/4)/(c + d*x^4), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ
[b*c - a*d, 0]

Rule 508

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = Denominato
r[p]}, Dist[k*(a^(p + (m + 1)/n)/n), Subst[Int[x^(k*((m + 1)/n) - 1)*((c - (b*c - a*d)*x^k)^q/(1 - b*x^k)^(p +
q + (m + 1)/n + 1)), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && Ration
alQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 0]

Rule 551

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1/(a*Sqr
t[c]*Sqrt[e]*Rt[-d/c, 2]))*EllipticPi[b*(c/(a*d)), ArcSin[Rt[-d/c, 2]*x], c*(f/(d*e))], x] /; FreeQ[{a, b, c,
d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-f/e, -d/c])

Rule 1227

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[Sqrt[-c],
Int[1/((d + e*x^2)*Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c,
0]

Rule 1254

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^4)^p, (d/
(d^2 - e^2*x^4) - e*(x^2/(d^2 - e^2*x^4)))^(-q), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&&  !IntegerQ[p] && ILtQ[q, 0]

Rule 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

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]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 6865

Int[(u_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k, Subst[Int[x^(k*(m + 1) - 1)*(u /. x -> x^k
), x], x, x^(1/k)], x]] /; FractionQ[m]

Rubi steps

\begin {align*} \int \frac {\left (1+x^2\right ) \sqrt [4]{x^3+x^4}}{x^2 \left (-1+x^2\right )} \, dx &=\frac {\sqrt [4]{x^3+x^4} \int \frac {\sqrt [4]{1+x} \left (1+x^2\right )}{x^{5/4} \left (-1+x^2\right )} \, dx}{x^{3/4} \sqrt [4]{1+x}}\\ &=\frac {\sqrt [4]{x^3+x^4} \int \frac {1+x^2}{(-1+x) x^{5/4} (1+x)^{3/4}} \, dx}{x^{3/4} \sqrt [4]{1+x}}\\ &=\frac {\left (4 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {1+x^8}{x^2 \left (-1+x^4\right ) \left (1+x^4\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{x^{3/4} \sqrt [4]{1+x}}\\ &=\frac {\left (4 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \left (-\frac {1}{x^2 \left (1+x^4\right )^{3/4}}+\frac {x^2}{\left (1+x^4\right )^{3/4}}+\frac {1}{\left (-1+x^2\right ) \left (1+x^4\right )^{3/4}}+\frac {1}{\left (1+x^2\right ) \left (1+x^4\right )^{3/4}}\right ) \, dx,x,\sqrt [4]{x}\right )}{x^{3/4} \sqrt [4]{1+x}}\\ &=-\frac {\left (4 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{x^2 \left (1+x^4\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{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 {\left (4 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{\left (-1+x^2\right ) \left (1+x^4\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{x^{3/4} \sqrt [4]{1+x}}+\frac {\left (4 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \left (1+x^4\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{x^{3/4} \sqrt [4]{1+x}}\\ &=\frac {4 \sqrt [4]{x^3+x^4}}{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 (4 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \left (\frac {1}{\left (1-x^4\right ) \left (1+x^4\right )^{3/4}}+\frac {x^2}{\left (-1+x^4\right ) \left (1+x^4\right )^{3/4}}\right ) \, dx,x,\sqrt [4]{x}\right )}{x^{3/4} \sqrt [4]{1+x}}+\frac {\left (4 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \left (\frac {1}{\left (-1+x^4\right ) \left (1+x^4\right )^{3/4}}+\frac {x^2}{\left (-1+x^4\right ) \left (1+x^4\right )^{3/4}}\right ) \, dx,x,\sqrt [4]{x}\right )}{x^{3/4} \sqrt [4]{1+x}}\\ &=\frac {4 \sqrt [4]{x^3+x^4}}{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 (4 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{\left (1-x^4\right ) \left (1+x^4\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{x^{3/4} \sqrt [4]{1+x}}+\frac {\left (4 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{\left (-1+x^4\right ) \left (1+x^4\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{x^{3/4} \sqrt [4]{1+x}}+2 \frac {\left (4 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {x^2}{\left (-1+x^4\right ) \left (1+x^4\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{x^{3/4} \sqrt [4]{1+x}}\\ &=\frac {4 \sqrt [4]{x^3+x^4}}{x}-\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 (2 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {\sqrt [4]{1+x^4}}{1-x^4} \, dx,x,\sqrt [4]{x}\right )}{x^{3/4} \sqrt [4]{1+x}}+\frac {\left (2 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {\sqrt [4]{1+x^4}}{-1+x^4} \, dx,x,\sqrt [4]{x}\right )}{x^{3/4} \sqrt [4]{1+x}}+2 \frac {\left (4 \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {x^2}{-1+2 x^4} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{x^{3/4} \sqrt [4]{1+x}}\\ &=\frac {4 \sqrt [4]{x^3+x^4}}{x}-\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}}+2 \left (-\frac {\left (\sqrt {2} \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{x^{3/4} \sqrt [4]{1+x}}+\frac {\left (\sqrt {2} \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x^2} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{x^{3/4} \sqrt [4]{1+x}}\right )+\frac {\left (2 \sqrt {\frac {1}{1+x}} \sqrt [4]{1+x} \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{\left (1-2 x^4\right ) \sqrt {1-x^4}} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{x^{3/4}}+\frac {\left (2 \sqrt {\frac {1}{1+x}} \sqrt [4]{1+x} \sqrt [4]{x^3+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^4} \left (-1+2 x^4\right )} \, dx,x,\frac {\sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{x^{3/4}}\\ &=\frac {4 \sqrt [4]{x^3+x^4}}{x}-\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}}+2 \left (\frac {\sqrt [4]{2} \sqrt [4]{x^3+x^4} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{x^{3/4} \sqrt [4]{1+x}}-\frac {\sqrt [4]{2} \sqrt [4]{x^3+x^4} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{1+x}}\right )}{x^{3/4} \sqrt [4]{1+x}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 129, normalized size = 1.28 \begin {gather*} \frac {2 x^2 (1+x)^{3/4} \left (2 \sqrt [4]{1+x}-\sqrt [4]{x} \text {ArcTan}\left (\sqrt [4]{\frac {x}{1+x}}\right )+\sqrt [4]{2} \sqrt [4]{x} \text {ArcTan}\left (\sqrt [4]{2} \sqrt [4]{\frac {x}{1+x}}\right )+\sqrt [4]{x} \tanh ^{-1}\left (\sqrt [4]{\frac {x}{1+x}}\right )-\sqrt [4]{2} \sqrt [4]{x} \tanh ^{-1}\left (\sqrt [4]{2} \sqrt [4]{\frac {x}{1+x}}\right )\right )}{\left (x^3 (1+x)\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 3.61, size = 376, normalized size = 3.72

method	result	size
trager	\(\frac {4 \left (x^{4}+x^{3}\right )^{\frac {1}{4}}}{x}+\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}+x^{3}}\, \RootOf \left (\textit {\_Z}^{2}+1\right ) x -2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}-\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}+2 \left (x^{4}+x^{3}\right )^{\frac {3}{4}}-2 \left (x^{4}+x^{3}\right )^{\frac {1}{4}} x^{2}}{x^{2}}\right )-\RootOf \left (\textit {\_Z}^{4}-2\right ) \ln \left (\frac {3 \RootOf \left (\textit {\_Z}^{4}-2\right )^{3} x^{3}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{3} x^{2}+4 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \left (x^{4}+x^{3}\right )^{\frac {1}{4}} x^{2}+4 \sqrt {x^{4}+x^{3}}\, \RootOf \left (\textit {\_Z}^{4}-2\right ) x +4 \left (x^{4}+x^{3}\right )^{\frac {3}{4}}}{\left (-1+x \right ) x^{2}}\right )-\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (-\frac {3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{3}+\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{2}+4 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \left (x^{4}+x^{3}\right )^{\frac {1}{4}} x^{2}-4 \sqrt {x^{4}+x^{3}}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) x -4 \left (x^{4}+x^{3}\right )^{\frac {3}{4}}}{\left (-1+x \right ) x^{2}}\right )+\ln \left (\frac {2 \left (x^{4}+x^{3}\right )^{\frac {3}{4}}+2 \sqrt {x^{4}+x^{3}}\, x +2 \left (x^{4}+x^{3}\right )^{\frac {1}{4}} x^{2}+2 x^{3}+x^{2}}{x^{2}}\right )\)	\(376\)
risch	\(\text {Expression too large to display}\)	\(974\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*(x^4+x^3)^(1/4)/x+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^4+x^3)^(1/4)*x^2)/x^2)-RootOf(_Z^4-2)*ln((3*RootOf(_Z^4-2)^3*x^3+RootOf(_Z^4-2)^3*x
^2+4*RootOf(_Z^4-2)^2*(x^4+x^3)^(1/4)*x^2+4*(x^4+x^3)^(1/2)*RootOf(_Z^4-2)*x+4*(x^4+x^3)^(3/4))/(-1+x)/x^2)-Ro
otOf(_Z^2+RootOf(_Z^4-2)^2)*ln(-(3*RootOf(_Z^2+RootOf(_Z^4-2)^2)*RootOf(_Z^4-2)^2*x^3+RootOf(_Z^2+RootOf(_Z^4-
2)^2)*RootOf(_Z^4-2)^2*x^2+4*RootOf(_Z^4-2)^2*(x^4+x^3)^(1/4)*x^2-4*(x^4+x^3)^(1/2)*RootOf(_Z^2+RootOf(_Z^4-2)
^2)*x-4*(x^4+x^3)^(3/4))/(-1+x)/x^2)+ln((2*(x^4+x^3)^(3/4)+2*(x^4+x^3)^(1/2)*x+2*(x^4+x^3)^(1/4)*x^2+2*x^3+x^2
)/x^2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (83) = 166\).
time = 0.35, size = 182, normalized size = 1.80 \begin {gather*} \frac {4 \cdot 2^{\frac {1}{4}} x \arctan \left (\frac {2^{\frac {3}{4}} x \sqrt {\frac {\sqrt {2} x^{2} + \sqrt {x^{4} + x^{3}}}{x^{2}}} - 2^{\frac {3}{4}} {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{2 \, x}\right ) - 2^{\frac {1}{4}} x \log \left (\frac {2^{\frac {1}{4}} x + {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + 2^{\frac {1}{4}} x \log \left (-\frac {2^{\frac {1}{4}} x - {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + 2 \, x \arctan \left (\frac {{\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + x \log \left (\frac {x + {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - x \log \left (-\frac {x - {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + 4 \, {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.45, size = 97, normalized size = 0.96 \begin {gather*} -2 \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) - 2^{\frac {1}{4}} \log \left (2^{\frac {1}{4}} + {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) + 2^{\frac {1}{4}} \log \left ({\left | -2^{\frac {1}{4}} + {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} \right |}\right ) + 4 \, {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 2 \, \arctan \left ({\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) + \log \left ({\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 1\right ) - \log \left ({\left | {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (x^4+x^3\right )}^{1/4}\,\left (x^2+1\right )}{x^2\,\left (x^2-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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