\(\int \frac {1}{(-1+x) \sqrt [4]{x+x^3}} \, dx\) [795]

Optimal result

Integrand size = 15, antiderivative size = 61 \[ \int \frac {1}{(-1+x) \sqrt [4]{x+x^3}} \, dx=\frac {\arctan \left (\frac {2^{3/4} \sqrt [4]{x+x^3}}{1+x}\right )}{2 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {2^{3/4} \sqrt [4]{x+x^3}}{1+x}\right )}{2 \sqrt [4]{2}} \]

[Out]

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

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 0.15 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.43, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2081, 973, 477, 524} \[ \int \frac {1}{(-1+x) \sqrt [4]{x+x^3}} \, dx=-\frac {4 \sqrt [4]{x^2+1} x^2 \operatorname {AppellF1}\left (\frac {7}{8},1,\frac {1}{4},\frac {15}{8},x^2,-x^2\right )}{7 \sqrt [4]{x^3+x}}-\frac {4 \sqrt [4]{x^2+1} x \operatorname {AppellF1}\left (\frac {3}{8},1,\frac {1}{4},\frac {11}{8},x^2,-x^2\right )}{3 \sqrt [4]{x^3+x}} \]

[In]

Int[1/((-1 + x)*(x + x^3)^(1/4)),x]

[Out]

(-4*x*(1 + x^2)^(1/4)*AppellF1[3/8, 1, 1/4, 11/8, x^2, -x^2])/(3*(x + x^3)^(1/4)) - (4*x^2*(1 + x^2)^(1/4)*App
ellF1[7/8, 1, 1/4, 15/8, x^2, -x^2])/(7*(x + x^3)^(1/4))

Rule 477

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 973

Int[(((g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[d*((g*x)^n/x^n), In
t[(x^n*(a + c*x^2)^p)/(d^2 - e^2*x^2), x], x] - Dist[e*((g*x)^n/x^n), Int[(x^(n + 1)*(a + c*x^2)^p)/(d^2 - e^2
*x^2), x], x] /; FreeQ[{a, c, d, e, g, n, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] &&  !IntegersQ[n, 2
*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}& = \frac {\left (\sqrt [4]{x} \sqrt [4]{1+x^2}\right ) \int \frac {1}{(-1+x) \sqrt [4]{x} \sqrt [4]{1+x^2}} \, dx}{\sqrt [4]{x+x^3}} \\ & = -\frac {\left (\sqrt [4]{x} \sqrt [4]{1+x^2}\right ) \int \frac {1}{\sqrt [4]{x} \left (1-x^2\right ) \sqrt [4]{1+x^2}} \, dx}{\sqrt [4]{x+x^3}}-\frac {\left (\sqrt [4]{x} \sqrt [4]{1+x^2}\right ) \int \frac {x^{3/4}}{\left (1-x^2\right ) \sqrt [4]{1+x^2}} \, dx}{\sqrt [4]{x+x^3}} \\ & = -\frac {\left (4 \sqrt [4]{x} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \frac {x^2}{\left (1-x^8\right ) \sqrt [4]{1+x^8}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x+x^3}}-\frac {\left (4 \sqrt [4]{x} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \frac {x^6}{\left (1-x^8\right ) \sqrt [4]{1+x^8}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x+x^3}} \\ & = -\frac {4 x \sqrt [4]{1+x^2} \operatorname {AppellF1}\left (\frac {3}{8},1,\frac {1}{4},\frac {11}{8},x^2,-x^2\right )}{3 \sqrt [4]{x+x^3}}-\frac {4 x^2 \sqrt [4]{1+x^2} \operatorname {AppellF1}\left (\frac {7}{8},1,\frac {1}{4},\frac {15}{8},x^2,-x^2\right )}{7 \sqrt [4]{x+x^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.36 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(-1+x) \sqrt [4]{x+x^3}} \, dx=\frac {\arctan \left (\frac {2^{3/4} \sqrt [4]{x+x^3}}{1+x}\right )-\text {arctanh}\left (\frac {2^{3/4} \sqrt [4]{x+x^3}}{1+x}\right )}{2 \sqrt [4]{2}} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 7.62 (sec) , antiderivative size = 507, normalized size of antiderivative = 8.31

method	result	size
trager	\(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) \ln \left (\frac {2 \sqrt {x^{3}+x}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{3} x^{2}+4 \sqrt {x^{3}+x}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{3} x +2 \left (x^{3}+x \right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{3}+2 \sqrt {x^{3}+x}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{3}+6 \left (x^{3}+x \right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{2}+x^{4} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )+16 \left (x^{3}+x \right )^{\frac {3}{4}} x +6 \left (x^{3}+x \right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x +12 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) x^{3}+16 \left (x^{3}+x \right )^{\frac {3}{4}}+2 \left (x^{3}+x \right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}+6 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) x^{2}+12 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )}{\left (x -1\right )^{4}}\right )}{8}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (\frac {-2 \sqrt {x^{3}+x}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{2}-4 \sqrt {x^{3}+x}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x -2 \left (x^{3}+x \right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{3}-2 \sqrt {x^{3}+x}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}-6 \left (x^{3}+x \right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{4}+16 \left (x^{3}+x \right )^{\frac {3}{4}} x -6 \left (x^{3}+x \right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x +12 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{3}+16 \left (x^{3}+x \right )^{\frac {3}{4}}-2 \left (x^{3}+x \right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}+6 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{2}+12 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right )}{\left (x -1\right )^{4}}\right )}{8}\)	\(507\)

[In]

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

[Out]

-1/8*RootOf(_Z^4-8)*ln((2*(x^3+x)^(1/2)*RootOf(_Z^4-8)^3*x^2+4*(x^3+x)^(1/2)*RootOf(_Z^4-8)^3*x+2*(x^3+x)^(1/4
)*RootOf(_Z^4-8)^2*x^3+2*(x^3+x)^(1/2)*RootOf(_Z^4-8)^3+6*(x^3+x)^(1/4)*RootOf(_Z^4-8)^2*x^2+x^4*RootOf(_Z^4-8
)+16*(x^3+x)^(3/4)*x+6*(x^3+x)^(1/4)*RootOf(_Z^4-8)^2*x+12*RootOf(_Z^4-8)*x^3+16*(x^3+x)^(3/4)+2*(x^3+x)^(1/4)
*RootOf(_Z^4-8)^2+6*RootOf(_Z^4-8)*x^2+12*RootOf(_Z^4-8)*x+RootOf(_Z^4-8))/(x-1)^4)-1/8*RootOf(_Z^2+RootOf(_Z^
4-8)^2)*ln((-2*(x^3+x)^(1/2)*RootOf(_Z^4-8)^2*RootOf(_Z^2+RootOf(_Z^4-8)^2)*x^2-4*(x^3+x)^(1/2)*RootOf(_Z^2+Ro
otOf(_Z^4-8)^2)*RootOf(_Z^4-8)^2*x-2*(x^3+x)^(1/4)*RootOf(_Z^4-8)^2*x^3-2*(x^3+x)^(1/2)*RootOf(_Z^2+RootOf(_Z^
4-8)^2)*RootOf(_Z^4-8)^2-6*(x^3+x)^(1/4)*RootOf(_Z^4-8)^2*x^2+RootOf(_Z^2+RootOf(_Z^4-8)^2)*x^4+16*(x^3+x)^(3/
4)*x-6*(x^3+x)^(1/4)*RootOf(_Z^4-8)^2*x+12*RootOf(_Z^2+RootOf(_Z^4-8)^2)*x^3+16*(x^3+x)^(3/4)-2*(x^3+x)^(1/4)*
RootOf(_Z^4-8)^2+6*RootOf(_Z^2+RootOf(_Z^4-8)^2)*x^2+12*RootOf(_Z^2+RootOf(_Z^4-8)^2)*x+RootOf(_Z^2+RootOf(_Z^
4-8)^2))/(x-1)^4)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.47 (sec) , antiderivative size = 438, normalized size of antiderivative = 7.18 \[ \int \frac {1}{(-1+x) \sqrt [4]{x+x^3}} \, dx=-\frac {1}{16} \cdot 2^{\frac {3}{4}} \log \left (\frac {2^{\frac {3}{4}} {\left (x^{4} + 12 \, x^{3} + 6 \, x^{2} + 12 \, x + 1\right )} + 4 \, \sqrt {2} {\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )} {\left (x^{3} + x\right )}^{\frac {1}{4}} + 8 \cdot 2^{\frac {1}{4}} \sqrt {x^{3} + x} {\left (x^{2} + 2 \, x + 1\right )} + 16 \, {\left (x^{3} + x\right )}^{\frac {3}{4}} {\left (x + 1\right )}}{x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1}\right ) + \frac {1}{16} \cdot 2^{\frac {3}{4}} \log \left (-\frac {2^{\frac {3}{4}} {\left (x^{4} + 12 \, x^{3} + 6 \, x^{2} + 12 \, x + 1\right )} - 4 \, \sqrt {2} {\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )} {\left (x^{3} + x\right )}^{\frac {1}{4}} + 8 \cdot 2^{\frac {1}{4}} \sqrt {x^{3} + x} {\left (x^{2} + 2 \, x + 1\right )} - 16 \, {\left (x^{3} + x\right )}^{\frac {3}{4}} {\left (x + 1\right )}}{x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1}\right ) - \frac {1}{16} i \cdot 2^{\frac {3}{4}} \log \left (\frac {2^{\frac {3}{4}} {\left (i \, x^{4} + 12 i \, x^{3} + 6 i \, x^{2} + 12 i \, x + i\right )} - 4 \, \sqrt {2} {\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )} {\left (x^{3} + x\right )}^{\frac {1}{4}} - 8 \cdot 2^{\frac {1}{4}} \sqrt {x^{3} + x} {\left (i \, x^{2} + 2 i \, x + i\right )} + 16 \, {\left (x^{3} + x\right )}^{\frac {3}{4}} {\left (x + 1\right )}}{x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1}\right ) + \frac {1}{16} i \cdot 2^{\frac {3}{4}} \log \left (\frac {2^{\frac {3}{4}} {\left (-i \, x^{4} - 12 i \, x^{3} - 6 i \, x^{2} - 12 i \, x - i\right )} - 4 \, \sqrt {2} {\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )} {\left (x^{3} + x\right )}^{\frac {1}{4}} - 8 \cdot 2^{\frac {1}{4}} \sqrt {x^{3} + x} {\left (-i \, x^{2} - 2 i \, x - i\right )} + 16 \, {\left (x^{3} + x\right )}^{\frac {3}{4}} {\left (x + 1\right )}}{x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1}\right ) \]

[In]

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

[Out]

-1/16*2^(3/4)*log((2^(3/4)*(x^4 + 12*x^3 + 6*x^2 + 12*x + 1) + 4*sqrt(2)*(x^3 + 3*x^2 + 3*x + 1)*(x^3 + x)^(1/
4) + 8*2^(1/4)*sqrt(x^3 + x)*(x^2 + 2*x + 1) + 16*(x^3 + x)^(3/4)*(x + 1))/(x^4 - 4*x^3 + 6*x^2 - 4*x + 1)) +
1/16*2^(3/4)*log(-(2^(3/4)*(x^4 + 12*x^3 + 6*x^2 + 12*x + 1) - 4*sqrt(2)*(x^3 + 3*x^2 + 3*x + 1)*(x^3 + x)^(1/
4) + 8*2^(1/4)*sqrt(x^3 + x)*(x^2 + 2*x + 1) - 16*(x^3 + x)^(3/4)*(x + 1))/(x^4 - 4*x^3 + 6*x^2 - 4*x + 1)) -
1/16*I*2^(3/4)*log((2^(3/4)*(I*x^4 + 12*I*x^3 + 6*I*x^2 + 12*I*x + I) - 4*sqrt(2)*(x^3 + 3*x^2 + 3*x + 1)*(x^3
+ x)^(1/4) - 8*2^(1/4)*sqrt(x^3 + x)*(I*x^2 + 2*I*x + I) + 16*(x^3 + x)^(3/4)*(x + 1))/(x^4 - 4*x^3 + 6*x^2 -
4*x + 1)) + 1/16*I*2^(3/4)*log((2^(3/4)*(-I*x^4 - 12*I*x^3 - 6*I*x^2 - 12*I*x - I) - 4*sqrt(2)*(x^3 + 3*x^2 +
3*x + 1)*(x^3 + x)^(1/4) - 8*2^(1/4)*sqrt(x^3 + x)*(-I*x^2 - 2*I*x - I) + 16*(x^3 + x)^(3/4)*(x + 1))/(x^4 -
4*x^3 + 6*x^2 - 4*x + 1))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/((x + x^3)^(1/4)*(x - 1)),x)

[Out]

int(1/((x + x^3)^(1/4)*(x - 1)), x)