3.11.1 \(\int \frac {\sqrt [4]{-1-x^4} (-1+x^4)}{x^6 (1+2 x^4)} \, dx\)

Optimal. Leaf size=76 \[ \frac {3}{2} \tan ^{-1}\left (\frac {x \left (-x^4-1\right )^{3/4}}{x^4+1}\right )-\frac {3}{2} \tanh ^{-1}\left (\frac {x \left (-x^4-1\right )^{3/4}}{x^4+1}\right )+\frac {\sqrt [4]{-x^4-1} \left (1-14 x^4\right )}{5 x^5} \]

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 73, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {580, 583, 12, 494, 298, 203, 206} \begin {gather*} -\frac {14 \sqrt [4]{-x^4-1}}{5 x}-\frac {3}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{-x^4-1}}\right )+\frac {3}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{-x^4-1}}\right )+\frac {\sqrt [4]{-x^4-1}}{5 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-1 - x^4)^(1/4)*(-1 + x^4))/(x^6*(1 + 2*x^4)),x]

[Out]

(-1 - x^4)^(1/4)/(5*x^5) - (14*(-1 - x^4)^(1/4))/(5*x) - (3*ArcTan[x/(-1 - x^4)^(1/4)])/2 + (3*ArcTanh[x/(-1 -
x^4)^(1/4)])/2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 494

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 580

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*g*(m + 1)), x] - Dist[1/(a*g^n*(m + 1
)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f)*(m + 1) + e*n*(b*c*(p + 1) + a*d*q)
+ d*((b*e - a*f)*(m + 1) + b*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && IGtQ[n
, 0] && GtQ[q, 0] && LtQ[m, -1] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\sqrt [4]{-1-x^4} \left (-1+x^4\right )}{x^6 \left (1+2 x^4\right )} \, dx &=\frac {\sqrt [4]{-1-x^4}}{5 x^5}+\frac {1}{5} \int \frac {-14-13 x^4}{x^2 \left (-1-x^4\right )^{3/4} \left (1+2 x^4\right )} \, dx\\ &=\frac {\sqrt [4]{-1-x^4}}{5 x^5}-\frac {14 \sqrt [4]{-1-x^4}}{5 x}+\frac {1}{5} \int \frac {15 x^2}{\left (-1-x^4\right )^{3/4} \left (1+2 x^4\right )} \, dx\\ &=\frac {\sqrt [4]{-1-x^4}}{5 x^5}-\frac {14 \sqrt [4]{-1-x^4}}{5 x}+3 \int \frac {x^2}{\left (-1-x^4\right )^{3/4} \left (1+2 x^4\right )} \, dx\\ &=\frac {\sqrt [4]{-1-x^4}}{5 x^5}-\frac {14 \sqrt [4]{-1-x^4}}{5 x}+3 \operatorname {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{-1-x^4}}\right )\\ &=\frac {\sqrt [4]{-1-x^4}}{5 x^5}-\frac {14 \sqrt [4]{-1-x^4}}{5 x}+\frac {3}{2} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt [4]{-1-x^4}}\right )-\frac {3}{2} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt [4]{-1-x^4}}\right )\\ &=\frac {\sqrt [4]{-1-x^4}}{5 x^5}-\frac {14 \sqrt [4]{-1-x^4}}{5 x}-\frac {3}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{-1-x^4}}\right )+\frac {3}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{-1-x^4}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.08, size = 78, normalized size = 1.03 \begin {gather*} \frac {\frac {5 \left (x^4+1\right )^{3/4} x^8 \, _2F_1\left (\frac {3}{4},\frac {3}{4};\frac {7}{4};\frac {x^4}{2 x^4+1}\right )}{\left (2 x^4+1\right )^{3/4}}+14 x^8+13 x^4-1}{5 x^5 \left (-x^4-1\right )^{3/4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((-1 - x^4)^(1/4)*(-1 + x^4))/(x^6*(1 + 2*x^4)),x]

[Out]

(-1 + 13*x^4 + 14*x^8 + (5*x^8*(1 + x^4)^(3/4)*Hypergeometric2F1[3/4, 3/4, 7/4, x^4/(1 + 2*x^4)])/(1 + 2*x^4)^
(3/4))/(5*x^5*(-1 - x^4)^(3/4))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.26, size = 76, normalized size = 1.00 \begin {gather*} \frac {\left (1-14 x^4\right ) \sqrt [4]{-1-x^4}}{5 x^5}+\frac {3}{2} \tan ^{-1}\left (\frac {x \left (-1-x^4\right )^{3/4}}{1+x^4}\right )-\frac {3}{2} \tanh ^{-1}\left (\frac {x \left (-1-x^4\right )^{3/4}}{1+x^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-1 - x^4)^(1/4)*(-1 + x^4))/(x^6*(1 + 2*x^4)),x]

[Out]

((1 - 14*x^4)*(-1 - x^4)^(1/4))/(5*x^5) + (3*ArcTan[(x*(-1 - x^4)^(3/4))/(1 + x^4)])/2 - (3*ArcTanh[(x*(-1 - x
^4)^(3/4))/(1 + x^4)])/2

________________________________________________________________________________________

fricas [C]  time = 3.43, size = 201, normalized size = 2.64 \begin {gather*} \frac {30 \, x^{5} \log \left (-\frac {2 \, {\left (2 \, {\left (-x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 2 \, \sqrt {-x^{4} - 1} x^{2} + 2 \, {\left (-x^{4} - 1\right )}^{\frac {3}{4}} x - 1\right )}}{2 \, x^{4} + 1}\right ) - 15 i \, x^{5} \log \left (-\frac {2 \, {\left (2 i \, {\left (-x^{4} - 1\right )}^{\frac {1}{4}} x^{3} - 2 \, \sqrt {-x^{4} - 1} x^{2} - 2 i \, {\left (-x^{4} - 1\right )}^{\frac {3}{4}} x - 1\right )}}{2 \, x^{4} + 1}\right ) + 15 i \, x^{5} \log \left (-\frac {2 \, {\left (-2 i \, {\left (-x^{4} - 1\right )}^{\frac {1}{4}} x^{3} - 2 \, \sqrt {-x^{4} - 1} x^{2} + 2 i \, {\left (-x^{4} - 1\right )}^{\frac {3}{4}} x - 1\right )}}{2 \, x^{4} + 1}\right ) - 8 \, {\left (14 \, x^{4} - 1\right )} {\left (-x^{4} - 1\right )}^{\frac {1}{4}}}{40 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4-1)^(1/4)*(x^4-1)/x^6/(2*x^4+1),x, algorithm="fricas")

[Out]

1/40*(30*x^5*log(-2*(2*(-x^4 - 1)^(1/4)*x^3 + 2*sqrt(-x^4 - 1)*x^2 + 2*(-x^4 - 1)^(3/4)*x - 1)/(2*x^4 + 1)) -
15*I*x^5*log(-2*(2*I*(-x^4 - 1)^(1/4)*x^3 - 2*sqrt(-x^4 - 1)*x^2 - 2*I*(-x^4 - 1)^(3/4)*x - 1)/(2*x^4 + 1)) +
15*I*x^5*log(-2*(-2*I*(-x^4 - 1)^(1/4)*x^3 - 2*sqrt(-x^4 - 1)*x^2 + 2*I*(-x^4 - 1)^(3/4)*x - 1)/(2*x^4 + 1)) -
8*(14*x^4 - 1)*(-x^4 - 1)^(1/4))/x^5

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} - 1\right )} {\left (-x^{4} - 1\right )}^{\frac {1}{4}}}{{\left (2 \, x^{4} + 1\right )} x^{6}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4-1)^(1/4)*(x^4-1)/x^6/(2*x^4+1),x, algorithm="giac")

[Out]

integrate((x^4 - 1)*(-x^4 - 1)^(1/4)/((2*x^4 + 1)*x^6), x)

________________________________________________________________________________________

maple [C]  time = 3.10, size = 152, normalized size = 2.00

method	result	size
trager	\(-\frac {\left (14 x^{4}-1\right ) \left (-x^{4}-1\right )^{\frac {1}{4}}}{5 x^{5}}-\frac {3 \ln \left (\frac {2 \left (-x^{4}-1\right )^{\frac {3}{4}} x -2 \sqrt {-x^{4}-1}\, x^{2}+2 \left (-x^{4}-1\right )^{\frac {1}{4}} x^{3}+1}{2 x^{4}+1}\right )}{4}-\frac {3 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-2 \sqrt {-x^{4}-1}\, \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}+2 \left (-x^{4}-1\right )^{\frac {3}{4}} x -2 \left (-x^{4}-1\right )^{\frac {1}{4}} x^{3}-\RootOf \left (\textit {\_Z}^{2}+1\right )}{2 x^{4}+1}\right )}{4}\)	\(152\)
risch	\(\frac {14 x^{8}+13 x^{4}-1}{5 x^{5} \left (-x^{4}-1\right )^{\frac {3}{4}}}+\frac {\left (-\frac {3 \ln \left (\frac {2 \left (-x^{12}-3 x^{8}-3 x^{4}-1\right )^{\frac {1}{4}} x^{9}+2 \sqrt {-x^{12}-3 x^{8}-3 x^{4}-1}\, x^{6}+x^{8}+2 \left (-x^{12}-3 x^{8}-3 x^{4}-1\right )^{\frac {3}{4}} x^{3}+4 \left (-x^{12}-3 x^{8}-3 x^{4}-1\right )^{\frac {1}{4}} x^{5}+2 \sqrt {-x^{12}-3 x^{8}-3 x^{4}-1}\, x^{2}+2 x^{4}+2 \left (-x^{12}-3 x^{8}-3 x^{4}-1\right )^{\frac {1}{4}} x +1}{\left (x^{4}+1\right )^{2} \left (2 x^{4}+1\right )}\right )}{4}-\frac {3 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (-x^{12}-3 x^{8}-3 x^{4}-1\right )^{\frac {1}{4}} x^{9}+2 \sqrt {-x^{12}-3 x^{8}-3 x^{4}-1}\, x^{6}-x^{8}+2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (-x^{12}-3 x^{8}-3 x^{4}-1\right )^{\frac {3}{4}} x^{3}-4 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (-x^{12}-3 x^{8}-3 x^{4}-1\right )^{\frac {1}{4}} x^{5}+2 \sqrt {-x^{12}-3 x^{8}-3 x^{4}-1}\, x^{2}-2 x^{4}-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (-x^{12}-3 x^{8}-3 x^{4}-1\right )^{\frac {1}{4}} x -1}{\left (x^{4}+1\right )^{2} \left (2 x^{4}+1\right )}\right )}{4}\right ) \left (-\left (x^{4}+1\right )^{3}\right )^{\frac {1}{4}}}{\left (-x^{4}-1\right )^{\frac {3}{4}}}\)	\(426\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^4-1)^(1/4)*(x^4-1)/x^6/(2*x^4+1),x,method=_RETURNVERBOSE)

[Out]

-1/5*(14*x^4-1)/x^5*(-x^4-1)^(1/4)-3/4*ln((2*(-x^4-1)^(3/4)*x-2*(-x^4-1)^(1/2)*x^2+2*(-x^4-1)^(1/4)*x^3+1)/(2*
x^4+1))-3/4*RootOf(_Z^2+1)*ln((-2*(-x^4-1)^(1/2)*RootOf(_Z^2+1)*x^2+2*(-x^4-1)^(3/4)*x-2*(-x^4-1)^(1/4)*x^3-Ro
otOf(_Z^2+1))/(2*x^4+1))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} - 1\right )} {\left (-x^{4} - 1\right )}^{\frac {1}{4}}}{{\left (2 \, x^{4} + 1\right )} x^{6}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4-1)^(1/4)*(x^4-1)/x^6/(2*x^4+1),x, algorithm="maxima")

[Out]

integrate((x^4 - 1)*(-x^4 - 1)^(1/4)/((2*x^4 + 1)*x^6), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (x^4-1\right )\,{\left (-x^4-1\right )}^{1/4}}{x^6\,\left (2\,x^4+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^4 - 1)*(- x^4 - 1)^(1/4))/(x^6*(2*x^4 + 1)),x)

[Out]

int(((x^4 - 1)*(- x^4 - 1)^(1/4))/(x^6*(2*x^4 + 1)), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \sqrt [4]{- x^{4} - 1}}{x^{6} \left (2 x^{4} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**4-1)**(1/4)*(x**4-1)/x**6/(2*x**4+1),x)

[Out]

Integral((x - 1)*(x + 1)*(x**2 + 1)*(-x**4 - 1)**(1/4)/(x**6*(2*x**4 + 1)), x)

________________________________________________________________________________________