Integrand size = 18, antiderivative size = 47 \[ \int \frac {\left (-1+x^4\right ) \sqrt [4]{1+x^4}}{x} \, dx=\frac {1}{5} \left (-4+x^4\right ) \sqrt [4]{1+x^4}+\frac {1}{2} \arctan \left (\sqrt [4]{1+x^4}\right )+\frac {1}{2} \text {arctanh}\left (\sqrt [4]{1+x^4}\right ) \]
Time = 0.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-1+x^4\right ) \sqrt [4]{1+x^4}}{x} \, dx=\frac {1}{5} \left (-4+x^4\right ) \sqrt [4]{1+x^4}+\frac {1}{2} \arctan \left (\sqrt [4]{1+x^4}\right )+\frac {1}{2} \text {arctanh}\left (\sqrt [4]{1+x^4}\right ) \]
Time = 0.18 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.28, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {948, 25, 90, 60, 73, 756, 216, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (x^4-1\right ) \sqrt [4]{x^4+1}}{x} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{4} \int -\frac {\left (1-x^4\right ) \sqrt [4]{x^4+1}}{x^4}dx^4\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} \int \frac {\left (1-x^4\right ) \sqrt [4]{x^4+1}}{x^4}dx^4\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{4} \left (\frac {4}{5} \left (x^4+1\right )^{5/4}-\int \frac {\sqrt [4]{x^4+1}}{x^4}dx^4\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{4} \left (-\int \frac {1}{x^4 \left (x^4+1\right )^{3/4}}dx^4+\frac {4}{5} \left (x^4+1\right )^{5/4}-4 \sqrt [4]{x^4+1}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (-4 \int \frac {1}{x^{16}-1}d\sqrt [4]{x^4+1}+\frac {4}{5} \left (x^4+1\right )^{5/4}-4 \sqrt [4]{x^4+1}\right )\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {1}{4} \left (-4 \left (-\frac {1}{2} \int \frac {1}{1-x^8}d\sqrt [4]{x^4+1}-\frac {1}{2} \int \frac {1}{x^8+1}d\sqrt [4]{x^4+1}\right )+\frac {4}{5} \left (x^4+1\right )^{5/4}-4 \sqrt [4]{x^4+1}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{4} \left (-4 \left (-\frac {1}{2} \int \frac {1}{1-x^8}d\sqrt [4]{x^4+1}-\frac {1}{2} \arctan \left (\sqrt [4]{x^4+1}\right )\right )+\frac {4}{5} \left (x^4+1\right )^{5/4}-4 \sqrt [4]{x^4+1}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (-4 \left (-\frac {1}{2} \arctan \left (\sqrt [4]{x^4+1}\right )-\frac {1}{2} \text {arctanh}\left (\sqrt [4]{x^4+1}\right )\right )+\frac {4}{5} \left (x^4+1\right )^{5/4}-4 \sqrt [4]{x^4+1}\right )\) |
(-4*(1 + x^4)^(1/4) + (4*(1 + x^4)^(5/4))/5 - 4*(-1/2*ArcTan[(1 + x^4)^(1/ 4)] - ArcTanh[(1 + x^4)^(1/4)]/2))/4
3.6.95.3.1 Defintions of rubi rules used
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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 3.56 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.21
method | result | size |
pseudoelliptic | \(\frac {x^{4} \left (x^{4}+1\right )^{\frac {1}{4}}}{5}-\frac {4 \left (x^{4}+1\right )^{\frac {1}{4}}}{5}-\frac {\ln \left (\left (x^{4}+1\right )^{\frac {1}{4}}-1\right )}{4}+\frac {\ln \left (\left (x^{4}+1\right )^{\frac {1}{4}}+1\right )}{4}+\frac {\arctan \left (\left (x^{4}+1\right )^{\frac {1}{4}}\right )}{2}\) | \(57\) |
meijerg | \(\frac {-\Gamma \left (\frac {3}{4}\right ) x^{4} \operatorname {hypergeom}\left (\left [\frac {3}{4}, 1, 1\right ], \left [2, 2\right ], -x^{4}\right )-4 \left (4-3 \ln \left (2\right )+\frac {\pi }{2}+4 \ln \left (x \right )\right ) \Gamma \left (\frac {3}{4}\right )}{16 \Gamma \left (\frac {3}{4}\right )}+\frac {x^{4} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, 1\right ], \left [2\right ], -x^{4}\right )}{4}\) | \(62\) |
trager | \(\left (\frac {x^{4}}{5}-\frac {4}{5}\right ) \left (x^{4}+1\right )^{\frac {1}{4}}+\frac {\ln \left (-\frac {x^{4}+2 \left (x^{4}+1\right )^{\frac {3}{4}}+2 \sqrt {x^{4}+1}+2 \left (x^{4}+1\right )^{\frac {1}{4}}+2}{x^{4}}\right )}{4}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{4}+2 \left (x^{4}+1\right )^{\frac {3}{4}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}+1}-2 \left (x^{4}+1\right )^{\frac {1}{4}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{x^{4}}\right )}{4}\) | \(123\) |
1/5*x^4*(x^4+1)^(1/4)-4/5*(x^4+1)^(1/4)-1/4*ln((x^4+1)^(1/4)-1)+1/4*ln((x^ 4+1)^(1/4)+1)+1/2*arctan((x^4+1)^(1/4))
Time = 0.25 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.04 \[ \int \frac {\left (-1+x^4\right ) \sqrt [4]{1+x^4}}{x} \, dx=\frac {1}{5} \, {\left (x^{4} + 1\right )}^{\frac {1}{4}} {\left (x^{4} - 4\right )} + \frac {1}{2} \, \arctan \left ({\left (x^{4} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{4} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{4} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{4}} - 1\right ) \]
1/5*(x^4 + 1)^(1/4)*(x^4 - 4) + 1/2*arctan((x^4 + 1)^(1/4)) + 1/4*log((x^4 + 1)^(1/4) + 1) - 1/4*log((x^4 + 1)^(1/4) - 1)
Time = 11.61 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.19 \[ \int \frac {\left (-1+x^4\right ) \sqrt [4]{1+x^4}}{x} \, dx=\frac {\left (x^{4} + 1\right )^{\frac {5}{4}}}{5} - \sqrt [4]{x^{4} + 1} - \frac {\log {\left (\sqrt [4]{x^{4} + 1} - 1 \right )}}{4} + \frac {\log {\left (\sqrt [4]{x^{4} + 1} + 1 \right )}}{4} + \frac {\operatorname {atan}{\left (\sqrt [4]{x^{4} + 1} \right )}}{2} \]
(x**4 + 1)**(5/4)/5 - (x**4 + 1)**(1/4) - log((x**4 + 1)**(1/4) - 1)/4 + l og((x**4 + 1)**(1/4) + 1)/4 + atan((x**4 + 1)**(1/4))/2
Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.13 \[ \int \frac {\left (-1+x^4\right ) \sqrt [4]{1+x^4}}{x} \, dx=\frac {1}{5} \, {\left (x^{4} + 1\right )}^{\frac {5}{4}} - {\left (x^{4} + 1\right )}^{\frac {1}{4}} + \frac {1}{2} \, \arctan \left ({\left (x^{4} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{4} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{4} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{4}} - 1\right ) \]
1/5*(x^4 + 1)^(5/4) - (x^4 + 1)^(1/4) + 1/2*arctan((x^4 + 1)^(1/4)) + 1/4* log((x^4 + 1)^(1/4) + 1) - 1/4*log((x^4 + 1)^(1/4) - 1)
Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.13 \[ \int \frac {\left (-1+x^4\right ) \sqrt [4]{1+x^4}}{x} \, dx=\frac {1}{5} \, {\left (x^{4} + 1\right )}^{\frac {5}{4}} - {\left (x^{4} + 1\right )}^{\frac {1}{4}} + \frac {1}{2} \, \arctan \left ({\left (x^{4} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{4} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{4} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{4}} - 1\right ) \]
1/5*(x^4 + 1)^(5/4) - (x^4 + 1)^(1/4) + 1/2*arctan((x^4 + 1)^(1/4)) + 1/4* log((x^4 + 1)^(1/4) + 1) - 1/4*log((x^4 + 1)^(1/4) - 1)
Time = 5.52 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int \frac {\left (-1+x^4\right ) \sqrt [4]{1+x^4}}{x} \, dx=\frac {\mathrm {atan}\left ({\left (x^4+1\right )}^{1/4}\right )}{2}+\frac {\mathrm {atanh}\left ({\left (x^4+1\right )}^{1/4}\right )}{2}-{\left (x^4+1\right )}^{1/4}+\frac {{\left (x^4+1\right )}^{5/4}}{5} \]