Integrand size = 18, antiderivative size = 54 \[ \int \frac {\left (-1+x^4\right ) \sqrt [4]{1+x^4}}{x^2} \, dx=\frac {\sqrt [4]{1+x^4} \left (4+x^4\right )}{4 x}+\frac {3}{8} \arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {3}{8} \text {arctanh}\left (\frac {x}{\sqrt [4]{1+x^4}}\right ) \]
Time = 0.17 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-1+x^4\right ) \sqrt [4]{1+x^4}}{x^2} \, dx=\frac {\sqrt [4]{1+x^4} \left (4+x^4\right )}{4 x}+\frac {3}{8} \arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {3}{8} \text {arctanh}\left (\frac {x}{\sqrt [4]{1+x^4}}\right ) \]
((1 + x^4)^(1/4)*(4 + x^4))/(4*x) + (3*ArcTan[x/(1 + x^4)^(1/4)])/8 - (3*A rcTanh[x/(1 + x^4)^(1/4)])/8
Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.30, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {955, 811, 854, 827, 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.
\(\displaystyle \int \frac {\left (x^4-1\right ) \sqrt [4]{x^4+1}}{x^2} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle \frac {\left (x^4+1\right )^{5/4}}{x}-3 \int x^2 \sqrt [4]{x^4+1}dx\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {\left (x^4+1\right )^{5/4}}{x}-3 \left (\frac {1}{4} \int \frac {x^2}{\left (x^4+1\right )^{3/4}}dx+\frac {1}{4} \sqrt [4]{x^4+1} x^3\right )\) |
\(\Big \downarrow \) 854 |
\(\displaystyle \frac {\left (x^4+1\right )^{5/4}}{x}-3 \left (\frac {1}{4} \int \frac {x^2}{\sqrt {x^4+1} \left (1-\frac {x^4}{x^4+1}\right )}d\frac {x}{\sqrt [4]{x^4+1}}+\frac {1}{4} \sqrt [4]{x^4+1} x^3\right )\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {\left (x^4+1\right )^{5/4}}{x}-3 \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {1}{1-\frac {x^2}{\sqrt {x^4+1}}}d\frac {x}{\sqrt [4]{x^4+1}}-\frac {1}{2} \int \frac {1}{\frac {x^2}{\sqrt {x^4+1}}+1}d\frac {x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{4} \sqrt [4]{x^4+1} x^3\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\left (x^4+1\right )^{5/4}}{x}-3 \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {1}{1-\frac {x^2}{\sqrt {x^4+1}}}d\frac {x}{\sqrt [4]{x^4+1}}-\frac {1}{2} \arctan \left (\frac {x}{\sqrt [4]{x^4+1}}\right )\right )+\frac {1}{4} \sqrt [4]{x^4+1} x^3\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\left (x^4+1\right )^{5/4}}{x}-3 \left (\frac {1}{4} \left (\frac {1}{2} \text {arctanh}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {1}{2} \arctan \left (\frac {x}{\sqrt [4]{x^4+1}}\right )\right )+\frac {1}{4} \sqrt [4]{x^4+1} x^3\right )\) |
(1 + x^4)^(5/4)/x - 3*((x^3*(1 + x^4)^(1/4))/4 + (-1/2*ArcTan[x/(1 + x^4)^ (1/4)] + ArcTanh[x/(1 + x^4)^(1/4)]/2)/4)
3.7.91.3.1 Defintions of rubi rules used
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 )) Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m , p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)) Int[(e *x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[p, -1]
Result contains higher order function than in optimal. Order 5 vs. order 3.
Time = 4.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.61
method | result | size |
meijerg | \(\frac {\operatorname {hypergeom}\left (\left [-\frac {1}{4}, -\frac {1}{4}\right ], \left [\frac {3}{4}\right ], -x^{4}\right )}{x}+\frac {x^{3} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -x^{4}\right )}{3}\) | \(33\) |
risch | \(\frac {x^{8}+5 x^{4}+4}{4 x \left (x^{4}+1\right )^{\frac {3}{4}}}-\frac {x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -x^{4}\right )}{4}\) | \(40\) |
pseudoelliptic | \(\frac {4 x^{4} \left (x^{4}+1\right )^{\frac {1}{4}}+3 \ln \left (\frac {\left (x^{4}+1\right )^{\frac {1}{4}}-x}{x}\right ) x -6 \arctan \left (\frac {\left (x^{4}+1\right )^{\frac {1}{4}}}{x}\right ) x -3 \ln \left (\frac {\left (x^{4}+1\right )^{\frac {1}{4}}+x}{x}\right ) x +16 \left (x^{4}+1\right )^{\frac {1}{4}}}{16 x}\) | \(79\) |
trager | \(\frac {\left (x^{4}+1\right )^{\frac {1}{4}} \left (x^{4}+4\right )}{4 x}+\frac {3 \ln \left (-2 \left (x^{4}+1\right )^{\frac {3}{4}} x +2 \sqrt {x^{4}+1}\, x^{2}-2 x^{3} \left (x^{4}+1\right )^{\frac {1}{4}}+2 x^{4}+1\right )}{16}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}+1}\, x^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{4}+2 \left (x^{4}+1\right )^{\frac {3}{4}} x -2 x^{3} \left (x^{4}+1\right )^{\frac {1}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right )}{16}\) | \(130\) |
Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (42) = 84\).
Time = 1.51 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.70 \[ \int \frac {\left (-1+x^4\right ) \sqrt [4]{1+x^4}}{x^2} \, dx=\frac {3 \, x \arctan \left (2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 2 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x\right ) + 3 \, x \log \left (-2 \, x^{4} + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 2 \, \sqrt {x^{4} + 1} x^{2} + 2 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x - 1\right ) + 4 \, {\left (x^{4} + 4\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{16 \, x} \]
1/16*(3*x*arctan(2*(x^4 + 1)^(1/4)*x^3 + 2*(x^4 + 1)^(3/4)*x) + 3*x*log(-2 *x^4 + 2*(x^4 + 1)^(1/4)*x^3 - 2*sqrt(x^4 + 1)*x^2 + 2*(x^4 + 1)^(3/4)*x - 1) + 4*(x^4 + 4)*(x^4 + 1)^(1/4))/x
Result contains complex when optimal does not.
Time = 1.40 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.20 \[ \int \frac {\left (-1+x^4\right ) \sqrt [4]{1+x^4}}{x^2} \, dx=\frac {x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} - \frac {\Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \]
x**3*gamma(3/4)*hyper((-1/4, 3/4), (7/4,), x**4*exp_polar(I*pi))/(4*gamma( 7/4)) - gamma(-1/4)*hyper((-1/4, -1/4), (3/4,), x**4*exp_polar(I*pi))/(4*x *gamma(3/4))
Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.54 \[ \int \frac {\left (-1+x^4\right ) \sqrt [4]{1+x^4}}{x^2} \, dx=\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} + \frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{4 \, x {\left (\frac {x^{4} + 1}{x^{4}} - 1\right )}} - \frac {3}{8} \, \arctan \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {3}{16} \, \log \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} + 1\right ) + \frac {3}{16} \, \log \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} - 1\right ) \]
(x^4 + 1)^(1/4)/x + 1/4*(x^4 + 1)^(1/4)/(x*((x^4 + 1)/x^4 - 1)) - 3/8*arct an((x^4 + 1)^(1/4)/x) - 3/16*log((x^4 + 1)^(1/4)/x + 1) + 3/16*log((x^4 + 1)^(1/4)/x - 1)
Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.30 \[ \int \frac {\left (-1+x^4\right ) \sqrt [4]{1+x^4}}{x^2} \, dx=\frac {1}{4} \, {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + \frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} - \frac {3}{8} \, \arctan \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {3}{16} \, \log \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} + 1\right ) + \frac {3}{16} \, \log \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} - 1\right ) \]
1/4*(x^4 + 1)^(1/4)*x^3 + (x^4 + 1)^(1/4)/x - 3/8*arctan((x^4 + 1)^(1/4)/x ) - 3/16*log((x^4 + 1)^(1/4)/x + 1) + 3/16*log((x^4 + 1)^(1/4)/x - 1)
Timed out. \[ \int \frac {\left (-1+x^4\right ) \sqrt [4]{1+x^4}}{x^2} \, dx=\int \frac {\left (x^4-1\right )\,{\left (x^4+1\right )}^{1/4}}{x^2} \,d x \]