Integrand size = 18, antiderivative size = 60 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{x^5} \, dx=\frac {\left (1-2 x^2\right ) \sqrt {1+x^4}}{4 x^4}+\frac {1}{2} \text {arctanh}\left (x^2+\sqrt {1+x^4}\right )+\frac {1}{2} \log \left (x^2+\sqrt {1+x^4}\right ) \]
Time = 0.17 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{x^5} \, dx=\frac {1}{4} \left (\frac {\left (1-2 x^2\right ) \sqrt {1+x^4}}{x^4}-2 \text {arctanh}\left (x^2-\sqrt {1+x^4}\right )-2 \log \left (-x^2+\sqrt {1+x^4}\right )\right ) \]
(((1 - 2*x^2)*Sqrt[1 + x^4])/x^4 - 2*ArcTanh[x^2 - Sqrt[1 + x^4]] - 2*Log[ -x^2 + Sqrt[1 + x^4]])/4
Time = 0.22 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.82, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1579, 25, 537, 25, 538, 222, 243, 73, 220}
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.
\(\displaystyle \int \frac {\left (x^2-1\right ) \sqrt {x^4+1}}{x^5} \, dx\) |
\(\Big \downarrow \) 1579 |
\(\displaystyle \frac {1}{2} \int -\frac {\left (1-x^2\right ) \sqrt {x^4+1}}{x^6}dx^2\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {\left (1-x^2\right ) \sqrt {x^4+1}}{x^6}dx^2\) |
\(\Big \downarrow \) 537 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int -\frac {1-2 x^2}{x^2 \sqrt {x^4+1}}dx^2+\frac {\sqrt {x^4+1} \left (1-2 x^2\right )}{2 x^4}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (1-2 x^2\right ) \sqrt {x^4+1}}{2 x^4}-\frac {1}{2} \int \frac {1-2 x^2}{x^2 \sqrt {x^4+1}}dx^2\right )\) |
\(\Big \downarrow \) 538 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (2 \int \frac {1}{\sqrt {x^4+1}}dx^2-\int \frac {1}{x^2 \sqrt {x^4+1}}dx^2\right )+\frac {\sqrt {x^4+1} \left (1-2 x^2\right )}{2 x^4}\right )\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (2 \text {arcsinh}\left (x^2\right )-\int \frac {1}{x^2 \sqrt {x^4+1}}dx^2\right )+\frac {\sqrt {x^4+1} \left (1-2 x^2\right )}{2 x^4}\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (2 \text {arcsinh}\left (x^2\right )-\frac {1}{2} \int \frac {1}{x^2 \sqrt {x^4+1}}dx^4\right )+\frac {\sqrt {x^4+1} \left (1-2 x^2\right )}{2 x^4}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (2 \text {arcsinh}\left (x^2\right )-\int \frac {1}{\sqrt {x^4+1}-1}d\sqrt {x^4+1}\right )+\frac {\sqrt {x^4+1} \left (1-2 x^2\right )}{2 x^4}\right )\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (2 \text {arcsinh}\left (x^2\right )+\text {arctanh}\left (\sqrt {x^4+1}\right )\right )+\frac {\sqrt {x^4+1} \left (1-2 x^2\right )}{2 x^4}\right )\) |
3.8.82.3.1 Defintions of rubi rules used
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_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*(c*(m + 2) + d*(m + 1)*x)*((a + b*x^2)^p/((m + 1)*(m + 2))), x] - Simp[2*b*(p/((m + 1)*(m + 2))) Int[x^(m + 2)*(c*(m + 2) + d*(m + 1) *x)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, -2] && GtQ[p, 0] && !ILtQ[m + 2*p + 3, 0] && IntegerQ[2*p]
Int[((c_) + (d_.)*(x_))/((x_)*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Simp [c Int[1/(x*Sqrt[a + b*x^2]), x], x] + Simp[d Int[1/Sqrt[a + b*x^2], x] , x] /; FreeQ[{a, b, c, d}, x]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]
Time = 0.94 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.70
method | result | size |
elliptic | \(\frac {\operatorname {arcsinh}\left (x^{2}\right )}{2}-\frac {\sqrt {x^{4}+1}}{2 x^{2}}+\frac {\sqrt {x^{4}+1}}{4 x^{4}}+\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{4}+1}}\right )}{4}\) | \(42\) |
risch | \(-\frac {2 x^{6}-x^{4}+2 x^{2}-1}{4 x^{4} \sqrt {x^{4}+1}}+\frac {\operatorname {arcsinh}\left (x^{2}\right )}{2}+\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{4}+1}}\right )}{4}\) | \(47\) |
pseudoelliptic | \(\frac {2 \,\operatorname {arcsinh}\left (x^{2}\right ) x^{4}+\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{4}+1}}\right ) x^{4}-2 \sqrt {x^{4}+1}\, x^{2}+\sqrt {x^{4}+1}}{4 x^{4}}\) | \(47\) |
default | \(-\frac {\left (x^{4}+1\right )^{\frac {3}{2}}}{2 x^{2}}+\frac {\sqrt {x^{4}+1}\, x^{2}}{2}+\frac {\operatorname {arcsinh}\left (x^{2}\right )}{2}+\frac {\left (x^{4}+1\right )^{\frac {3}{2}}}{4 x^{4}}-\frac {\sqrt {x^{4}+1}}{4}+\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{4}+1}}\right )}{4}\) | \(63\) |
trager | \(-\frac {\left (2 x^{2}-1\right ) \sqrt {x^{4}+1}}{4 x^{4}}+\frac {\ln \left (-\frac {2 x^{6}+2 x^{4} \sqrt {x^{4}+1}+2 x^{4}+2 \sqrt {x^{4}+1}\, x^{2}+2 x^{2}+\sqrt {x^{4}+1}+1}{x^{2}}\right )}{4}\) | \(77\) |
meijerg | \(\frac {-\frac {\sqrt {\pi }\, \left (4 x^{4}+8\right )}{4 x^{4}}+\frac {2 \sqrt {\pi }\, \sqrt {x^{4}+1}}{x^{4}}+2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {x^{4}+1}}{2}\right )-\left (-2 \ln \left (2\right )-1+4 \ln \left (x \right )\right ) \sqrt {\pi }+\frac {2 \sqrt {\pi }}{x^{4}}}{8 \sqrt {\pi }}-\frac {\frac {4 \sqrt {\pi }\, \sqrt {x^{4}+1}}{x^{2}}-4 \sqrt {\pi }\, \operatorname {arcsinh}\left (x^{2}\right )}{8 \sqrt {\pi }}\) | \(108\) |
Time = 0.26 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.42 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{x^5} \, dx=\frac {x^{4} \log \left (-x^{2} + \sqrt {x^{4} + 1} + 1\right ) - 2 \, x^{4} \log \left (-x^{2} + \sqrt {x^{4} + 1}\right ) - x^{4} \log \left (-x^{2} + \sqrt {x^{4} + 1} - 1\right ) - 2 \, x^{4} - \sqrt {x^{4} + 1} {\left (2 \, x^{2} - 1\right )}}{4 \, x^{4}} \]
1/4*(x^4*log(-x^2 + sqrt(x^4 + 1) + 1) - 2*x^4*log(-x^2 + sqrt(x^4 + 1)) - x^4*log(-x^2 + sqrt(x^4 + 1) - 1) - 2*x^4 - sqrt(x^4 + 1)*(2*x^2 - 1))/x^ 4
Time = 2.76 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{x^5} \, dx=- \frac {x^{2}}{2 \sqrt {x^{4} + 1}} + \frac {\operatorname {asinh}{\left (\frac {1}{x^{2}} \right )}}{4} + \frac {\operatorname {asinh}{\left (x^{2} \right )}}{2} + \frac {\sqrt {1 + \frac {1}{x^{4}}}}{4 x^{2}} - \frac {1}{2 x^{2} \sqrt {x^{4} + 1}} \]
-x**2/(2*sqrt(x**4 + 1)) + asinh(x**(-2))/4 + asinh(x**2)/2 + sqrt(1 + x** (-4))/(4*x**2) - 1/(2*x**2*sqrt(x**4 + 1))
Time = 0.27 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.35 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{x^5} \, dx=-\frac {\sqrt {x^{4} + 1}}{2 \, x^{2}} + \frac {\sqrt {x^{4} + 1}}{4 \, x^{4}} + \frac {1}{8} \, \log \left (\sqrt {x^{4} + 1} + 1\right ) - \frac {1}{8} \, \log \left (\sqrt {x^{4} + 1} - 1\right ) + \frac {1}{4} \, \log \left (\frac {\sqrt {x^{4} + 1}}{x^{2}} + 1\right ) - \frac {1}{4} \, \log \left (\frac {\sqrt {x^{4} + 1}}{x^{2}} - 1\right ) \]
-1/2*sqrt(x^4 + 1)/x^2 + 1/4*sqrt(x^4 + 1)/x^4 + 1/8*log(sqrt(x^4 + 1) + 1 ) - 1/8*log(sqrt(x^4 + 1) - 1) + 1/4*log(sqrt(x^4 + 1)/x^2 + 1) - 1/4*log( sqrt(x^4 + 1)/x^2 - 1)
Leaf count of result is larger than twice the leaf count of optimal. 118 vs. \(2 (48) = 96\).
Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.97 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{x^5} \, dx=-\frac {{\left (x^{2} - \sqrt {x^{4} + 1}\right )}^{3} - 2 \, {\left (x^{2} - \sqrt {x^{4} + 1}\right )}^{2} + x^{2} - \sqrt {x^{4} + 1} + 2}{2 \, {\left ({\left (x^{2} - \sqrt {x^{4} + 1}\right )}^{2} - 1\right )}^{2}} - \frac {1}{4} \, \log \left (x^{2} - \sqrt {x^{4} + 1} + 1\right ) + \frac {1}{4} \, \log \left (-x^{2} + \sqrt {x^{4} + 1} + 1\right ) - \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 1}\right ) \]
-1/2*((x^2 - sqrt(x^4 + 1))^3 - 2*(x^2 - sqrt(x^4 + 1))^2 + x^2 - sqrt(x^4 + 1) + 2)/((x^2 - sqrt(x^4 + 1))^2 - 1)^2 - 1/4*log(x^2 - sqrt(x^4 + 1) + 1) + 1/4*log(-x^2 + sqrt(x^4 + 1) + 1) - 1/2*log(-x^2 + sqrt(x^4 + 1))
Time = 6.29 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.75 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{x^5} \, dx=\frac {\mathrm {asinh}\left (x^2\right )}{2}-\frac {\sqrt {x^4+1}}{2\,x^2}+\frac {\sqrt {x^4+1}}{4\,x^4}-\frac {\mathrm {atan}\left (\sqrt {x^4+1}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4} \]