Integrand size = 18, antiderivative size = 60 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{x^3} \, dx=\frac {\left (1+x^2\right ) \sqrt {1+x^4}}{2 x^2}+\text {arctanh}\left (1-2 x^2-2 \sqrt {1+x^4}\right )-\frac {1}{2} \log \left (1+x^2+\sqrt {1+x^4}\right ) \]
Time = 0.10 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{x^3} \, dx=\frac {\left (1+x^2\right ) \sqrt {1+x^4}}{2 x^2}+\text {arctanh}\left (x^2-\sqrt {1+x^4}\right )+\frac {1}{2} \log \left (-x^2+\sqrt {1+x^4}\right ) \]
((1 + x^2)*Sqrt[1 + x^4])/(2*x^2) + ArcTanh[x^2 - Sqrt[1 + x^4]] + Log[-x^ 2 + Sqrt[1 + x^4]]/2
Time = 0.20 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.68, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {1579, 25, 536, 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.
\(\displaystyle \int \frac {\left (x^2-1\right ) \sqrt {x^4+1}}{x^3} \, dx\) |
\(\Big \downarrow \) 1579 |
\(\displaystyle \frac {1}{2} \int -\frac {\left (1-x^2\right ) \sqrt {x^4+1}}{x^4}dx^2\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {\left (1-x^2\right ) \sqrt {x^4+1}}{x^4}dx^2\) |
\(\Big \downarrow \) 536 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (x^2+1\right ) \sqrt {x^4+1}}{x^2}-\int \frac {x^2-1}{x^2 \sqrt {x^4+1}}dx^2\right )\) |
\(\Big \downarrow \) 538 |
\(\displaystyle \frac {1}{2} \left (-\int \frac {1}{\sqrt {x^4+1}}dx^2+\int \frac {1}{x^2 \sqrt {x^4+1}}dx^2+\frac {\sqrt {x^4+1} \left (x^2+1\right )}{x^2}\right )\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{2} \left (\int \frac {1}{x^2 \sqrt {x^4+1}}dx^2-\text {arcsinh}\left (x^2\right )+\frac {\sqrt {x^4+1} \left (x^2+1\right )}{x^2}\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2 \sqrt {x^4+1}}dx^4-\text {arcsinh}\left (x^2\right )+\frac {\sqrt {x^4+1} \left (x^2+1\right )}{x^2}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\int \frac {1}{\sqrt {x^4+1}-1}d\sqrt {x^4+1}-\text {arcsinh}\left (x^2\right )+\frac {\sqrt {x^4+1} \left (x^2+1\right )}{x^2}\right )\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {1}{2} \left (-\text {arcsinh}\left (x^2\right )-\text {arctanh}\left (\sqrt {x^4+1}\right )+\frac {\sqrt {x^4+1} \left (x^2+1\right )}{x^2}\right )\) |
3.8.83.3.1 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[(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_))/(x_)^2, x_Symbol] :> S imp[(-(2*c*p - d*x))*((a + b*x^2)^p/(2*p*x)), x] + Int[(a*d + 2*b*c*p*x)*(( a + b*x^2)^(p - 1)/x), x] /; FreeQ[{a, b, c, d}, x] && GtQ[p, 0] && Integer Q[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 = 0.90 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.63
method | result | size |
trager | \(\frac {\left (x^{2}+1\right ) \sqrt {x^{4}+1}}{2 x^{2}}-\ln \left (\frac {1+x^{2}+\sqrt {x^{4}+1}}{x}\right )\) | \(38\) |
risch | \(\frac {\sqrt {x^{4}+1}}{2 x^{2}}+\frac {\sqrt {x^{4}+1}}{2}-\frac {\operatorname {arcsinh}\left (x^{2}\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{4}+1}}\right )}{2}\) | \(39\) |
elliptic | \(\frac {\sqrt {x^{4}+1}}{2 x^{2}}+\frac {\sqrt {x^{4}+1}}{2}-\frac {\operatorname {arcsinh}\left (x^{2}\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{4}+1}}\right )}{2}\) | \(39\) |
pseudoelliptic | \(\frac {\sqrt {x^{4}+1}\, x^{2}-\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{4}+1}}\right ) x^{2}-x^{2} \operatorname {arcsinh}\left (x^{2}\right )+\sqrt {x^{4}+1}}{2 x^{2}}\) | \(47\) |
default | \(\frac {\sqrt {x^{4}+1}}{2}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{4}+1}}\right )}{2}+\frac {\left (x^{4}+1\right )^{\frac {3}{2}}}{2 x^{2}}-\frac {\sqrt {x^{4}+1}\, x^{2}}{2}-\frac {\operatorname {arcsinh}\left (x^{2}\right )}{2}\) | \(51\) |
meijerg | \(\frac {\frac {4 \sqrt {\pi }\, \sqrt {x^{4}+1}}{x^{2}}-4 \sqrt {\pi }\, \operatorname {arcsinh}\left (x^{2}\right )}{8 \sqrt {\pi }}-\frac {4 \sqrt {\pi }-4 \sqrt {\pi }\, \sqrt {x^{4}+1}+4 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {x^{4}+1}}{2}\right )-2 \left (2-2 \ln \left (2\right )+4 \ln \left (x \right )\right ) \sqrt {\pi }}{8 \sqrt {\pi }}\) | \(87\) |
Time = 0.25 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.22 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{x^3} \, dx=\frac {x^{2} \log \left (2 \, x^{4} + x^{2} - \sqrt {x^{4} + 1} {\left (2 \, x^{2} + 1\right )} + 1\right ) - x^{2} \log \left (-x^{2} + \sqrt {x^{4} + 1} + 1\right ) + x^{2} + \sqrt {x^{4} + 1} {\left (x^{2} + 1\right )}}{2 \, x^{2}} \]
1/2*(x^2*log(2*x^4 + x^2 - sqrt(x^4 + 1)*(2*x^2 + 1) + 1) - x^2*log(-x^2 + sqrt(x^4 + 1) + 1) + x^2 + sqrt(x^4 + 1)*(x^2 + 1))/x^2
Time = 3.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.25 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{x^3} \, dx=\frac {x^{2}}{2 \sqrt {x^{4} + 1}} + \frac {x^{2}}{2 \sqrt {1 + \frac {1}{x^{4}}}} - \frac {\operatorname {asinh}{\left (\frac {1}{x^{2}} \right )}}{2} - \frac {\operatorname {asinh}{\left (x^{2} \right )}}{2} + \frac {1}{2 x^{2} \sqrt {x^{4} + 1}} + \frac {1}{2 x^{2} \sqrt {1 + \frac {1}{x^{4}}}} \]
x**2/(2*sqrt(x**4 + 1)) + x**2/(2*sqrt(1 + x**(-4))) - asinh(x**(-2))/2 - asinh(x**2)/2 + 1/(2*x**2*sqrt(x**4 + 1)) + 1/(2*x**2*sqrt(1 + x**(-4)))
Time = 0.28 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.30 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{x^3} \, dx=\frac {1}{2} \, \sqrt {x^{4} + 1} + \frac {\sqrt {x^{4} + 1}}{2 \, x^{2}} - \frac {1}{4} \, \log \left (\sqrt {x^{4} + 1} + 1\right ) + \frac {1}{4} \, \log \left (\sqrt {x^{4} + 1} - 1\right ) - \frac {1}{4} \, \log \left (\frac {\sqrt {x^{4} + 1}}{x^{2}} + 1\right ) + \frac {1}{4} \, \log \left (\frac {\sqrt {x^{4} + 1}}{x^{2}} - 1\right ) \]
1/2*sqrt(x^4 + 1) + 1/2*sqrt(x^4 + 1)/x^2 - 1/4*log(sqrt(x^4 + 1) + 1) + 1 /4*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)
Time = 0.27 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.35 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{x^3} \, dx=\frac {1}{2} \, \sqrt {x^{4} + 1} - \frac {1}{{\left (x^{2} - \sqrt {x^{4} + 1}\right )}^{2} - 1} + \frac {1}{2} \, \log \left (x^{2} - \sqrt {x^{4} + 1} + 1\right ) - \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 1} + 1\right ) + \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 1}\right ) \]
1/2*sqrt(x^4 + 1) - 1/((x^2 - sqrt(x^4 + 1))^2 - 1) + 1/2*log(x^2 - sqrt(x ^4 + 1) + 1) - 1/2*log(-x^2 + sqrt(x^4 + 1) + 1) + 1/2*log(-x^2 + sqrt(x^4 + 1))
Time = 6.43 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.70 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{x^3} \, dx=\frac {\sqrt {x^4+1}}{2}-\frac {\mathrm {asinh}\left (x^2\right )}{2}+\frac {\sqrt {x^4+1}}{2\,x^2}+\frac {\mathrm {atan}\left (\sqrt {x^4+1}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} \]