Integrand size = 15, antiderivative size = 70 \[ \int \frac {x^3 \sec ^{-1}(x)}{\sqrt {-1+x^4}} \, dx=-\frac {\sqrt {-1+x^4}}{2 \sqrt {1-\frac {1}{x^2}} x}+\frac {1}{2} \sqrt {-1+x^4} \sec ^{-1}(x)+\frac {1}{2} \text {arctanh}\left (\frac {\sqrt {1-\frac {1}{x^2}} x}{\sqrt {-1+x^4}}\right ) \]
1/2*arctanh(x*(1-1/x^2)^(1/2)/(x^4-1)^(1/2))+1/2*arcsec(x)*(x^4-1)^(1/2)-1 /2*(x^4-1)^(1/2)/x/(1-1/x^2)^(1/2)
Time = 0.06 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.26 \[ \int \frac {x^3 \sec ^{-1}(x)}{\sqrt {-1+x^4}} \, dx=\frac {1}{2} \left (-\frac {\sqrt {1-\frac {1}{x^2}} x \sqrt {-1+x^4}}{-1+x^2}+\sqrt {-1+x^4} \sec ^{-1}(x)-\log \left (x-x^3\right )+\log \left (1-x^2-\sqrt {1-\frac {1}{x^2}} x \sqrt {-1+x^4}\right )\right ) \]
(-((Sqrt[1 - x^(-2)]*x*Sqrt[-1 + x^4])/(-1 + x^2)) + Sqrt[-1 + x^4]*ArcSec [x] - Log[x - x^3] + Log[1 - x^2 - Sqrt[1 - x^(-2)]*x*Sqrt[-1 + x^4]])/2
Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.17, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {5769, 27, 1896, 1396, 243, 60, 73, 217}
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.
\(\displaystyle \int \frac {x^3 \sec ^{-1}(x)}{\sqrt {x^4-1}} \, dx\) |
\(\Big \downarrow \) 5769 |
\(\displaystyle \frac {1}{2} \sqrt {x^4-1} \sec ^{-1}(x)-\int \frac {\sqrt {x^4-1}}{2 \sqrt {1-\frac {1}{x^2}} x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \sqrt {x^4-1} \sec ^{-1}(x)-\frac {1}{2} \int \frac {\sqrt {x^4-1}}{\sqrt {1-\frac {1}{x^2}} x^2}dx\) |
\(\Big \downarrow \) 1896 |
\(\displaystyle \frac {1}{2} \sqrt {x^4-1} \sec ^{-1}(x)-\frac {\sqrt {1-x^2} \int \frac {\sqrt {x^4-1}}{x \sqrt {1-x^2}}dx}{2 \sqrt {1-\frac {1}{x^2}} x}\) |
\(\Big \downarrow \) 1396 |
\(\displaystyle \frac {1}{2} \sqrt {x^4-1} \sec ^{-1}(x)-\frac {\sqrt {x^4-1} \int \frac {\sqrt {-x^2-1}}{x}dx}{2 \sqrt {1-\frac {1}{x^2}} x \sqrt {-x^2-1}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \sqrt {x^4-1} \sec ^{-1}(x)-\frac {\sqrt {x^4-1} \int \frac {\sqrt {-x^2-1}}{x^2}dx^2}{4 \sqrt {1-\frac {1}{x^2}} x \sqrt {-x^2-1}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \sqrt {x^4-1} \sec ^{-1}(x)-\frac {\sqrt {x^4-1} \left (2 \sqrt {-x^2-1}-\int \frac {1}{x^2 \sqrt {-x^2-1}}dx^2\right )}{4 \sqrt {1-\frac {1}{x^2}} x \sqrt {-x^2-1}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \sqrt {x^4-1} \sec ^{-1}(x)-\frac {\sqrt {x^4-1} \left (2 \int \frac {1}{-x^4-1}d\sqrt {-x^2-1}+2 \sqrt {-x^2-1}\right )}{4 \sqrt {1-\frac {1}{x^2}} x \sqrt {-x^2-1}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \sqrt {x^4-1} \sec ^{-1}(x)-\frac {\sqrt {x^4-1} \left (2 \sqrt {-x^2-1}-2 \arctan \left (\sqrt {-x^2-1}\right )\right )}{4 \sqrt {1-\frac {1}{x^2}} x \sqrt {-x^2-1}}\) |
(Sqrt[-1 + x^4]*ArcSec[x])/2 - (Sqrt[-1 + x^4]*(2*Sqrt[-1 - x^2] - 2*ArcTa n[Sqrt[-1 - x^2]]))/(4*Sqrt[1 - x^(-2)]*x*Sqrt[-1 - x^2])
3.1.22.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
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[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_)*((d_) + (e_.)*(x_)^(n_))^(q_.), x _Symbol] :> Simp[(a + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + c*(x^n/e))^FracPart[p]) Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a* e^2, 0] && !IntegerQ[p] && !(EqQ[q, 1] && EqQ[n, 2])
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^(mn_.))^(q_)*((a_) + (c_.)*(x_)^(n2_.))^( p_.), x_Symbol] :> Simp[(e^IntPart[q]*((d + e*x^mn)^FracPart[q]/(1 + d*(1/( x^mn*e)))^FracPart[q]))/x^(mn*FracPart[q]) Int[x^(m + mn*q)*(1 + d*(1/(x^ mn*e)))^q*(a + c*x^n2)^p, x], x] /; FreeQ[{a, c, d, e, m, mn, p, q}, x] && EqQ[n2, -2*mn] && !IntegerQ[p] && !IntegerQ[q] && PosQ[n2]
Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*(u_), x_Symbol] :> With[{v = IntHide [u, x]}, Simp[(a + b*ArcSec[c*x]) v, x] - Simp[b/c Int[SimplifyIntegran d[v/(x^2*Sqrt[1 - 1/(c^2*x^2)]), x], x], x] /; InverseFunctionFreeQ[v, x]] /; FreeQ[{a, b, c}, x]
\[\int \frac {x^{3} \operatorname {arcsec}\left (x \right )}{\sqrt {x^{4}-1}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (54) = 108\).
Time = 0.26 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.57 \[ \int \frac {x^3 \sec ^{-1}(x)}{\sqrt {-1+x^4}} \, dx=\frac {{\left (x^{2} - 1\right )} \log \left (\frac {x^{2} + \sqrt {x^{4} - 1} \sqrt {x^{2} - 1} - 1}{x^{2} - 1}\right ) - {\left (x^{2} - 1\right )} \log \left (-\frac {x^{2} - \sqrt {x^{4} - 1} \sqrt {x^{2} - 1} - 1}{x^{2} - 1}\right ) + 2 \, \sqrt {x^{4} - 1} {\left ({\left (x^{2} - 1\right )} \operatorname {arcsec}\left (x\right ) - \sqrt {x^{2} - 1}\right )}}{4 \, {\left (x^{2} - 1\right )}} \]
1/4*((x^2 - 1)*log((x^2 + sqrt(x^4 - 1)*sqrt(x^2 - 1) - 1)/(x^2 - 1)) - (x ^2 - 1)*log(-(x^2 - sqrt(x^4 - 1)*sqrt(x^2 - 1) - 1)/(x^2 - 1)) + 2*sqrt(x ^4 - 1)*((x^2 - 1)*arcsec(x) - sqrt(x^2 - 1)))/(x^2 - 1)
\[ \int \frac {x^3 \sec ^{-1}(x)}{\sqrt {-1+x^4}} \, dx=\int \frac {x^{3} \operatorname {asec}{\left (x \right )}}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}}\, dx \]
\[ \int \frac {x^3 \sec ^{-1}(x)}{\sqrt {-1+x^4}} \, dx=\int { \frac {x^{3} \operatorname {arcsec}\left (x\right )}{\sqrt {x^{4} - 1}} \,d x } \]
1/2*sqrt(x^2 + 1)*sqrt(x + 1)*sqrt(x - 1)*arctan(sqrt(x + 1)*sqrt(x - 1)) - integrate((2*(x^3*e^(3/2*log(x + 1) + 3/2*log(x - 1)) + x^3*e^(1/2*log(x + 1) + 1/2*log(x - 1)))*sqrt(x^2 + 1)*log(x) + (x^3 + x)*e^(1/2*log(x^2 + 1) + 3/2*log(x + 1) + 3/2*log(x - 1)))/((x^2 + 1)*(e^(2*log(x + 1) + 2*lo g(x - 1)) + e^(log(x + 1) + log(x - 1)))), x)
Time = 0.30 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.74 \[ \int \frac {x^3 \sec ^{-1}(x)}{\sqrt {-1+x^4}} \, dx=\frac {1}{2} \, \sqrt {x^{4} - 1} \arccos \left (\frac {1}{x}\right ) - \frac {2 \, \sqrt {x^{2} + 1} - \log \left (\sqrt {x^{2} + 1} + 1\right ) + \log \left (\sqrt {x^{2} + 1} - 1\right )}{4 \, \mathrm {sgn}\left (x\right )} \]
1/2*sqrt(x^4 - 1)*arccos(1/x) - 1/4*(2*sqrt(x^2 + 1) - log(sqrt(x^2 + 1) + 1) + log(sqrt(x^2 + 1) - 1))/sgn(x)
Timed out. \[ \int \frac {x^3 \sec ^{-1}(x)}{\sqrt {-1+x^4}} \, dx=\int \frac {x^3\,\mathrm {acos}\left (\frac {1}{x}\right )}{\sqrt {x^4-1}} \,d x \]
\[ \int \frac {x^3 \sec ^{-1}(x)}{\sqrt {-1+x^4}} \, dx=\int \frac {\sqrt {x^{4}-1}\, \mathit {asec} \left (x \right ) x^{3}}{x^{4}-1}d x \]