\(\int \frac {x^3 \sec ^{-1}(x)}{\sqrt {-1+x^4}} \, dx\) [22]

Optimal result

Integrand size = 15, antiderivative size = 70 \[ \int \frac {x^3 \sec ^{-1}(x)}{\sqrt {-1+x^4}} \, dx=-\frac {\sqrt {-1+x^4}}{2 \sqrt {1-\frac {1}{x^2}} x}+\frac {1}{2} \sqrt {-1+x^4} \sec ^{-1}(x)+\frac {1}{2} \text {arctanh}\left (\frac {\sqrt {1-\frac {1}{x^2}} x}{\sqrt {-1+x^4}}\right ) \] Output:

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)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.26 \[ \int \frac {x^3 \sec ^{-1}(x)}{\sqrt {-1+x^4}} \, dx=\frac {1}{2} \left (-\frac {\sqrt {1-\frac {1}{x^2}} x \sqrt {-1+x^4}}{-1+x^2}+\sqrt {-1+x^4} \sec ^{-1}(x)-\log \left (x-x^3\right )+\log \left (1-x^2-\sqrt {1-\frac {1}{x^2}} x \sqrt {-1+x^4}\right )\right ) \] Input:

Integrate[(x^3*ArcSec[x])/Sqrt[-1 + x^4],x]
 

Output:

(-((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
 

Rubi [A] (verified)

Time = 0.29 (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

\(\displaystyle \frac {1}{2} \sqrt {x^4-1} \sec ^{-1}(x)-\frac {\sqrt {x^4-1} \left (2 \sqrt {-x^2-1}-2 \arctan \left (\sqrt {-x^2-1}\right )\right )}{4 \sqrt {1-\frac {1}{x^2}} x \sqrt {-x^2-1}}\)

Input:

Int[(x^3*ArcSec[x])/Sqrt[-1 + x^4],x]
 

Output:

(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])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 60

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]
 

rule 73

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]
 

rule 217

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])
 

rule 243

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]
 

rule 1396

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])
 

rule 1896

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]
 

rule 5769

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]
 

Maple [F]

\[\int \frac {x^{3} \operatorname {arcsec}\left (x \right )}{\sqrt {x^{4}-1}}d x\]

Input:

int(x^3*arcsec(x)/(x^4-1)^(1/2),x)
 

Output:

int(x^3*arcsec(x)/(x^4-1)^(1/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (54) = 108\).

Time = 0.09 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.57 \[ \int \frac {x^3 \sec ^{-1}(x)}{\sqrt {-1+x^4}} \, dx=\frac {{\left (x^{2} - 1\right )} \log \left (\frac {x^{2} + \sqrt {x^{4} - 1} \sqrt {x^{2} - 1} - 1}{x^{2} - 1}\right ) - {\left (x^{2} - 1\right )} \log \left (-\frac {x^{2} - \sqrt {x^{4} - 1} \sqrt {x^{2} - 1} - 1}{x^{2} - 1}\right ) + 2 \, \sqrt {x^{4} - 1} {\left ({\left (x^{2} - 1\right )} \operatorname {arcsec}\left (x\right ) - \sqrt {x^{2} - 1}\right )}}{4 \, {\left (x^{2} - 1\right )}} \] Input:

integrate(x^3*arcsec(x)/(x^4-1)^(1/2),x, algorithm="fricas")
 

Output:

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)
 

Sympy [F]

\[ \int \frac {x^3 \sec ^{-1}(x)}{\sqrt {-1+x^4}} \, dx=\int \frac {x^{3} \operatorname {asec}{\left (x \right )}}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}}\, dx \] Input:

integrate(x**3*asec(x)/(x**4-1)**(1/2),x)
 

Output:

Integral(x**3*asec(x)/sqrt((x - 1)*(x + 1)*(x**2 + 1)), x)
 

Maxima [F]

\[ \int \frac {x^3 \sec ^{-1}(x)}{\sqrt {-1+x^4}} \, dx=\int { \frac {x^{3} \operatorname {arcsec}\left (x\right )}{\sqrt {x^{4} - 1}} \,d x } \] Input:

integrate(x^3*arcsec(x)/(x^4-1)^(1/2),x, algorithm="maxima")
 

Output:

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)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.74 \[ \int \frac {x^3 \sec ^{-1}(x)}{\sqrt {-1+x^4}} \, dx=\frac {1}{2} \, \sqrt {x^{4} - 1} \arccos \left (\frac {1}{x}\right ) - \frac {2 \, \sqrt {x^{2} + 1} - \log \left (\sqrt {x^{2} + 1} + 1\right ) + \log \left (\sqrt {x^{2} + 1} - 1\right )}{4 \, \mathrm {sgn}\left (x\right )} \] Input:

integrate(x^3*arcsec(x)/(x^4-1)^(1/2),x, algorithm="giac")
 

Output:

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)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \sec ^{-1}(x)}{\sqrt {-1+x^4}} \, dx=\int \frac {x^3\,\mathrm {acos}\left (\frac {1}{x}\right )}{\sqrt {x^4-1}} \,d x \] Input:

int((x^3*acos(1/x))/(x^4 - 1)^(1/2),x)
 

Output:

int((x^3*acos(1/x))/(x^4 - 1)^(1/2), x)
 

Reduce [F]

\[ \int \frac {x^3 \sec ^{-1}(x)}{\sqrt {-1+x^4}} \, dx=\int \frac {\sqrt {x^{4}-1}\, \mathit {asec} \left (x \right ) x^{3}}{x^{4}-1}d x \] Input:

int(x^3*asec(x)/(x^4-1)^(1/2),x)
 

Output:

int((sqrt(x**4 - 1)*asec(x)*x**3)/(x**4 - 1),x)