3.7.89 \(\int \frac {x^6 \sec ^{-1}(x)}{(-1+x^2)^{5/2}} \, dx\) [689]

Optimal. Leaf size=175 \[ \frac {\sqrt {x^2} \left (2-3 x^2\right )}{6 \left (-1+x^2\right )}-\frac {13}{6} \coth ^{-1}\left (\sqrt {x^2}\right )-\frac {5 x^3 \sec ^{-1}(x)}{6 \left (-1+x^2\right )^{3/2}}+\frac {x^5 \sec ^{-1}(x)}{2 \left (-1+x^2\right )^{3/2}}-\frac {5 x \sec ^{-1}(x)}{2 \sqrt {-1+x^2}}-\frac {5 i \sqrt {x^2} \sec ^{-1}(x) \tan ^{-1}\left (e^{i \sec ^{-1}(x)}\right )}{x}+\frac {5 i \sqrt {x^2} \text {Li}_2\left (-i e^{i \sec ^{-1}(x)}\right )}{2 x}-\frac {5 i \sqrt {x^2} \text {Li}_2\left (i e^{i \sec ^{-1}(x)}\right )}{2 x} \]

[Out]

-13/6*arccoth((x^2)^(1/2))-5/6*x^3*arcsec(x)/(x^2-1)^(3/2)+1/2*x^5*arcsec(x)/(x^2-1)^(3/2)+1/6*(-3*x^2+2)*(x^2
)^(1/2)/(x^2-1)-5*I*arcsec(x)*arctan(1/x+I*(1-1/x^2)^(1/2))*(x^2)^(1/2)/x+5/2*I*polylog(2,-I*(1/x+I*(1-1/x^2)^
(1/2)))*(x^2)^(1/2)/x-5/2*I*polylog(2,I*(1/x+I*(1-1/x^2)^(1/2)))*(x^2)^(1/2)/x-5/2*x*arcsec(x)/(x^2-1)^(1/2)

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Rubi [A]
time = 0.22, antiderivative size = 232, normalized size of antiderivative = 1.33, number of steps used = 16, number of rules used = 11, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.733, Rules used = {5350, 4790, 4794, 4804, 4266, 2317, 2438, 212, 205, 296, 331} \begin {gather*} \frac {5 i \sqrt {x^2} \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(x)}\right )}{2 x}-\frac {5 i \sqrt {x^2} \text {PolyLog}\left (2,i e^{i \sec ^{-1}(x)}\right )}{2 x}-\frac {5 i \sqrt {x^2} \sec ^{-1}(x) \text {ArcTan}\left (e^{i \sec ^{-1}(x)}\right )}{x}+\frac {\sqrt {x^2}}{4 \left (1-\frac {1}{x^2}\right )}-\frac {3 \sqrt {x^2}}{4}-\frac {5}{12 \left (1-\frac {1}{x^2}\right ) \sqrt {x^2}}+\frac {x \sqrt {x^2} \sec ^{-1}(x)}{2 \left (1-\frac {1}{x^2}\right )^{3/2}}-\frac {5 \sqrt {x^2} \sec ^{-1}(x)}{2 \sqrt {1-\frac {1}{x^2}} x}-\frac {5 \sqrt {x^2} \sec ^{-1}(x)}{6 \left (1-\frac {1}{x^2}\right )^{3/2} x}-\frac {13 \sqrt {x^2} \coth ^{-1}(x)}{6 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^6*ArcSec[x])/(-1 + x^2)^(5/2),x]

[Out]

-5/(12*(1 - x^(-2))*Sqrt[x^2]) - (3*Sqrt[x^2])/4 + Sqrt[x^2]/(4*(1 - x^(-2))) - (13*Sqrt[x^2]*ArcCoth[x])/(6*x
) - (5*Sqrt[x^2]*ArcSec[x])/(6*(1 - x^(-2))^(3/2)*x) - (5*Sqrt[x^2]*ArcSec[x])/(2*Sqrt[1 - x^(-2)]*x) + (x*Sqr
t[x^2]*ArcSec[x])/(2*(1 - x^(-2))^(3/2)) - ((5*I)*Sqrt[x^2]*ArcSec[x]*ArcTan[E^(I*ArcSec[x])])/x + (((5*I)/2)*
Sqrt[x^2]*PolyLog[2, (-I)*E^(I*ArcSec[x])])/x - (((5*I)/2)*Sqrt[x^2]*PolyLog[2, I*E^(I*ArcSec[x])])/x

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4266

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))]/f), x] + (-Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 4790

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
f*x)^(m + 1)*(d + e*x^2)^(p + 1)*((a + b*ArcCos[c*x])^n/(d*f*(m + 1))), x] + (Dist[c^2*((m + 2*p + 3)/(f^2*(m
+ 1))), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcCos[c*x])^n, x], x] + Dist[b*c*(n/(f*(m + 1)))*Simp[(d + e*x
^2)^p/(1 - c^2*x^2)^p], Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcCos[c*x])^(n - 1), x], x]) /; Free
Q[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && ILtQ[m, -1]

Rule 4794

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
-(f*x)^(m + 1))*(d + e*x^2)^(p + 1)*((a + b*ArcCos[c*x])^n/(2*d*f*(p + 1))), x] + (Dist[(m + 2*p + 3)/(2*d*(p
+ 1)), Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcCos[c*x])^n, x], x] - Dist[b*c*(n/(2*f*(p + 1)))*Simp[(d + e*
x^2)^p/(1 - c^2*x^2)^p], Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcCos[c*x])^(n - 1), x], x]) /; Fre
eQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ[m, 1] && (IntegerQ[m] ||
IntegerQ[p] || EqQ[n, 1])

Rule 4804

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[(-(c^(m
+ 1))^(-1))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]], Subst[Int[(a + b*x)^n*Cos[x]^m, x], x, ArcCos[c*x]], x] /
; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 5350

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Dist[-Sqrt[x
^2]/x, Subst[Int[(e + d*x^2)^p*((a + b*ArcCos[x/c])^n/x^(m + 2*(p + 1))), x], x, 1/x], x] /; FreeQ[{a, b, c, d
, e, n}, x] && IGtQ[n, 0] && EqQ[c^2*d + e, 0] && IntegerQ[m] && IntegerQ[p + 1/2] && GtQ[e, 0] && LtQ[d, 0]

Rubi steps

\begin {align*} \int \frac {x^6 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx &=-\frac {\sqrt {x^2} \text {Subst}\left (\int \frac {\cos ^{-1}(x)}{x^3 \left (1-x^2\right )^{5/2}} \, dx,x,\frac {1}{x}\right )}{x}\\ &=\frac {x \sqrt {x^2} \sec ^{-1}(x)}{2 \left (1-\frac {1}{x^2}\right )^{3/2}}+\frac {\sqrt {x^2} \text {Subst}\left (\int \frac {1}{x^2 \left (1-x^2\right )^2} \, dx,x,\frac {1}{x}\right )}{2 x}-\frac {\left (5 \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {\cos ^{-1}(x)}{x \left (1-x^2\right )^{5/2}} \, dx,x,\frac {1}{x}\right )}{2 x}\\ &=\frac {\sqrt {x^2}}{4 \left (1-\frac {1}{x^2}\right )}-\frac {5 \sqrt {x^2} \sec ^{-1}(x)}{6 \left (1-\frac {1}{x^2}\right )^{3/2} x}+\frac {x \sqrt {x^2} \sec ^{-1}(x)}{2 \left (1-\frac {1}{x^2}\right )^{3/2}}+\frac {\left (3 \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1}{x^2 \left (1-x^2\right )} \, dx,x,\frac {1}{x}\right )}{4 x}-\frac {\left (5 \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^2} \, dx,x,\frac {1}{x}\right )}{6 x}-\frac {\left (5 \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {\cos ^{-1}(x)}{x \left (1-x^2\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{2 x}\\ &=-\frac {5}{12 \left (1-\frac {1}{x^2}\right ) \sqrt {x^2}}-\frac {3 \sqrt {x^2}}{4}+\frac {\sqrt {x^2}}{4 \left (1-\frac {1}{x^2}\right )}-\frac {5 \sqrt {x^2} \sec ^{-1}(x)}{6 \left (1-\frac {1}{x^2}\right )^{3/2} x}-\frac {5 \sqrt {x^2} \sec ^{-1}(x)}{2 \sqrt {1-\frac {1}{x^2}} x}+\frac {x \sqrt {x^2} \sec ^{-1}(x)}{2 \left (1-\frac {1}{x^2}\right )^{3/2}}-\frac {\left (5 \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {1}{x}\right )}{12 x}+\frac {\left (3 \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {1}{x}\right )}{4 x}-\frac {\left (5 \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {1}{x}\right )}{2 x}-\frac {\left (5 \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {\cos ^{-1}(x)}{x \sqrt {1-x^2}} \, dx,x,\frac {1}{x}\right )}{2 x}\\ &=-\frac {5}{12 \left (1-\frac {1}{x^2}\right ) \sqrt {x^2}}-\frac {3 \sqrt {x^2}}{4}+\frac {\sqrt {x^2}}{4 \left (1-\frac {1}{x^2}\right )}-\frac {13 \sqrt {x^2} \coth ^{-1}(x)}{6 x}-\frac {5 \sqrt {x^2} \sec ^{-1}(x)}{6 \left (1-\frac {1}{x^2}\right )^{3/2} x}-\frac {5 \sqrt {x^2} \sec ^{-1}(x)}{2 \sqrt {1-\frac {1}{x^2}} x}+\frac {x \sqrt {x^2} \sec ^{-1}(x)}{2 \left (1-\frac {1}{x^2}\right )^{3/2}}+\frac {\left (5 \sqrt {x^2}\right ) \text {Subst}\left (\int x \sec (x) \, dx,x,\sec ^{-1}(x)\right )}{2 x}\\ &=-\frac {5}{12 \left (1-\frac {1}{x^2}\right ) \sqrt {x^2}}-\frac {3 \sqrt {x^2}}{4}+\frac {\sqrt {x^2}}{4 \left (1-\frac {1}{x^2}\right )}-\frac {13 \sqrt {x^2} \coth ^{-1}(x)}{6 x}-\frac {5 \sqrt {x^2} \sec ^{-1}(x)}{6 \left (1-\frac {1}{x^2}\right )^{3/2} x}-\frac {5 \sqrt {x^2} \sec ^{-1}(x)}{2 \sqrt {1-\frac {1}{x^2}} x}+\frac {x \sqrt {x^2} \sec ^{-1}(x)}{2 \left (1-\frac {1}{x^2}\right )^{3/2}}-\frac {5 i \sqrt {x^2} \sec ^{-1}(x) \tan ^{-1}\left (e^{i \sec ^{-1}(x)}\right )}{x}-\frac {\left (5 \sqrt {x^2}\right ) \text {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(x)\right )}{2 x}+\frac {\left (5 \sqrt {x^2}\right ) \text {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sec ^{-1}(x)\right )}{2 x}\\ &=-\frac {5}{12 \left (1-\frac {1}{x^2}\right ) \sqrt {x^2}}-\frac {3 \sqrt {x^2}}{4}+\frac {\sqrt {x^2}}{4 \left (1-\frac {1}{x^2}\right )}-\frac {13 \sqrt {x^2} \coth ^{-1}(x)}{6 x}-\frac {5 \sqrt {x^2} \sec ^{-1}(x)}{6 \left (1-\frac {1}{x^2}\right )^{3/2} x}-\frac {5 \sqrt {x^2} \sec ^{-1}(x)}{2 \sqrt {1-\frac {1}{x^2}} x}+\frac {x \sqrt {x^2} \sec ^{-1}(x)}{2 \left (1-\frac {1}{x^2}\right )^{3/2}}-\frac {5 i \sqrt {x^2} \sec ^{-1}(x) \tan ^{-1}\left (e^{i \sec ^{-1}(x)}\right )}{x}+\frac {\left (5 i \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \sec ^{-1}(x)}\right )}{2 x}-\frac {\left (5 i \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \sec ^{-1}(x)}\right )}{2 x}\\ &=-\frac {5}{12 \left (1-\frac {1}{x^2}\right ) \sqrt {x^2}}-\frac {3 \sqrt {x^2}}{4}+\frac {\sqrt {x^2}}{4 \left (1-\frac {1}{x^2}\right )}-\frac {13 \sqrt {x^2} \coth ^{-1}(x)}{6 x}-\frac {5 \sqrt {x^2} \sec ^{-1}(x)}{6 \left (1-\frac {1}{x^2}\right )^{3/2} x}-\frac {5 \sqrt {x^2} \sec ^{-1}(x)}{2 \sqrt {1-\frac {1}{x^2}} x}+\frac {x \sqrt {x^2} \sec ^{-1}(x)}{2 \left (1-\frac {1}{x^2}\right )^{3/2}}-\frac {5 i \sqrt {x^2} \sec ^{-1}(x) \tan ^{-1}\left (e^{i \sec ^{-1}(x)}\right )}{x}+\frac {5 i \sqrt {x^2} \text {Li}_2\left (-i e^{i \sec ^{-1}(x)}\right )}{2 x}-\frac {5 i \sqrt {x^2} \text {Li}_2\left (i e^{i \sec ^{-1}(x)}\right )}{2 x}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(383\) vs. \(2(175)=350\).
time = 1.25, size = 383, normalized size = 2.19 \begin {gather*} -\frac {x^5 \left (22 \sec ^{-1}(x)+40 \sec ^{-1}(x) \cos \left (2 \sec ^{-1}(x)\right )-30 \sec ^{-1}(x) \cos \left (4 \sec ^{-1}(x)\right )-30 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x) \log \left (1-i e^{i \sec ^{-1}(x)}\right )+30 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x) \log \left (1+i e^{i \sec ^{-1}(x)}\right )+26 \sqrt {1-\frac {1}{x^2}} \log \left (\cos \left (\frac {1}{2} \sec ^{-1}(x)\right )\right )-26 \sqrt {1-\frac {1}{x^2}} \log \left (\sin \left (\frac {1}{2} \sec ^{-1}(x)\right )\right )+16 \sin \left (2 \sec ^{-1}(x)\right )-60 i \sqrt {1-\frac {1}{x^2}} \text {Li}_2\left (-i e^{i \sec ^{-1}(x)}\right ) \sin ^2\left (2 \sec ^{-1}(x)\right )+60 i \sqrt {1-\frac {1}{x^2}} \text {Li}_2\left (i e^{i \sec ^{-1}(x)}\right ) \sin ^2\left (2 \sec ^{-1}(x)\right )-15 \sec ^{-1}(x) \log \left (1-i e^{i \sec ^{-1}(x)}\right ) \sin \left (3 \sec ^{-1}(x)\right )+15 \sec ^{-1}(x) \log \left (1+i e^{i \sec ^{-1}(x)}\right ) \sin \left (3 \sec ^{-1}(x)\right )+13 \log \left (\cos \left (\frac {1}{2} \sec ^{-1}(x)\right )\right ) \sin \left (3 \sec ^{-1}(x)\right )-13 \log \left (\sin \left (\frac {1}{2} \sec ^{-1}(x)\right )\right ) \sin \left (3 \sec ^{-1}(x)\right )-4 \sin \left (4 \sec ^{-1}(x)\right )+15 \sec ^{-1}(x) \log \left (1-i e^{i \sec ^{-1}(x)}\right ) \sin \left (5 \sec ^{-1}(x)\right )-15 \sec ^{-1}(x) \log \left (1+i e^{i \sec ^{-1}(x)}\right ) \sin \left (5 \sec ^{-1}(x)\right )-13 \log \left (\cos \left (\frac {1}{2} \sec ^{-1}(x)\right )\right ) \sin \left (5 \sec ^{-1}(x)\right )+13 \log \left (\sin \left (\frac {1}{2} \sec ^{-1}(x)\right )\right ) \sin \left (5 \sec ^{-1}(x)\right )\right )}{96 \left (-1+x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^6*ArcSec[x])/(-1 + x^2)^(5/2),x]

[Out]

-1/96*(x^5*(22*ArcSec[x] + 40*ArcSec[x]*Cos[2*ArcSec[x]] - 30*ArcSec[x]*Cos[4*ArcSec[x]] - 30*Sqrt[1 - x^(-2)]
*ArcSec[x]*Log[1 - I*E^(I*ArcSec[x])] + 30*Sqrt[1 - x^(-2)]*ArcSec[x]*Log[1 + I*E^(I*ArcSec[x])] + 26*Sqrt[1 -
x^(-2)]*Log[Cos[ArcSec[x]/2]] - 26*Sqrt[1 - x^(-2)]*Log[Sin[ArcSec[x]/2]] + 16*Sin[2*ArcSec[x]] - (60*I)*Sqrt
[1 - x^(-2)]*PolyLog[2, (-I)*E^(I*ArcSec[x])]*Sin[2*ArcSec[x]]^2 + (60*I)*Sqrt[1 - x^(-2)]*PolyLog[2, I*E^(I*A
rcSec[x])]*Sin[2*ArcSec[x]]^2 - 15*ArcSec[x]*Log[1 - I*E^(I*ArcSec[x])]*Sin[3*ArcSec[x]] + 15*ArcSec[x]*Log[1
+ I*E^(I*ArcSec[x])]*Sin[3*ArcSec[x]] + 13*Log[Cos[ArcSec[x]/2]]*Sin[3*ArcSec[x]] - 13*Log[Sin[ArcSec[x]/2]]*S
in[3*ArcSec[x]] - 4*Sin[4*ArcSec[x]] + 15*ArcSec[x]*Log[1 - I*E^(I*ArcSec[x])]*Sin[5*ArcSec[x]] - 15*ArcSec[x]
*Log[1 + I*E^(I*ArcSec[x])]*Sin[5*ArcSec[x]] - 13*Log[Cos[ArcSec[x]/2]]*Sin[5*ArcSec[x]] + 13*Log[Sin[ArcSec[x
]/2]]*Sin[5*ArcSec[x]]))/(-1 + x^2)^(3/2)

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Maple [A]
time = 0.84, size = 240, normalized size = 1.37

method	result	size
default	\(\frac {\sqrt {x^{2}-1}\, x \left (3 x^{4} \mathrm {arcsec}\left (x \right )-3 \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{3}-20 \,\mathrm {arcsec}\left (x \right ) x^{2}+2 \sqrt {\frac {x^{2}-1}{x^{2}}}\, x +15 \,\mathrm {arcsec}\left (x \right )\right )}{6 x^{4}-12 x^{2}+6}-\frac {i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x \left (15 i \mathrm {arcsec}\left (x \right ) \ln \left (1-i \left (\frac {1}{x}+i \sqrt {1-\frac {1}{x^{2}}}\right )\right )-15 i \mathrm {arcsec}\left (x \right ) \ln \left (1+i \left (\frac {1}{x}+i \sqrt {1-\frac {1}{x^{2}}}\right )\right )+13 i \ln \left (\frac {1}{x}+i \sqrt {1-\frac {1}{x^{2}}}-1\right )-13 i \ln \left (\frac {1}{x}+i \sqrt {1-\frac {1}{x^{2}}}+1\right )+15 \dilog \left (1-i \left (\frac {1}{x}+i \sqrt {1-\frac {1}{x^{2}}}\right )\right )-15 \dilog \left (1+i \left (\frac {1}{x}+i \sqrt {1-\frac {1}{x^{2}}}\right )\right )\right )}{6 \sqrt {x^{2}-1}}\)	\(240\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*arcsec(x)/(x^2-1)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/6*(x^2-1)^(1/2)*x/(x^4-2*x^2+1)*(3*x^4*arcsec(x)-3*((x^2-1)/x^2)^(1/2)*x^3-20*arcsec(x)*x^2+2*((x^2-1)/x^2)^
(1/2)*x+15*arcsec(x))-1/6*I/(x^2-1)^(1/2)*((x^2-1)/x^2)^(1/2)*x*(15*I*arcsec(x)*ln(1-I*(1/x+I*(1-1/x^2)^(1/2))
)-15*I*arcsec(x)*ln(1+I*(1/x+I*(1-1/x^2)^(1/2)))+13*I*ln(1/x+I*(1-1/x^2)^(1/2)-1)-13*I*ln(1/x+I*(1-1/x^2)^(1/2
)+1)+15*dilog(1-I*(1/x+I*(1-1/x^2)^(1/2)))-15*dilog(1+I*(1/x+I*(1-1/x^2)^(1/2))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*arcsec(x)/(x^2-1)^(5/2),x, algorithm="maxima")

[Out]

integrate(x^6*arcsec(x)/(x^2 - 1)^(5/2), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*arcsec(x)/(x^2-1)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^2 - 1)*x^6*arcsec(x)/(x^6 - 3*x^4 + 3*x^2 - 1), x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6*asec(x)/(x**2-1)**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6190 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*arcsec(x)/(x^2-1)^(5/2),x, algorithm="giac")

[Out]

integrate(x^6*arcsec(x)/(x^2 - 1)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^6\,\mathrm {acos}\left (\frac {1}{x}\right )}{{\left (x^2-1\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6*acos(1/x))/(x^2 - 1)^(5/2),x)

[Out]

int((x^6*acos(1/x))/(x^2 - 1)^(5/2), x)

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