3.689 \(\int \frac {x^6 \sec ^{-1}(x)}{(-1+x^2)^{5/2}} \, dx\)

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

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Rubi [A]  time = 0.32, 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 = {5242, 4702, 4706, 4710, 4181, 2279, 2391, 206, 199, 290, 325} \[ \frac {5 i \sqrt {x^2} \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(x)}\right )}{2 x}-\frac {5 i \sqrt {x^2} \text {PolyLog}\left (2,i e^{i \sec ^{-1}(x)}\right )}{2 x}+\frac {\sqrt {x^2}}{4 \left (1-\frac {1}{x^2}\right )}-\frac {3 \sqrt {x^2}}{4}-\frac {5}{12 \left (1-\frac {1}{x^2}\right ) \sqrt {x^2}}+\frac {x \sqrt {x^2} \sec ^{-1}(x)}{2 \left (1-\frac {1}{x^2}\right )^{3/2}}-\frac {5 \sqrt {x^2} \sec ^{-1}(x)}{2 \sqrt {1-\frac {1}{x^2}} x}-\frac {5 \sqrt {x^2} \sec ^{-1}(x)}{6 \left (1-\frac {1}{x^2}\right )^{3/2} x}-\frac {13 \sqrt {x^2} \coth ^{-1}(x)}{6 x}-\frac {5 i \sqrt {x^2} \sec ^{-1}(x) \tan ^{-1}\left (e^{i \sec ^{-1}(x)}\right )}{x} \]

Warning: Unable to verify antiderivative.

[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 199

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] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 290

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] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

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 2279

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 2391

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 4181

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 4702

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*d^IntPart[p]*(d + e*x^2)^F
racPart[p])/(f*(m + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcCos[c*x
])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[m, -1] && Inte
gerQ[m]

Rule 4706

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*d^IntPart[p]*(d + e*x^2)^Frac
Part[p])/(2*f*(p + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcCos[c*x]
)^(n - 1), x], x]) /; FreeQ[{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 4710

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> -Dist[(c^(m +
1)*Sqrt[d])^(-1), 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] && GtQ[d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 5242

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} \operatorname {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} \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {1}{x}\right )}{12 x}+\frac {\left (3 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {1}{x}\right )}{4 x}-\frac {\left (5 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {1}{x}\right )}{2 x}-\frac {\left (5 \sqrt {x^2}\right ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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]  time = 1.99, size = 383, normalized size = 2.19 \[ -\frac {x^5 \left (-60 i \sqrt {1-\frac {1}{x^2}} \sin ^2\left (2 \sec ^{-1}(x)\right ) \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(x)}\right )+60 i \sqrt {1-\frac {1}{x^2}} \sin ^2\left (2 \sec ^{-1}(x)\right ) \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(x)}\right )-30 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x) \log \left (1-i e^{i \sec ^{-1}(x)}\right )+30 \sqrt {1-\frac {1}{x^2}} \sec ^{-1}(x) \log \left (1+i e^{i \sec ^{-1}(x)}\right )-26 \sqrt {1-\frac {1}{x^2}} \log \left (\sin \left (\frac {1}{2} \sec ^{-1}(x)\right )\right )+26 \sqrt {1-\frac {1}{x^2}} \log \left (\cos \left (\frac {1}{2} \sec ^{-1}(x)\right )\right )+22 \sec ^{-1}(x)+40 \sec ^{-1}(x) \cos \left (2 \sec ^{-1}(x)\right )-30 \sec ^{-1}(x) \cos \left (4 \sec ^{-1}(x)\right )+16 \sin \left (2 \sec ^{-1}(x)\right )-4 \sin \left (4 \sec ^{-1}(x)\right )-15 \sec ^{-1}(x) \log \left (1-i e^{i \sec ^{-1}(x)}\right ) \sin \left (3 \sec ^{-1}(x)\right )+15 \sec ^{-1}(x) \log \left (1+i e^{i \sec ^{-1}(x)}\right ) \sin \left (3 \sec ^{-1}(x)\right )-13 \sin \left (3 \sec ^{-1}(x)\right ) \log \left (\sin \left (\frac {1}{2} \sec ^{-1}(x)\right )\right )+15 \sec ^{-1}(x) \log \left (1-i e^{i \sec ^{-1}(x)}\right ) \sin \left (5 \sec ^{-1}(x)\right )-15 \sec ^{-1}(x) \log \left (1+i e^{i \sec ^{-1}(x)}\right ) \sin \left (5 \sec ^{-1}(x)\right )+13 \sin \left (5 \sec ^{-1}(x)\right ) \log \left (\sin \left (\frac {1}{2} \sec ^{-1}(x)\right )\right )+13 \sin \left (3 \sec ^{-1}(x)\right ) \log \left (\cos \left (\frac {1}{2} \sec ^{-1}(x)\right )\right )-13 \sin \left (5 \sec ^{-1}(x)\right ) \log \left (\cos \left (\frac {1}{2} \sec ^{-1}(x)\right )\right )\right )}{96 \left (x^2-1\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^6 \sec ^{-1}(x)}{\left (-1+x^2\right )^{5/2}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{2} - 1} x^{6} \operatorname {arcsec}\relax (x)}{x^{6} - 3 \, x^{4} + 3 \, x^{2} - 1}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6} \operatorname {arcsec}\relax (x)}{{\left (x^{2} - 1\right )}^{\frac {5}{2}}}\,{d x} \]

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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maple [A]  time = 0.89, size = 240, normalized size = 1.37

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

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 \[ \int \frac {x^{6} \operatorname {arcsec}\relax (x)}{{\left (x^{2} - 1\right )}^{\frac {5}{2}}}\,{d x} \]

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

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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