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

Optimal. Leaf size=70 \[ \frac {1}{2} \sqrt {x^4-1} \sec ^{-1}(x)-\frac {\sqrt {x^4-1}}{2 \sqrt {1-\frac {1}{x^2}} x}+\frac {1}{2} \tanh ^{-1}\left (\frac {\sqrt {1-\frac {1}{x^2}} x}{\sqrt {x^4-1}}\right ) \]

[Out]

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)

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Rubi [A]  time = 0.13, antiderivative size = 94, normalized size of antiderivative = 1.34, number of steps used = 7, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {261, 5246, 12, 1572, 1252, 865, 875, 203} \[ -\frac {\sqrt {x^4-1}}{2 \sqrt {1-\frac {1}{x^2}} x}+\frac {\sqrt {1-x^2} \tan ^{-1}\left (\frac {\sqrt {x^4-1}}{\sqrt {1-x^2}}\right )}{2 \sqrt {1-\frac {1}{x^2}} x}+\frac {1}{2} \sqrt {x^4-1} \sec ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcSec[x])/Sqrt[-1 + x^4],x]

[Out]

-Sqrt[-1 + x^4]/(2*Sqrt[1 - x^(-2)]*x) + (Sqrt[-1 + x^4]*ArcSec[x])/2 + (Sqrt[1 - x^2]*ArcTan[Sqrt[-1 + x^4]/S
qrt[1 - x^2]])/(2*Sqrt[1 - x^(-2)]*x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 203

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

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 865

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*
x)^m*(f + g*x)^(n + 1)*(a + c*x^2)^p)/(g*(m - n - 1)), x] - Dist[(c*m*(e*f + d*g))/(e^2*g*(m - n - 1)), Int[(d
+ e*x)^(m + 1)*(f + g*x)^n*(a + c*x^2)^(p - 1), x], x] /; FreeQ[{a, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0
] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0] && GtQ[p, 0] && NeQ[m - n - 1, 0] &&  !IGtQ[n, 0]
&&  !(IntegerQ[n + p] && LtQ[n + p + 2, 0]) && RationalQ[n]

Rule 875

Int[Sqrt[(d_) + (e_.)*(x_)]/(((f_.) + (g_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e^2, Subst[I
nt[1/(c*(e*f + d*g) + e^2*g*x^2), x], x, Sqrt[a + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x] &&
NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0]

Rule 1252

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[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]

Rule 1572

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^(mn_.))^(q_)*((a_) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Dist[(e^IntPart[
q]*(d + e*x^mn)^FracPart[q])/(x^(mn*FracPart[q])*(1 + d/(x^mn*e))^FracPart[q]), Int[x^(m + mn*q)*(1 + d/(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] &&  !In
tegerQ[q] && PosQ[n2]

Rule 5246

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*(u_), x_Symbol] :> With[{v = IntHide[u, x]}, Dist[a + b*ArcSec[c*x], v,
x] - Dist[b/c, Int[SimplifyIntegrand[v/(x^2*Sqrt[1 - 1/(c^2*x^2)]), x], x], x] /; InverseFunctionFreeQ[v, x]]
/; FreeQ[{a, b, c}, x]

Rubi steps

\begin {align*} \int \frac {x^3 \sec ^{-1}(x)}{\sqrt {-1+x^4}} \, dx &=\frac {1}{2} \sqrt {-1+x^4} \sec ^{-1}(x)-\int \frac {\sqrt {-1+x^4}}{2 \sqrt {1-\frac {1}{x^2}} x^2} \, dx\\ &=\frac {1}{2} \sqrt {-1+x^4} \sec ^{-1}(x)-\frac {1}{2} \int \frac {\sqrt {-1+x^4}}{\sqrt {1-\frac {1}{x^2}} x^2} \, dx\\ &=\frac {1}{2} \sqrt {-1+x^4} \sec ^{-1}(x)-\frac {\sqrt {1-x^2} \int \frac {\sqrt {-1+x^4}}{x \sqrt {1-x^2}} \, dx}{2 \sqrt {1-\frac {1}{x^2}} x}\\ &=\frac {1}{2} \sqrt {-1+x^4} \sec ^{-1}(x)-\frac {\sqrt {1-x^2} \operatorname {Subst}\left (\int \frac {\sqrt {-1+x^2}}{\sqrt {1-x} x} \, dx,x,x^2\right )}{4 \sqrt {1-\frac {1}{x^2}} x}\\ &=-\frac {\sqrt {-1+x^4}}{2 \sqrt {1-\frac {1}{x^2}} x}+\frac {1}{2} \sqrt {-1+x^4} \sec ^{-1}(x)+\frac {\sqrt {1-x^2} \operatorname {Subst}\left (\int \frac {\sqrt {1-x}}{x \sqrt {-1+x^2}} \, dx,x,x^2\right )}{4 \sqrt {1-\frac {1}{x^2}} x}\\ &=-\frac {\sqrt {-1+x^4}}{2 \sqrt {1-\frac {1}{x^2}} x}+\frac {1}{2} \sqrt {-1+x^4} \sec ^{-1}(x)+\frac {\sqrt {1-x^2} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt {-1+x^4}}{\sqrt {1-x^2}}\right )}{2 \sqrt {1-\frac {1}{x^2}} x}\\ &=-\frac {\sqrt {-1+x^4}}{2 \sqrt {1-\frac {1}{x^2}} x}+\frac {1}{2} \sqrt {-1+x^4} \sec ^{-1}(x)+\frac {\sqrt {1-x^2} \tan ^{-1}\left (\frac {\sqrt {-1+x^4}}{\sqrt {1-x^2}}\right )}{2 \sqrt {1-\frac {1}{x^2}} x}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 88, normalized size = 1.26 \[ \frac {1}{2} \left (\sqrt {x^4-1} \sec ^{-1}(x)-\log \left (x-x^3\right )-\frac {\sqrt {1-\frac {1}{x^2}} \sqrt {x^4-1} x}{x^2-1}+\log \left (-x^2-\sqrt {1-\frac {1}{x^2}} \sqrt {x^4-1} x+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((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 - S
qrt[1 - x^(-2)]*x*Sqrt[-1 + x^4]])/2

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fricas [B]  time = 0.48, size = 110, normalized size = 1.57 \[ \frac {{\left (x^{2} - 1\right )} \log \left (\frac {x^{2} + \sqrt {x^{4} - 1} \sqrt {x^{2} - 1} - 1}{x^{2} - 1}\right ) - {\left (x^{2} - 1\right )} \log \left (-\frac {x^{2} - \sqrt {x^{4} - 1} \sqrt {x^{2} - 1} - 1}{x^{2} - 1}\right ) + 2 \, \sqrt {x^{4} - 1} {\left ({\left (x^{2} - 1\right )} \operatorname {arcsec}\relax (x) - \sqrt {x^{2} - 1}\right )}}{4 \, {\left (x^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)*sq
rt(x^2 - 1) - 1)/(x^2 - 1)) + 2*sqrt(x^4 - 1)*((x^2 - 1)*arcsec(x) - sqrt(x^2 - 1)))/(x^2 - 1)

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giac [A]  time = 1.22, size = 52, normalized size = 0.74 \[ \frac {1}{2} \, \sqrt {x^{4} - 1} \arccos \left (\frac {1}{x}\right ) - \frac {2 \, \sqrt {x^{2} + 1} - \log \left (\sqrt {x^{2} + 1} + 1\right ) + \log \left (\sqrt {x^{2} + 1} - 1\right )}{4 \, \mathrm {sgn}\relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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maple [F]  time = 0.74, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \mathrm {arcsec}\relax (x )}{\sqrt {x^{4}-1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, \sqrt {x^{2} + 1} \sqrt {x + 1} \sqrt {x - 1} \arctan \left (\sqrt {x + 1} \sqrt {x - 1}\right ) - \int \frac {2 \, {\left (x^{3} e^{\left (\frac {3}{2} \, \log \left (x + 1\right ) + \frac {3}{2} \, \log \left (x - 1\right )\right )} + x^{3} e^{\left (\frac {1}{2} \, \log \left (x + 1\right ) + \frac {1}{2} \, \log \left (x - 1\right )\right )}\right )} \sqrt {x^{2} + 1} \log \relax (x) + {\left (x^{3} + x\right )} e^{\left (\frac {1}{2} \, \log \left (x^{2} + 1\right ) + \frac {3}{2} \, \log \left (x + 1\right ) + \frac {3}{2} \, \log \left (x - 1\right )\right )}}{{\left (x^{2} + 1\right )} {\left (e^{\left (2 \, \log \left (x + 1\right ) + 2 \, \log \left (x - 1\right )\right )} + e^{\left (\log \left (x + 1\right ) + \log \left (x - 1\right )\right )}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*log(x - 1)) + e^(log(x + 1) + log(x
- 1)))), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\mathrm {acos}\left (\frac {1}{x}\right )}{\sqrt {x^4-1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \operatorname {asec}{\relax (x )}}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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