3.1.17 \(\int \frac {\text {sech}^{-1}(a x)^3}{x^3} \, dx\) [17]

Optimal. Leaf size=137 \[ \frac {3 \sqrt {\frac {1-a x}{1+a x}} (1+a x)}{8 x^2}-\frac {3}{8} a^2 \text {sech}^{-1}(a x)-\frac {3 (1-a x) (1+a x) \text {sech}^{-1}(a x)}{4 x^2}+\frac {3 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{4 x^2}-\frac {1}{4} a^2 \text {sech}^{-1}(a x)^3-\frac {(1-a x) (1+a x) \text {sech}^{-1}(a x)^3}{2 x^2} \]

[Out]

-3/8*a^2*arcsech(a*x)-3/4*(-a*x+1)*(a*x+1)*arcsech(a*x)/x^2-1/4*a^2*arcsech(a*x)^3-1/2*(-a*x+1)*(a*x+1)*arcsec
h(a*x)^3/x^2+3/8*(a*x+1)*((-a*x+1)/(a*x+1))^(1/2)/x^2+3/4*(a*x+1)*arcsech(a*x)^2*((-a*x+1)/(a*x+1))^(1/2)/x^2

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Rubi [A]
time = 0.06, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6420, 5480, 3392, 30, 2715, 8} \begin {gather*} -\frac {1}{4} a^2 \text {sech}^{-1}(a x)^3-\frac {3}{8} a^2 \text {sech}^{-1}(a x)+\frac {3 \sqrt {\frac {1-a x}{a x+1}} (a x+1)}{8 x^2}-\frac {(1-a x) (a x+1) \text {sech}^{-1}(a x)^3}{2 x^2}+\frac {3 \sqrt {\frac {1-a x}{a x+1}} (a x+1) \text {sech}^{-1}(a x)^2}{4 x^2}-\frac {3 (1-a x) (a x+1) \text {sech}^{-1}(a x)}{4 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSech[a*x]^3/x^3,x]

[Out]

(3*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x))/(8*x^2) - (3*a^2*ArcSech[a*x])/8 - (3*(1 - a*x)*(1 + a*x)*ArcSech[a*x]
)/(4*x^2) + (3*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x]^2)/(4*x^2) - (a^2*ArcSech[a*x]^3)/4 - ((1 - a*
x)*(1 + a*x)*ArcSech[a*x]^3)/(2*x^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 3392

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*m*(c + d*x)^(m - 1)*((
b*Sin[e + f*x])^n/(f^2*n^2)), x] + (Dist[b^2*((n - 1)/n), Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[d^2*m*((m - 1)/(f^2*n^2)), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[b*(c + d*x)^m*Cos[e + f
*x]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 5480

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[x^(m - n
+ 1)*(Sinh[a + b*x^n]^(p + 1)/(b*n*(p + 1))), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Sinh[a + b*x
^n]^(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rule 6420

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[-(c^(m + 1))^(-1), Subst[Int[(a + b
*x)^n*Sech[x]^(m + 1)*Tanh[x], x], x, ArcSech[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] &
& (GtQ[n, 0] || LtQ[m, -1])

Rubi steps

\begin {align*} \int \frac {\text {sech}^{-1}(a x)^3}{x^3} \, dx &=-\left (a^2 \text {Subst}\left (\int x^3 \cosh (x) \sinh (x) \, dx,x,\text {sech}^{-1}(a x)\right )\right )\\ &=-\frac {(1-a x) (1+a x) \text {sech}^{-1}(a x)^3}{2 x^2}+\frac {1}{2} \left (3 a^2\right ) \text {Subst}\left (\int x^2 \sinh ^2(x) \, dx,x,\text {sech}^{-1}(a x)\right )\\ &=-\frac {3 (1-a x) (1+a x) \text {sech}^{-1}(a x)}{4 x^2}+\frac {3 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{4 x^2}-\frac {(1-a x) (1+a x) \text {sech}^{-1}(a x)^3}{2 x^2}-\frac {1}{4} \left (3 a^2\right ) \text {Subst}\left (\int x^2 \, dx,x,\text {sech}^{-1}(a x)\right )+\frac {1}{4} \left (3 a^2\right ) \text {Subst}\left (\int \sinh ^2(x) \, dx,x,\text {sech}^{-1}(a x)\right )\\ &=\frac {3 \sqrt {\frac {1-a x}{1+a x}} (1+a x)}{8 x^2}-\frac {3 (1-a x) (1+a x) \text {sech}^{-1}(a x)}{4 x^2}+\frac {3 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{4 x^2}-\frac {1}{4} a^2 \text {sech}^{-1}(a x)^3-\frac {(1-a x) (1+a x) \text {sech}^{-1}(a x)^3}{2 x^2}-\frac {1}{8} \left (3 a^2\right ) \text {Subst}\left (\int 1 \, dx,x,\text {sech}^{-1}(a x)\right )\\ &=\frac {3 \sqrt {\frac {1-a x}{1+a x}} (1+a x)}{8 x^2}-\frac {3}{8} a^2 \text {sech}^{-1}(a x)-\frac {3 (1-a x) (1+a x) \text {sech}^{-1}(a x)}{4 x^2}+\frac {3 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{4 x^2}-\frac {1}{4} a^2 \text {sech}^{-1}(a x)^3-\frac {(1-a x) (1+a x) \text {sech}^{-1}(a x)^3}{2 x^2}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 147, normalized size = 1.07 \begin {gather*} \frac {3 \sqrt {\frac {1-a x}{1+a x}} (1+a x)-6 \text {sech}^{-1}(a x)+6 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2+2 \left (-2+a^2 x^2\right ) \text {sech}^{-1}(a x)^3-3 a^2 x^2 \log (x)+3 a^2 x^2 \log \left (1+\sqrt {\frac {1-a x}{1+a x}}+a x \sqrt {\frac {1-a x}{1+a x}}\right )}{8 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSech[a*x]^3/x^3,x]

[Out]

(3*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x) - 6*ArcSech[a*x] + 6*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x]^2
+ 2*(-2 + a^2*x^2)*ArcSech[a*x]^3 - 3*a^2*x^2*Log[x] + 3*a^2*x^2*Log[1 + Sqrt[(1 - a*x)/(1 + a*x)] + a*x*Sqrt
[(1 - a*x)/(1 + a*x)]])/(8*x^2)

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Maple [A]
time = 0.15, size = 126, normalized size = 0.92

method	result	size
derivativedivides	\(a^{2} \left (-\frac {\mathrm {arcsech}\left (a x \right )^{3}}{2 a^{2} x^{2}}+\frac {3 \mathrm {arcsech}\left (a x \right )^{2} \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}}{4 a x}+\frac {\mathrm {arcsech}\left (a x \right )^{3}}{4}-\frac {3 \,\mathrm {arcsech}\left (a x \right )}{4 a^{2} x^{2}}+\frac {3 \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}}{8 a x}+\frac {3 \,\mathrm {arcsech}\left (a x \right )}{8}\right )\)	\(126\)
default	\(a^{2} \left (-\frac {\mathrm {arcsech}\left (a x \right )^{3}}{2 a^{2} x^{2}}+\frac {3 \mathrm {arcsech}\left (a x \right )^{2} \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}}{4 a x}+\frac {\mathrm {arcsech}\left (a x \right )^{3}}{4}-\frac {3 \,\mathrm {arcsech}\left (a x \right )}{4 a^{2} x^{2}}+\frac {3 \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}}{8 a x}+\frac {3 \,\mathrm {arcsech}\left (a x \right )}{8}\right )\)	\(126\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsech(a*x)^3/x^3,x,method=_RETURNVERBOSE)

[Out]

a^2*(-1/2*arcsech(a*x)^3/a^2/x^2+3/4*arcsech(a*x)^2/a/x*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(1/2)+1/4*arcsech(a
*x)^3-3/4/a^2/x^2*arcsech(a*x)+3/8/a/x*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(1/2)+3/8*arcsech(a*x))

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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(arcsech(a*x)^3/x^3,x, algorithm="maxima")

[Out]

integrate(arcsech(a*x)^3/x^3, x)

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Fricas [A]
time = 0.39, size = 174, normalized size = 1.27 \begin {gather*} \frac {6 \, a x \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} \log \left (\frac {a x \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} + 1}{a x}\right )^{2} + 2 \, {\left (a^{2} x^{2} - 2\right )} \log \left (\frac {a x \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} + 1}{a x}\right )^{3} + 3 \, a x \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} + 3 \, {\left (a^{2} x^{2} - 2\right )} \log \left (\frac {a x \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} + 1}{a x}\right )}{8 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsech(a*x)^3/x^3,x, algorithm="fricas")

[Out]

1/8*(6*a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2))*log((a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2)) + 1)/(a*x))^2 + 2*(a^2*x^2 -
2)*log((a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2)) + 1)/(a*x))^3 + 3*a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2)) + 3*(a^2*x^2
- 2)*log((a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2)) + 1)/(a*x)))/x^2

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {asech}^{3}{\left (a x \right )}}{x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asech(a*x)**3/x**3,x)

[Out]

Integral(asech(a*x)**3/x**3, x)

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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(arcsech(a*x)^3/x^3,x, algorithm="giac")

[Out]

integrate(arcsech(a*x)^3/x^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {acosh}\left (\frac {1}{a\,x}\right )}^3}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acosh(1/(a*x))^3/x^3,x)

[Out]

int(acosh(1/(a*x))^3/x^3, x)

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