3.1.9 \(\int \frac {\text {sech}^{-1}(a x)^2}{x^4} \, dx\) [9]

Optimal. Leaf size=102 \[ -\frac {2}{27 x^3}-\frac {4 a^2}{9 x}+\frac {2 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{9 x^3}+\frac {4 a^2 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{9 x}-\frac {\text {sech}^{-1}(a x)^2}{3 x^3} \]

[Out]

-2/27/x^3-4/9*a^2/x-1/3*arcsech(a*x)^2/x^3+2/9*(a*x+1)*arcsech(a*x)*((-a*x+1)/(a*x+1))^(1/2)/x^3+4/9*a^2*(a*x+
1)*arcsech(a*x)*((-a*x+1)/(a*x+1))^(1/2)/x

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Rubi [A]
time = 0.05, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6420, 5481, 3391, 3377, 2718} \begin {gather*} -\frac {4 a^2}{9 x}+\frac {4 a^2 \sqrt {\frac {1-a x}{a x+1}} (a x+1) \text {sech}^{-1}(a x)}{9 x}-\frac {\text {sech}^{-1}(a x)^2}{3 x^3}+\frac {2 \sqrt {\frac {1-a x}{a x+1}} (a x+1) \text {sech}^{-1}(a x)}{9 x^3}-\frac {2}{27 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSech[a*x]^2/x^4,x]

[Out]

-2/(27*x^3) - (4*a^2)/(9*x) + (2*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x])/(9*x^3) + (4*a^2*Sqrt[(1 -
a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x])/(9*x) - ArcSech[a*x]^2/(3*x^3)

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3391

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

Rule 5481

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[x^(m - n
+ 1)*(Cosh[a + b*x^n]^(p + 1)/(b*n*(p + 1))), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Cosh[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)^2}{x^4} \, dx &=-\left (a^3 \text {Subst}\left (\int x^2 \cosh ^2(x) \sinh (x) \, dx,x,\text {sech}^{-1}(a x)\right )\right )\\ &=-\frac {\text {sech}^{-1}(a x)^2}{3 x^3}+\frac {1}{3} \left (2 a^3\right ) \text {Subst}\left (\int x \cosh ^3(x) \, dx,x,\text {sech}^{-1}(a x)\right )\\ &=-\frac {2}{27 x^3}+\frac {2 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{9 x^3}-\frac {\text {sech}^{-1}(a x)^2}{3 x^3}+\frac {1}{9} \left (4 a^3\right ) \text {Subst}\left (\int x \cosh (x) \, dx,x,\text {sech}^{-1}(a x)\right )\\ &=-\frac {2}{27 x^3}+\frac {2 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{9 x^3}+\frac {4 a^2 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{9 x}-\frac {\text {sech}^{-1}(a x)^2}{3 x^3}-\frac {1}{9} \left (4 a^3\right ) \text {Subst}\left (\int \sinh (x) \, dx,x,\text {sech}^{-1}(a x)\right )\\ &=-\frac {2}{27 x^3}-\frac {4 a^2}{9 x}+\frac {2 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{9 x^3}+\frac {4 a^2 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{9 x}-\frac {\text {sech}^{-1}(a x)^2}{3 x^3}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 73, normalized size = 0.72 \begin {gather*} \frac {-2 \left (1+6 a^2 x^2\right )+6 \sqrt {\frac {1-a x}{1+a x}} \left (1+a x+2 a^2 x^2+2 a^3 x^3\right ) \text {sech}^{-1}(a x)-9 \text {sech}^{-1}(a x)^2}{27 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSech[a*x]^2/x^4,x]

[Out]

(-2*(1 + 6*a^2*x^2) + 6*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x + 2*a^2*x^2 + 2*a^3*x^3)*ArcSech[a*x] - 9*ArcSech[a
*x]^2)/(27*x^3)

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Maple [A]
time = 0.23, size = 112, normalized size = 1.10

method	result	size
derivativedivides	\(a^{3} \left (-\frac {\mathrm {arcsech}\left (a x \right )^{2}}{3 a^{3} x^{3}}+\frac {4 \,\mathrm {arcsech}\left (a x \right ) \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}}{9}+\frac {2 \,\mathrm {arcsech}\left (a x \right ) \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}}{9 a^{2} x^{2}}-\frac {4}{9 a x}-\frac {2}{27 a^{3} x^{3}}\right )\)	\(112\)
default	\(a^{3} \left (-\frac {\mathrm {arcsech}\left (a x \right )^{2}}{3 a^{3} x^{3}}+\frac {4 \,\mathrm {arcsech}\left (a x \right ) \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}}{9}+\frac {2 \,\mathrm {arcsech}\left (a x \right ) \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}}{9 a^{2} x^{2}}-\frac {4}{9 a x}-\frac {2}{27 a^{3} x^{3}}\right )\)	\(112\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsech(a*x)^2/x^4,x,method=_RETURNVERBOSE)

[Out]

a^3*(-1/3*arcsech(a*x)^2/a^3/x^3+4/9*arcsech(a*x)*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(1/2)+2/9*arcsech(a*x)/a^
2/x^2*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(1/2)-4/9/a/x-2/27/a^3/x^3)

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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)^2/x^4,x, algorithm="maxima")

[Out]

integrate(arcsech(a*x)^2/x^4, x)

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Fricas [A]
time = 0.35, size = 116, normalized size = 1.14 \begin {gather*} -\frac {12 \, a^{2} x^{2} - 6 \, {\left (2 \, a^{3} x^{3} + a x\right )} \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 ) + 9 \, \log \left (\frac {a x \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} + 1}{a x}\right )^{2} + 2}{27 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsech(a*x)^2/x^4,x, algorithm="fricas")

[Out]

-1/27*(12*a^2*x^2 - 6*(2*a^3*x^3 + 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)) + 9*log((a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2)) + 1)/(a*x))^2 + 2)/x^3

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {asech}^{2}{\left (a x \right )}}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asech(a*x)**2/x**4,x)

[Out]

Integral(asech(a*x)**2/x**4, 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)^2/x^4,x, algorithm="giac")

[Out]

integrate(arcsech(a*x)^2/x^4, 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 )}^2}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acosh(1/(a*x))^2/x^4,x)

[Out]

int(acosh(1/(a*x))^2/x^4, x)

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