3.1.6 \(\int \frac {\text {sech}^{-1}(a+b x)}{x^2} \, dx\) [6]

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

[Out]

-b*arcsech(b*x+a)/a-arcsech(b*x+a)/x+2*b*arctanh((1+a)^(1/2)*tanh(1/2*arcsech(b*x+a))/(1-a)^(1/2))/a/(-a^2+1)^
(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 70, 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 = {6456, 5576, 3868, 2738, 214} \begin {gather*} \frac {2 b \tanh ^{-1}\left (\frac {\sqrt {a+1} \tanh \left (\frac {1}{2} \text {sech}^{-1}(a+b x)\right )}{\sqrt {1-a}}\right )}{a \sqrt {1-a^2}}-\frac {b \text {sech}^{-1}(a+b x)}{a}-\frac {\text {sech}^{-1}(a+b x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSech[a + b*x]/x^2,x]

[Out]

-((b*ArcSech[a + b*x])/a) - ArcSech[a + b*x]/x + (2*b*ArcTanh[(Sqrt[1 + a]*Tanh[ArcSech[a + b*x]/2])/Sqrt[1 -
a]])/(a*Sqrt[1 - a^2])

Rule 214

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

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 3868

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Dist[1/a, Int[1/(1 + (a/b)*Sin[c
+ d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 5576

Int[((e_.) + (f_.)*(x_))^(m_.)*Sech[(c_.) + (d_.)*(x_)]*((a_) + (b_.)*Sech[(c_.) + (d_.)*(x_)])^(n_.)*Tanh[(c_
.) + (d_.)*(x_)], x_Symbol] :> Simp[(-(e + f*x)^m)*((a + b*Sech[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Dist[f*
(m/(b*d*(n + 1))), Int[(e + f*x)^(m - 1)*(a + b*Sech[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n},
x] && IGtQ[m, 0] && NeQ[n, -1]

Rule 6456

Int[((a_.) + ArcSech[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[-(d^(m + 1)
)^(-1), Subst[Int[(a + b*x)^p*Sech[x]*Tanh[x]*(d*e - c*f + f*Sech[x])^m, x], x, ArcSech[c + d*x]], x] /; FreeQ
[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && IntegerQ[m]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(244\) vs. \(2(70)=140\).
time = 0.17, size = 244, normalized size = 3.49 \begin {gather*} -\frac {\text {sech}^{-1}(a+b x)}{x}+\frac {b \left (-\log (x)+\sqrt {1-a^2} \log (a+b x)-\sqrt {1-a^2} \log \left (1+\sqrt {-\frac {-1+a+b x}{1+a+b x}}+a \sqrt {-\frac {-1+a+b x}{1+a+b x}}+b x \sqrt {-\frac {-1+a+b x}{1+a+b x}}\right )+\log \left (1-a^2-a b x+\sqrt {1-a^2} \sqrt {-\frac {-1+a+b x}{1+a+b x}}+a \sqrt {1-a^2} \sqrt {-\frac {-1+a+b x}{1+a+b x}}+\sqrt {1-a^2} b x \sqrt {-\frac {-1+a+b x}{1+a+b x}}\right )\right )}{a \sqrt {1-a^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSech[a + b*x]/x^2,x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(170\) vs. \(2(62)=124\).
time = 0.44, size = 171, normalized size = 2.44

method	result	size
derivativedivides	\(b \left (-\frac {\mathrm {arcsech}\left (b x +a \right )}{b x}-\frac {\sqrt {-\frac {b x +a -1}{b x +a}}\, \left (b x +a \right ) \sqrt {\frac {b x +a +1}{b x +a}}\, \left (\arctanh \left (\frac {1}{\sqrt {-\left (b x +a \right )^{2}+1}}\right ) a^{2}+\sqrt {-a^{2}+1}\, \ln \left (\frac {2 \sqrt {-a^{2}+1}\, \sqrt {-\left (b x +a \right )^{2}+1}-2 a \left (b x +a \right )+2}{b x}\right )-\arctanh \left (\frac {1}{\sqrt {-\left (b x +a \right )^{2}+1}}\right )\right )}{\sqrt {-\left (b x +a \right )^{2}+1}\, a \left (-1+a \right ) \left (1+a \right )}\right )\)	\(171\)
default	\(b \left (-\frac {\mathrm {arcsech}\left (b x +a \right )}{b x}-\frac {\sqrt {-\frac {b x +a -1}{b x +a}}\, \left (b x +a \right ) \sqrt {\frac {b x +a +1}{b x +a}}\, \left (\arctanh \left (\frac {1}{\sqrt {-\left (b x +a \right )^{2}+1}}\right ) a^{2}+\sqrt {-a^{2}+1}\, \ln \left (\frac {2 \sqrt {-a^{2}+1}\, \sqrt {-\left (b x +a \right )^{2}+1}-2 a \left (b x +a \right )+2}{b x}\right )-\arctanh \left (\frac {1}{\sqrt {-\left (b x +a \right )^{2}+1}}\right )\right )}{\sqrt {-\left (b x +a \right )^{2}+1}\, a \left (-1+a \right ) \left (1+a \right )}\right )\)	\(171\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*(-1/b/x*arcsech(b*x+a)-(-(b*x+a-1)/(b*x+a))^(1/2)*(b*x+a)*((b*x+a+1)/(b*x+a))^(1/2)*(arctanh(1/(-(b*x+a)^2+1
)^(1/2))*a^2+(-a^2+1)^(1/2)*ln(2*((-a^2+1)^(1/2)*(-(b*x+a)^2+1)^(1/2)-a*(b*x+a)+1)/b/x)-arctanh(1/(-(b*x+a)^2+
1)^(1/2)))/(-(b*x+a)^2+1)^(1/2)/a/(-1+a)/(1+a))

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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(b*x+a)/x^2,x, algorithm="maxima")

[Out]

b*log(x)/(a^3 - a) - 1/2*((a^2*b - a*b)*x*log(b*x + a + 1) + (a^2*b + a*b)*x*log(-b*x - a + 1) + 2*(a^3 - a)*l
og(sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*b*x + sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*a + b*x + a) - 2*(a^3 + (a^
2*b - b)*x - a)*log(b*x + a) - 2*(a^3 - a)*log(b*x + a))/((a^3 - a)*x) - integrate((b^2*x + a*b)/(b^2*x^3 + 2*
a*b*x^2 + (a^2 - 1)*x + (b^2*x^3 + 2*a*b*x^2 + (a^2 - 1)*x)*e^(1/2*log(b*x + a + 1) + 1/2*log(-b*x - a + 1))),
x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (62) = 124\).
time = 0.40, size = 651, normalized size = 9.30 \begin {gather*} \left [-\frac {{\left (a^{2} - 1\right )} b x \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{x}\right ) - {\left (a^{2} - 1\right )} b x \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - 1}{x}\right ) + \sqrt {-a^{2} + 1} b x \log \left (\frac {{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \, {\left (a^{3} - a\right )} b x - 4 \, a^{2} - 2 \, {\left (a b^{2} x^{2} + a^{3} + {\left (2 \, a^{2} - 1\right )} b x - a\right )} \sqrt {-a^{2} + 1} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 2}{x^{2}}\right ) + 2 \, {\left (a^{3} - a\right )} \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{b x + a}\right )}{2 \, {\left (a^{3} - a\right )} x}, -\frac {{\left (a^{2} - 1\right )} b x \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{x}\right ) - {\left (a^{2} - 1\right )} b x \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - 1}{x}\right ) + 2 \, \sqrt {a^{2} - 1} b x \arctan \left (\frac {{\left (a b^{2} x^{2} + a^{3} + {\left (2 \, a^{2} - 1\right )} b x - a\right )} \sqrt {a^{2} - 1} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{{\left (a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 2 \, {\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}\right ) + 2 \, {\left (a^{3} - a\right )} \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{b x + a}\right )}{2 \, {\left (a^{3} - a\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((a^2 - 1)*b*x*log(((b*x + a)*sqrt(-(b^2*x^2 + 2*a*b*x + a^2 - 1)/(b^2*x^2 + 2*a*b*x + a^2)) + 1)/x) - (
a^2 - 1)*b*x*log(((b*x + a)*sqrt(-(b^2*x^2 + 2*a*b*x + a^2 - 1)/(b^2*x^2 + 2*a*b*x + a^2)) - 1)/x) + sqrt(-a^2
+ 1)*b*x*log(((2*a^2 - 1)*b^2*x^2 + 2*a^4 + 4*(a^3 - a)*b*x - 4*a^2 - 2*(a*b^2*x^2 + a^3 + (2*a^2 - 1)*b*x -
a)*sqrt(-a^2 + 1)*sqrt(-(b^2*x^2 + 2*a*b*x + a^2 - 1)/(b^2*x^2 + 2*a*b*x + a^2)) + 2)/x^2) + 2*(a^3 - a)*log((
(b*x + a)*sqrt(-(b^2*x^2 + 2*a*b*x + a^2 - 1)/(b^2*x^2 + 2*a*b*x + a^2)) + 1)/(b*x + a)))/((a^3 - a)*x), -1/2*
((a^2 - 1)*b*x*log(((b*x + a)*sqrt(-(b^2*x^2 + 2*a*b*x + a^2 - 1)/(b^2*x^2 + 2*a*b*x + a^2)) + 1)/x) - (a^2 -
1)*b*x*log(((b*x + a)*sqrt(-(b^2*x^2 + 2*a*b*x + a^2 - 1)/(b^2*x^2 + 2*a*b*x + a^2)) - 1)/x) + 2*sqrt(a^2 - 1)
*b*x*arctan((a*b^2*x^2 + a^3 + (2*a^2 - 1)*b*x - a)*sqrt(a^2 - 1)*sqrt(-(b^2*x^2 + 2*a*b*x + a^2 - 1)/(b^2*x^2
+ 2*a*b*x + a^2))/((a^2 - 1)*b^2*x^2 + a^4 + 2*(a^3 - a)*b*x - 2*a^2 + 1)) + 2*(a^3 - a)*log(((b*x + a)*sqrt(
-(b^2*x^2 + 2*a*b*x + a^2 - 1)/(b^2*x^2 + 2*a*b*x + a^2)) + 1)/(b*x + a)))/((a^3 - a)*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asech(b*x+a)/x**2,x)

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acosh(1/(a + b*x))/x^2,x)

[Out]

int(acosh(1/(a + b*x))/x^2, x)

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