3.6 \(\int \frac{\text{sech}^{-1}(a+b x)}{x^2} \, dx\)

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

[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])

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Rubi [A]  time = 0.105885, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6321, 5468, 3783, 2659, 208} \[ \frac{2 b \tanh ^{-1}\left (\frac{\sqrt{a+1} \tanh \left (\frac{1}{2} \text{sech}^{-1}(a+b x)\right )}{\sqrt{1-a}}\right )}{a \sqrt{1-a^2}}-\frac{b \text{sech}^{-1}(a+b x)}{a}-\frac{\text{sech}^{-1}(a+b x)}{x} \]

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 6321

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]

Rule 5468

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 3783

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

Rule 2659

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 208

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

Rubi steps

\begin{align*} \int \frac{\text{sech}^{-1}(a+b x)}{x^2} \, dx &=-\left (b \operatorname{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 \operatorname{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 \operatorname{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) \operatorname{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*}

Mathematica [B]  time = 0.215134, size = 244, normalized size = 3.49 \[ \frac{b \left (\sqrt{1-a^2} \log (a+b x)-\sqrt{1-a^2} \log \left (a \sqrt{-\frac{a+b x-1}{a+b x+1}}+b x \sqrt{-\frac{a+b x-1}{a+b x+1}}+\sqrt{-\frac{a+b x-1}{a+b x+1}}+1\right )+\log \left (\sqrt{1-a^2} a \sqrt{-\frac{a+b x-1}{a+b x+1}}+\sqrt{1-a^2} b x \sqrt{-\frac{a+b x-1}{a+b x+1}}+\sqrt{1-a^2} \sqrt{-\frac{a+b x-1}{a+b x+1}}-a^2-a b x+1\right )-\log (x)\right )}{a \sqrt{1-a^2}}-\frac{\text{sech}^{-1}(a+b x)}{x} \]

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]  time = 0.252, size = 542, normalized size = 7.7 \begin{align*} -{\frac{{\rm arcsech} \left (bx+a\right )}{x}}-{\frac{a{b}^{2}x}{ \left ( 1+a \right ) \left ( a-1 \right ) }\sqrt{-{\frac{bx+a-1}{bx+a}}}\sqrt{{\frac{bx+a+1}{bx+a}}}{\it Artanh} \left ({\frac{1}{\sqrt{1- \left ( bx+a \right ) ^{2}}}} \right ){\frac{1}{\sqrt{1- \left ( bx+a \right ) ^{2}}}}}-{\frac{{a}^{2}b}{ \left ( 1+a \right ) \left ( a-1 \right ) }\sqrt{-{\frac{bx+a-1}{bx+a}}}\sqrt{{\frac{bx+a+1}{bx+a}}}{\it Artanh} \left ({\frac{1}{\sqrt{1- \left ( bx+a \right ) ^{2}}}} \right ){\frac{1}{\sqrt{1- \left ( bx+a \right ) ^{2}}}}}-{\frac{{b}^{2}x}{a \left ( a-1 \right ) \left ( 1+a \right ) }\sqrt{-{\frac{bx+a-1}{bx+a}}}\sqrt{{\frac{bx+a+1}{bx+a}}}\ln \left ( 2\,{\frac{\sqrt{-{a}^{2}+1}\sqrt{1- \left ( bx+a \right ) ^{2}}-a \left ( bx+a \right ) +1}{bx}} \right ) \sqrt{-{a}^{2}+1}{\frac{1}{\sqrt{1- \left ( bx+a \right ) ^{2}}}}}+{\frac{{b}^{2}x}{a \left ( a-1 \right ) \left ( 1+a \right ) }\sqrt{-{\frac{bx+a-1}{bx+a}}}\sqrt{{\frac{bx+a+1}{bx+a}}}{\it Artanh} \left ({\frac{1}{\sqrt{1- \left ( bx+a \right ) ^{2}}}} \right ){\frac{1}{\sqrt{1- \left ( bx+a \right ) ^{2}}}}}-{\frac{b}{ \left ( 1+a \right ) \left ( a-1 \right ) }\sqrt{-{\frac{bx+a-1}{bx+a}}}\sqrt{{\frac{bx+a+1}{bx+a}}}\ln \left ( 2\,{\frac{\sqrt{-{a}^{2}+1}\sqrt{1- \left ( bx+a \right ) ^{2}}-a \left ( bx+a \right ) +1}{bx}} \right ) \sqrt{-{a}^{2}+1}{\frac{1}{\sqrt{1- \left ( bx+a \right ) ^{2}}}}}+{\frac{b}{ \left ( 1+a \right ) \left ( a-1 \right ) }\sqrt{-{\frac{bx+a-1}{bx+a}}}\sqrt{{\frac{bx+a+1}{bx+a}}}{\it Artanh} \left ({\frac{1}{\sqrt{1- \left ( bx+a \right ) ^{2}}}} \right ){\frac{1}{\sqrt{1- \left ( bx+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{b \log \left (x\right )}{a^{3} - a} - \frac{{\left (a^{2} b - a b\right )} x \log \left (b x + a + 1\right ) +{\left (a^{2} b + a b\right )} x \log \left (-b x - a + 1\right ) + 2 \,{\left (a^{3} - a\right )} \log \left (\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\right ) - 2 \,{\left (a^{3} +{\left (a^{2} b - b\right )} x - a\right )} \log \left (b x + a\right ) - 2 \,{\left (a^{3} - a\right )} \log \left (b x + a\right )}{2 \,{\left (a^{3} - a\right )} x} - \int \frac{b^{2} x + a b}{b^{2} x^{3} + 2 \, a b x^{2} +{\left (a^{2} - 1\right )} x +{\left (b^{2} x^{3} + 2 \, a b x^{2} +{\left (a^{2} - 1\right )} x\right )} e^{\left (\frac{1}{2} \, \log \left (b x + a + 1\right ) + \frac{1}{2} \, \log \left (-b x - a + 1\right )\right )}}\,{d x} \end{align*}

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]  time = 2.19925, size = 1462, normalized size = 20.89 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asech}{\left (a + b x \right )}}{x^{2}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arsech}\left (b x + a\right )}{x^{2}}\,{d x} \end{align*}

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)