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(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 = 0.11, 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 = {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 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]

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 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 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 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]

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*}

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Mathematica [B]  time = 0.26, 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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fricas [B]  time = 0.71, size = 651, normalized size = 9.30 \[ \left [-\frac {{\left (a^{2} - 1\right )} b x \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{x}\right ) - {\left (a^{2} - 1\right )} b x \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - 1}{x}\right ) + \sqrt {-a^{2} + 1} b x \log \left (\frac {{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \, {\left (a^{3} - a\right )} b x - 4 \, a^{2} - 2 \, {\left (a b^{2} x^{2} + a^{3} + {\left (2 \, a^{2} - 1\right )} b x - a\right )} \sqrt {-a^{2} + 1} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 2}{x^{2}}\right ) + 2 \, {\left (a^{3} - a\right )} \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{b x + a}\right )}{2 \, {\left (a^{3} - a\right )} x}, -\frac {{\left (a^{2} - 1\right )} b x \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{x}\right ) - {\left (a^{2} - 1\right )} b x \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - 1}{x}\right ) + 2 \, \sqrt {a^{2} - 1} b x \arctan \left (\frac {{\left (a b^{2} x^{2} + a^{3} + {\left (2 \, a^{2} - 1\right )} b x - a\right )} \sqrt {a^{2} - 1} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{{\left (a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 2 \, {\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}\right ) + 2 \, {\left (a^{3} - a\right )} \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{b x + a}\right )}{2 \, {\left (a^{3} - a\right )} x}\right ] \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arsech}\left (b x + a\right )}{x^{2}}\,{d x} \]

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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maple [B]  time = 0.08, size = 542, normalized size = 7.74 \[ -\frac {\mathrm {arcsech}\left (b x +a \right )}{x}-\frac {\sqrt {-\frac {b x +a -1}{b x +a}}\, \sqrt {\frac {b x +a +1}{b x +a}}\, a \arctanh \left (\frac {1}{\sqrt {1-\left (b x +a \right )^{2}}}\right ) x \,b^{2}}{\sqrt {1-\left (b x +a \right )^{2}}\, \left (a -1\right ) \left (1+a \right )}-\frac {\sqrt {-\frac {b x +a -1}{b x +a}}\, \sqrt {\frac {b x +a +1}{b x +a}}\, \sqrt {-a^{2}+1}\, \ln \left (\frac {2 \sqrt {-a^{2}+1}\, \sqrt {1-\left (b x +a \right )^{2}}-2 a \left (b x +a \right )+2}{b x}\right ) x \,b^{2}}{\sqrt {1-\left (b x +a \right )^{2}}\, a \left (a -1\right ) \left (1+a \right )}-\frac {b \sqrt {-\frac {b x +a -1}{b x +a}}\, \sqrt {\frac {b x +a +1}{b x +a}}\, a^{2} \arctanh \left (\frac {1}{\sqrt {1-\left (b x +a \right )^{2}}}\right )}{\sqrt {1-\left (b x +a \right )^{2}}\, \left (a -1\right ) \left (1+a \right )}-\frac {b \sqrt {-\frac {b x +a -1}{b x +a}}\, \sqrt {\frac {b x +a +1}{b x +a}}\, \sqrt {-a^{2}+1}\, \ln \left (\frac {2 \sqrt {-a^{2}+1}\, \sqrt {1-\left (b x +a \right )^{2}}-2 a \left (b x +a \right )+2}{b x}\right )}{\sqrt {1-\left (b x +a \right )^{2}}\, \left (a -1\right ) \left (1+a \right )}+\frac {\sqrt {-\frac {b x +a -1}{b x +a}}\, \sqrt {\frac {b x +a +1}{b x +a}}\, \arctanh \left (\frac {1}{\sqrt {1-\left (b x +a \right )^{2}}}\right ) x \,b^{2}}{\sqrt {1-\left (b x +a \right )^{2}}\, a \left (a -1\right ) \left (1+a \right )}+\frac {b \sqrt {-\frac {b x +a -1}{b x +a}}\, \sqrt {\frac {b x +a +1}{b x +a}}\, \arctanh \left (\frac {1}{\sqrt {1-\left (b x +a \right )^{2}}}\right )}{\sqrt {1-\left (b x +a \right )^{2}}\, \left (a -1\right ) \left (1+a \right )} \]

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

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

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acosh}\left (\frac {1}{a+b\,x}\right )}{x^2} \,d x \]

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

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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