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

Optimal. Leaf size=44 \[ \frac{(a+b x) \text{sech}^{-1}(a+b x)}{b}-\frac{2 \tan ^{-1}\left (\sqrt{\frac{-a-b x+1}{a+b x+1}}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.0556131, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {6313, 1961, 12, 203} \[ \frac{(a+b x) \text{sech}^{-1}(a+b x)}{b}-\frac{2 \tan ^{-1}\left (\sqrt{\frac{-a-b x+1}{a+b x+1}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcSech[a + b*x],x]

[Out]

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

Rule 6313

Int[ArcSech[(c_) + (d_.)*(x_)], x_Symbol] :> Simp[((c + d*x)*ArcSech[c + d*x])/d, x] + Int[Sqrt[(1 - c - d*x)/
(1 + c + d*x)]/(1 - c - d*x), x] /; FreeQ[{c, d}, x]

Rule 1961

Int[(u_)^(r_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> With[{q = Den
ominator[p]}, Dist[(q*e*(b*c - a*d))/n, Subst[Int[SimplifyIntegrand[(x^(q*(p + 1) - 1)*(-(a*e) + c*x^q)^(1/n -
1)*(u /. x -> (-(a*e) + c*x^q)^(1/n)/(b*e - d*x^q)^(1/n))^r)/(b*e - d*x^q)^(1/n + 1), x], x], x, ((e*(a + b*x
^n))/(c + d*x^n))^(1/q)], x]] /; FreeQ[{a, b, c, d, e}, x] && PolynomialQ[u, x] && FractionQ[p] && IntegerQ[1/
n] && IntegerQ[r]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [B]  time = 0.195596, size = 95, normalized size = 2.16 \[ x \text{sech}^{-1}(a+b x)+\frac{2 \sqrt{-\frac{a+b x-1}{a+b x+1}} \left (a \tan ^{-1}\left (\sqrt{\frac{a+b x-1}{a+b x+1}}\right )-\sinh ^{-1}\left (\frac{\sqrt{a+b x-1}}{\sqrt{2}}\right )\right )}{b \sqrt{\frac{a+b x-1}{a+b x+1}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSech[a + b*x],x]

[Out]

x*ArcSech[a + b*x] + (2*Sqrt[-((-1 + a + b*x)/(1 + a + b*x))]*(-ArcSinh[Sqrt[-1 + a + b*x]/Sqrt[2]] + a*ArcTan
[Sqrt[(-1 + a + b*x)/(1 + a + b*x)]]))/(b*Sqrt[(-1 + a + b*x)/(1 + a + b*x)])

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Maple [A]  time = 0.213, size = 50, normalized size = 1.1 \begin{align*} x{\rm arcsech} \left (bx+a\right )+{\frac{{\rm arcsech} \left (bx+a\right )a}{b}}-{\frac{1}{b}\arctan \left ( \sqrt{ \left ( bx+a \right ) ^{-1}-1}\sqrt{ \left ( bx+a \right ) ^{-1}+1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsech(b*x+a),x)

[Out]

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

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Maxima [A]  time = 0.9777, size = 42, normalized size = 0.95 \begin{align*} \frac{{\left (b x + a\right )} \operatorname{arsech}\left (b x + a\right ) - \arctan \left (\sqrt{\frac{1}{{\left (b x + a\right )}^{2}} - 1}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsech(b*x+a),x, algorithm="maxima")

[Out]

((b*x + a)*arcsech(b*x + a) - arctan(sqrt(1/(b*x + a)^2 - 1)))/b

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Fricas [B]  time = 2.1473, size = 567, normalized size = 12.89 \begin{align*} \frac{2 \, 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}{b x + a}\right ) + a \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 ) - a \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 \, \arctan \left (\frac{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt{-\frac{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*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)/(b*x + a)) + a*l
og(((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*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*arctan((b^2*x^2 + 2*a*b*x + a^2)*sqrt(-(b
^2*x^2 + 2*a*b*x + a^2 - 1)/(b^2*x^2 + 2*a*b*x + a^2))/(b^2*x^2 + 2*a*b*x + a^2 - 1)))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asech(a + b*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{arsech}\left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsech(b*x+a),x, algorithm="giac")

[Out]

integrate(arcsech(b*x + a), x)