3.1.4 \(\int \text {sech}^{-1}(a+b x) \, dx\) [4]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 44, normalized size of antiderivative = 1.00, 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 = {6448, 1983, 12, 209} \begin {gather*} \frac {(a+b x) \text {sech}^{-1}(a+b x)}{b}-\frac {2 \text {ArcTan}\left (\sqrt {\frac {-a-b x+1}{a+b x+1}}\right )}{b} \end {gather*}

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 12

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

Rule 209

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

Rule 1983

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)/(b*e - d*x^q)^(1/n + 1))*(u /. x -> ((-a)*e + c*x^q)^(1/n)/(b*e - d*x^q)^(1/n))^r, 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 6448

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]

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) \text {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 \text {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] Leaf count is larger than twice the leaf count of optimal. \(97\) vs. \(2(44)=88\).
time = 0.18, size = 97, normalized size = 2.20 \begin {gather*} x \text {sech}^{-1}(a+b x)-\frac {2 \sqrt {-\frac {-1+a+b x}{1+a+b x}} \left (-a \text {ArcTan}\left (\sqrt {\frac {-1+a+b x}{1+a+b x}}\right )+\tanh ^{-1}\left (\sqrt {\frac {-1+a+b x}{1+a+b x}}\right )\right )}{b \sqrt {\frac {-1+a+b x}{1+a+b x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSech[a + b*x],x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.04, size = 44, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {\left (b x +a \right ) \mathrm {arcsech}\left (b x +a \right )-\arctan \left (\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )}{b}\)	\(44\)
default	\(\frac {\left (b x +a \right ) \mathrm {arcsech}\left (b x +a \right )-\arctan \left (\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )}{b}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.25, size = 31, normalized size = 0.70 \begin {gather*} \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 {gather*}

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

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (40) = 80\).
time = 0.44, size = 253, normalized size = 5.75 \begin {gather*} \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 {gather*}

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asech(a + b*x), x)

________________________________________________________________________________________

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, algorithm="giac")

[Out]

integrate(arcsech(b*x + a), x)

________________________________________________________________________________________

Mupad [B]
time = 2.16, size = 43, normalized size = 0.98 \begin {gather*} \frac {\mathrm {atan}\left (\frac {1}{\sqrt {\frac {1}{a+b\,x}-1}\,\sqrt {\frac {1}{a+b\,x}+1}}\right )+\mathrm {acosh}\left (\frac {1}{a+b\,x}\right )\,\left (a+b\,x\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acosh(1/(a + b*x)),x)

[Out]

(atan(1/((1/(a + b*x) - 1)^(1/2)*(1/(a + b*x) + 1)^(1/2))) + acosh(1/(a + b*x))*(a + b*x))/b

________________________________________________________________________________________