3.22 \(\int \sec ^{-1}(a+b x) \, dx\)

Optimal. Leaf size=37 \[ \frac{(a+b x) \sec ^{-1}(a+b x)}{b}-\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{(a+b x)^2}}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.0230683, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.833, Rules used = {5250, 372, 266, 63, 206} \[ \frac{(a+b x) \sec ^{-1}(a+b x)}{b}-\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{(a+b x)^2}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcSec[a + b*x],x]

[Out]

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

Rule 5250

Int[ArcSec[(c_) + (d_.)*(x_)], x_Symbol] :> Simp[((c + d*x)*ArcSec[c + d*x])/d, x] - Int[1/((c + d*x)*Sqrt[1 -
1/(c + d*x)^2]), x] /; FreeQ[{c, d}, x]

Rule 372

Int[(u_)^(m_.)*((a_) + (b_.)*(v_)^(n_))^(p_.), x_Symbol] :> Dist[u^m/(Coefficient[v, x, 1]*v^m), Subst[Int[x^m
*(a + b*x^n)^p, x], x, v], x] /; FreeQ[{a, b, m, n, p}, x] && LinearPairQ[u, v, x]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

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

Rubi steps

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

Mathematica [B]  time = 0.139065, size = 115, normalized size = 3.11 \[ x \sec ^{-1}(a+b x)-\frac{(a+b x) \sqrt{\frac{a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}} \left (\tanh ^{-1}\left (\frac{a+b x}{\sqrt{a^2+2 a b x+b^2 x^2-1}}\right )-a \tan ^{-1}\left (\sqrt{(a+b x)^2-1}\right )\right )}{b \sqrt{a^2+2 a b x+b^2 x^2-1}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSec[a + b*x],x]

[Out]

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

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Maple [A]  time = 0.203, size = 51, normalized size = 1.4 \begin{align*} x{\rm arcsec} \left (bx+a\right )+{\frac{{\rm arcsec} \left (bx+a\right )a}{b}}-{\frac{1}{b}\ln \left ( bx+a+ \left ( bx+a \right ) \sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.945117, size = 74, normalized size = 2. \begin{align*} \frac{2 \,{\left (b x + a\right )} \operatorname{arcsec}\left (b x + a\right ) - \log \left (\sqrt{-\frac{1}{{\left (b x + a\right )}^{2}} + 1} + 1\right ) + \log \left (-\sqrt{-\frac{1}{{\left (b x + a\right )}^{2}} + 1} + 1\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(b*x + a)*arcsec(b*x + a) - log(sqrt(-1/(b*x + a)^2 + 1) + 1) + log(-sqrt(-1/(b*x + a)^2 + 1) + 1))/b

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Fricas [B]  time = 2.54098, size = 184, normalized size = 4.97 \begin{align*} \frac{b x \operatorname{arcsec}\left (b x + a\right ) + 2 \, a \arctan \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x*arcsec(b*x + a) + 2*a*arctan(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)) + log(-b*x - a + sqrt(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{asec}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asec(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsec(b*x + a), x)