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

Optimal. Leaf size=78 \[ -\frac{a^2 \sec ^{-1}(a+b x)}{2 b^2}-\frac{(a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{2 b^2}+\frac{a \tanh ^{-1}\left (\sqrt{1-\frac{1}{(a+b x)^2}}\right )}{b^2}+\frac{1}{2} x^2 \sec ^{-1}(a+b x) \]

[Out]

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

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Rubi [A]  time = 0.0530349, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {5258, 4426, 3773, 3770, 3767, 8} \[ -\frac{a^2 \sec ^{-1}(a+b x)}{2 b^2}-\frac{(a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{2 b^2}+\frac{a \tanh ^{-1}\left (\sqrt{1-\frac{1}{(a+b x)^2}}\right )}{b^2}+\frac{1}{2} x^2 \sec ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x*ArcSec[a + b*x],x]

[Out]

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

Rule 5258

Int[((a_.) + ArcSec[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d^(m + 1),
Subst[Int[(a + b*x)^p*Sec[x]*Tan[x]*(d*e - c*f + f*Sec[x])^m, x], x, ArcSec[c + d*x]], x] /; FreeQ[{a, b, c,
d, e, f}, x] && IGtQ[p, 0] && IntegerQ[m]

Rule 4426

Int[((e_.) + (f_.)*(x_))^(m_.)*Sec[(c_.) + (d_.)*(x_)]*((a_) + (b_.)*Sec[(c_.) + (d_.)*(x_)])^(n_.)*Tan[(c_.)
+ (d_.)*(x_)], x_Symbol] :> Simp[((e + f*x)^m*(a + b*Sec[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*Sec[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && I
GtQ[m, 0] && NeQ[n, -1]

Rule 3773

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

Rule 3770

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.117792, size = 110, normalized size = 1.41 \[ \frac{-(a+b x) \sqrt{\frac{a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}}+2 a \log \left ((a+b x) \left (\sqrt{\frac{a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}}+1\right )\right )+a^2 \sin ^{-1}\left (\frac{1}{a+b x}\right )+b^2 x^2 \sec ^{-1}(a+b x)}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcSec[a + b*x],x]

[Out]

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

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Maple [A]  time = 0.208, size = 126, normalized size = 1.6 \begin{align*}{\frac{{x}^{2}{\rm arcsec} \left (bx+a\right )}{2}}-{\frac{{a}^{2}{\rm arcsec} \left (bx+a\right )}{2\,{b}^{2}}}+{\frac{a}{{b}^{2} \left ( bx+a \right ) }\sqrt{-1+ \left ( bx+a \right ) ^{2}}\ln \left ( bx+a+\sqrt{-1+ \left ( bx+a \right ) ^{2}} \right ){\frac{1}{\sqrt{{\frac{-1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}}}-{\frac{-1+ \left ( bx+a \right ) ^{2}}{2\,{b}^{2} \left ( bx+a \right ) }{\frac{1}{\sqrt{{\frac{-1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) - integrate(1/2*(b^2*x^3 + a*b*x^2)*e^(1/2*log(b*x + a + 1
) + 1/2*log(b*x + a - 1))/(b^2*x^2 + 2*a*b*x + a^2 + (b^2*x^2 + 2*a*b*x + a^2 - 1)*e^(log(b*x + a + 1) + log(b
*x + a - 1)) - 1), x)

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Fricas [A]  time = 2.43836, size = 254, normalized size = 3.26 \begin{align*} \frac{b^{2} x^{2} \operatorname{arcsec}\left (b x + a\right ) - 2 \, a^{2} \arctan \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - 2 \, a \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*asec(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arcsec(b*x + a), x)