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

Optimal. Leaf size=107 \[ -\frac {a^2 \text {sech}^{-1}(a+b x)}{2 b^2}-\frac {\sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1)}{2 b^2}+\frac {a \tan ^{-1}\left (\frac {\sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1)}{a+b x}\right )}{b^2}+\frac {1}{2} x^2 \text {sech}^{-1}(a+b x) \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6321, 5468, 3773, 3770, 3767, 8} \[ -\frac {a^2 \text {sech}^{-1}(a+b x)}{2 b^2}-\frac {\sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1)}{2 b^2}+\frac {a \tan ^{-1}\left (\frac {\sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1)}{a+b x}\right )}{b^2}+\frac {1}{2} x^2 \text {sech}^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x*ArcSech[a + b*x],x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 3770

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

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

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Mathematica [C]  time = 0.17, size = 176, normalized size = 1.64 \[ \frac {a^2 \log (a+b x)-a^2 \log \left (a \sqrt {-\frac {a+b x-1}{a+b x+1}}+b x \sqrt {-\frac {a+b x-1}{a+b x+1}}+\sqrt {-\frac {a+b x-1}{a+b x+1}}+1\right )+b^2 x^2 \text {sech}^{-1}(a+b x)-\sqrt {-\frac {a+b x-1}{a+b x+1}} (a+b x+1)-2 i a \log \left (2 \sqrt {-\frac {a+b x-1}{a+b x+1}} (a+b x+1)-2 i (a+b x)\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcSech[a + b*x],x]

[Out]

(-(Sqrt[-((-1 + a + b*x)/(1 + a + b*x))]*(1 + a + b*x)) + b^2*x^2*ArcSech[a + b*x] + a^2*Log[a + b*x] - a^2*Lo
g[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))]] - (2*I)*a*Log[(-2*I)*(a + b*x) + 2*Sqrt[-((-1 + a + b*x)/(1 + a + b*x))]*(1 + a + b*x)])/
(2*b^2)

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fricas [B]  time = 1.39, size = 308, normalized size = 2.88 \[ \frac {2 \, b^{2} x^{2} \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{b x + a}\right ) - a^{2} \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{x}\right ) + a^{2} \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - 1}{x}\right ) + 4 \, a \arctan \left (\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - 2 \, {\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{4 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*b^2*x^2*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^2*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*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) + 4*a*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)) - 2*(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)))/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arcsech(b*x + a), x)

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maple [A]  time = 0.07, size = 111, normalized size = 1.04 \[ \frac {\frac {\mathrm {arcsech}\left (b x +a \right ) \left (b x +a \right )^{2}}{2}-\mathrm {arcsech}\left (b x +a \right ) a \left (b x +a \right )-\frac {\sqrt {-\frac {b x +a -1}{b x +a}}\, \left (b x +a \right ) \sqrt {\frac {b x +a +1}{b x +a}}\, \left (2 a \arcsin \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )}{2 \sqrt {1-\left (b x +a \right )^{2}}}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {2 \, b^{2} x^{2} \log \left (\sqrt {b x + a + 1} \sqrt {-b x - a + 1} b x + \sqrt {b x + a + 1} \sqrt {-b x - a + 1} a + b x + a\right ) - 2 \, b^{2} x^{2} \log \left (b x + a\right ) - {\left (a^{2} + 2 \, a + 1\right )} \log \left (b x + a + 1\right ) - 2 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (b x + a\right ) - {\left (a^{2} - 2 \, a + 1\right )} \log \left (-b x - a + 1\right )}{4 \, b^{2}} + \int \frac {b^{2} x^{3} + a b x^{2}}{2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )} e^{\left (\frac {1}{2} \, \log \left (b x + a + 1\right ) + \frac {1}{2} \, \log \left (-b x - a + 1\right )\right )} - 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*b^2*x^2*log(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*b^2*x^2*log(b*x + a) - (a^2 + 2*a + 1)*log(b*x + a + 1) - 2*(b^2*x^2 - a^2)*log(b*x + a) - (a^2 - 2*a +
1)*log(-b*x - a + 1))/b^2 + integrate(1/2*(b^2*x^3 + a*b*x^2)/(b^2*x^2 + 2*a*b*x + a^2 + (b^2*x^2 + 2*a*b*x +
a^2 - 1)*e^(1/2*log(b*x + a + 1) + 1/2*log(-b*x - a + 1)) - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*acosh(1/(a + b*x)),x)

[Out]

int(x*acosh(1/(a + b*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*asech(a + b*x), x)

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