3.2 \(\int x^2 \text{sech}^{-1}(a+b x) \, dx\)

Optimal. Leaf size=153 \[ -\frac{\left (6 a^2+1\right ) \tan ^{-1}\left (\frac{\sqrt{\frac{-a-b x+1}{a+b x+1}} (a+b x+1)}{a+b x}\right )}{6 b^3}+\frac{a^3 \text{sech}^{-1}(a+b x)}{3 b^3}+\frac{5 a \sqrt{\frac{-a-b x+1}{a+b x+1}} (a+b x+1)}{6 b^3}-\frac{x \sqrt{\frac{-a-b x+1}{a+b x+1}} (a+b x+1)}{6 b^2}+\frac{1}{3} x^3 \text{sech}^{-1}(a+b x) \]

[Out]

(5*a*Sqrt[(1 - a - b*x)/(1 + a + b*x)]*(1 + a + b*x))/(6*b^3) - (x*Sqrt[(1 - a - b*x)/(1 + a + b*x)]*(1 + a +
b*x))/(6*b^2) + (a^3*ArcSech[a + b*x])/(3*b^3) + (x^3*ArcSech[a + b*x])/3 - ((1 + 6*a^2)*ArcTan[(Sqrt[(1 - a -
b*x)/(1 + a + b*x)]*(1 + a + b*x))/(a + b*x)])/(6*b^3)

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Rubi [A]  time = 0.0985505, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {6321, 5468, 3782, 3770, 3767, 8} \[ -\frac{\left (6 a^2+1\right ) \tan ^{-1}\left (\frac{\sqrt{\frac{-a-b x+1}{a+b x+1}} (a+b x+1)}{a+b x}\right )}{6 b^3}+\frac{a^3 \text{sech}^{-1}(a+b x)}{3 b^3}+\frac{5 a \sqrt{\frac{-a-b x+1}{a+b x+1}} (a+b x+1)}{6 b^3}-\frac{x \sqrt{\frac{-a-b x+1}{a+b x+1}} (a+b x+1)}{6 b^2}+\frac{1}{3} x^3 \text{sech}^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcSech[a + b*x],x]

[Out]

(5*a*Sqrt[(1 - a - b*x)/(1 + a + b*x)]*(1 + a + b*x))/(6*b^3) - (x*Sqrt[(1 - a - b*x)/(1 + a + b*x)]*(1 + a +
b*x))/(6*b^2) + (a^3*ArcSech[a + b*x])/(3*b^3) + (x^3*ArcSech[a + b*x])/3 - ((1 + 6*a^2)*ArcTan[(Sqrt[(1 - a -
b*x)/(1 + a + b*x)]*(1 + a + b*x))/(a + b*x)])/(6*b^3)

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]

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 3782

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n
- 2))/(d*(n - 1)), x] + Dist[1/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 3)*Simp[a^3*(n - 1) + (b*(b^2*(n - 2) +
3*a^2*(n - 1)))*Csc[c + d*x] + (a*b^2*(3*n - 4))*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && GtQ[n, 2] && IntegerQ[2*n]

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^2 \text{sech}^{-1}(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int x \text{sech}(x) (-a+\text{sech}(x))^2 \tanh (x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b^3}\\ &=\frac{1}{3} x^3 \text{sech}^{-1}(a+b x)-\frac{\operatorname{Subst}\left (\int (-a+\text{sech}(x))^3 \, dx,x,\text{sech}^{-1}(a+b x)\right )}{3 b^3}\\ &=-\frac{x \sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x)}{6 b^2}+\frac{1}{3} x^3 \text{sech}^{-1}(a+b x)-\frac{\operatorname{Subst}\left (\int \left (-2 a^3+\left (1+6 a^2\right ) \text{sech}(x)-5 a \text{sech}^2(x)\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{6 b^3}\\ &=-\frac{x \sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x)}{6 b^2}+\frac{a^3 \text{sech}^{-1}(a+b x)}{3 b^3}+\frac{1}{3} x^3 \text{sech}^{-1}(a+b x)+\frac{(5 a) \operatorname{Subst}\left (\int \text{sech}^2(x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{6 b^3}-\frac{\left (1+6 a^2\right ) \operatorname{Subst}\left (\int \text{sech}(x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{6 b^3}\\ &=-\frac{x \sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x)}{6 b^2}+\frac{a^3 \text{sech}^{-1}(a+b x)}{3 b^3}+\frac{1}{3} x^3 \text{sech}^{-1}(a+b x)-\frac{\left (1+6 a^2\right ) \tan ^{-1}\left (\frac{\sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x)}{a+b x}\right )}{6 b^3}+\frac{(5 i a) \operatorname{Subst}\left (\int 1 \, dx,x,-i \sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x)\right )}{6 b^3}\\ &=\frac{5 a \sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x)}{6 b^3}-\frac{x \sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x)}{6 b^2}+\frac{a^3 \text{sech}^{-1}(a+b x)}{3 b^3}+\frac{1}{3} x^3 \text{sech}^{-1}(a+b x)-\frac{\left (1+6 a^2\right ) \tan ^{-1}\left (\frac{\sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x)}{a+b x}\right )}{6 b^3}\\ \end{align*}

Mathematica [C]  time = 0.258412, size = 200, normalized size = 1.31 \[ \frac{\sqrt{-\frac{a+b x-1}{a+b x+1}} \left (5 a^2+a (4 b x+5)-b x (b x+1)\right )-2 a^3 \log (a+b x)+2 a^3 \log \left (a \sqrt{-\frac{a+b x-1}{a+b x+1}}+b x \sqrt{-\frac{a+b x-1}{a+b x+1}}+\sqrt{-\frac{a+b x-1}{a+b x+1}}+1\right )+i \left (6 a^2+1\right ) \log \left (2 \sqrt{-\frac{a+b x-1}{a+b x+1}} (a+b x+1)-2 i (a+b x)\right )+2 b^3 x^3 \text{sech}^{-1}(a+b x)}{6 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcSech[a + b*x],x]

[Out]

(Sqrt[-((-1 + a + b*x)/(1 + a + b*x))]*(5*a^2 - b*x*(1 + b*x) + a*(5 + 4*b*x)) + 2*b^3*x^3*ArcSech[a + b*x] -
2*a^3*Log[a + b*x] + 2*a^3*Log[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))]] + I*(1 + 6*a^2)*Log[(-2*I)*(a + b*x) + 2*Sqrt[-((-1 + a + b*
x)/(1 + a + b*x))]*(1 + a + b*x)])/(6*b^3)

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Maple [A]  time = 0.244, size = 189, normalized size = 1.2 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{ \left ( bx+a \right ) ^{3}{\rm arcsech} \left (bx+a\right )}{3}}-{\rm arcsech} \left (bx+a\right ) \left ( bx+a \right ) ^{2}a+{\rm arcsech} \left (bx+a\right ) \left ( bx+a \right ){a}^{2}-{\frac{{\rm arcsech} \left (bx+a\right ){a}^{3}}{3}}+{\frac{bx+a}{6}\sqrt{-{\frac{bx+a-1}{bx+a}}}\sqrt{{\frac{bx+a+1}{bx+a}}} \left ( 2\,{a}^{3}{\it Artanh} \left ({\frac{1}{\sqrt{1- \left ( bx+a \right ) ^{2}}}} \right ) +6\,{a}^{2}\arcsin \left ( bx+a \right ) - \left ( bx+a \right ) \sqrt{1- \left ( bx+a \right ) ^{2}}+6\,a\sqrt{1- \left ( bx+a \right ) ^{2}}+\arcsin \left ( bx+a \right ) \right ){\frac{1}{\sqrt{1- \left ( bx+a \right ) ^{2}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arcsech(b*x+a),x)

[Out]

1/b^3*(1/3*(b*x+a)^3*arcsech(b*x+a)-arcsech(b*x+a)*(b*x+a)^2*a+arcsech(b*x+a)*(b*x+a)*a^2-1/3*arcsech(b*x+a)*a
^3+1/6*(-(b*x+a-1)/(b*x+a))^(1/2)*(b*x+a)*((b*x+a+1)/(b*x+a))^(1/2)*(2*a^3*arctanh(1/(1-(b*x+a)^2)^(1/2))+6*a^
2*arcsin(b*x+a)-(b*x+a)*(1-(b*x+a)^2)^(1/2)+6*a*(1-(b*x+a)^2)^(1/2)+arcsin(b*x+a))/(1-(b*x+a)^2)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \, b^{3} x^{3} \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^{3} x^{3} \log \left (b x + a\right ) - 2 \, b x +{\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} \log \left (b x + a + 1\right ) - 2 \,{\left (b^{3} x^{3} + a^{3}\right )} \log \left (b x + a\right ) +{\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} \log \left (-b x - a + 1\right )}{6 \, b^{3}} + \int \frac{b^{2} x^{4} + a b x^{3}}{3 \,{\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arcsech(b*x+a),x, algorithm="maxima")

[Out]

1/6*(2*b^3*x^3*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^3*x^3*log(b*x + a) - 2*b*x + (a^3 + 3*a^2 + 3*a + 1)*log(b*x + a + 1) - 2*(b^3*x^3 + a^3)*log(b*x + a)
+ (a^3 - 3*a^2 + 3*a - 1)*log(-b*x - a + 1))/b^3 + integrate(1/3*(b^2*x^4 + a*b*x^3)/(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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Fricas [B]  time = 2.35117, size = 720, normalized size = 4.71 \begin{align*} \frac{2 \, b^{3} x^{3} \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^{3} \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^{3} \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 ) -{\left (6 \, a^{2} + 1\right )} \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 ) -{\left (b^{2} x^{2} - 4 \, a b x - 5 \, 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}}}}{6 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arcsech(b*x+a),x, algorithm="fricas")

[Out]

1/6*(2*b^3*x^3*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^3*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^3*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) - (6*a^2 + 1)*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^2
*x^2 - 4*a*b*x - 5*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^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{asech}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*asech(b*x+a),x)

[Out]

Integral(x**2*asech(a + b*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{arsech}\left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arcsech(b*x+a),x, algorithm="giac")

[Out]

integrate(x^2*arcsech(b*x + a), x)