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

Optimal. Leaf size=65 \[ -\frac{b \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )}{c^2}+\frac{1}{2} x^2 \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{b^2 \log (x)}{c^2} \]

[Out]

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

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Rubi [A]  time = 0.0752577, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6285, 5451, 4184, 3475} \[ -\frac{b \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )}{c^2}+\frac{1}{2} x^2 \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{b^2 \log (x)}{c^2} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcSech[c*x])^2,x]

[Out]

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

Rule 6285

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Dist[(c^(m + 1))^(-1), Subst[Int[(a + b
*x)^n*Sech[x]^(m + 1)*Tanh[x], x], x, ArcSech[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] &
& (GtQ[n, 0] || LtQ[m, -1])

Rule 5451

Int[((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.)*Tanh[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> -Si
mp[((c + d*x)^m*Sech[a + b*x]^n)/(b*n), x] + Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Sech[a + b*x]^n, x], x] /
; FreeQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3475

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

Rubi steps

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

Mathematica [A]  time = 0.208577, size = 112, normalized size = 1.72 \[ \frac{a \left (a c^2 x^2-2 b \sqrt{\frac{1-c x}{c x+1}} (c x+1)\right )-2 b \text{sech}^{-1}(c x) \left (b \sqrt{\frac{1-c x}{c x+1}} (c x+1)-a c^2 x^2\right )+b^2 c^2 x^2 \text{sech}^{-1}(c x)^2-2 b^2 \log (c x)}{2 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*ArcSech[c*x])^2,x]

[Out]

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

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Maple [B]  time = 0.253, size = 168, normalized size = 2.6 \begin{align*}{\frac{{a}^{2}{x}^{2}}{2}}-{\frac{{b}^{2}{\rm arcsech} \left (cx\right )}{{c}^{2}}}-{\frac{{b}^{2}{\rm arcsech} \left (cx\right )x}{c}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}}+{\frac{{x}^{2}{b}^{2} \left ({\rm arcsech} \left (cx\right ) \right ) ^{2}}{2}}+{\frac{{b}^{2}}{{c}^{2}}\ln \left ( 1+ \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) ^{2} \right ) }+ab{\rm arcsech} \left (cx\right ){x}^{2}-{\frac{xab}{c}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01994, size = 113, normalized size = 1.74 \begin{align*} \frac{1}{2} \, b^{2} x^{2} \operatorname{arsech}\left (c x\right )^{2} + \frac{1}{2} \, a^{2} x^{2} +{\left (x^{2} \operatorname{arsech}\left (c x\right ) - \frac{x \sqrt{\frac{1}{c^{2} x^{2}} - 1}}{c}\right )} a b -{\left (\frac{x \sqrt{\frac{1}{c^{2} x^{2}} - 1} \operatorname{arsech}\left (c x\right )}{c} + \frac{\log \left (x\right )}{c^{2}}\right )} b^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.06572, size = 446, normalized size = 6.86 \begin{align*} \frac{b^{2} c^{2} x^{2} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )^{2} + a^{2} c^{2} x^{2} - 2 \, a b c^{2} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) - 2 \, a b c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - 2 \, b^{2} \log \left (x\right ) + 2 \,{\left (a b c^{2} x^{2} - b^{2} c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - a b c^{2}\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )}{2 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.94712, size = 99, normalized size = 1.52 \begin{align*} \begin{cases} \frac{a^{2} x^{2}}{2} + a b x^{2} \operatorname{asech}{\left (c x \right )} - \frac{a b \sqrt{- c^{2} x^{2} + 1}}{c^{2}} + \frac{b^{2} x^{2} \operatorname{asech}^{2}{\left (c x \right )}}{2} - \frac{b^{2} \sqrt{- c^{2} x^{2} + 1} \operatorname{asech}{\left (c x \right )}}{c^{2}} - \frac{b^{2} \log{\left (x \right )}}{c^{2}} & \text{for}\: c \neq 0 \\\frac{x^{2} \left (a + \infty b\right )^{2}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*x**2/2 + a*b*x**2*asech(c*x) - a*b*sqrt(-c**2*x**2 + 1)/c**2 + b**2*x**2*asech(c*x)**2/2 - b**
2*sqrt(-c**2*x**2 + 1)*asech(c*x)/c**2 - b**2*log(x)/c**2, Ne(c, 0)), (x**2*(a + oo*b)**2/2, True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsech(c*x) + a)^2*x, x)