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

Optimal. Leaf size=78 \[ \frac{1}{3} x^3 \left (a+b \text{sech}^{-1}(c x)\right )-\frac{b x \sqrt{1-c x}}{6 c^2 \sqrt{\frac{1}{c x+1}}}+\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sin ^{-1}(c x)}{6 c^3} \]

[Out]

-(b*x*Sqrt[1 - c*x])/(6*c^2*Sqrt[(1 + c*x)^(-1)]) + (x^3*(a + b*ArcSech[c*x]))/3 + (b*Sqrt[(1 + c*x)^(-1)]*Sqr
t[1 + c*x]*ArcSin[c*x])/(6*c^3)

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Rubi [A]  time = 0.0274582, antiderivative size = 78, 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 = {6283, 90, 41, 216} \[ \frac{1}{3} x^3 \left (a+b \text{sech}^{-1}(c x)\right )-\frac{b x \sqrt{1-c x}}{6 c^2 \sqrt{\frac{1}{c x+1}}}+\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sin ^{-1}(c x)}{6 c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcSech[c*x]),x]

[Out]

-(b*x*Sqrt[1 - c*x])/(6*c^2*Sqrt[(1 + c*x)^(-1)]) + (x^3*(a + b*ArcSech[c*x]))/3 + (b*Sqrt[(1 + c*x)^(-1)]*Sqr
t[1 + c*x]*ArcSin[c*x])/(6*c^3)

Rule 6283

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSech[c*
x]))/(d*(m + 1)), x] + Dist[(b*Sqrt[1 + c*x]*Sqrt[1/(1 + c*x)])/(m + 1), Int[(d*x)^m/(Sqrt[1 - c*x]*Sqrt[1 + c
*x]), x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1]

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 216

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

Rubi steps

\begin{align*} \int x^2 \left (a+b \text{sech}^{-1}(c x)\right ) \, dx &=\frac{1}{3} x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{3} \left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{x^2}{\sqrt{1-c x} \sqrt{1+c x}} \, dx\\ &=-\frac{b x \sqrt{1-c x}}{6 c^2 \sqrt{\frac{1}{1+c x}}}+\frac{1}{3} x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{1}{\sqrt{1-c x} \sqrt{1+c x}} \, dx}{6 c^2}\\ &=-\frac{b x \sqrt{1-c x}}{6 c^2 \sqrt{\frac{1}{1+c x}}}+\frac{1}{3} x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{1}{\sqrt{1-c^2 x^2}} \, dx}{6 c^2}\\ &=-\frac{b x \sqrt{1-c x}}{6 c^2 \sqrt{\frac{1}{1+c x}}}+\frac{1}{3} x^3 \left (a+b \text{sech}^{-1}(c x)\right )+\frac{b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sin ^{-1}(c x)}{6 c^3}\\ \end{align*}

Mathematica [C]  time = 0.0865029, size = 103, normalized size = 1.32 \[ \frac{a x^3}{3}+b \sqrt{\frac{1-c x}{c x+1}} \left (-\frac{x}{6 c^2}-\frac{x^2}{6 c}\right )+\frac{i b \log \left (2 \sqrt{\frac{1-c x}{c x+1}} (c x+1)-2 i c x\right )}{6 c^3}+\frac{1}{3} b x^3 \text{sech}^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*ArcSech[c*x]),x]

[Out]

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

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Maple [A]  time = 0.192, size = 96, normalized size = 1.2 \begin{align*}{\frac{1}{{c}^{3}} \left ({\frac{{c}^{3}{x}^{3}a}{3}}+b \left ({\frac{{c}^{3}{x}^{3}{\rm arcsech} \left (cx\right )}{3}}+{\frac{cx}{6}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}} \left ( -cx\sqrt{-{c}^{2}{x}^{2}+1}+\arcsin \left ( cx \right ) \right ){\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.53006, size = 99, normalized size = 1.27 \begin{align*} \frac{1}{3} \, a x^{3} + \frac{1}{6} \,{\left (2 \, x^{3} \operatorname{arsech}\left (c x\right ) - \frac{\frac{\sqrt{\frac{1}{c^{2} x^{2}} - 1}}{c^{2}{\left (\frac{1}{c^{2} x^{2}} - 1\right )} + c^{2}} + \frac{\arctan \left (\sqrt{\frac{1}{c^{2} x^{2}} - 1}\right )}{c^{2}}}{c}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*x^3 + 1/6*(2*x^3*arcsech(c*x) - (sqrt(1/(c^2*x^2) - 1)/(c^2*(1/(c^2*x^2) - 1) + c^2) + arctan(sqrt(1/(c^
2*x^2) - 1))/c^2)/c)*b

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Fricas [B]  time = 2.12044, size = 352, normalized size = 4.51 \begin{align*} \frac{2 \, a c^{3} x^{3} - b c^{2} x^{2} \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - 2 \, b c^{3} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) - 2 \, b \arctan \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{c x}\right ) + 2 \,{\left (b c^{3} x^{3} - b c^{3}\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )}{6 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a*c^3*x^3 - b*c^2*x^2*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 2*b*c^3*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2))
- 1)/x) - 2*b*arctan((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 1)/(c*x)) + 2*(b*c^3*x^3 - b*c^3)*log((c*x*sqrt(-(c
^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)))/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*(a + b*asech(c*x)), x)

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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 )} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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