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

Optimal. Leaf size=140 \[ \frac{i b^2 \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(c x)}\right )}{3 c^3}-\frac{i b^2 \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(c x)}\right )}{3 c^3}-\frac{b x \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )}{3 c^2}-\frac{2 b \tan ^{-1}\left (e^{\text{sech}^{-1}(c x)}\right ) \left (a+b \text{sech}^{-1}(c x)\right )}{3 c^3}+\frac{1}{3} x^3 \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{b^2 x}{3 c^2} \]

[Out]

-(b^2*x)/(3*c^2) - (b*x*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x]))/(3*c^2) + (x^3*(a + b*ArcSec
h[c*x])^2)/3 - (2*b*(a + b*ArcSech[c*x])*ArcTan[E^ArcSech[c*x]])/(3*c^3) + ((I/3)*b^2*PolyLog[2, (-I)*E^ArcSec
h[c*x]])/c^3 - ((I/3)*b^2*PolyLog[2, I*E^ArcSech[c*x]])/c^3

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Rubi [A]  time = 0.12288, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {6285, 5451, 4185, 4180, 2279, 2391} \[ \frac{i b^2 \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(c x)}\right )}{3 c^3}-\frac{i b^2 \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(c x)}\right )}{3 c^3}-\frac{b x \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )}{3 c^2}-\frac{2 b \tan ^{-1}\left (e^{\text{sech}^{-1}(c x)}\right ) \left (a+b \text{sech}^{-1}(c x)\right )}{3 c^3}+\frac{1}{3} x^3 \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{b^2 x}{3 c^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcSech[c*x])^2,x]

[Out]

-(b^2*x)/(3*c^2) - (b*x*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x]))/(3*c^2) + (x^3*(a + b*ArcSec
h[c*x])^2)/3 - (2*b*(a + b*ArcSech[c*x])*ArcTan[E^ArcSech[c*x]])/(3*c^3) + ((I/3)*b^2*PolyLog[2, (-I)*E^ArcSec
h[c*x]])/c^3 - ((I/3)*b^2*PolyLog[2, I*E^ArcSech[c*x]])/c^3

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 4185

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> -Simp[(b^2*(c + d*x)*Cot[e + f*x]*
(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2
), x], x] - Simp[(b^2*d*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[{b, c, d, e, f}, x] && G
tQ[n, 1] && NeQ[n, 2]

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int x^2 \left (a+b \text{sech}^{-1}(c x)\right )^2 \, dx &=-\frac{\operatorname{Subst}\left (\int (a+b x)^2 \text{sech}^3(x) \tanh (x) \, dx,x,\text{sech}^{-1}(c x)\right )}{c^3}\\ &=\frac{1}{3} x^3 \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \text{sech}^3(x) \, dx,x,\text{sech}^{-1}(c x)\right )}{3 c^3}\\ &=-\frac{b^2 x}{3 c^2}-\frac{b x \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{3 c^2}+\frac{1}{3} x^3 \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{b \operatorname{Subst}\left (\int (a+b x) \text{sech}(x) \, dx,x,\text{sech}^{-1}(c x)\right )}{3 c^3}\\ &=-\frac{b^2 x}{3 c^2}-\frac{b x \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{3 c^2}+\frac{1}{3} x^3 \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{2 b \left (a+b \text{sech}^{-1}(c x)\right ) \tan ^{-1}\left (e^{\text{sech}^{-1}(c x)}\right )}{3 c^3}+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\text{sech}^{-1}(c x)\right )}{3 c^3}-\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\text{sech}^{-1}(c x)\right )}{3 c^3}\\ &=-\frac{b^2 x}{3 c^2}-\frac{b x \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{3 c^2}+\frac{1}{3} x^3 \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{2 b \left (a+b \text{sech}^{-1}(c x)\right ) \tan ^{-1}\left (e^{\text{sech}^{-1}(c x)}\right )}{3 c^3}+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{\text{sech}^{-1}(c x)}\right )}{3 c^3}-\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{\text{sech}^{-1}(c x)}\right )}{3 c^3}\\ &=-\frac{b^2 x}{3 c^2}-\frac{b x \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{3 c^2}+\frac{1}{3} x^3 \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{2 b \left (a+b \text{sech}^{-1}(c x)\right ) \tan ^{-1}\left (e^{\text{sech}^{-1}(c x)}\right )}{3 c^3}+\frac{i b^2 \text{Li}_2\left (-i e^{\text{sech}^{-1}(c x)}\right )}{3 c^3}-\frac{i b^2 \text{Li}_2\left (i e^{\text{sech}^{-1}(c x)}\right )}{3 c^3}\\ \end{align*}

Mathematica [A]  time = 1.15445, size = 224, normalized size = 1.6 \[ \frac{1}{3} \left (\frac{b^2 \left (i \text{PolyLog}\left (2,-i e^{-\text{sech}^{-1}(c x)}\right )-i \text{PolyLog}\left (2,i e^{-\text{sech}^{-1}(c x)}\right )+c^3 x^3 \text{sech}^{-1}(c x)^2-c x-c x \sqrt{\frac{1-c x}{c x+1}} (c x+1) \text{sech}^{-1}(c x)+i \text{sech}^{-1}(c x) \log \left (1-i e^{-\text{sech}^{-1}(c x)}\right )-i \text{sech}^{-1}(c x) \log \left (1+i e^{-\text{sech}^{-1}(c x)}\right )\right )}{c^3}+a^2 x^3+a b \left (2 x^3 \text{sech}^{-1}(c x)-\frac{\sqrt{\frac{1-c x}{c x+1}} \left (c^3 x^3+\sqrt{1-c^2 x^2} \sin ^{-1}(c x)-c x\right )}{c^3 (c x-1)}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2*(a + b*ArcSech[c*x])^2,x]

[Out]

(a^2*x^3 + a*b*(2*x^3*ArcSech[c*x] - (Sqrt[(1 - c*x)/(1 + c*x)]*(-(c*x) + c^3*x^3 + Sqrt[1 - c^2*x^2]*ArcSin[c
*x]))/(c^3*(-1 + c*x))) + (b^2*(-(c*x) - c*x*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*ArcSech[c*x] + c^3*x^3*ArcSec
h[c*x]^2 + I*ArcSech[c*x]*Log[1 - I/E^ArcSech[c*x]] - I*ArcSech[c*x]*Log[1 + I/E^ArcSech[c*x]] + I*PolyLog[2,
(-I)/E^ArcSech[c*x]] - I*PolyLog[2, I/E^ArcSech[c*x]]))/c^3)/3

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Maple [A]  time = 0.278, size = 372, normalized size = 2.7 \begin{align*}{\frac{{x}^{3}{a}^{2}}{3}}-{\frac{{b}^{2}{\rm arcsech} \left (cx\right ){x}^{2}}{3\,c}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}}+{\frac{{x}^{3}{b}^{2} \left ({\rm arcsech} \left (cx\right ) \right ) ^{2}}{3}}-{\frac{{b}^{2}x}{3\,{c}^{2}}}+{\frac{{\frac{i}{3}}{b}^{2}{\rm arcsech} \left (cx\right )}{{c}^{3}}\ln \left ( 1+i \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) \right ) }-{\frac{{\frac{i}{3}}{b}^{2}{\rm arcsech} \left (cx\right )}{{c}^{3}}\ln \left ( 1-i \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) \right ) }+{\frac{{\frac{i}{3}}{b}^{2}}{{c}^{3}}{\it dilog} \left ( 1+i \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) \right ) }-{\frac{{\frac{i}{3}}{b}^{2}}{{c}^{3}}{\it dilog} \left ( 1-i \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) \right ) }+{\frac{2\,ab{x}^{3}{\rm arcsech} \left (cx\right )}{3}}-{\frac{{x}^{2}ab}{3\,c}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}}+{\frac{xab\arcsin \left ( cx \right ) }{3\,{c}^{2}}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}{\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*(a + b*asech(c*x))**2, 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 )}^{2} 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))^2,x, algorithm="giac")

[Out]

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