3.44 \(\int x (a+b \text{sech}^{-1}(c x))^3 \, dx\)

Optimal. Leaf size=126 \[ \frac{3 b^3 \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(c x)}\right )}{2 c^2}+\frac{3 b^2 \log \left (e^{2 \text{sech}^{-1}(c x)}+1\right ) \left (a+b \text{sech}^{-1}(c x)\right )}{c^2}-\frac{3 b \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )^2}{2 c^2}-\frac{3 b \left (a+b \text{sech}^{-1}(c x)\right )^2}{2 c^2}+\frac{1}{2} x^2 \left (a+b \text{sech}^{-1}(c x)\right )^3 \]

[Out]

(-3*b*(a + b*ArcSech[c*x])^2)/(2*c^2) - (3*b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x])^2)/(2*c^
2) + (x^2*(a + b*ArcSech[c*x])^3)/2 + (3*b^2*(a + b*ArcSech[c*x])*Log[1 + E^(2*ArcSech[c*x])])/c^2 + (3*b^3*Po
lyLog[2, -E^(2*ArcSech[c*x])])/(2*c^2)

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Rubi [A]  time = 0.156583, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {6285, 5451, 4184, 3718, 2190, 2279, 2391} \[ \frac{3 b^3 \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(c x)}\right )}{2 c^2}+\frac{3 b^2 \log \left (e^{2 \text{sech}^{-1}(c x)}+1\right ) \left (a+b \text{sech}^{-1}(c x)\right )}{c^2}-\frac{3 b \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )^2}{2 c^2}-\frac{3 b \left (a+b \text{sech}^{-1}(c x)\right )^2}{2 c^2}+\frac{1}{2} x^2 \left (a+b \text{sech}^{-1}(c x)\right )^3 \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcSech[c*x])^3,x]

[Out]

(-3*b*(a + b*ArcSech[c*x])^2)/(2*c^2) - (3*b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x])^2)/(2*c^
2) + (x^2*(a + b*ArcSech[c*x])^3)/2 + (3*b^2*(a + b*ArcSech[c*x])*Log[1 + E^(2*ArcSech[c*x])])/c^2 + (3*b^3*Po
lyLog[2, -E^(2*ArcSech[c*x])])/(2*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 3718

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

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && 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 \left (a+b \text{sech}^{-1}(c x)\right )^3 \, dx &=-\frac{\operatorname{Subst}\left (\int (a+b x)^3 \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 )^3-\frac{(3 b) \operatorname{Subst}\left (\int (a+b x)^2 \text{sech}^2(x) \, dx,x,\text{sech}^{-1}(c x)\right )}{2 c^2}\\ &=-\frac{3 b \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )^2}{2 c^2}+\frac{1}{2} x^2 \left (a+b \text{sech}^{-1}(c x)\right )^3+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int (a+b x) \tanh (x) \, dx,x,\text{sech}^{-1}(c x)\right )}{c^2}\\ &=-\frac{3 b \left (a+b \text{sech}^{-1}(c x)\right )^2}{2 c^2}-\frac{3 b \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )^2}{2 c^2}+\frac{1}{2} x^2 \left (a+b \text{sech}^{-1}(c x)\right )^3+\frac{\left (6 b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 x} (a+b x)}{1+e^{2 x}} \, dx,x,\text{sech}^{-1}(c x)\right )}{c^2}\\ &=-\frac{3 b \left (a+b \text{sech}^{-1}(c x)\right )^2}{2 c^2}-\frac{3 b \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )^2}{2 c^2}+\frac{1}{2} x^2 \left (a+b \text{sech}^{-1}(c x)\right )^3+\frac{3 b^2 \left (a+b \text{sech}^{-1}(c x)\right ) \log \left (1+e^{2 \text{sech}^{-1}(c x)}\right )}{c^2}-\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\text{sech}^{-1}(c x)\right )}{c^2}\\ &=-\frac{3 b \left (a+b \text{sech}^{-1}(c x)\right )^2}{2 c^2}-\frac{3 b \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )^2}{2 c^2}+\frac{1}{2} x^2 \left (a+b \text{sech}^{-1}(c x)\right )^3+\frac{3 b^2 \left (a+b \text{sech}^{-1}(c x)\right ) \log \left (1+e^{2 \text{sech}^{-1}(c x)}\right )}{c^2}-\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \text{sech}^{-1}(c x)}\right )}{2 c^2}\\ &=-\frac{3 b \left (a+b \text{sech}^{-1}(c x)\right )^2}{2 c^2}-\frac{3 b \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )^2}{2 c^2}+\frac{1}{2} x^2 \left (a+b \text{sech}^{-1}(c x)\right )^3+\frac{3 b^2 \left (a+b \text{sech}^{-1}(c x)\right ) \log \left (1+e^{2 \text{sech}^{-1}(c x)}\right )}{c^2}+\frac{3 b^3 \text{Li}_2\left (-e^{2 \text{sech}^{-1}(c x)}\right )}{2 c^2}\\ \end{align*}

Mathematica [A]  time = 0.867704, size = 219, normalized size = 1.74 \[ \frac{-3 b^3 \text{PolyLog}\left (2,-e^{-2 \text{sech}^{-1}(c x)}\right )+a \left (a \left (a c^2 x^2-3 b \sqrt{\frac{1-c x}{c x+1}} (c x+1)\right )+6 b^2 \log \left (\frac{1}{c x}\right )\right )-3 b^2 \text{sech}^{-1}(c x)^2 \left (b \left (c x \sqrt{\frac{1-c x}{c x+1}}+\sqrt{\frac{1-c x}{c x+1}}-1\right )-a c^2 x^2\right )+3 b \text{sech}^{-1}(c x) \left (a \left (a c^2 x^2-2 b \sqrt{\frac{1-c x}{c x+1}} (c x+1)\right )+2 b^2 \log \left (e^{-2 \text{sech}^{-1}(c x)}+1\right )\right )+b^3 c^2 x^2 \text{sech}^{-1}(c x)^3}{2 c^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x*(a + b*ArcSech[c*x])^3,x]

[Out]

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

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Maple [B]  time = 0.302, size = 343, normalized size = 2.7 \begin{align*}{\frac{{x}^{2}{a}^{3}}{2}}-{\frac{3\,{b}^{3} \left ({\rm arcsech} \left (cx\right ) \right ) ^{2}x}{2\,c}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}}+{\frac{{x}^{2}{b}^{3} \left ({\rm arcsech} \left (cx\right ) \right ) ^{3}}{2}}-{\frac{3\,{b}^{3} \left ({\rm arcsech} \left (cx\right ) \right ) ^{2}}{2\,{c}^{2}}}+3\,{\frac{{b}^{3}{\rm arcsech} \left (cx\right )}{{c}^{2}}\ln \left ( 1+ \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) ^{2} \right ) }+{\frac{3\,{b}^{3}}{2\,{c}^{2}}{\it polylog} \left ( 2,- \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) ^{2} \right ) }-3\,{\frac{a{b}^{2}{\rm arcsech} \left (cx\right )}{{c}^{2}}}-3\,{\frac{a{b}^{2}{\rm arcsech} \left (cx\right )x}{c}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}}+{\frac{3\,{x}^{2}a{b}^{2} \left ({\rm arcsech} \left (cx\right ) \right ) ^{2}}{2}}+3\,{\frac{a{b}^{2}}{{c}^{2}}\ln \left ( 1+ \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) ^{2} \right ) }+{\frac{3\,{x}^{2}{a}^{2}b{\rm arcsech} \left (cx\right )}{2}}-{\frac{3\,{a}^{2}bx}{2\,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))^3,x)

[Out]

1/2*x^2*a^3-3/2/c*b^3*arcsech(c*x)^2*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*x+1/2*x^2*b^3*arcsech(c*x)^3-3/2
/c^2*b^3*arcsech(c*x)^2+3/c^2*b^3*arcsech(c*x)*ln(1+(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)+3/2*b^3*polylo
g(2,-(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)/c^2-3/c^2*a*b^2*arcsech(c*x)-3/c*a*b^2*(-(c*x-1)/c/x)^(1/2)*(
(c*x+1)/c/x)^(1/2)*arcsech(c*x)*x+3/2*x^2*a*b^2*arcsech(c*x)^2+3/c^2*a*b^2*ln(1+(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c
/x)^(1/2))^2)+3/2*x^2*a^2*b*arcsech(c*x)-3/2/c*a^2*b*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsech(c*x))^3,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b^{3} x \operatorname{arsech}\left (c x\right )^{3} + 3 \, a b^{2} x \operatorname{arsech}\left (c x\right )^{2} + 3 \, a^{2} b x \operatorname{arsech}\left (c x\right ) + a^{3} x, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsech(c*x))^3,x, algorithm="fricas")

[Out]

integral(b^3*x*arcsech(c*x)^3 + 3*a*b^2*x*arcsech(c*x)^2 + 3*a^2*b*x*arcsech(c*x) + a^3*x, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \left (a + b \operatorname{asech}{\left (c x \right )}\right )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*asech(c*x))**3,x)

[Out]

Integral(x*(a + b*asech(c*x))**3, 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 )}^{3} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsech(c*x))^3,x, algorithm="giac")

[Out]

integrate((b*arcsech(c*x) + a)^3*x, x)