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

Optimal. Leaf size=223 \[ \frac {b^2 \log \left (e^{2 \text {sech}^{-1}(c x)}+1\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{c^4}-\frac {b^2 x^2 \left (a+b \text {sech}^{-1}(c x)\right )}{4 c^2}-\frac {b \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )^2}{2 c^4}-\frac {b \left (a+b \text {sech}^{-1}(c x)\right )^2}{2 c^4}-\frac {b x^2 \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )^2}{4 c^2}+\frac {1}{4} x^4 \left (a+b \text {sech}^{-1}(c x)\right )^3+\frac {b^3 \text {Li}_2\left (-e^{2 \text {sech}^{-1}(c x)}\right )}{2 c^4}+\frac {b^3 \sqrt {\frac {1-c x}{c x+1}} (c x+1)}{4 c^4} \]

[Out]

-1/4*b^2*x^2*(a+b*arcsech(c*x))/c^2-1/2*b*(a+b*arcsech(c*x))^2/c^4+1/4*x^4*(a+b*arcsech(c*x))^3+b^2*(a+b*arcse
ch(c*x))*ln(1+(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)/c^4+1/2*b^3*polylog(2,-(1/c/x+(-1+1/c/x)^(1/2)*(1+1/
c/x)^(1/2))^2)/c^4+1/4*b^3*(c*x+1)*((-c*x+1)/(c*x+1))^(1/2)/c^4-1/2*b*(c*x+1)*(a+b*arcsech(c*x))^2*((-c*x+1)/(
c*x+1))^(1/2)/c^4-1/4*b*x^2*(c*x+1)*(a+b*arcsech(c*x))^2*((-c*x+1)/(c*x+1))^(1/2)/c^2

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Rubi [A]  time = 0.24, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {6285, 5451, 4186, 3767, 8, 4184, 3718, 2190, 2279, 2391} \[ \frac {b^3 \text {PolyLog}\left (2,-e^{2 \text {sech}^{-1}(c x)}\right )}{2 c^4}-\frac {b^2 x^2 \left (a+b \text {sech}^{-1}(c x)\right )}{4 c^2}+\frac {b^2 \log \left (e^{2 \text {sech}^{-1}(c x)}+1\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{c^4}-\frac {b x^2 \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )^2}{4 c^2}-\frac {b \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )^2}{2 c^4}-\frac {b \left (a+b \text {sech}^{-1}(c x)\right )^2}{2 c^4}+\frac {1}{4} x^4 \left (a+b \text {sech}^{-1}(c x)\right )^3+\frac {b^3 \sqrt {\frac {1-c x}{c x+1}} (c x+1)}{4 c^4} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcSech[c*x])^3,x]

[Out]

(b^3*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x))/(4*c^4) - (b^2*x^2*(a + b*ArcSech[c*x]))/(4*c^2) - (b*(a + b*ArcSech
[c*x])^2)/(2*c^4) - (b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x])^2)/(2*c^4) - (b*x^2*Sqrt[(1 -
c*x)/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x])^2)/(4*c^2) + (x^4*(a + b*ArcSech[c*x])^3)/4 + (b^2*(a + b*ArcSe
ch[c*x])*Log[1 + E^(2*ArcSech[c*x])])/c^4 + (b^3*PolyLog[2, -E^(2*ArcSech[c*x])])/(2*c^4)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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]

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 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 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 4186

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

Rubi steps

\begin {align*} \int x^3 \left (a+b \text {sech}^{-1}(c x)\right )^3 \, dx &=-\frac {\operatorname {Subst}\left (\int (a+b x)^3 \text {sech}^4(x) \tanh (x) \, dx,x,\text {sech}^{-1}(c x)\right )}{c^4}\\ &=\frac {1}{4} x^4 \left (a+b \text {sech}^{-1}(c x)\right )^3-\frac {(3 b) \operatorname {Subst}\left (\int (a+b x)^2 \text {sech}^4(x) \, dx,x,\text {sech}^{-1}(c x)\right )}{4 c^4}\\ &=-\frac {b^2 x^2 \left (a+b \text {sech}^{-1}(c x)\right )}{4 c^2}-\frac {b x^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )^2}{4 c^2}+\frac {1}{4} x^4 \left (a+b \text {sech}^{-1}(c x)\right )^3-\frac {b \operatorname {Subst}\left (\int (a+b x)^2 \text {sech}^2(x) \, dx,x,\text {sech}^{-1}(c x)\right )}{2 c^4}+\frac {b^3 \operatorname {Subst}\left (\int \text {sech}^2(x) \, dx,x,\text {sech}^{-1}(c x)\right )}{4 c^4}\\ &=-\frac {b^2 x^2 \left (a+b \text {sech}^{-1}(c x)\right )}{4 c^2}-\frac {b \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )^2}{2 c^4}-\frac {b x^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )^2}{4 c^2}+\frac {1}{4} x^4 \left (a+b \text {sech}^{-1}(c x)\right )^3+\frac {b^2 \operatorname {Subst}\left (\int (a+b x) \tanh (x) \, dx,x,\text {sech}^{-1}(c x)\right )}{c^4}+\frac {\left (i b^3\right ) \operatorname {Subst}\left (\int 1 \, dx,x,-i \sqrt {\frac {1-c x}{1+c x}} (1+c x)\right )}{4 c^4}\\ &=\frac {b^3 \sqrt {\frac {1-c x}{1+c x}} (1+c x)}{4 c^4}-\frac {b^2 x^2 \left (a+b \text {sech}^{-1}(c x)\right )}{4 c^2}-\frac {b \left (a+b \text {sech}^{-1}(c x)\right )^2}{2 c^4}-\frac {b \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )^2}{2 c^4}-\frac {b x^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )^2}{4 c^2}+\frac {1}{4} x^4 \left (a+b \text {sech}^{-1}(c x)\right )^3+\frac {\left (2 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^4}\\ &=\frac {b^3 \sqrt {\frac {1-c x}{1+c x}} (1+c x)}{4 c^4}-\frac {b^2 x^2 \left (a+b \text {sech}^{-1}(c x)\right )}{4 c^2}-\frac {b \left (a+b \text {sech}^{-1}(c x)\right )^2}{2 c^4}-\frac {b \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )^2}{2 c^4}-\frac {b x^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )^2}{4 c^2}+\frac {1}{4} x^4 \left (a+b \text {sech}^{-1}(c x)\right )^3+\frac {b^2 \left (a+b \text {sech}^{-1}(c x)\right ) \log \left (1+e^{2 \text {sech}^{-1}(c x)}\right )}{c^4}-\frac {b^3 \operatorname {Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\text {sech}^{-1}(c x)\right )}{c^4}\\ &=\frac {b^3 \sqrt {\frac {1-c x}{1+c x}} (1+c x)}{4 c^4}-\frac {b^2 x^2 \left (a+b \text {sech}^{-1}(c x)\right )}{4 c^2}-\frac {b \left (a+b \text {sech}^{-1}(c x)\right )^2}{2 c^4}-\frac {b \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )^2}{2 c^4}-\frac {b x^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )^2}{4 c^2}+\frac {1}{4} x^4 \left (a+b \text {sech}^{-1}(c x)\right )^3+\frac {b^2 \left (a+b \text {sech}^{-1}(c x)\right ) \log \left (1+e^{2 \text {sech}^{-1}(c x)}\right )}{c^4}-\frac {b^3 \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \text {sech}^{-1}(c x)}\right )}{2 c^4}\\ &=\frac {b^3 \sqrt {\frac {1-c x}{1+c x}} (1+c x)}{4 c^4}-\frac {b^2 x^2 \left (a+b \text {sech}^{-1}(c x)\right )}{4 c^2}-\frac {b \left (a+b \text {sech}^{-1}(c x)\right )^2}{2 c^4}-\frac {b \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )^2}{2 c^4}-\frac {b x^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )^2}{4 c^2}+\frac {1}{4} x^4 \left (a+b \text {sech}^{-1}(c x)\right )^3+\frac {b^2 \left (a+b \text {sech}^{-1}(c x)\right ) \log \left (1+e^{2 \text {sech}^{-1}(c x)}\right )}{c^4}+\frac {b^3 \text {Li}_2\left (-e^{2 \text {sech}^{-1}(c x)}\right )}{2 c^4}\\ \end {align*}

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Mathematica [A]  time = 2.15, size = 337, normalized size = 1.51 \[ \frac {1}{4} \left (a^3 x^4+a^2 b \left (3 x^4 \text {sech}^{-1}(c x)-\frac {\sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (c^2 x^2+2\right )}{c^4}\right )+\frac {a b^2 \left (3 c^4 x^4 \text {sech}^{-1}(c x)^2-c^2 x^2-2 \sqrt {\frac {1-c x}{c x+1}} \left (c^3 x^3+c^2 x^2+2 c x+2\right ) \text {sech}^{-1}(c x)+4 \log \left (\frac {1}{c x}\right )\right )}{c^4}-\frac {b^3 \left (\text {sech}^{-1}(c x) \left (c^2 x^2-4 \log \left (e^{-2 \text {sech}^{-1}(c x)}+1\right )\right )+\left (c^3 x^3 \sqrt {\frac {1-c x}{c x+1}}+c^2 x^2 \sqrt {\frac {1-c x}{c x+1}}+2 c x \sqrt {\frac {1-c x}{c x+1}}+2 \sqrt {\frac {1-c x}{c x+1}}-2\right ) \text {sech}^{-1}(c x)^2+2 \text {Li}_2\left (-e^{-2 \text {sech}^{-1}(c x)}\right )-\sqrt {\frac {1-c x}{c x+1}} (c x+1)\right )}{c^4}+b^3 x^4 \text {sech}^{-1}(c x)^3\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3*(a + b*ArcSech[c*x])^3,x]

[Out]

(a^3*x^4 + b^3*x^4*ArcSech[c*x]^3 + a^2*b*(-((Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(2 + c^2*x^2))/c^4) + 3*x^4*
ArcSech[c*x]) + (a*b^2*(-(c^2*x^2) - 2*Sqrt[(1 - c*x)/(1 + c*x)]*(2 + 2*c*x + c^2*x^2 + c^3*x^3)*ArcSech[c*x]
+ 3*c^4*x^4*ArcSech[c*x]^2 + 4*Log[1/(c*x)]))/c^4 - (b^3*(-(Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)) + (-2 + 2*Sqr
t[(1 - c*x)/(1 + c*x)] + 2*c*x*Sqrt[(1 - c*x)/(1 + c*x)] + c^2*x^2*Sqrt[(1 - c*x)/(1 + c*x)] + c^3*x^3*Sqrt[(1
- c*x)/(1 + c*x)])*ArcSech[c*x]^2 + ArcSech[c*x]*(c^2*x^2 - 4*Log[1 + E^(-2*ArcSech[c*x])]) + 2*PolyLog[2, -E
^(-2*ArcSech[c*x])]))/c^4)/4

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{3} x^{3} \operatorname {arsech}\left (c x\right )^{3} + 3 \, a b^{2} x^{3} \operatorname {arsech}\left (c x\right )^{2} + 3 \, a^{2} b x^{3} \operatorname {arsech}\left (c x\right ) + a^{3} x^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \operatorname {arsech}\left (c x\right ) + a\right )}^{3} x^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.96, size = 546, normalized size = 2.45 \[ \frac {x^{4} a^{3}}{4}+\frac {b^{3} \mathrm {arcsech}\left (c x \right )^{3} x^{4}}{4}-\frac {b^{3} \mathrm {arcsech}\left (c x \right )^{2} \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, x^{3}}{4 c}-\frac {b^{3} \mathrm {arcsech}\left (c x \right )^{2} \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, x}{2 c^{3}}-\frac {b^{3} \mathrm {arcsech}\left (c x \right ) x^{2}}{4 c^{2}}+\frac {b^{3} \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, x}{4 c^{3}}-\frac {b^{3} \mathrm {arcsech}\left (c x \right )^{2}}{2 c^{4}}-\frac {b^{3}}{4 c^{4}}+\frac {b^{3} \mathrm {arcsech}\left (c x \right ) \ln \left (1+\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )}{c^{4}}+\frac {b^{3} \polylog \left (2, -\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )}{2 c^{4}}-\frac {a \,b^{2} \mathrm {arcsech}\left (c x \right )}{c^{4}}+\frac {3 a \,b^{2} \mathrm {arcsech}\left (c x \right )^{2} x^{4}}{4}-\frac {a \,b^{2} \mathrm {arcsech}\left (c x \right ) \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, x^{3}}{2 c}-\frac {a \,b^{2} \mathrm {arcsech}\left (c x \right ) \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, x}{c^{3}}-\frac {x^{2} a \,b^{2}}{4 c^{2}}+\frac {a \,b^{2} \ln \left (1+\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )}{c^{4}}+\frac {3 a^{2} b \,x^{4} \mathrm {arcsech}\left (c x \right )}{4}-\frac {a^{2} b \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, x^{3}}{4 c}-\frac {a^{2} b \sqrt {-\frac {c x -1}{c x}}\, x \sqrt {\frac {c x +1}{c x}}}{2 c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*a^3+1/4*b^3*arcsech(c*x)^3*x^4-1/4/c*b^3*arcsech(c*x)^2*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*x^3-1
/2/c^3*b^3*arcsech(c*x)^2*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*x-1/4/c^2*b^3*arcsech(c*x)*x^2+1/4/c^3*b^3*
(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*x-1/2/c^4*b^3*arcsech(c*x)^2-1/4/c^4*b^3+1/c^4*b^3*arcsech(c*x)*ln(1+
(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)+1/2*b^3*polylog(2,-(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)/c^4
-1/c^4*a*b^2*arcsech(c*x)+3/4*a*b^2*arcsech(c*x)^2*x^4-1/2/c*a*b^2*arcsech(c*x)*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/
c/x)^(1/2)*x^3-1/c^3*a*b^2*arcsech(c*x)*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*x-1/4/c^2*x^2*a*b^2+1/c^4*a*b
^2*ln(1+(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)+3/4*a^2*b*x^4*arcsech(c*x)-1/4/c*a^2*b*(-(c*x-1)/c/x)^(1/2
)*((c*x+1)/c/x)^(1/2)*x^3-1/2/c^3*a^2*b*(-(c*x-1)/c/x)^(1/2)*x*((c*x+1)/c/x)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, a^{3} x^{4} + \frac {1}{4} \, {\left (3 \, x^{4} \operatorname {arsech}\left (c x\right ) + \frac {c^{2} x^{3} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} - 3 \, x \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c^{3}}\right )} a^{2} b + \int b^{3} x^{3} \log \left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )^{3} + 3 \, a b^{2} x^{3} \log \left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^3*x^4 + 1/4*(3*x^4*arcsech(c*x) + (c^2*x^3*(1/(c^2*x^2) - 1)^(3/2) - 3*x*sqrt(1/(c^2*x^2) - 1))/c^3)*a^2
*b + integrate(b^3*x^3*log(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))^3 + 3*a*b^2*x^3*log(sqrt(1/(c*x) + 1
)*sqrt(1/(c*x) - 1) + 1/(c*x))^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^3\,{\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*acosh(1/(c*x)))^3,x)

[Out]

int(x^3*(a + b*acosh(1/(c*x)))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \left (a + b \operatorname {asech}{\left (c x \right )}\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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