3.1.43 \(\int x^2 (a+b \text {sech}^{-1}(c x))^3 \, dx\) [43]

Optimal. Leaf size=242 \[ -\frac {b^2 x \left (a+b \text {sech}^{-1}(c x)\right )}{c^2}-\frac {b x \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )^2}{2 c^2}+\frac {1}{3} x^3 \left (a+b \text {sech}^{-1}(c x)\right )^3-\frac {b \left (a+b \text {sech}^{-1}(c x)\right )^2 \text {ArcTan}\left (e^{\text {sech}^{-1}(c x)}\right )}{c^3}+\frac {b^3 \text {ArcTan}\left (\frac {\sqrt {\frac {1-c x}{1+c x}} (1+c x)}{c x}\right )}{c^3}+\frac {i b^2 \left (a+b \text {sech}^{-1}(c x)\right ) \text {PolyLog}\left (2,-i e^{\text {sech}^{-1}(c x)}\right )}{c^3}-\frac {i b^2 \left (a+b \text {sech}^{-1}(c x)\right ) \text {PolyLog}\left (2,i e^{\text {sech}^{-1}(c x)}\right )}{c^3}-\frac {i b^3 \text {PolyLog}\left (3,-i e^{\text {sech}^{-1}(c x)}\right )}{c^3}+\frac {i b^3 \text {PolyLog}\left (3,i e^{\text {sech}^{-1}(c x)}\right )}{c^3} \]

[Out]

-b^2*x*(a+b*arcsech(c*x))/c^2+1/3*x^3*(a+b*arcsech(c*x))^3-b*(a+b*arcsech(c*x))^2*arctan(1/c/x+(-1+1/c/x)^(1/2
)*(1+1/c/x)^(1/2))/c^3+b^3*arctan((c*x+1)*((-c*x+1)/(c*x+1))^(1/2)/c/x)/c^3+I*b^2*(a+b*arcsech(c*x))*polylog(2
,-I*(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2)))/c^3-I*b^2*(a+b*arcsech(c*x))*polylog(2,I*(1/c/x+(-1+1/c/x)^(1/2)
*(1+1/c/x)^(1/2)))/c^3-I*b^3*polylog(3,-I*(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2)))/c^3+I*b^3*polylog(3,I*(1/c
/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2)))/c^3-1/2*b*x*(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 = 0.14, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {6420, 5559, 4271, 3855, 4265, 2611, 2320, 6724} \begin {gather*} -\frac {b \text {ArcTan}\left (e^{\text {sech}^{-1}(c x)}\right ) \left (a+b \text {sech}^{-1}(c x)\right )^2}{c^3}+\frac {i b^2 \text {Li}_2\left (-i e^{\text {sech}^{-1}(c x)}\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{c^3}-\frac {i b^2 \text {Li}_2\left (i e^{\text {sech}^{-1}(c x)}\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{c^3}-\frac {b^2 x \left (a+b \text {sech}^{-1}(c x)\right )}{c^2}-\frac {b x \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 {1}{3} x^3 \left (a+b \text {sech}^{-1}(c x)\right )^3+\frac {b^3 \text {ArcTan}\left (\frac {\sqrt {\frac {1-c x}{c x+1}} (c x+1)}{c x}\right )}{c^3}-\frac {i b^3 \text {Li}_3\left (-i e^{\text {sech}^{-1}(c x)}\right )}{c^3}+\frac {i b^3 \text {Li}_3\left (i e^{\text {sech}^{-1}(c x)}\right )}{c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcSech[c*x])^3,x]

[Out]

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

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3855

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

Rule 4265

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 4271

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]) /; Free
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]

Rule 5559

Int[((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.)*Tanh[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Sim
p[(-(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 6420

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 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int x^2 \left (a+b \text {sech}^{-1}(c x)\right )^3 \, dx &=-\frac {\text {Subst}\left (\int (a+b x)^3 \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 )^3-\frac {b \text {Subst}\left (\int (a+b x)^2 \text {sech}^3(x) \, dx,x,\text {sech}^{-1}(c x)\right )}{c^3}\\ &=-\frac {b^2 x \left (a+b \text {sech}^{-1}(c x)\right )}{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 )^2}{2 c^2}+\frac {1}{3} x^3 \left (a+b \text {sech}^{-1}(c x)\right )^3-\frac {b \text {Subst}\left (\int (a+b x)^2 \text {sech}(x) \, dx,x,\text {sech}^{-1}(c x)\right )}{2 c^3}+\frac {b^3 \text {Subst}\left (\int \text {sech}(x) \, dx,x,\text {sech}^{-1}(c x)\right )}{c^3}\\ &=-\frac {b^2 x \left (a+b \text {sech}^{-1}(c x)\right )}{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 )^2}{2 c^2}+\frac {1}{3} x^3 \left (a+b \text {sech}^{-1}(c x)\right )^3-\frac {b \left (a+b \text {sech}^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\text {sech}^{-1}(c x)}\right )}{c^3}+\frac {b^3 \tan ^{-1}\left (\frac {\sqrt {\frac {1-c x}{1+c x}} (1+c x)}{c x}\right )}{c^3}+\frac {\left (i b^2\right ) \text {Subst}\left (\int (a+b x) \log \left (1-i e^x\right ) \, dx,x,\text {sech}^{-1}(c x)\right )}{c^3}-\frac {\left (i b^2\right ) \text {Subst}\left (\int (a+b x) \log \left (1+i e^x\right ) \, dx,x,\text {sech}^{-1}(c x)\right )}{c^3}\\ &=-\frac {b^2 x \left (a+b \text {sech}^{-1}(c x)\right )}{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 )^2}{2 c^2}+\frac {1}{3} x^3 \left (a+b \text {sech}^{-1}(c x)\right )^3-\frac {b \left (a+b \text {sech}^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\text {sech}^{-1}(c x)}\right )}{c^3}+\frac {b^3 \tan ^{-1}\left (\frac {\sqrt {\frac {1-c x}{1+c x}} (1+c x)}{c x}\right )}{c^3}+\frac {i b^2 \left (a+b \text {sech}^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\text {sech}^{-1}(c x)}\right )}{c^3}-\frac {i b^2 \left (a+b \text {sech}^{-1}(c x)\right ) \text {Li}_2\left (i e^{\text {sech}^{-1}(c x)}\right )}{c^3}-\frac {\left (i b^3\right ) \text {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\text {sech}^{-1}(c x)\right )}{c^3}+\frac {\left (i b^3\right ) \text {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\text {sech}^{-1}(c x)\right )}{c^3}\\ &=-\frac {b^2 x \left (a+b \text {sech}^{-1}(c x)\right )}{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 )^2}{2 c^2}+\frac {1}{3} x^3 \left (a+b \text {sech}^{-1}(c x)\right )^3-\frac {b \left (a+b \text {sech}^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\text {sech}^{-1}(c x)}\right )}{c^3}+\frac {b^3 \tan ^{-1}\left (\frac {\sqrt {\frac {1-c x}{1+c x}} (1+c x)}{c x}\right )}{c^3}+\frac {i b^2 \left (a+b \text {sech}^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\text {sech}^{-1}(c x)}\right )}{c^3}-\frac {i b^2 \left (a+b \text {sech}^{-1}(c x)\right ) \text {Li}_2\left (i e^{\text {sech}^{-1}(c x)}\right )}{c^3}-\frac {\left (i b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\text {sech}^{-1}(c x)}\right )}{c^3}+\frac {\left (i b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\text {sech}^{-1}(c x)}\right )}{c^3}\\ &=-\frac {b^2 x \left (a+b \text {sech}^{-1}(c x)\right )}{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 )^2}{2 c^2}+\frac {1}{3} x^3 \left (a+b \text {sech}^{-1}(c x)\right )^3-\frac {b \left (a+b \text {sech}^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\text {sech}^{-1}(c x)}\right )}{c^3}+\frac {b^3 \tan ^{-1}\left (\frac {\sqrt {\frac {1-c x}{1+c x}} (1+c x)}{c x}\right )}{c^3}+\frac {i b^2 \left (a+b \text {sech}^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\text {sech}^{-1}(c x)}\right )}{c^3}-\frac {i b^2 \left (a+b \text {sech}^{-1}(c x)\right ) \text {Li}_2\left (i e^{\text {sech}^{-1}(c x)}\right )}{c^3}-\frac {i b^3 \text {Li}_3\left (-i e^{\text {sech}^{-1}(c x)}\right )}{c^3}+\frac {i b^3 \text {Li}_3\left (i e^{\text {sech}^{-1}(c x)}\right )}{c^3}\\ \end {align*}

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Mathematica [A]
time = 0.75, size = 440, normalized size = 1.82 \begin {gather*} \frac {2 a^3 c^3 x^3-3 a^2 b c x \sqrt {\frac {1-c x}{1+c x}} (1+c x)+6 a^2 b c^3 x^3 \text {sech}^{-1}(c x)+3 i a^2 b \log \left (-2 i c x+2 \sqrt {\frac {1-c x}{1+c x}} (1+c x)\right )-6 a b^2 \left (c x+c x \sqrt {\frac {1-c x}{1+c x}} (1+c x) \text {sech}^{-1}(c x)-c^3 x^3 \text {sech}^{-1}(c x)^2-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 )-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 )\right )-b^3 \left (6 c x \text {sech}^{-1}(c x)+3 c x \sqrt {\frac {1-c x}{1+c x}} (1+c x) \text {sech}^{-1}(c x)^2-2 c^3 x^3 \text {sech}^{-1}(c x)^3-3 i \left (-4 i \text {ArcTan}\left (\tanh \left (\frac {1}{2} \text {sech}^{-1}(c x)\right )\right )+\text {sech}^{-1}(c x)^2 \log \left (1-i e^{-\text {sech}^{-1}(c x)}\right )-\text {sech}^{-1}(c x)^2 \log \left (1+i e^{-\text {sech}^{-1}(c x)}\right )+2 \text {sech}^{-1}(c x) \text {PolyLog}\left (2,-i e^{-\text {sech}^{-1}(c x)}\right )-2 \text {sech}^{-1}(c x) \text {PolyLog}\left (2,i e^{-\text {sech}^{-1}(c x)}\right )+2 \text {PolyLog}\left (3,-i e^{-\text {sech}^{-1}(c x)}\right )-2 \text {PolyLog}\left (3,i e^{-\text {sech}^{-1}(c x)}\right )\right )\right )}{6 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*ArcSech[c*x])^3,x]

[Out]

(2*a^3*c^3*x^3 - 3*a^2*b*c*x*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x) + 6*a^2*b*c^3*x^3*ArcSech[c*x] + (3*I)*a^2*b*
Log[(-2*I)*c*x + 2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)] - 6*a*b^2*(c*x + c*x*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*
x)*ArcSech[c*x] - c^3*x^3*ArcSech[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]]) - b^3*(6*c*x*ArcSech[c*
x] + 3*c*x*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*ArcSech[c*x]^2 - 2*c^3*x^3*ArcSech[c*x]^3 - (3*I)*((-4*I)*ArcTa
n[Tanh[ArcSech[c*x]/2]] + ArcSech[c*x]^2*Log[1 - I/E^ArcSech[c*x]] - ArcSech[c*x]^2*Log[1 + I/E^ArcSech[c*x]]
+ 2*ArcSech[c*x]*PolyLog[2, (-I)/E^ArcSech[c*x]] - 2*ArcSech[c*x]*PolyLog[2, I/E^ArcSech[c*x]] + 2*PolyLog[3,
(-I)/E^ArcSech[c*x]] - 2*PolyLog[3, I/E^ArcSech[c*x]])))/(6*c^3)

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int x^{2} \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )^{3}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsech(c*x))^3,x, algorithm="maxima")

[Out]

1/3*a^3*x^3 + 1/2*(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^2*b + integrate(b^3*x^2*log(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))^3 + 3*a*b^
2*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsech(c*x))^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \left (a + b \operatorname {asech}{\left (c x \right )}\right )^{3}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*asech(c*x))**3,x)

[Out]

Integral(x**2*(a + b*asech(c*x))**3, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsech(c*x))^3,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^2\,{\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*acosh(1/(c*x)))^3,x)

[Out]

int(x^2*(a + b*acosh(1/(c*x)))^3, x)

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