3.1.45 \(\int (a+b \text {sech}^{-1}(c x))^3 \, dx\) [45]

Optimal. Leaf size=140 \[ x \left (a+b \text {sech}^{-1}(c x)\right )^3-\frac {6 b \left (a+b \text {sech}^{-1}(c x)\right )^2 \text {ArcTan}\left (e^{\text {sech}^{-1}(c x)}\right )}{c}+\frac {6 i b^2 \left (a+b \text {sech}^{-1}(c x)\right ) \text {PolyLog}\left (2,-i e^{\text {sech}^{-1}(c x)}\right )}{c}-\frac {6 i b^2 \left (a+b \text {sech}^{-1}(c x)\right ) \text {PolyLog}\left (2,i e^{\text {sech}^{-1}(c x)}\right )}{c}-\frac {6 i b^3 \text {PolyLog}\left (3,-i e^{\text {sech}^{-1}(c x)}\right )}{c}+\frac {6 i b^3 \text {PolyLog}\left (3,i e^{\text {sech}^{-1}(c x)}\right )}{c} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6414, 5559, 4265, 2611, 2320, 6724} \begin {gather*} -\frac {6 b \text {ArcTan}\left (e^{\text {sech}^{-1}(c x)}\right ) \left (a+b \text {sech}^{-1}(c x)\right )^2}{c}+\frac {6 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}-\frac {6 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}+x \left (a+b \text {sech}^{-1}(c x)\right )^3-\frac {6 i b^3 \text {Li}_3\left (-i e^{\text {sech}^{-1}(c x)}\right )}{c}+\frac {6 i b^3 \text {Li}_3\left (i e^{\text {sech}^{-1}(c x)}\right )}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])^3,x]

[Out]

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

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 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 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 6414

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[-c^(-1), Subst[Int[(a + b*x)^n*Sech[x]*Tanh[x]
, x], x, ArcSech[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[n, 0]

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 \left (a+b \text {sech}^{-1}(c x)\right )^3 \, dx &=-\frac {\text {Subst}\left (\int (a+b x)^3 \text {sech}(x) \tanh (x) \, dx,x,\text {sech}^{-1}(c x)\right )}{c}\\ &=x \left (a+b \text {sech}^{-1}(c x)\right )^3-\frac {(3 b) \text {Subst}\left (\int (a+b x)^2 \text {sech}(x) \, dx,x,\text {sech}^{-1}(c x)\right )}{c}\\ &=x \left (a+b \text {sech}^{-1}(c x)\right )^3-\frac {6 b \left (a+b \text {sech}^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\text {sech}^{-1}(c x)}\right )}{c}+\frac {\left (6 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}-\frac {\left (6 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}\\ &=x \left (a+b \text {sech}^{-1}(c x)\right )^3-\frac {6 b \left (a+b \text {sech}^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\text {sech}^{-1}(c x)}\right )}{c}+\frac {6 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}-\frac {6 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}-\frac {\left (6 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}+\frac {\left (6 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}\\ &=x \left (a+b \text {sech}^{-1}(c x)\right )^3-\frac {6 b \left (a+b \text {sech}^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\text {sech}^{-1}(c x)}\right )}{c}+\frac {6 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}-\frac {6 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}-\frac {\left (6 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}+\frac {\left (6 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}\\ &=x \left (a+b \text {sech}^{-1}(c x)\right )^3-\frac {6 b \left (a+b \text {sech}^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\text {sech}^{-1}(c x)}\right )}{c}+\frac {6 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}-\frac {6 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}-\frac {6 i b^3 \text {Li}_3\left (-i e^{\text {sech}^{-1}(c x)}\right )}{c}+\frac {6 i b^3 \text {Li}_3\left (i e^{\text {sech}^{-1}(c x)}\right )}{c}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(282\) vs. \(2(140)=280\).
time = 0.28, size = 282, normalized size = 2.01 \begin {gather*} a^3 x+3 a^2 b x \text {sech}^{-1}(c x)-\frac {3 a^2 b \text {ArcTan}\left (\frac {c x \sqrt {\frac {1-c x}{1+c x}}}{-1+c x}\right )}{c}+\frac {3 i a b^2 \left (\text {sech}^{-1}(c x) \left (-i c x \text {sech}^{-1}(c x)+2 \log \left (1-i e^{-\text {sech}^{-1}(c x)}\right )-2 \log \left (1+i e^{-\text {sech}^{-1}(c x)}\right )\right )+2 \text {PolyLog}\left (2,-i e^{-\text {sech}^{-1}(c x)}\right )-2 \text {PolyLog}\left (2,i e^{-\text {sech}^{-1}(c x)}\right )\right )}{c}+\frac {b^3 \left (c x \text {sech}^{-1}(c x)^3-3 i \left (-\text {sech}^{-1}(c x)^2 \left (\log \left (1-i e^{-\text {sech}^{-1}(c x)}\right )-\log \left (1+i e^{-\text {sech}^{-1}(c x)}\right )\right )-2 \text {sech}^{-1}(c x) \left (\text {PolyLog}\left (2,-i e^{-\text {sech}^{-1}(c x)}\right )-\text {PolyLog}\left (2,i e^{-\text {sech}^{-1}(c x)}\right )\right )-2 \left (\text {PolyLog}\left (3,-i e^{-\text {sech}^{-1}(c x)}\right )-\text {PolyLog}\left (3,i e^{-\text {sech}^{-1}(c x)}\right )\right )\right )\right )}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSech[c*x])^3,x]

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \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((a+b*asech(c*x))**3,x)

[Out]

Integral((a + b*asech(c*x))**3, x)

________________________________________________________________________________________

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

[Out]

integrate((b*arcsech(c*x) + a)^3, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\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((a + b*acosh(1/(c*x)))^3,x)

[Out]

int((a + b*acosh(1/(c*x)))^3, x)

________________________________________________________________________________________