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

Optimal. Leaf size=260 \[ -\frac {a^2 \text {sech}^{-1}(a+b x)^3}{2 b^2}-\frac {6 i a \text {sech}^{-1}(a+b x) \text {Li}_2\left (-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \text {sech}^{-1}(a+b x) \text {Li}_2\left (i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {3 \text {Li}_2\left (-e^{2 \text {sech}^{-1}(a+b x)}\right )}{2 b^2}+\frac {6 i a \text {Li}_3\left (-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \text {Li}_3\left (i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {3 \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)^2}{2 b^2}-\frac {3 \text {sech}^{-1}(a+b x)^2}{2 b^2}+\frac {3 \text {sech}^{-1}(a+b x) \log \left (e^{2 \text {sech}^{-1}(a+b x)}+1\right )}{b^2}+\frac {6 a \text {sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {1}{2} x^2 \text {sech}^{-1}(a+b x)^3 \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.200, Rules used = {6321, 5468, 4190, 4180, 2531, 2282, 6589, 4184, 3718, 2190, 2279, 2391} \[ -\frac {6 i a \text {sech}^{-1}(a+b x) \text {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \text {sech}^{-1}(a+b x) \text {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {3 \text {PolyLog}\left (2,-e^{2 \text {sech}^{-1}(a+b x)}\right )}{2 b^2}+\frac {6 i a \text {PolyLog}\left (3,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \text {PolyLog}\left (3,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {a^2 \text {sech}^{-1}(a+b x)^3}{2 b^2}-\frac {3 \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)^2}{2 b^2}-\frac {3 \text {sech}^{-1}(a+b x)^2}{2 b^2}+\frac {3 \text {sech}^{-1}(a+b x) \log \left (e^{2 \text {sech}^{-1}(a+b x)}+1\right )}{b^2}+\frac {6 a \text {sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {1}{2} x^2 \text {sech}^{-1}(a+b x)^3 \]

Antiderivative was successfully verified.

[In]

Int[x*ArcSech[a + b*x]^3,x]

[Out]

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

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 2282

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 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 2531

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 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 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 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 4190

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 5468

Int[((e_.) + (f_.)*(x_))^(m_.)*Sech[(c_.) + (d_.)*(x_)]*((a_) + (b_.)*Sech[(c_.) + (d_.)*(x_)])^(n_.)*Tanh[(c_
.) + (d_.)*(x_)], x_Symbol] :> -Simp[((e + f*x)^m*(a + b*Sech[c + d*x])^(n + 1))/(b*d*(n + 1)), x] + Dist[(f*m
)/(b*d*(n + 1)), Int[(e + f*x)^(m - 1)*(a + b*Sech[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x
] && IGtQ[m, 0] && NeQ[n, -1]

Rule 6321

Int[((a_.) + ArcSech[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Dist[(d^(m + 1)
)^(-1), Subst[Int[(a + b*x)^p*Sech[x]*Tanh[x]*(d*e - c*f + f*Sech[x])^m, x], x, ArcSech[c + d*x]], x] /; FreeQ
[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && IntegerQ[m]

Rule 6589

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 \text {sech}^{-1}(a+b x)^3 \, dx &=-\frac {\operatorname {Subst}\left (\int x^3 \text {sech}(x) (-a+\text {sech}(x)) \tanh (x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^2}\\ &=\frac {1}{2} x^2 \text {sech}^{-1}(a+b x)^3-\frac {3 \operatorname {Subst}\left (\int x^2 (-a+\text {sech}(x))^2 \, dx,x,\text {sech}^{-1}(a+b x)\right )}{2 b^2}\\ &=\frac {1}{2} x^2 \text {sech}^{-1}(a+b x)^3-\frac {3 \operatorname {Subst}\left (\int \left (a^2 x^2-2 a x^2 \text {sech}(x)+x^2 \text {sech}^2(x)\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{2 b^2}\\ &=-\frac {a^2 \text {sech}^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \text {sech}^{-1}(a+b x)^3-\frac {3 \operatorname {Subst}\left (\int x^2 \text {sech}^2(x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{2 b^2}+\frac {(3 a) \operatorname {Subst}\left (\int x^2 \text {sech}(x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^2}\\ &=-\frac {3 \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)^2}{2 b^2}-\frac {a^2 \text {sech}^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \text {sech}^{-1}(a+b x)^3+\frac {6 a \text {sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {3 \operatorname {Subst}\left (\int x \tanh (x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^2}-\frac {(6 i a) \operatorname {Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^2}+\frac {(6 i a) \operatorname {Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^2}\\ &=-\frac {3 \text {sech}^{-1}(a+b x)^2}{2 b^2}-\frac {3 \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)^2}{2 b^2}-\frac {a^2 \text {sech}^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \text {sech}^{-1}(a+b x)^3+\frac {6 a \text {sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \text {sech}^{-1}(a+b x) \text {Li}_2\left (-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \text {sech}^{-1}(a+b x) \text {Li}_2\left (i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {6 \operatorname {Subst}\left (\int \frac {e^{2 x} x}{1+e^{2 x}} \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^2}+\frac {(6 i a) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^2}-\frac {(6 i a) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^2}\\ &=-\frac {3 \text {sech}^{-1}(a+b x)^2}{2 b^2}-\frac {3 \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)^2}{2 b^2}-\frac {a^2 \text {sech}^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \text {sech}^{-1}(a+b x)^3+\frac {6 a \text {sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {3 \text {sech}^{-1}(a+b x) \log \left (1+e^{2 \text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \text {sech}^{-1}(a+b x) \text {Li}_2\left (-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \text {sech}^{-1}(a+b x) \text {Li}_2\left (i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {3 \operatorname {Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^2}+\frac {(6 i a) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {(6 i a) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}\\ &=-\frac {3 \text {sech}^{-1}(a+b x)^2}{2 b^2}-\frac {3 \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)^2}{2 b^2}-\frac {a^2 \text {sech}^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \text {sech}^{-1}(a+b x)^3+\frac {6 a \text {sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {3 \text {sech}^{-1}(a+b x) \log \left (1+e^{2 \text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \text {sech}^{-1}(a+b x) \text {Li}_2\left (-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \text {sech}^{-1}(a+b x) \text {Li}_2\left (i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \text {Li}_3\left (-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \text {Li}_3\left (i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {3 \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \text {sech}^{-1}(a+b x)}\right )}{2 b^2}\\ &=-\frac {3 \text {sech}^{-1}(a+b x)^2}{2 b^2}-\frac {3 \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)^2}{2 b^2}-\frac {a^2 \text {sech}^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \text {sech}^{-1}(a+b x)^3+\frac {6 a \text {sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {3 \text {sech}^{-1}(a+b x) \log \left (1+e^{2 \text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \text {sech}^{-1}(a+b x) \text {Li}_2\left (-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \text {sech}^{-1}(a+b x) \text {Li}_2\left (i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {3 \text {Li}_2\left (-e^{2 \text {sech}^{-1}(a+b x)}\right )}{2 b^2}+\frac {6 i a \text {Li}_3\left (-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \text {Li}_3\left (i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}\\ \end {align*}

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Mathematica [A]  time = 0.54, size = 254, normalized size = 0.98 \[ \frac {-3 \text {Li}_2\left (-e^{-2 \text {sech}^{-1}(a+b x)}\right )+6 i a \left (-2 \text {sech}^{-1}(a+b x) \left (\text {Li}_2\left (-i e^{-\text {sech}^{-1}(a+b x)}\right )-\text {Li}_2\left (i e^{-\text {sech}^{-1}(a+b x)}\right )\right )-2 \text {Li}_3\left (-i e^{-\text {sech}^{-1}(a+b x)}\right )+2 \text {Li}_3\left (i e^{-\text {sech}^{-1}(a+b x)}\right )-\left (\text {sech}^{-1}(a+b x)^2 \left (\log \left (1-i e^{-\text {sech}^{-1}(a+b x)}\right )-\log \left (1+i e^{-\text {sech}^{-1}(a+b x)}\right )\right )\right )\right )+(a+b x)^2 \text {sech}^{-1}(a+b x)^3-2 a (a+b x) \text {sech}^{-1}(a+b x)^3-3 \sqrt {-\frac {a+b x-1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)^2+3 \text {sech}^{-1}(a+b x) \left (\text {sech}^{-1}(a+b x)+2 \log \left (e^{-2 \text {sech}^{-1}(a+b x)}+1\right )\right )}{2 b^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x*ArcSech[a + b*x]^3,x]

[Out]

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

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x \operatorname {arsech}\left (b x + a\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*arcsech(b*x + a)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arcsech(b*x + a)^3, x)

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maple [F]  time = 1.88, size = 0, normalized size = 0.00 \[ \int x \mathrm {arcsech}\left (b x +a \right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, x^{2} \log \left (\sqrt {b x + a + 1} \sqrt {-b x - a + 1} b x + \sqrt {b x + a + 1} \sqrt {-b x - a + 1} a + b x + a\right )^{3} - \int \frac {16 \, {\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} b - b\right )} x^{2} + {\left (a^{3} - a\right )} x\right )} \sqrt {b x + a + 1} \sqrt {-b x - a + 1} \log \left (b x + a\right )^{3} + 16 \, {\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} b - b\right )} x^{2} + {\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right )^{3} + 3 \, {\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} + {\left (a^{2} b - b\right )} x^{2} + 4 \, {\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} b - b\right )} x^{2} + {\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right ) + {\left (2 \, {\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} b - b\right )} x^{2} + {\left (a^{3} - a\right )} x\right )} \sqrt {b x + a + 1} \log \left (b x + a\right ) + {\left (2 \, b^{3} x^{4} + 4 \, a b^{2} x^{3} + {\left (2 \, a^{2} b - b\right )} x^{2} + 2 \, {\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} b - b\right )} x^{2} + {\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right )\right )} \sqrt {b x + a + 1}\right )} \sqrt {-b x - a + 1}\right )} \log \left (\sqrt {b x + a + 1} \sqrt {-b x - a + 1} b x + \sqrt {b x + a + 1} \sqrt {-b x - a + 1} a + b x + a\right )^{2} - 24 \, {\left ({\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} b - b\right )} x^{2} + {\left (a^{3} - a\right )} x\right )} \sqrt {b x + a + 1} \sqrt {-b x - a + 1} \log \left (b x + a\right )^{2} + {\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} b - b\right )} x^{2} + {\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right )^{2}\right )} \log \left (\sqrt {b x + a + 1} \sqrt {-b x - a + 1} b x + \sqrt {b x + a + 1} \sqrt {-b x - a + 1} a + b x + a\right )}{2 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} b - b\right )} x - a\right )} \sqrt {b x + a + 1} \sqrt {-b x - a + 1} + {\left (3 \, a^{2} b - b\right )} x - a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*log(sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*b*x + sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*a + b*x + a)^3 - i
ntegrate(1/2*(16*(b^3*x^4 + 3*a*b^2*x^3 + (3*a^2*b - b)*x^2 + (a^3 - a)*x)*sqrt(b*x + a + 1)*sqrt(-b*x - a + 1
)*log(b*x + a)^3 + 16*(b^3*x^4 + 3*a*b^2*x^3 + (3*a^2*b - b)*x^2 + (a^3 - a)*x)*log(b*x + a)^3 + 3*(b^3*x^4 +
2*a*b^2*x^3 + (a^2*b - b)*x^2 + 4*(b^3*x^4 + 3*a*b^2*x^3 + (3*a^2*b - b)*x^2 + (a^3 - a)*x)*log(b*x + a) + (2*
(b^3*x^4 + 3*a*b^2*x^3 + (3*a^2*b - b)*x^2 + (a^3 - a)*x)*sqrt(b*x + a + 1)*log(b*x + a) + (2*b^3*x^4 + 4*a*b^
2*x^3 + (2*a^2*b - b)*x^2 + 2*(b^3*x^4 + 3*a*b^2*x^3 + (3*a^2*b - b)*x^2 + (a^3 - a)*x)*log(b*x + a))*sqrt(b*x
+ a + 1))*sqrt(-b*x - a + 1))*log(sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*b*x + sqrt(b*x + a + 1)*sqrt(-b*x - a
+ 1)*a + b*x + a)^2 - 24*((b^3*x^4 + 3*a*b^2*x^3 + (3*a^2*b - b)*x^2 + (a^3 - a)*x)*sqrt(b*x + a + 1)*sqrt(-b*
x - a + 1)*log(b*x + a)^2 + (b^3*x^4 + 3*a*b^2*x^3 + (3*a^2*b - b)*x^2 + (a^3 - a)*x)*log(b*x + a)^2)*log(sqrt
(b*x + a + 1)*sqrt(-b*x - a + 1)*b*x + sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*a + b*x + a))/(b^3*x^3 + 3*a*b^2*x
^2 + a^3 + (b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b - b)*x - a)*sqrt(b*x + a + 1)*sqrt(-b*x - a + 1) + (3*a^2*b
- b)*x - a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*acosh(1/(a + b*x))^3,x)

[Out]

int(x*acosh(1/(a + b*x))^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*asech(a + b*x)**3, x)

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