3.34 \(\int x \sec ^{-1}(a+b x)^3 \, dx\)

Optimal. Leaf size=278 \[ \frac{6 i a \sec ^{-1}(a+b x) \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac{6 i a \sec ^{-1}(a+b x) \text{PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac{3 i \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )}{2 b^2}-\frac{6 a \text{PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac{6 a \text{PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac{a^2 \sec ^{-1}(a+b x)^3}{2 b^2}-\frac{3 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^2}+\frac{3 i \sec ^{-1}(a+b x)^2}{2 b^2}-\frac{3 \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac{6 i a \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac{1}{2} x^2 \sec ^{-1}(a+b x)^3 \]

[Out]

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

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Rubi [A]  time = 0.259244, antiderivative size = 278, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 12, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.2, Rules used = {5258, 4426, 4190, 4181, 2531, 2282, 6589, 4184, 3719, 2190, 2279, 2391} \[ \frac{6 i a \sec ^{-1}(a+b x) \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac{6 i a \sec ^{-1}(a+b x) \text{PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac{3 i \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )}{2 b^2}-\frac{6 a \text{PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac{6 a \text{PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac{a^2 \sec ^{-1}(a+b x)^3}{2 b^2}-\frac{3 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^2}+\frac{3 i \sec ^{-1}(a+b x)^2}{2 b^2}-\frac{3 \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac{6 i a \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac{1}{2} x^2 \sec ^{-1}(a+b x)^3 \]

Antiderivative was successfully verified.

[In]

Int[x*ArcSec[a + b*x]^3,x]

[Out]

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

Rule 5258

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

Rule 4426

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

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 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

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 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 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]

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 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, 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 \sec ^{-1}(a+b x)^3 \, dx &=\frac{\operatorname{Subst}\left (\int x^3 \sec (x) (-a+\sec (x)) \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}\\ &=\frac{1}{2} x^2 \sec ^{-1}(a+b x)^3-\frac{3 \operatorname{Subst}\left (\int x^2 (-a+\sec (x))^2 \, dx,x,\sec ^{-1}(a+b x)\right )}{2 b^2}\\ &=\frac{1}{2} x^2 \sec ^{-1}(a+b x)^3-\frac{3 \operatorname{Subst}\left (\int \left (a^2 x^2-2 a x^2 \sec (x)+x^2 \sec ^2(x)\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{2 b^2}\\ &=-\frac{a^2 \sec ^{-1}(a+b x)^3}{2 b^2}+\frac{1}{2} x^2 \sec ^{-1}(a+b x)^3-\frac{3 \operatorname{Subst}\left (\int x^2 \sec ^2(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{2 b^2}+\frac{(3 a) \operatorname{Subst}\left (\int x^2 \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}\\ &=-\frac{3 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^2}-\frac{a^2 \sec ^{-1}(a+b x)^3}{2 b^2}+\frac{1}{2} x^2 \sec ^{-1}(a+b x)^3-\frac{6 i a \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac{3 \operatorname{Subst}\left (\int x \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}-\frac{(6 a) \operatorname{Subst}\left (\int x \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}+\frac{(6 a) \operatorname{Subst}\left (\int x \log \left (1+i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}\\ &=\frac{3 i \sec ^{-1}(a+b x)^2}{2 b^2}-\frac{3 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^2}-\frac{a^2 \sec ^{-1}(a+b x)^3}{2 b^2}+\frac{1}{2} x^2 \sec ^{-1}(a+b x)^3-\frac{6 i a \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac{6 i a \sec ^{-1}(a+b x) \text{Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac{6 i a \sec ^{-1}(a+b x) \text{Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac{(6 i) \operatorname{Subst}\left (\int \frac{e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}-\frac{(6 i a) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}+\frac{(6 i a) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}\\ &=\frac{3 i \sec ^{-1}(a+b x)^2}{2 b^2}-\frac{3 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^2}-\frac{a^2 \sec ^{-1}(a+b x)^3}{2 b^2}+\frac{1}{2} x^2 \sec ^{-1}(a+b x)^3-\frac{6 i a \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac{3 \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac{6 i a \sec ^{-1}(a+b x) \text{Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac{6 i a \sec ^{-1}(a+b x) \text{Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac{3 \operatorname{Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^2}-\frac{(6 a) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac{(6 a) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^2}\\ &=\frac{3 i \sec ^{-1}(a+b x)^2}{2 b^2}-\frac{3 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^2}-\frac{a^2 \sec ^{-1}(a+b x)^3}{2 b^2}+\frac{1}{2} x^2 \sec ^{-1}(a+b x)^3-\frac{6 i a \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac{3 \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac{6 i a \sec ^{-1}(a+b x) \text{Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac{6 i a \sec ^{-1}(a+b x) \text{Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac{6 a \text{Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac{6 a \text{Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac{(3 i) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \sec ^{-1}(a+b x)}\right )}{2 b^2}\\ &=\frac{3 i \sec ^{-1}(a+b x)^2}{2 b^2}-\frac{3 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^2}-\frac{a^2 \sec ^{-1}(a+b x)^3}{2 b^2}+\frac{1}{2} x^2 \sec ^{-1}(a+b x)^3-\frac{6 i a \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac{3 \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac{6 i a \sec ^{-1}(a+b x) \text{Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac{6 i a \sec ^{-1}(a+b x) \text{Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac{3 i \text{Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )}{2 b^2}-\frac{6 a \text{Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac{6 a \text{Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}\\ \end{align*}

Mathematica [A]  time = 0.423032, size = 248, normalized size = 0.89 \[ \frac{1}{2} \left (x^2 \sec ^{-1}(a+b x)^3-\frac{3 \left (4 i a \sec ^{-1}(a+b x) \text{PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )-i \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )+4 a \left (\text{PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )-i \sec ^{-1}(a+b x) \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )\right )-4 a \text{PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )+\frac{1}{3} a^2 \sec ^{-1}(a+b x)^3+(a+b x) \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2-i \sec ^{-1}(a+b x)^2+2 \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )+4 i a \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )\right )}{b^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcSec[a + b*x]^3,x]

[Out]

(x^2*ArcSec[a + b*x]^3 - (3*((-I)*ArcSec[a + b*x]^2 + (a + b*x)*Sqrt[1 - (a + b*x)^(-2)]*ArcSec[a + b*x]^2 + (
a^2*ArcSec[a + b*x]^3)/3 + (4*I)*a*ArcSec[a + b*x]^2*ArcTan[E^(I*ArcSec[a + b*x])] + 2*ArcSec[a + b*x]*Log[1 +
E^((2*I)*ArcSec[a + b*x])] + (4*I)*a*ArcSec[a + b*x]*PolyLog[2, I*E^(I*ArcSec[a + b*x])] - I*PolyLog[2, -E^((
2*I)*ArcSec[a + b*x])] + 4*a*((-I)*ArcSec[a + b*x]*PolyLog[2, (-I)*E^(I*ArcSec[a + b*x])] + PolyLog[3, (-I)*E^
(I*ArcSec[a + b*x])]) - 4*a*PolyLog[3, I*E^(I*ArcSec[a + b*x])]))/b^2)/2

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Maple [A]  time = 0.476, size = 429, normalized size = 1.5 \begin{align*}{\frac{{x}^{2} \left ({\rm arcsec} \left (bx+a\right ) \right ) ^{3}}{2}}-{\frac{{a}^{2} \left ({\rm arcsec} \left (bx+a\right ) \right ) ^{3}}{2\,{b}^{2}}}-{\frac{3\,x \left ({\rm arcsec} \left (bx+a\right ) \right ) ^{2}}{2\,b}\sqrt{{\frac{-1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}-{\frac{3\, \left ({\rm arcsec} \left (bx+a\right ) \right ) ^{2}a}{2\,{b}^{2}}\sqrt{{\frac{-1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}+{\frac{{\frac{3\,i}{2}} \left ({\rm arcsec} \left (bx+a\right ) \right ) ^{2}}{{b}^{2}}}-3\,{\frac{ \left ({\rm arcsec} \left (bx+a\right ) \right ) ^{2}a}{{b}^{2}}\ln \left ( 1+i \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) \right ) }+{\frac{6\,ia{\rm arcsec} \left (bx+a\right )}{{b}^{2}}{\it polylog} \left ( 2,-i \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) \right ) }-6\,{\frac{a}{{b}^{2}}{\it polylog} \left ( 3,-i \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) \right ) }-{\frac{6\,ia{\rm arcsec} \left (bx+a\right )}{{b}^{2}}{\it polylog} \left ( 2,i \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) \right ) }+3\,{\frac{ \left ({\rm arcsec} \left (bx+a\right ) \right ) ^{2}a}{{b}^{2}}\ln \left ( 1-i \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) \right ) }+6\,{\frac{a}{{b}^{2}}{\it polylog} \left ( 3,i \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) \right ) }-3\,{\frac{{\rm arcsec} \left (bx+a\right )}{{b}^{2}}\ln \left ( 1+ \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) ^{2} \right ) }+{\frac{{\frac{3\,i}{2}}}{{b}^{2}}{\it polylog} \left ( 2,- \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, x^{2} \arctan \left (\sqrt{b x + a + 1} \sqrt{b x + a - 1}\right )^{3} - \frac{3}{8} \, x^{2} \arctan \left (\sqrt{b x + a + 1} \sqrt{b x + a - 1}\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2} - \int \frac{3 \,{\left ({\left (4 \, b x^{2} \arctan \left (\sqrt{b x + a + 1} \sqrt{b x + a - 1}\right )^{2} - b x^{2} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2}\right )} \sqrt{b x + a + 1} \sqrt{b x + a - 1} + 4 \,{\left (2 \,{\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} +{\left (3 \, a^{2} - 1\right )} b x^{2} +{\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right )^{2} -{\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} +{\left (a^{2} - 1\right )} b x^{2} + 2 \,{\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} +{\left (3 \, a^{2} - 1\right )} b x^{2} +{\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right )\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )\right )} \arctan \left (\sqrt{b x + a + 1} \sqrt{b x + a - 1}\right )\right )}}{8 \,{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} - 1\right )} b x - a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^3 - 3/8*x^2*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))*lo
g(b^2*x^2 + 2*a*b*x + a^2)^2 - integrate(3/8*((4*b*x^2*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^2 - b*x^2*l
og(b^2*x^2 + 2*a*b*x + a^2)^2)*sqrt(b*x + a + 1)*sqrt(b*x + a - 1) + 4*(2*(b^3*x^4 + 3*a*b^2*x^3 + (3*a^2 - 1)
*b*x^2 + (a^3 - a)*x)*log(b*x + a)^2 - (b^3*x^4 + 2*a*b^2*x^3 + (a^2 - 1)*b*x^2 + 2*(b^3*x^4 + 3*a*b^2*x^3 + (
3*a^2 - 1)*b*x^2 + (a^3 - a)*x)*log(b*x + a))*log(b^2*x^2 + 2*a*b*x + a^2))*arctan(sqrt(b*x + a + 1)*sqrt(b*x
+ a - 1)))/(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2 - 1)*b*x - a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*arcsec(b*x + a)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*asec(a + b*x)**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arcsec(b*x + a)^3, x)