3.33 \(\int x^2 \sec ^{-1}(a+b x)^3 \, dx\)

Optimal. Leaf size=494 \[ -\frac{6 i a^2 \sec ^{-1}(a+b x) \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{6 i a^2 \sec ^{-1}(a+b x) \text{PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{6 a^2 \text{PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{6 a^2 \text{PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{i \sec ^{-1}(a+b x) \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{i \sec ^{-1}(a+b x) \text{PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{3 i a \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{\text{PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{\text{PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{a^3 \sec ^{-1}(a+b x)^3}{3 b^3}+\frac{6 i a^2 \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{(a+b x)^2 \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^3}+\frac{3 a (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{b^3}-\frac{3 i a \sec ^{-1}(a+b x)^2}{b^3}+\frac{(a+b x) \sec ^{-1}(a+b x)}{b^3}-\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{(a+b x)^2}}\right )}{b^3}+\frac{6 a \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{1}{3} x^3 \sec ^{-1}(a+b x)^3 \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.396809, antiderivative size = 494, normalized size of antiderivative = 1., number of steps used = 25, number of rules used = 14, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.167, Rules used = {5258, 4426, 4190, 4181, 2531, 2282, 6589, 4184, 3719, 2190, 2279, 2391, 4186, 3770} \[ -\frac{6 i a^2 \sec ^{-1}(a+b x) \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{6 i a^2 \sec ^{-1}(a+b x) \text{PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{6 a^2 \text{PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{6 a^2 \text{PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{i \sec ^{-1}(a+b x) \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{i \sec ^{-1}(a+b x) \text{PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{3 i a \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{\text{PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{\text{PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{a^3 \sec ^{-1}(a+b x)^3}{3 b^3}+\frac{6 i a^2 \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{(a+b x)^2 \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^3}+\frac{3 a (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{b^3}-\frac{3 i a \sec ^{-1}(a+b x)^2}{b^3}+\frac{(a+b x) \sec ^{-1}(a+b x)}{b^3}-\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{(a+b x)^2}}\right )}{b^3}+\frac{6 a \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{1}{3} x^3 \sec ^{-1}(a+b x)^3 \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcSec[a + b*x]^3,x]

[Out]

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

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]

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 3770

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

Rubi steps

\begin{align*} \int x^2 \sec ^{-1}(a+b x)^3 \, dx &=\frac{\operatorname{Subst}\left (\int x^3 \sec (x) (-a+\sec (x))^2 \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}\\ &=\frac{1}{3} x^3 \sec ^{-1}(a+b x)^3-\frac{\operatorname{Subst}\left (\int x^2 (-a+\sec (x))^3 \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}\\ &=\frac{1}{3} x^3 \sec ^{-1}(a+b x)^3-\frac{\operatorname{Subst}\left (\int \left (-a^3 x^2+3 a^2 x^2 \sec (x)-3 a x^2 \sec ^2(x)+x^2 \sec ^3(x)\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}\\ &=\frac{a^3 \sec ^{-1}(a+b x)^3}{3 b^3}+\frac{1}{3} x^3 \sec ^{-1}(a+b x)^3-\frac{\operatorname{Subst}\left (\int x^2 \sec ^3(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}+\frac{(3 a) \operatorname{Subst}\left (\int x^2 \sec ^2(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}-\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int x^2 \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}\\ &=\frac{(a+b x) \sec ^{-1}(a+b x)}{b^3}+\frac{3 a (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{b^3}-\frac{(a+b x)^2 \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^3}+\frac{a^3 \sec ^{-1}(a+b x)^3}{3 b^3}+\frac{1}{3} x^3 \sec ^{-1}(a+b x)^3+\frac{6 i a^2 \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{\operatorname{Subst}\left (\int x^2 \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{2 b^3}-\frac{\operatorname{Subst}\left (\int \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}-\frac{(6 a) \operatorname{Subst}\left (\int x \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}+\frac{\left (6 a^2\right ) \operatorname{Subst}\left (\int x \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}-\frac{\left (6 a^2\right ) \operatorname{Subst}\left (\int x \log \left (1+i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}\\ &=\frac{(a+b x) \sec ^{-1}(a+b x)}{b^3}-\frac{3 i a \sec ^{-1}(a+b x)^2}{b^3}+\frac{3 a (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{b^3}-\frac{(a+b x)^2 \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^3}+\frac{a^3 \sec ^{-1}(a+b x)^3}{3 b^3}+\frac{1}{3} x^3 \sec ^{-1}(a+b x)^3+\frac{i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{6 i a^2 \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{(a+b x)^2}}\right )}{b^3}-\frac{6 i a^2 \sec ^{-1}(a+b x) \text{Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{6 i a^2 \sec ^{-1}(a+b x) \text{Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{\operatorname{Subst}\left (\int x \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}-\frac{\operatorname{Subst}\left (\int x \log \left (1+i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}+\frac{(12 i a) \operatorname{Subst}\left (\int \frac{e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}+\frac{\left (6 i a^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}-\frac{\left (6 i a^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}\\ &=\frac{(a+b x) \sec ^{-1}(a+b x)}{b^3}-\frac{3 i a \sec ^{-1}(a+b x)^2}{b^3}+\frac{3 a (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{b^3}-\frac{(a+b x)^2 \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^3}+\frac{a^3 \sec ^{-1}(a+b x)^3}{3 b^3}+\frac{1}{3} x^3 \sec ^{-1}(a+b x)^3+\frac{i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{6 i a^2 \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{(a+b x)^2}}\right )}{b^3}+\frac{6 a \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{i \sec ^{-1}(a+b x) \text{Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{6 i a^2 \sec ^{-1}(a+b x) \text{Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{i \sec ^{-1}(a+b x) \text{Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{6 i a^2 \sec ^{-1}(a+b x) \text{Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{i \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}-\frac{i \operatorname{Subst}\left (\int \text{Li}_2\left (i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}-\frac{(6 a) \operatorname{Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^3}+\frac{\left (6 a^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{\left (6 a^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^3}\\ &=\frac{(a+b x) \sec ^{-1}(a+b x)}{b^3}-\frac{3 i a \sec ^{-1}(a+b x)^2}{b^3}+\frac{3 a (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{b^3}-\frac{(a+b x)^2 \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^3}+\frac{a^3 \sec ^{-1}(a+b x)^3}{3 b^3}+\frac{1}{3} x^3 \sec ^{-1}(a+b x)^3+\frac{i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{6 i a^2 \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{(a+b x)^2}}\right )}{b^3}+\frac{6 a \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{i \sec ^{-1}(a+b x) \text{Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{6 i a^2 \sec ^{-1}(a+b x) \text{Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{i \sec ^{-1}(a+b x) \text{Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{6 i a^2 \sec ^{-1}(a+b x) \text{Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{6 a^2 \text{Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{6 a^2 \text{Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{(3 i a) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}\\ &=\frac{(a+b x) \sec ^{-1}(a+b x)}{b^3}-\frac{3 i a \sec ^{-1}(a+b x)^2}{b^3}+\frac{3 a (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{b^3}-\frac{(a+b x)^2 \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^3}+\frac{a^3 \sec ^{-1}(a+b x)^3}{3 b^3}+\frac{1}{3} x^3 \sec ^{-1}(a+b x)^3+\frac{i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{6 i a^2 \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{(a+b x)^2}}\right )}{b^3}+\frac{6 a \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{i \sec ^{-1}(a+b x) \text{Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{6 i a^2 \sec ^{-1}(a+b x) \text{Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{i \sec ^{-1}(a+b x) \text{Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{6 i a^2 \sec ^{-1}(a+b x) \text{Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{3 i a \text{Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{\text{Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac{6 a^2 \text{Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{\text{Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac{6 a^2 \text{Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}\\ \end{align*}

Mathematica [A]  time = 0.490697, size = 442, normalized size = 0.89 \[ \frac{6 a^2 \left (\text{PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )-i \sec ^{-1}(a+b x) \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )\right )+6 i a^2 \left (\sec ^{-1}(a+b x) \text{PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )+i \text{PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )\right )-i \sec ^{-1}(a+b x) \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )+i \sec ^{-1}(a+b x) \text{PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )+\text{PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )-\text{PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )-3 i a \left (\text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )+\sec ^{-1}(a+b x) \left (\sec ^{-1}(a+b x)+2 i \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )\right )\right )+\frac{1}{3} a^3 \sec ^{-1}(a+b x)^3+6 i a^2 \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )+\frac{1}{3} b^3 x^3 \sec ^{-1}(a+b x)^3-\frac{1}{2} (a+b x)^2 \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2+3 a (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2+(a+b x) \sec ^{-1}(a+b x)-\tanh ^{-1}\left (\sqrt{1-\frac{1}{(a+b x)^2}}\right )+i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcSec[a + b*x]^3,x]

[Out]

((a + b*x)*ArcSec[a + b*x] + 3*a*(a + b*x)*Sqrt[1 - (a + b*x)^(-2)]*ArcSec[a + b*x]^2 - ((a + b*x)^2*Sqrt[1 -
(a + b*x)^(-2)]*ArcSec[a + b*x]^2)/2 + (a^3*ArcSec[a + b*x]^3)/3 + (b^3*x^3*ArcSec[a + b*x]^3)/3 + I*ArcSec[a
+ b*x]^2*ArcTan[E^(I*ArcSec[a + b*x])] + (6*I)*a^2*ArcSec[a + b*x]^2*ArcTan[E^(I*ArcSec[a + b*x])] - ArcTanh[S
qrt[1 - (a + b*x)^(-2)]] - I*ArcSec[a + b*x]*PolyLog[2, (-I)*E^(I*ArcSec[a + b*x])] + I*ArcSec[a + b*x]*PolyLo
g[2, I*E^(I*ArcSec[a + b*x])] - (3*I)*a*(ArcSec[a + b*x]*(ArcSec[a + b*x] + (2*I)*Log[1 + E^((2*I)*ArcSec[a +
b*x])]) + PolyLog[2, -E^((2*I)*ArcSec[a + b*x])]) + PolyLog[3, (-I)*E^(I*ArcSec[a + b*x])] + 6*a^2*((-I)*ArcSe
c[a + b*x]*PolyLog[2, (-I)*E^(I*ArcSec[a + b*x])] + PolyLog[3, (-I)*E^(I*ArcSec[a + b*x])]) + (6*I)*a^2*(ArcSe
c[a + b*x]*PolyLog[2, I*E^(I*ArcSec[a + b*x])] + I*PolyLog[3, I*E^(I*ArcSec[a + b*x])]) - PolyLog[3, I*E^(I*Ar
cSec[a + b*x])])/b^3

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Maple [A]  time = 0.59, size = 770, normalized size = 1.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arcsec(b*x+a)^3-1/2/b*arcsec(b*x+a)^2*((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)*x^2+1/3*a^3*arcsec(b*x+a)^3/b^3
+2/b^2*arcsec(b*x+a)^2*((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)*x*a+5/2/b^3*arcsec(b*x+a)^2*((-1+(b*x+a)^2)/(b*x+a)^2)
^(1/2)*a^2+6*I*a^2*arcsec(b*x+a)*polylog(2,I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b^3+1/b^2*arcsec(b*x+a)*x+1/
b^3*arcsec(b*x+a)*a-3*I*a*arcsec(b*x+a)^2/b^3-3*I*a*polylog(2,-(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))^2)/b^3+3/b^
3*a^2*arcsec(b*x+a)^2*ln(1+I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))+6*a^2*polylog(3,-I*(1/(b*x+a)+I*(1-1/(b*x+a)
^2)^(1/2)))/b^3+I*arcsec(b*x+a)*polylog(2,I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b^3-I*arcsec(b*x+a)*polylog(2
,-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b^3-3/b^3*a^2*arcsec(b*x+a)^2*ln(1-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/
2)))-6*a^2*polylog(3,I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b^3+6*a*arcsec(b*x+a)*ln(1+(1/(b*x+a)+I*(1-1/(b*x+
a)^2)^(1/2))^2)/b^3-6*I*a^2*arcsec(b*x+a)*polylog(2,-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b^3+1/2/b^3*arcsec
(b*x+a)^2*ln(1+I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))+2*I/b^3*arctan(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))+polylo
g(3,-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b^3-1/2/b^3*arcsec(b*x+a)^2*ln(1-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1
/2)))-polylog(3,I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{3} \, x^{3} \arctan \left (\sqrt{b x + a + 1} \sqrt{b x + a - 1}\right )^{3} - \frac{1}{4} \, x^{3} \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{{\left (4 \, b x^{3} \arctan \left (\sqrt{b x + a + 1} \sqrt{b x + a - 1}\right )^{2} - b x^{3} \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 (3 \,{\left (b^{3} x^{5} + 3 \, a b^{2} x^{4} +{\left (3 \, a^{2} - 1\right )} b x^{3} +{\left (a^{3} - a\right )} x^{2}\right )} \log \left (b x + a\right )^{2} -{\left (b^{3} x^{5} + 2 \, a b^{2} x^{4} +{\left (a^{2} - 1\right )} b x^{3} + 3 \,{\left (b^{3} x^{5} + 3 \, a b^{2} x^{4} +{\left (3 \, a^{2} - 1\right )} b x^{3} +{\left (a^{3} - a\right )} x^{2}\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 )}{4 \,{\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^2*arcsec(b*x+a)^3,x, algorithm="maxima")

[Out]

1/3*x^3*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^3 - 1/4*x^3*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(1/4*((4*b*x^3*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^2 - b*x^3*l
og(b^2*x^2 + 2*a*b*x + a^2)^2)*sqrt(b*x + a + 1)*sqrt(b*x + a - 1) + 4*(3*(b^3*x^5 + 3*a*b^2*x^4 + (3*a^2 - 1)
*b*x^3 + (a^3 - a)*x^2)*log(b*x + a)^2 - (b^3*x^5 + 2*a*b^2*x^4 + (a^2 - 1)*b*x^3 + 3*(b^3*x^5 + 3*a*b^2*x^4 +
(3*a^2 - 1)*b*x^3 + (a^3 - a)*x^2)*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^{2} \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^2*arcsec(b*x+a)^3,x, algorithm="fricas")

[Out]

integral(x^2*arcsec(b*x + a)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*asec(b*x+a)**3,x)

[Out]

Integral(x**2*asec(a + b*x)**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \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^2*arcsec(b*x+a)^3,x, algorithm="giac")

[Out]

integrate(x^2*arcsec(b*x + a)^3, x)