3.1.33 \(\int x^2 \sec ^{-1}(a+b x)^3 \, dx\) [33]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.31, antiderivative size = 494, normalized size of antiderivative = 1.00, 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 = {5366, 4511, 4275, 4266, 2611, 2320, 6724, 4269, 3800, 2221, 2317, 2438, 4271, 3855} \begin {gather*} \frac {a^3 \sec ^{-1}(a+b x)^3}{3 b^3}+\frac {6 i a^2 \sec ^{-1}(a+b x)^2 \text {ArcTan}\left (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 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 {i \sec ^{-1}(a+b x)^2 \text {ArcTan}\left (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 {i \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 {\text {Li}_3\left (i 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 {1}{3} x^3 \sec ^{-1}(a+b x)^3 \end {gather*}

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 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 2438

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 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 3800

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 3855

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

Rule 4266

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 4269

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 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 4275

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 4511

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 5366

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

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Mathematica [A]
time = 0.36, size = 442, normalized size = 0.89 \begin {gather*} \frac {(a+b x) \sec ^{-1}(a+b x)+3 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2-\frac {1}{2} (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2+\frac {1}{3} a^3 \sec ^{-1}(a+b x)^3+\frac {1}{3} b^3 x^3 \sec ^{-1}(a+b x)^3+i \sec ^{-1}(a+b x)^2 \text {ArcTan}\left (e^{i \sec ^{-1}(a+b x)}\right )+6 i a^2 \sec ^{-1}(a+b x)^2 \text {ArcTan}\left (e^{i \sec ^{-1}(a+b x)}\right )-\tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\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 )-3 i a \left (\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 )+\text {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )\right )+\text {PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )+6 a^2 \left (-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 )\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 )-\text {PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3} \end {gather*}

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 = 1.10, size = 716, normalized size = 1.45 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^3*(1/6*arcsec(b*x+a)*(6*arcsec(b*x+a)^2*a^2*(b*x+a)-6*arcsec(b*x+a)^2*a*(b*x+a)^2+2*arcsec(b*x+a)^2*(b*x+a
)^3+18*arcsec(b*x+a)*(((b*x+a)^2-1)/(b*x+a)^2)^(1/2)*a*(b*x+a)-3*arcsec(b*x+a)*(((b*x+a)^2-1)/(b*x+a)^2)^(1/2)
*(b*x+a)^2+18*I*a*arcsec(b*x+a)+6*b*x+6*a)-I*arcsec(b*x+a)*polylog(2,-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))+2
*I*arctan(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))+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)))-6*I*arcsec(b*x+a)^2*a-6*I*a^2*arcsec(b*x+a)*polylog(2
,-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))-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)))+6*a*arcsec(b*x+a)*ln(1+(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))
^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)))+I*arcsec(b*x+a)*polylog(2,I*(1/(b*x
+a)+I*(1-1/(b*x+a)^2)^(1/2)))+1/2*arcsec(b*x+a)^2*ln(1+I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))-3*I*a*polylog(2,
-(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))^2)+polylog(3,-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))-1/2*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))))

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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*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 = 0.00, 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*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \operatorname {asec}^{3}{\left (a + b x \right )}\, dx \end {gather*}

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 = 0.00, 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*arcsec(b*x+a)^3,x, algorithm="giac")

[Out]

integrate(x^2*arcsec(b*x + a)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*acos(1/(a + b*x))^3,x)

[Out]

int(x^2*acos(1/(a + b*x))^3, x)

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