3.1.34 \(\int x \sec ^{-1}(a+b x)^3 \, dx\) [34]

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

[Out]

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

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Rubi [A]
time = 0.20, antiderivative size = 278, 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 = {5366, 4511, 4275, 4266, 2611, 2320, 6724, 4269, 3800, 2221, 2317, 2438} \begin {gather*} -\frac {a^2 \sec ^{-1}(a+b x)^3}{2 b^2}-\frac {6 i a \sec ^{-1}(a+b x)^2 \text {ArcTan}\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 {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}-\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 {1}{2} x^2 \sec ^{-1}(a+b x)^3 \end {gather*}

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

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Mathematica [A]
time = 0.29, size = 248, normalized size = 0.89 \begin {gather*} \frac {1}{2} \left (x^2 \sec ^{-1}(a+b x)^3-\frac {3 \left (-i \sec ^{-1}(a+b x)^2+(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2+\frac {1}{3} a^2 \sec ^{-1}(a+b x)^3+4 i a \sec ^{-1}(a+b x)^2 \text {ArcTan}\left (e^{i \sec ^{-1}(a+b x)}\right )+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) \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 (-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 )-4 a \text {PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )\right )}{b^2}\right ) \end {gather*}

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 = 1.01, size = 379, normalized size = 1.36

method	result	size
derivativedivides	\(\frac {-\frac {\mathrm {arcsec}\left (b x +a \right )^{2} \left (2 \,\mathrm {arcsec}\left (b x +a \right ) a \left (b x +a \right )-\mathrm {arcsec}\left (b x +a \right ) \left (b x +a \right )^{2}+3 \sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )+3 i\right )}{2}+6 i a \,\mathrm {arcsec}\left (b x +a \right ) \polylog \left (2, -i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-3 a \mathrm {arcsec}\left (b x +a \right )^{2} \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-6 a \polylog \left (3, -i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-6 i a \,\mathrm {arcsec}\left (b x +a \right ) \polylog \left (2, i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+3 a \mathrm {arcsec}\left (b x +a \right )^{2} \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+6 a \polylog \left (3, i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+3 i \mathrm {arcsec}\left (b x +a \right )^{2}-3 \,\mathrm {arcsec}\left (b x +a \right ) \ln \left (1+\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )^{2}\right )+\frac {3 i \polylog \left (2, -\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )^{2}\right )}{2}}{b^{2}}\)	\(379\)
default	\(\frac {-\frac {\mathrm {arcsec}\left (b x +a \right )^{2} \left (2 \,\mathrm {arcsec}\left (b x +a \right ) a \left (b x +a \right )-\mathrm {arcsec}\left (b x +a \right ) \left (b x +a \right )^{2}+3 \sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )+3 i\right )}{2}+6 i a \,\mathrm {arcsec}\left (b x +a \right ) \polylog \left (2, -i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-3 a \mathrm {arcsec}\left (b x +a \right )^{2} \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-6 a \polylog \left (3, -i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-6 i a \,\mathrm {arcsec}\left (b x +a \right ) \polylog \left (2, i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+3 a \mathrm {arcsec}\left (b x +a \right )^{2} \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+6 a \polylog \left (3, i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+3 i \mathrm {arcsec}\left (b x +a \right )^{2}-3 \,\mathrm {arcsec}\left (b x +a \right ) \ln \left (1+\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )^{2}\right )+\frac {3 i \polylog \left (2, -\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )^{2}\right )}{2}}{b^{2}}\)	\(379\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^2*(-1/2*arcsec(b*x+a)^2*(2*arcsec(b*x+a)*a*(b*x+a)-arcsec(b*x+a)*(b*x+a)^2+3*(((b*x+a)^2-1)/(b*x+a)^2)^(1/
2)*(b*x+a)+3*I)+6*I*a*arcsec(b*x+a)*polylog(2,-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))-3*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)))-6*I*a*arcsec(b*x
+a)*polylog(2,I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))+3*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)))+3*I*arcsec(b*x+a)^2-3*arcsec(b*x+a)*ln(1+(1/(b*x+a
)+I*(1-1/(b*x+a)^2)^(1/2))^2)+3/2*I*polylog(2,-(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))^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*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 = 0.00, 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*arcsec(b*x+a)^3,x, algorithm="fricas")

[Out]

integral(x*arcsec(b*x + a)^3, x)

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

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

[Out]

integrate(x*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\,{\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*acos(1/(a + b*x))^3,x)

[Out]

int(x*acos(1/(a + b*x))^3, x)

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