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

Optimal. Leaf size=278 \[ -\frac {a^2 \sec ^{-1}(a+b x)^3}{2 b^2}+\frac {6 i a \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {3 i \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )}{2 b^2}-\frac {6 a \text {Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {6 a \text {Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {3 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2}{2 b^2}+\frac {3 i \sec ^{-1}(a+b x)^2}{2 b^2}-\frac {3 \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^3 \]

[Out]

3/2*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

________________________________________________________________________________________

Rubi [A]  time = 0.26, 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 = {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 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 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 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 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 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 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 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 \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 = 0.44, size = 248, normalized size = 0.89 \[ \frac {1}{2} \left (x^2 \sec ^{-1}(a+b x)^3-\frac {3 \left (\frac {1}{3} a^2 \sec ^{-1}(a+b x)^3+4 i a \sec ^{-1}(a+b x) \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )-i \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )+4 a \left (\text {Li}_3\left (-i e^{i \sec ^{-1}(a+b x)}\right )-i \sec ^{-1}(a+b x) \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )\right )-4 a \text {Li}_3\left (i e^{i \sec ^{-1}(a+b x)}\right )+(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^2-i \sec ^{-1}(a+b x)^2+2 \sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )+4 i a \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )\right )}{b^2}\right ) \]

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

________________________________________________________________________________________

fricas [F]  time = 1.19, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x \operatorname {arcsec}\left (b x + a\right )^{3}, x\right ) \]

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

maple [A]  time = 1.53, size = 429, normalized size = 1.54 \[ \frac {x^{2} \mathrm {arcsec}\left (b x +a \right )^{3}}{2}-\frac {3 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right )^{2} x}{2 b}-\frac {a^{2} \mathrm {arcsec}\left (b x +a \right )^{3}}{2 b^{2}}-\frac {3 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right )^{2} a}{2 b^{2}}+\frac {3 i \mathrm {arcsec}\left (b x +a \right )^{2}}{2 b^{2}}+\frac {3 a \mathrm {arcsec}\left (b x +a \right )^{2} \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{2}}-\frac {6 i a \,\mathrm {arcsec}\left (b x +a \right ) \polylog \left (2, i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{2}}+\frac {6 a \polylog \left (3, i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{2}}+\frac {6 i a \,\mathrm {arcsec}\left (b x +a \right ) \polylog \left (2, -i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{2}}-\frac {3 a \mathrm {arcsec}\left (b x +a \right )^{2} \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{2}}-\frac {6 a \polylog \left (3, -i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{2}}-\frac {3 \,\mathrm {arcsec}\left (b x +a \right ) \ln \left (1+\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )^{2}\right )}{b^{2}}+\frac {3 i \polylog \left (2, -\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )^{2}\right )}{2 b^{2}} \]

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)^3-3/2/b*((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)*arcsec(b*x+a)^2*x-1/2*a^2*arcsec(b*x+a)^3/b^2-3
/2/b^2*((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)*arcsec(b*x+a)^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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, x^{2} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{3} - \frac {3}{8} \, x^{2} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2} - \int \frac {3 \, {\left ({\left (4 \, b x^{2} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{2} - b x^{2} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2}\right )} \sqrt {b x + a + 1} \sqrt {b x + a - 1} + 4 \, {\left (2 \, {\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} - 1\right )} b x^{2} + {\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right )^{2} - {\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} + {\left (a^{2} - 1\right )} b x^{2} + 2 \, {\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} - 1\right )} b x^{2} + {\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right )\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )\right )} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )\right )}}{8 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a\right )}}\,{d x} \]

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________