\(\int (a+b \sec ^{-1}(c x))^3 \, dx\) [27]

Optimal result

Integrand size = 10, antiderivative size = 158 \[ \int \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=x \left (a+b \sec ^{-1}(c x)\right )^3+\frac {6 i b \left (a+b \sec ^{-1}(c x)\right )^2 \arctan \left (e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right )}{c}+\frac {6 b^3 \operatorname {PolyLog}\left (3,-i e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {6 b^3 \operatorname {PolyLog}\left (3,i e^{i \sec ^{-1}(c x)}\right )}{c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5324, 4494, 4266, 2611, 2320, 6724} \[ \int \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=\frac {6 i b \arctan \left (e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2}{c}-\frac {6 i b^2 \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c}+\frac {6 i b^2 \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c}+x \left (a+b \sec ^{-1}(c x)\right )^3+\frac {6 b^3 \operatorname {PolyLog}\left (3,-i e^{i \sec ^{-1}(c x)}\right )}{c}-\frac {6 b^3 \operatorname {PolyLog}\left (3,i e^{i \sec ^{-1}(c x)}\right )}{c} \]

[In]

Int[(a + b*ArcSec[c*x])^3,x]

[Out]

x*(a + b*ArcSec[c*x])^3 + ((6*I)*b*(a + b*ArcSec[c*x])^2*ArcTan[E^(I*ArcSec[c*x])])/c - ((6*I)*b^2*(a + b*ArcS
ec[c*x])*PolyLog[2, (-I)*E^(I*ArcSec[c*x])])/c + ((6*I)*b^2*(a + b*ArcSec[c*x])*PolyLog[2, I*E^(I*ArcSec[c*x])
])/c + (6*b^3*PolyLog[3, (-I)*E^(I*ArcSec[c*x])])/c - (6*b^3*PolyLog[3, I*E^(I*ArcSec[c*x])])/c

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 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 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 4494

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Simp[
(c + d*x)^m*(Sec[a + b*x]^n/(b*n)), x] - Dist[d*(m/(b*n)), Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; Fre
eQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rule 5324

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[1/c, Subst[Int[(a + b*x)^n*Sec[x]*Tan[x], x], x
, ArcSec[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[n, 0]

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.83 \[ \int \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=\frac {a^3 c x+3 a^2 b c x \sec ^{-1}(c x)+3 a b^2 c x \sec ^{-1}(c x)^2+b^3 c x \sec ^{-1}(c x)^3-6 a b^2 \sec ^{-1}(c x) \log \left (1-i e^{i \sec ^{-1}(c x)}\right )-3 b^3 \sec ^{-1}(c x)^2 \log \left (1-i e^{i \sec ^{-1}(c x)}\right )+6 a b^2 \sec ^{-1}(c x) \log \left (1+i e^{i \sec ^{-1}(c x)}\right )+3 b^3 \sec ^{-1}(c x)^2 \log \left (1+i e^{i \sec ^{-1}(c x)}\right )-3 a^2 b \log \left (c \left (1+\sqrt {1-\frac {1}{c^2 x^2}}\right ) x\right )-6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \operatorname {PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right )+6 i b^2 \left (a+b \sec ^{-1}(c x)\right ) \operatorname {PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right )+6 b^3 \operatorname {PolyLog}\left (3,-i e^{i \sec ^{-1}(c x)}\right )-6 b^3 \operatorname {PolyLog}\left (3,i e^{i \sec ^{-1}(c x)}\right )}{c} \]

[In]

Integrate[(a + b*ArcSec[c*x])^3,x]

[Out]

(a^3*c*x + 3*a^2*b*c*x*ArcSec[c*x] + 3*a*b^2*c*x*ArcSec[c*x]^2 + b^3*c*x*ArcSec[c*x]^3 - 6*a*b^2*ArcSec[c*x]*L
og[1 - I*E^(I*ArcSec[c*x])] - 3*b^3*ArcSec[c*x]^2*Log[1 - I*E^(I*ArcSec[c*x])] + 6*a*b^2*ArcSec[c*x]*Log[1 + I
*E^(I*ArcSec[c*x])] + 3*b^3*ArcSec[c*x]^2*Log[1 + I*E^(I*ArcSec[c*x])] - 3*a^2*b*Log[c*(1 + Sqrt[1 - 1/(c^2*x^
2)])*x] - (6*I)*b^2*(a + b*ArcSec[c*x])*PolyLog[2, (-I)*E^(I*ArcSec[c*x])] + (6*I)*b^2*(a + b*ArcSec[c*x])*Pol
yLog[2, I*E^(I*ArcSec[c*x])] + 6*b^3*PolyLog[3, (-I)*E^(I*ArcSec[c*x])] - 6*b^3*PolyLog[3, I*E^(I*ArcSec[c*x])
])/c

Maple [F]

\[\int \left (a +b \,\operatorname {arcsec}\left (c x \right )\right )^{3}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=\int { {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}^{3} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=\int \left (a + b \operatorname {asec}{\left (c x \right )}\right )^{3}\, dx \]

[In]

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

[Out]

Integral((a + b*asec(c*x))**3, x)

Maxima [F]

\[ \int \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=\int { {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}^{3} \,d x } \]

[In]

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

[Out]

-3/2*a*b^2*c^2*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3)*log(c)^2 - 12*b^3*c^2*integrate(1/4*x^2*arctan(
sqrt(c*x + 1)*sqrt(c*x - 1))/(c^2*x^2 - 1), x)*log(c)^2 + b^3*x*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^3 - 3/4*b^
3*x*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)^2 + 12*b^3*c^2*integrate(1/4*x^2*arctan(sqrt(c*x + 1)*sqr
t(c*x - 1))*log(c^2*x^2)/(c^2*x^2 - 1), x)*log(c) - 24*b^3*c^2*integrate(1/4*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x
- 1))*log(x)/(c^2*x^2 - 1), x)*log(c) + 12*a*b^2*c^2*integrate(1/4*x^2*log(c^2*x^2)/(c^2*x^2 - 1), x)*log(c)
- 24*a*b^2*c^2*integrate(1/4*x^2*log(x)/(c^2*x^2 - 1), x)*log(c) + 12*b^3*c^2*integrate(1/4*x^2*arctan(sqrt(c*
x + 1)*sqrt(c*x - 1))*log(c^2*x^2)*log(x)/(c^2*x^2 - 1), x) - 12*b^3*c^2*integrate(1/4*x^2*arctan(sqrt(c*x + 1
)*sqrt(c*x - 1))*log(x)^2/(c^2*x^2 - 1), x) + 12*a*b^2*c^2*integrate(1/4*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1
))^2/(c^2*x^2 - 1), x) + 12*b^3*c^2*integrate(1/4*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)/(c^2*x^
2 - 1), x) - 3*a*b^2*c^2*integrate(1/4*x^2*log(c^2*x^2)^2/(c^2*x^2 - 1), x) + 12*a*b^2*c^2*integrate(1/4*x^2*l
og(c^2*x^2)*log(x)/(c^2*x^2 - 1), x) - 12*a*b^2*c^2*integrate(1/4*x^2*log(x)^2/(c^2*x^2 - 1), x) - 3/2*a*b^2*(
log(c*x + 1)/c - log(c*x - 1)/c)*log(c)^2 + 12*b^3*integrate(1/4*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))/(c^2*x^2
- 1), x)*log(c)^2 - 12*b^3*integrate(1/4*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)/(c^2*x^2 - 1), x)*lo
g(c) + 24*b^3*integrate(1/4*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(x)/(c^2*x^2 - 1), x)*log(c) - 12*a*b^2*int
egrate(1/4*log(c^2*x^2)/(c^2*x^2 - 1), x)*log(c) + 24*a*b^2*integrate(1/4*log(x)/(c^2*x^2 - 1), x)*log(c) + a^
3*x - 12*b^3*integrate(1/4*sqrt(c*x + 1)*sqrt(c*x - 1)*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2/(c^2*x^2 - 1), x)
+ 3*b^3*integrate(1/4*sqrt(c*x + 1)*sqrt(c*x - 1)*log(c^2*x^2)^2/(c^2*x^2 - 1), x) - 12*b^3*integrate(1/4*arc
tan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)*log(x)/(c^2*x^2 - 1), x) + 12*b^3*integrate(1/4*arctan(sqrt(c*x
+ 1)*sqrt(c*x - 1))*log(x)^2/(c^2*x^2 - 1), x) - 12*a*b^2*integrate(1/4*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2/
(c^2*x^2 - 1), x) - 12*b^3*integrate(1/4*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)/(c^2*x^2 - 1), x) +
3*a*b^2*integrate(1/4*log(c^2*x^2)^2/(c^2*x^2 - 1), x) - 12*a*b^2*integrate(1/4*log(c^2*x^2)*log(x)/(c^2*x^2 -
1), x) + 12*a*b^2*integrate(1/4*log(x)^2/(c^2*x^2 - 1), x) + 3/2*(2*c*x*arcsec(c*x) - log(sqrt(-1/(c^2*x^2) +
1) + 1) + log(-sqrt(-1/(c^2*x^2) + 1) + 1))*a^2*b/c

Giac [F]

\[ \int \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=\int { {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}^{3} \,d x } \]

[In]

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

[Out]

integrate((b*arcsec(c*x) + a)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=\int {\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}^3 \,d x \]

[In]

int((a + b*acos(1/(c*x)))^3,x)

[Out]

int((a + b*acos(1/(c*x)))^3, x)