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

Optimal. Leaf size=158 \[ -\frac{6 i b^2 \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c}+\frac{6 i b^2 \text{PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c}+\frac{6 b^3 \text{PolyLog}\left (3,-i e^{i \sec ^{-1}(c x)}\right )}{c}-\frac{6 b^3 \text{PolyLog}\left (3,i e^{i \sec ^{-1}(c x)}\right )}{c}+x \left (a+b \sec ^{-1}(c x)\right )^3+\frac{6 i b \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2}{c} \]

[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

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Rubi [A]  time = 0.118292, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {5216, 4409, 4181, 2531, 2282, 6589} \[ -\frac{6 i b^2 \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c}+\frac{6 i b^2 \text{PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c}+\frac{6 b^3 \text{PolyLog}\left (3,-i e^{i \sec ^{-1}(c x)}\right )}{c}-\frac{6 b^3 \text{PolyLog}\left (3,i e^{i \sec ^{-1}(c x)}\right )}{c}+x \left (a+b \sec ^{-1}(c x)\right )^3+\frac{6 i b \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2}{c} \]

Antiderivative was successfully verified.

[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 5216

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 4409

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 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]

Rubi steps

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

Mathematica [A]  time = 0.236383, size = 289, normalized size = 1.83 \[ \frac{-6 i b^2 \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )+6 i b^2 \text{PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )+6 b^3 \text{PolyLog}\left (3,-i e^{i \sec ^{-1}(c x)}\right )-6 b^3 \text{PolyLog}\left (3,i e^{i \sec ^{-1}(c x)}\right )-3 a^2 b \log \left (c x \left (\sqrt{1-\frac{1}{c^2 x^2}}+1\right )\right )+3 a^2 b c x \sec ^{-1}(c x)+a^3 c x+3 a b^2 c x \sec ^{-1}(c x)^2-6 a b^2 \sec ^{-1}(c x) \log \left (1-i e^{i \sec ^{-1}(c x)}\right )+6 a b^2 \sec ^{-1}(c x) \log \left (1+i e^{i \sec ^{-1}(c x)}\right )+b^3 c x \sec ^{-1}(c x)^3-3 b^3 \sec ^{-1}(c x)^2 \log \left (1-i e^{i \sec ^{-1}(c x)}\right )+3 b^3 \sec ^{-1}(c x)^2 \log \left (1+i e^{i \sec ^{-1}(c x)}\right )}{c} \]

Warning: Unable to verify antiderivative.

[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

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Maple [F]  time = 0.508, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\rm arcsec} \left (cx\right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b^{3} \operatorname{arcsec}\left (c x\right )^{3} + 3 \, a b^{2} \operatorname{arcsec}\left (c x\right )^{2} + 3 \, a^{2} b \operatorname{arcsec}\left (c x\right ) + a^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asec(c*x))**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{arcsec}\left (c x\right ) + a\right )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsec(c*x) + a)^3, x)