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

Optimal. Leaf size=154 \[ -\frac{6 i \sec ^{-1}(a+b x) \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac{6 i \sec ^{-1}(a+b x) \text{PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac{6 \text{PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )}{b}-\frac{6 \text{PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac{(a+b x) \sec ^{-1}(a+b x)^3}{b}+\frac{6 i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.103838, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.875, Rules used = {5252, 5216, 3757, 4181, 2531, 2282, 6589} \[ -\frac{6 i \sec ^{-1}(a+b x) \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac{6 i \sec ^{-1}(a+b x) \text{PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac{6 \text{PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )}{b}-\frac{6 \text{PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac{(a+b x) \sec ^{-1}(a+b x)^3}{b}+\frac{6 i \sec ^{-1}(a+b x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcSec[a + b*x]^3,x]

[Out]

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

Rule 5252

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

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 3757

Int[(x_)^(m_.)*Sec[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*Tan[(a_.) + (b_.)*(x_)^(n_.)]^(q_.), x_Symbol] :> Simp[(x^(
m - n + 1)*Sec[a + b*x^n]^p)/(b*n*p), x] - Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Sec[a + b*x^n]^p, x], x] /;
FreeQ[{a, b, p}, x] && IntegerQ[n] && GeQ[m, n] && EqQ[q, 1]

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

Mathematica [A]  time = 0.0926012, size = 160, normalized size = 1.04 \[ \frac{-6 i \sec ^{-1}(a+b x) \left (\text{PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )-\text{PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )\right )+6 \left (\text{PolyLog}\left (3,-i e^{i \sec ^{-1}(a+b x)}\right )-\text{PolyLog}\left (3,i e^{i \sec ^{-1}(a+b x)}\right )\right )+(a+b x) \sec ^{-1}(a+b x)^3-3 \sec ^{-1}(a+b x)^2 \left (\log \left (1-i e^{i \sec ^{-1}(a+b x)}\right )-\log \left (1+i e^{i \sec ^{-1}(a+b x)}\right )\right )}{b} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSec[a + b*x]^3,x]

[Out]

((a + b*x)*ArcSec[a + b*x]^3 - 3*ArcSec[a + b*x]^2*(Log[1 - I*E^(I*ArcSec[a + b*x])] - Log[1 + I*E^(I*ArcSec[a
+ b*x])]) - (6*I)*ArcSec[a + b*x]*(PolyLog[2, (-I)*E^(I*ArcSec[a + b*x])] - PolyLog[2, I*E^(I*ArcSec[a + b*x]
)]) + 6*(PolyLog[3, (-I)*E^(I*ArcSec[a + b*x])] - PolyLog[3, I*E^(I*ArcSec[a + b*x])]))/b

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Maple [F]  time = 0.454, size = 0, normalized size = 0. \begin{align*} \int \left ({\rm arcsec} \left (bx+a\right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} x \arctan \left (\sqrt{b x + a + 1} \sqrt{b x + a - 1}\right )^{3} - \frac{3}{4} \, x \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 \arctan \left (\sqrt{b x + a + 1} \sqrt{b x + a - 1}\right )^{2} - b x \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 ({\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} - 1\right )} b x - a\right )} \log \left (b x + a\right )^{2} -{\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} +{\left (a^{2} - 1\right )} b x +{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} - 1\right )} b x - a\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 )}}{4 \,{\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^3 - 3/4*x*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))*log(b^2*x^
2 + 2*a*b*x + a^2)^2 - integrate(3/4*((4*b*x*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^2 - b*x*log(b^2*x^2 +
2*a*b*x + a^2)^2)*sqrt(b*x + a + 1)*sqrt(b*x + a - 1) + 4*((b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2 - 1)*b*x - a
)*log(b*x + a)^2 - (b^3*x^3 + 2*a*b^2*x^2 + (a^2 - 1)*b*x + (b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2 - 1)*b*x - a
)*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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\operatorname{arcsec}\left (b x + a\right )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsec(b*x + a)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asec(a + b*x)**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsec(b*x + a)^3, x)