3.37 \(\int \frac{\sec ^{-1}(a+b x)^3}{x^2} \, dx\)

Optimal. Leaf size=362 \[ -\frac{6 b \sec ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{6 b \sec ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )}{a \sqrt{1-a^2}}-\frac{6 i b \text{PolyLog}\left (3,\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{6 i b \text{PolyLog}\left (3,\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )}{a \sqrt{1-a^2}}-\frac{3 i b \sec ^{-1}(a+b x)^2 \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{3 i b \sec ^{-1}(a+b x)^2 \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )}{a \sqrt{1-a^2}}-\frac{b \sec ^{-1}(a+b x)^3}{a}-\frac{\sec ^{-1}(a+b x)^3}{x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.59477, antiderivative size = 362, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {5258, 4426, 4191, 3321, 2264, 2190, 2531, 2282, 6589} \[ -\frac{6 b \sec ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{6 b \sec ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )}{a \sqrt{1-a^2}}-\frac{6 i b \text{PolyLog}\left (3,\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{6 i b \text{PolyLog}\left (3,\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )}{a \sqrt{1-a^2}}-\frac{3 i b \sec ^{-1}(a+b x)^2 \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{3 i b \sec ^{-1}(a+b x)^2 \log \left (1-\frac{a e^{i \sec ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )}{a \sqrt{1-a^2}}-\frac{b \sec ^{-1}(a+b x)^3}{a}-\frac{\sec ^{-1}(a+b x)^3}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcSec[a + b*x]^3/x^2,x]

[Out]

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

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 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 4191

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] &
& IGtQ[m, 0]

Rule 3321

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c
+ d*x)^m*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)))/(b + 2*a*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)) - b*E^(2*I*k*Pi)*E^(
2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[2*k] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

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

Mathematica [F]  time = 180.003, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

Integrate[ArcSec[a + b*x]^3/x^2,x]

[Out]

$Aborted

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Maple [F]  time = 0.884, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ({\rm arcsec} \left (bx+a\right ) \right ) ^{3}}{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsec(b*x+a)^3/x^2,x)

[Out]

int(arcsec(b*x+a)^3/x^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{4 \, \arctan \left (\sqrt{b x + a + 1} \sqrt{b x + a - 1}\right )^{3} - 3 \, \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} - 3 \, x \int \frac{{\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 )}{b^{3} x^{5} + 3 \, a b^{2} x^{4} +{\left (3 \, a^{2} - 1\right )} b x^{3} +{\left (a^{3} - a\right )} x^{2}}\,{d x}}{4 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(b*x+a)^3/x^2,x, algorithm="maxima")

[Out]

-1/4*(4*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^3 - 3*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))*log(b^2*
x^2 + 2*a*b*x + a^2)^2 - 4*x*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^5 + 3*
a*b^2*x^4 + (3*a^2 - 1)*b*x^3 + (a^3 - a)*x^2), x))/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(b*x+a)^3/x^2,x, algorithm="fricas")

[Out]

integral(arcsec(b*x + a)^3/x^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asec}^{3}{\left (a + b x \right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asec(b*x+a)**3/x**2,x)

[Out]

Integral(asec(a + b*x)**3/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcsec}\left (b x + a\right )^{3}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(b*x+a)^3/x^2,x, algorithm="giac")

[Out]

integrate(arcsec(b*x + a)^3/x^2, x)