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

Optimal. Leaf size=378 \[ -\frac{6 b \csc ^{-1}(a+b x) \text{PolyLog}\left (2,-\frac{i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{6 b \csc ^{-1}(a+b x) \text{PolyLog}\left (2,-\frac{i a e^{i \csc ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )}{a \sqrt{1-a^2}}-\frac{6 i b \text{PolyLog}\left (3,-\frac{i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{6 i b \text{PolyLog}\left (3,-\frac{i a e^{i \csc ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )}{a \sqrt{1-a^2}}-\frac{3 i b \csc ^{-1}(a+b x)^2 \log \left (1+\frac{i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{3 i b \csc ^{-1}(a+b x)^2 \log \left (1+\frac{i a e^{i \csc ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )}{a \sqrt{1-a^2}}-\frac{b \csc ^{-1}(a+b x)^3}{a}-\frac{\csc ^{-1}(a+b x)^3}{x} \]

[Out]

-((b*ArcCsc[a + b*x]^3)/a) - ArcCsc[a + b*x]^3/x - ((3*I)*b*ArcCsc[a + b*x]^2*Log[1 + (I*a*E^(I*ArcCsc[a + b*x
]))/(1 - Sqrt[1 - a^2])])/(a*Sqrt[1 - a^2]) + ((3*I)*b*ArcCsc[a + b*x]^2*Log[1 + (I*a*E^(I*ArcCsc[a + b*x]))/(
1 + Sqrt[1 - a^2])])/(a*Sqrt[1 - a^2]) - (6*b*ArcCsc[a + b*x]*PolyLog[2, ((-I)*a*E^(I*ArcCsc[a + b*x]))/(1 - S
qrt[1 - a^2])])/(a*Sqrt[1 - a^2]) + (6*b*ArcCsc[a + b*x]*PolyLog[2, ((-I)*a*E^(I*ArcCsc[a + b*x]))/(1 + Sqrt[1
- a^2])])/(a*Sqrt[1 - a^2]) - ((6*I)*b*PolyLog[3, ((-I)*a*E^(I*ArcCsc[a + b*x]))/(1 - Sqrt[1 - a^2])])/(a*Sqr
t[1 - a^2]) + ((6*I)*b*PolyLog[3, ((-I)*a*E^(I*ArcCsc[a + b*x]))/(1 + Sqrt[1 - a^2])])/(a*Sqrt[1 - a^2])

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Rubi [A]  time = 0.616608, antiderivative size = 378, 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 = {5259, 4427, 4191, 3323, 2264, 2190, 2531, 2282, 6589} \[ -\frac{6 b \csc ^{-1}(a+b x) \text{PolyLog}\left (2,-\frac{i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{6 b \csc ^{-1}(a+b x) \text{PolyLog}\left (2,-\frac{i a e^{i \csc ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )}{a \sqrt{1-a^2}}-\frac{6 i b \text{PolyLog}\left (3,-\frac{i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{6 i b \text{PolyLog}\left (3,-\frac{i a e^{i \csc ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )}{a \sqrt{1-a^2}}-\frac{3 i b \csc ^{-1}(a+b x)^2 \log \left (1+\frac{i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{3 i b \csc ^{-1}(a+b x)^2 \log \left (1+\frac{i a e^{i \csc ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )}{a \sqrt{1-a^2}}-\frac{b \csc ^{-1}(a+b x)^3}{a}-\frac{\csc ^{-1}(a+b x)^3}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcCsc[a + b*x]^3/x^2,x]

[Out]

-((b*ArcCsc[a + b*x]^3)/a) - ArcCsc[a + b*x]^3/x - ((3*I)*b*ArcCsc[a + b*x]^2*Log[1 + (I*a*E^(I*ArcCsc[a + b*x
]))/(1 - Sqrt[1 - a^2])])/(a*Sqrt[1 - a^2]) + ((3*I)*b*ArcCsc[a + b*x]^2*Log[1 + (I*a*E^(I*ArcCsc[a + b*x]))/(
1 + Sqrt[1 - a^2])])/(a*Sqrt[1 - a^2]) - (6*b*ArcCsc[a + b*x]*PolyLog[2, ((-I)*a*E^(I*ArcCsc[a + b*x]))/(1 - S
qrt[1 - a^2])])/(a*Sqrt[1 - a^2]) + (6*b*ArcCsc[a + b*x]*PolyLog[2, ((-I)*a*E^(I*ArcCsc[a + b*x]))/(1 + Sqrt[1
- a^2])])/(a*Sqrt[1 - a^2]) - ((6*I)*b*PolyLog[3, ((-I)*a*E^(I*ArcCsc[a + b*x]))/(1 - Sqrt[1 - a^2])])/(a*Sqr
t[1 - a^2]) + ((6*I)*b*PolyLog[3, ((-I)*a*E^(I*ArcCsc[a + b*x]))/(1 + Sqrt[1 - a^2])])/(a*Sqrt[1 - a^2])

Rule 5259

Int[((a_.) + ArcCsc[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Dist[(d^(m + 1))
^(-1), Subst[Int[(a + b*x)^p*Csc[x]*Cot[x]*(d*e - c*f + f*Csc[x])^m, x], x, ArcCsc[c + d*x]], x] /; FreeQ[{a,
b, c, d, e, f}, x] && IGtQ[p, 0] && IntegerQ[m]

Rule 4427

Int[Cot[(c_.) + (d_.)*(x_)]*Csc[(c_.) + (d_.)*(x_)]*(Csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.)*((e_.) + (f_.
)*(x_))^(m_.), x_Symbol] :> -Simp[((e + f*x)^m*(a + b*Csc[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*Csc[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
IGtQ[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 3323

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c + d*x)^m*E
^(I*(e + f*x)))/(I*b + 2*a*E^(I*(e + f*x)) - I*b*E^(2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
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{\csc ^{-1}(a+b x)^3}{x^2} \, dx &=-\left (b \operatorname{Subst}\left (\int \frac{x^3 \cot (x) \csc (x)}{(-a+\csc (x))^2} \, dx,x,\csc ^{-1}(a+b x)\right )\right )\\ &=-\frac{\csc ^{-1}(a+b x)^3}{x}+(3 b) \operatorname{Subst}\left (\int \frac{x^2}{-a+\csc (x)} \, dx,x,\csc ^{-1}(a+b x)\right )\\ &=-\frac{\csc ^{-1}(a+b x)^3}{x}+(3 b) \operatorname{Subst}\left (\int \left (-\frac{x^2}{a}+\frac{x^2}{a (1-a \sin (x))}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )\\ &=-\frac{b \csc ^{-1}(a+b x)^3}{a}-\frac{\csc ^{-1}(a+b x)^3}{x}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{x^2}{1-a \sin (x)} \, dx,x,\csc ^{-1}(a+b x)\right )}{a}\\ &=-\frac{b \csc ^{-1}(a+b x)^3}{a}-\frac{\csc ^{-1}(a+b x)^3}{x}+\frac{(6 b) \operatorname{Subst}\left (\int \frac{e^{i x} x^2}{-i a+2 e^{i x}+i a e^{2 i x}} \, dx,x,\csc ^{-1}(a+b x)\right )}{a}\\ &=-\frac{b \csc ^{-1}(a+b x)^3}{a}-\frac{\csc ^{-1}(a+b x)^3}{x}+\frac{(6 i b) \operatorname{Subst}\left (\int \frac{e^{i x} x^2}{2-2 \sqrt{1-a^2}+2 i a e^{i x}} \, dx,x,\csc ^{-1}(a+b x)\right )}{\sqrt{1-a^2}}-\frac{(6 i b) \operatorname{Subst}\left (\int \frac{e^{i x} x^2}{2+2 \sqrt{1-a^2}+2 i a e^{i x}} \, dx,x,\csc ^{-1}(a+b x)\right )}{\sqrt{1-a^2}}\\ &=-\frac{b \csc ^{-1}(a+b x)^3}{a}-\frac{\csc ^{-1}(a+b x)^3}{x}-\frac{3 i b \csc ^{-1}(a+b x)^2 \log \left (1+\frac{i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{3 i b \csc ^{-1}(a+b x)^2 \log \left (1+\frac{i a e^{i \csc ^{-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 i a e^{i x}}{2-2 \sqrt{1-a^2}}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )}{a \sqrt{1-a^2}}-\frac{(6 i b) \operatorname{Subst}\left (\int x \log \left (1+\frac{2 i a e^{i x}}{2+2 \sqrt{1-a^2}}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )}{a \sqrt{1-a^2}}\\ &=-\frac{b \csc ^{-1}(a+b x)^3}{a}-\frac{\csc ^{-1}(a+b x)^3}{x}-\frac{3 i b \csc ^{-1}(a+b x)^2 \log \left (1+\frac{i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{3 i b \csc ^{-1}(a+b x)^2 \log \left (1+\frac{i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}-\frac{6 b \csc ^{-1}(a+b x) \text{Li}_2\left (-\frac{i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{6 b \csc ^{-1}(a+b x) \text{Li}_2\left (-\frac{i a e^{i \csc ^{-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 i a e^{i x}}{2-2 \sqrt{1-a^2}}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )}{a \sqrt{1-a^2}}-\frac{(6 b) \operatorname{Subst}\left (\int \text{Li}_2\left (-\frac{2 i a e^{i x}}{2+2 \sqrt{1-a^2}}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )}{a \sqrt{1-a^2}}\\ &=-\frac{b \csc ^{-1}(a+b x)^3}{a}-\frac{\csc ^{-1}(a+b x)^3}{x}-\frac{3 i b \csc ^{-1}(a+b x)^2 \log \left (1+\frac{i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{3 i b \csc ^{-1}(a+b x)^2 \log \left (1+\frac{i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}-\frac{6 b \csc ^{-1}(a+b x) \text{Li}_2\left (-\frac{i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{6 b \csc ^{-1}(a+b x) \text{Li}_2\left (-\frac{i a e^{i \csc ^{-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{i a x}{-1+\sqrt{1-a^2}}\right )}{x} \, dx,x,e^{i \csc ^{-1}(a+b x)}\right )}{a \sqrt{1-a^2}}+\frac{(6 i b) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{i a x}{1+\sqrt{1-a^2}}\right )}{x} \, dx,x,e^{i \csc ^{-1}(a+b x)}\right )}{a \sqrt{1-a^2}}\\ &=-\frac{b \csc ^{-1}(a+b x)^3}{a}-\frac{\csc ^{-1}(a+b x)^3}{x}-\frac{3 i b \csc ^{-1}(a+b x)^2 \log \left (1+\frac{i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{3 i b \csc ^{-1}(a+b x)^2 \log \left (1+\frac{i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}-\frac{6 b \csc ^{-1}(a+b x) \text{Li}_2\left (-\frac{i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{6 b \csc ^{-1}(a+b x) \text{Li}_2\left (-\frac{i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}-\frac{6 i b \text{Li}_3\left (-\frac{i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}+\frac{6 i b \text{Li}_3\left (-\frac{i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )}{a \sqrt{1-a^2}}\\ \end{align*}

Mathematica [A]  time = 0.531832, size = 289, normalized size = 0.76 \[ -\frac{\frac{(a+b x) \csc ^{-1}(a+b x)^3}{x}+\frac{3 i b \left (-2 i \csc ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{i a e^{i \csc ^{-1}(a+b x)}}{\sqrt{1-a^2}-1}\right )+2 i \csc ^{-1}(a+b x) \text{PolyLog}\left (2,-\frac{i a e^{i \csc ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )+2 \text{PolyLog}\left (3,\frac{i a e^{i \csc ^{-1}(a+b x)}}{\sqrt{1-a^2}-1}\right )-2 \text{PolyLog}\left (3,-\frac{i a e^{i \csc ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )+\csc ^{-1}(a+b x)^2 \log \left (1-\frac{i a e^{i \csc ^{-1}(a+b x)}}{\sqrt{1-a^2}-1}\right )-\csc ^{-1}(a+b x)^2 \log \left (1+\frac{i a e^{i \csc ^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )\right )}{\sqrt{1-a^2}}}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCsc[a + b*x]^3/x^2,x]

[Out]

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

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Maple [F]  time = 1.043, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ({\rm arccsc} \left (bx+a\right ) \right ) ^{3}}{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(arccsc(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 (1, \sqrt{b x + a + 1} \sqrt{b x + a - 1}\right )^{3} - 3 \, \arctan \left (1, \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{4 \,{\left (b^{3} x^{3} \arctan \left (1, \sqrt{b x + a + 1} \sqrt{b x + a - 1}\right ) + 3 \, a b^{2} x^{2} \arctan \left (1, \sqrt{b x + a + 1} \sqrt{b x + a - 1}\right ) + a^{3} \arctan \left (1, \sqrt{b x + a + 1} \sqrt{b x + a - 1}\right ) +{\left (3 \, a^{2} \arctan \left (1, \sqrt{b x + a + 1} \sqrt{b x + a - 1}\right ) - \arctan \left (1, \sqrt{b x + a + 1} \sqrt{b x + a - 1}\right )\right )} b x - a \arctan \left (1, \sqrt{b x + a + 1} \sqrt{b x + a - 1}\right )\right )} \log \left (b x + a\right )^{2} +{\left (4 \, b x \arctan \left (1, \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 (b^{3} x^{3} \arctan \left (1, \sqrt{b x + a + 1} \sqrt{b x + a - 1}\right ) + 2 \, a b^{2} x^{2} \arctan \left (1, \sqrt{b x + a + 1} \sqrt{b x + a - 1}\right ) +{\left (a^{2} \arctan \left (1, \sqrt{b x + a + 1} \sqrt{b x + a - 1}\right ) - \arctan \left (1, \sqrt{b x + a + 1} \sqrt{b x + a - 1}\right )\right )} b x -{\left (b^{3} x^{3} \arctan \left (1, \sqrt{b x + a + 1} \sqrt{b x + a - 1}\right ) + 3 \, a b^{2} x^{2} \arctan \left (1, \sqrt{b x + a + 1} \sqrt{b x + a - 1}\right ) + a^{3} \arctan \left (1, \sqrt{b x + a + 1} \sqrt{b x + a - 1}\right ) +{\left (3 \, a^{2} \arctan \left (1, \sqrt{b x + a + 1} \sqrt{b x + a - 1}\right ) - \arctan \left (1, \sqrt{b x + a + 1} \sqrt{b x + a - 1}\right )\right )} b x - a \arctan \left (1, \sqrt{b x + a + 1} \sqrt{b x + a - 1}\right )\right )} \log \left (b x + a\right )\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\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(arccsc(b*x+a)^3/x^2,x, algorithm="maxima")

[Out]

-1/4*(4*arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^3 - 3*arctan2(1, 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^3*x^3*arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x + a - 1)
) + 3*a*b^2*x^2*arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) + a^3*arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x +
a - 1)) + (3*a^2*arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) - arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x + a -
1)))*b*x - a*arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x + a - 1)))*log(b*x + a)^2 + (4*b*x*arctan2(1, 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*arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) + 2*a*b^2*x^2*arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x + a -
1)) + (a^2*arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) - arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x + a - 1)))*
b*x - (b^3*x^3*arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) + 3*a*b^2*x^2*arctan2(1, sqrt(b*x + a + 1)*sqrt
(b*x + a - 1)) + a^3*arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) + (3*a^2*arctan2(1, sqrt(b*x + a + 1)*sqr
t(b*x + a - 1)) - arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x + a - 1)))*b*x - a*arctan2(1, sqrt(b*x + a + 1)*sqrt(b
*x + a - 1)))*log(b*x + a))*log(b^2*x^2 + 2*a*b*x + a^2))/(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{arccsc}\left (b x + a\right )^{3}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccsc(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{acsc}^{3}{\left (a + b x \right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acsc(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{arccsc}\left (b x + a\right )^{3}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccsc(b*x + a)^3/x^2, x)