3.103 \(\int \frac{x^5 (a+b \csc ^{-1}(c x))}{(d+e x^2)^2} \, dx\)

Optimal. Leaf size=628 \[ \frac{i b d \text{PolyLog}\left (2,-\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}+\frac{i b d \text{PolyLog}\left (2,\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}+\frac{i b d \text{PolyLog}\left (2,-\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{e^3}+\frac{i b d \text{PolyLog}\left (2,\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{e^3}-\frac{i b d \text{PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right )}{e^3}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1-\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1+\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1-\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{e^3}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1+\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{e^3}+\frac{d \left (a+b \csc ^{-1}(c x)\right )}{2 e^2 \left (\frac{d}{x^2}+e\right )}+\frac{2 d \log \left (1-e^{2 i \csc ^{-1}(c x)}\right ) \left (a+b \csc ^{-1}(c x)\right )}{e^3}+\frac{x^2 \left (a+b \csc ^{-1}(c x)\right )}{2 e^2}-\frac{b d \tan ^{-1}\left (\frac{\sqrt{c^2 d+e}}{c \sqrt{e} x \sqrt{1-\frac{1}{c^2 x^2}}}\right )}{2 e^{5/2} \sqrt{c^2 d+e}}+\frac{b x \sqrt{1-\frac{1}{c^2 x^2}}}{2 c e^2} \]

[Out]

(b*Sqrt[1 - 1/(c^2*x^2)]*x)/(2*c*e^2) + (d*(a + b*ArcCsc[c*x]))/(2*e^2*(e + d/x^2)) + (x^2*(a + b*ArcCsc[c*x])
)/(2*e^2) - (b*d*ArcTan[Sqrt[c^2*d + e]/(c*Sqrt[e]*Sqrt[1 - 1/(c^2*x^2)]*x)])/(2*e^(5/2)*Sqrt[c^2*d + e]) - (d
*(a + b*ArcCsc[c*x])*Log[1 - (I*c*Sqrt[-d]*E^(I*ArcCsc[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e])])/e^3 - (d*(a + b*Ar
cCsc[c*x])*Log[1 + (I*c*Sqrt[-d]*E^(I*ArcCsc[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e])])/e^3 - (d*(a + b*ArcCsc[c*x])
*Log[1 - (I*c*Sqrt[-d]*E^(I*ArcCsc[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e])])/e^3 - (d*(a + b*ArcCsc[c*x])*Log[1 + (
I*c*Sqrt[-d]*E^(I*ArcCsc[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e])])/e^3 + (2*d*(a + b*ArcCsc[c*x])*Log[1 - E^((2*I)*
ArcCsc[c*x])])/e^3 + (I*b*d*PolyLog[2, ((-I)*c*Sqrt[-d]*E^(I*ArcCsc[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e])])/e^3 +
(I*b*d*PolyLog[2, (I*c*Sqrt[-d]*E^(I*ArcCsc[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e])])/e^3 + (I*b*d*PolyLog[2, ((-I
)*c*Sqrt[-d]*E^(I*ArcCsc[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e])])/e^3 + (I*b*d*PolyLog[2, (I*c*Sqrt[-d]*E^(I*ArcCs
c[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e])])/e^3 - (I*b*d*PolyLog[2, E^((2*I)*ArcCsc[c*x])])/e^3

________________________________________________________________________________________

Rubi [A]  time = 1.32061, antiderivative size = 628, normalized size of antiderivative = 1., number of steps used = 31, number of rules used = 14, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {5241, 4733, 4627, 264, 4625, 3717, 2190, 2279, 2391, 4729, 377, 205, 4741, 4519} \[ \frac{i b d \text{PolyLog}\left (2,-\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}+\frac{i b d \text{PolyLog}\left (2,\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}+\frac{i b d \text{PolyLog}\left (2,-\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{e^3}+\frac{i b d \text{PolyLog}\left (2,\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{e^3}-\frac{i b d \text{PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right )}{e^3}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1-\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1+\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1-\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{e^3}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1+\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{e^3}+\frac{d \left (a+b \csc ^{-1}(c x)\right )}{2 e^2 \left (\frac{d}{x^2}+e\right )}+\frac{2 d \log \left (1-e^{2 i \csc ^{-1}(c x)}\right ) \left (a+b \csc ^{-1}(c x)\right )}{e^3}+\frac{x^2 \left (a+b \csc ^{-1}(c x)\right )}{2 e^2}-\frac{b d \tan ^{-1}\left (\frac{\sqrt{c^2 d+e}}{c \sqrt{e} x \sqrt{1-\frac{1}{c^2 x^2}}}\right )}{2 e^{5/2} \sqrt{c^2 d+e}}+\frac{b x \sqrt{1-\frac{1}{c^2 x^2}}}{2 c e^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*ArcCsc[c*x]))/(d + e*x^2)^2,x]

[Out]

(b*Sqrt[1 - 1/(c^2*x^2)]*x)/(2*c*e^2) + (d*(a + b*ArcCsc[c*x]))/(2*e^2*(e + d/x^2)) + (x^2*(a + b*ArcCsc[c*x])
)/(2*e^2) - (b*d*ArcTan[Sqrt[c^2*d + e]/(c*Sqrt[e]*Sqrt[1 - 1/(c^2*x^2)]*x)])/(2*e^(5/2)*Sqrt[c^2*d + e]) - (d
*(a + b*ArcCsc[c*x])*Log[1 - (I*c*Sqrt[-d]*E^(I*ArcCsc[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e])])/e^3 - (d*(a + b*Ar
cCsc[c*x])*Log[1 + (I*c*Sqrt[-d]*E^(I*ArcCsc[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e])])/e^3 - (d*(a + b*ArcCsc[c*x])
*Log[1 - (I*c*Sqrt[-d]*E^(I*ArcCsc[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e])])/e^3 - (d*(a + b*ArcCsc[c*x])*Log[1 + (
I*c*Sqrt[-d]*E^(I*ArcCsc[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e])])/e^3 + (2*d*(a + b*ArcCsc[c*x])*Log[1 - E^((2*I)*
ArcCsc[c*x])])/e^3 + (I*b*d*PolyLog[2, ((-I)*c*Sqrt[-d]*E^(I*ArcCsc[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e])])/e^3 +
(I*b*d*PolyLog[2, (I*c*Sqrt[-d]*E^(I*ArcCsc[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e])])/e^3 + (I*b*d*PolyLog[2, ((-I
)*c*Sqrt[-d]*E^(I*ArcCsc[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e])])/e^3 + (I*b*d*PolyLog[2, (I*c*Sqrt[-d]*E^(I*ArcCs
c[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e])])/e^3 - (I*b*d*PolyLog[2, E^((2*I)*ArcCsc[c*x])])/e^3

Rule 5241

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> -Subst[Int[
((e + d*x^2)^p*(a + b*ArcSin[x/c])^n)/x^(m + 2*(p + 1)), x], x, 1/x] /; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[n
, 0] && IntegerQ[m] && IntegerQ[p]

Rule 4733

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Int[
ExpandIntegrand[(a + b*ArcSin[c*x])^n, (f*x)^m*(d + e*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[c^
2*d + e, 0] && IGtQ[n, 0] && IntegerQ[p] && IntegerQ[m]

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 4625

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

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && 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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4729

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^(p + 1
)*(a + b*ArcSin[c*x]))/(2*e*(p + 1)), x] - Dist[(b*c)/(2*e*(p + 1)), Int[(d + e*x^2)^(p + 1)/Sqrt[1 - c^2*x^2]
, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[c^2*d + e, 0] && NeQ[p, -1]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 4741

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Subst[Int[((a + b*x)^n*Cos[x])/
(c*d + e*Sin[x]), x], x, ArcSin[c*x]] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[n, 0]

Rule 4519

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(I*(e + f*x)^(m + 1))/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(I*(c + d*x)))/(a - Rt[a^2 - b^2, 2] - I*b
*E^(I*(c + d*x))), x] + Int[((e + f*x)^m*E^(I*(c + d*x)))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x))), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

Rubi steps

\begin{align*} \int \frac{x^5 \left (a+b \csc ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx &=-\operatorname{Subst}\left (\int \frac{a+b \sin ^{-1}\left (\frac{x}{c}\right )}{x^3 \left (e+d x^2\right )^2} \, dx,x,\frac{1}{x}\right )\\ &=-\operatorname{Subst}\left (\int \left (\frac{a+b \sin ^{-1}\left (\frac{x}{c}\right )}{e^2 x^3}-\frac{2 d \left (a+b \sin ^{-1}\left (\frac{x}{c}\right )\right )}{e^3 x}+\frac{d^2 x \left (a+b \sin ^{-1}\left (\frac{x}{c}\right )\right )}{e^2 \left (e+d x^2\right )^2}+\frac{2 d^2 x \left (a+b \sin ^{-1}\left (\frac{x}{c}\right )\right )}{e^3 \left (e+d x^2\right )}\right ) \, dx,x,\frac{1}{x}\right )\\ &=\frac{(2 d) \operatorname{Subst}\left (\int \frac{a+b \sin ^{-1}\left (\frac{x}{c}\right )}{x} \, dx,x,\frac{1}{x}\right )}{e^3}-\frac{\left (2 d^2\right ) \operatorname{Subst}\left (\int \frac{x \left (a+b \sin ^{-1}\left (\frac{x}{c}\right )\right )}{e+d x^2} \, dx,x,\frac{1}{x}\right )}{e^3}-\frac{\operatorname{Subst}\left (\int \frac{a+b \sin ^{-1}\left (\frac{x}{c}\right )}{x^3} \, dx,x,\frac{1}{x}\right )}{e^2}-\frac{d^2 \operatorname{Subst}\left (\int \frac{x \left (a+b \sin ^{-1}\left (\frac{x}{c}\right )\right )}{\left (e+d x^2\right )^2} \, dx,x,\frac{1}{x}\right )}{e^2}\\ &=\frac{d \left (a+b \csc ^{-1}(c x)\right )}{2 e^2 \left (e+\frac{d}{x^2}\right )}+\frac{x^2 \left (a+b \csc ^{-1}(c x)\right )}{2 e^2}+\frac{(2 d) \operatorname{Subst}\left (\int (a+b x) \cot (x) \, dx,x,\csc ^{-1}(c x)\right )}{e^3}-\frac{\left (2 d^2\right ) \operatorname{Subst}\left (\int \left (-\frac{\sqrt{-d} \left (a+b \sin ^{-1}\left (\frac{x}{c}\right )\right )}{2 d \left (\sqrt{e}-\sqrt{-d} x\right )}+\frac{\sqrt{-d} \left (a+b \sin ^{-1}\left (\frac{x}{c}\right )\right )}{2 d \left (\sqrt{e}+\sqrt{-d} x\right )}\right ) \, dx,x,\frac{1}{x}\right )}{e^3}-\frac{b \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{2 c e^2}-\frac{(b d) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{c^2}} \left (e+d x^2\right )} \, dx,x,\frac{1}{x}\right )}{2 c e^2}\\ &=\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x}{2 c e^2}+\frac{d \left (a+b \csc ^{-1}(c x)\right )}{2 e^2 \left (e+\frac{d}{x^2}\right )}+\frac{x^2 \left (a+b \csc ^{-1}(c x)\right )}{2 e^2}-\frac{i d \left (a+b \csc ^{-1}(c x)\right )^2}{b e^3}-\frac{(-d)^{3/2} \operatorname{Subst}\left (\int \frac{a+b \sin ^{-1}\left (\frac{x}{c}\right )}{\sqrt{e}-\sqrt{-d} x} \, dx,x,\frac{1}{x}\right )}{e^3}+\frac{(-d)^{3/2} \operatorname{Subst}\left (\int \frac{a+b \sin ^{-1}\left (\frac{x}{c}\right )}{\sqrt{e}+\sqrt{-d} x} \, dx,x,\frac{1}{x}\right )}{e^3}-\frac{(4 i d) \operatorname{Subst}\left (\int \frac{e^{2 i x} (a+b x)}{1-e^{2 i x}} \, dx,x,\csc ^{-1}(c x)\right )}{e^3}-\frac{(b d) \operatorname{Subst}\left (\int \frac{1}{e-\left (-d-\frac{e}{c^2}\right ) x^2} \, dx,x,\frac{1}{\sqrt{1-\frac{1}{c^2 x^2}} x}\right )}{2 c e^2}\\ &=\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x}{2 c e^2}+\frac{d \left (a+b \csc ^{-1}(c x)\right )}{2 e^2 \left (e+\frac{d}{x^2}\right )}+\frac{x^2 \left (a+b \csc ^{-1}(c x)\right )}{2 e^2}-\frac{i d \left (a+b \csc ^{-1}(c x)\right )^2}{b e^3}-\frac{b d \tan ^{-1}\left (\frac{\sqrt{c^2 d+e}}{c \sqrt{e} \sqrt{1-\frac{1}{c^2 x^2}} x}\right )}{2 e^{5/2} \sqrt{c^2 d+e}}+\frac{2 d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )}{e^3}-\frac{(-d)^{3/2} \operatorname{Subst}\left (\int \frac{(a+b x) \cos (x)}{\frac{\sqrt{e}}{c}-\sqrt{-d} \sin (x)} \, dx,x,\csc ^{-1}(c x)\right )}{e^3}+\frac{(-d)^{3/2} \operatorname{Subst}\left (\int \frac{(a+b x) \cos (x)}{\frac{\sqrt{e}}{c}+\sqrt{-d} \sin (x)} \, dx,x,\csc ^{-1}(c x)\right )}{e^3}-\frac{(2 b d) \operatorname{Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\csc ^{-1}(c x)\right )}{e^3}\\ &=\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x}{2 c e^2}+\frac{d \left (a+b \csc ^{-1}(c x)\right )}{2 e^2 \left (e+\frac{d}{x^2}\right )}+\frac{x^2 \left (a+b \csc ^{-1}(c x)\right )}{2 e^2}-\frac{b d \tan ^{-1}\left (\frac{\sqrt{c^2 d+e}}{c \sqrt{e} \sqrt{1-\frac{1}{c^2 x^2}} x}\right )}{2 e^{5/2} \sqrt{c^2 d+e}}+\frac{2 d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )}{e^3}+\frac{(-d)^{3/2} \operatorname{Subst}\left (\int \frac{e^{i x} (a+b x)}{\frac{\sqrt{e}}{c}-\frac{\sqrt{c^2 d+e}}{c}-i \sqrt{-d} e^{i x}} \, dx,x,\csc ^{-1}(c x)\right )}{e^3}+\frac{(-d)^{3/2} \operatorname{Subst}\left (\int \frac{e^{i x} (a+b x)}{\frac{\sqrt{e}}{c}+\frac{\sqrt{c^2 d+e}}{c}-i \sqrt{-d} e^{i x}} \, dx,x,\csc ^{-1}(c x)\right )}{e^3}-\frac{(-d)^{3/2} \operatorname{Subst}\left (\int \frac{e^{i x} (a+b x)}{\frac{\sqrt{e}}{c}-\frac{\sqrt{c^2 d+e}}{c}+i \sqrt{-d} e^{i x}} \, dx,x,\csc ^{-1}(c x)\right )}{e^3}-\frac{(-d)^{3/2} \operatorname{Subst}\left (\int \frac{e^{i x} (a+b x)}{\frac{\sqrt{e}}{c}+\frac{\sqrt{c^2 d+e}}{c}+i \sqrt{-d} e^{i x}} \, dx,x,\csc ^{-1}(c x)\right )}{e^3}+\frac{(i b d) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 i \csc ^{-1}(c x)}\right )}{e^3}\\ &=\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x}{2 c e^2}+\frac{d \left (a+b \csc ^{-1}(c x)\right )}{2 e^2 \left (e+\frac{d}{x^2}\right )}+\frac{x^2 \left (a+b \csc ^{-1}(c x)\right )}{2 e^2}-\frac{b d \tan ^{-1}\left (\frac{\sqrt{c^2 d+e}}{c \sqrt{e} \sqrt{1-\frac{1}{c^2 x^2}} x}\right )}{2 e^{5/2} \sqrt{c^2 d+e}}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1-\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1+\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1-\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}+\sqrt{c^2 d+e}}\right )}{e^3}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1+\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}+\sqrt{c^2 d+e}}\right )}{e^3}+\frac{2 d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )}{e^3}-\frac{i b d \text{Li}_2\left (e^{2 i \csc ^{-1}(c x)}\right )}{e^3}+\frac{(b d) \operatorname{Subst}\left (\int \log \left (1-\frac{i \sqrt{-d} e^{i x}}{\frac{\sqrt{e}}{c}-\frac{\sqrt{c^2 d+e}}{c}}\right ) \, dx,x,\csc ^{-1}(c x)\right )}{e^3}+\frac{(b d) \operatorname{Subst}\left (\int \log \left (1+\frac{i \sqrt{-d} e^{i x}}{\frac{\sqrt{e}}{c}-\frac{\sqrt{c^2 d+e}}{c}}\right ) \, dx,x,\csc ^{-1}(c x)\right )}{e^3}+\frac{(b d) \operatorname{Subst}\left (\int \log \left (1-\frac{i \sqrt{-d} e^{i x}}{\frac{\sqrt{e}}{c}+\frac{\sqrt{c^2 d+e}}{c}}\right ) \, dx,x,\csc ^{-1}(c x)\right )}{e^3}+\frac{(b d) \operatorname{Subst}\left (\int \log \left (1+\frac{i \sqrt{-d} e^{i x}}{\frac{\sqrt{e}}{c}+\frac{\sqrt{c^2 d+e}}{c}}\right ) \, dx,x,\csc ^{-1}(c x)\right )}{e^3}\\ &=\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x}{2 c e^2}+\frac{d \left (a+b \csc ^{-1}(c x)\right )}{2 e^2 \left (e+\frac{d}{x^2}\right )}+\frac{x^2 \left (a+b \csc ^{-1}(c x)\right )}{2 e^2}-\frac{b d \tan ^{-1}\left (\frac{\sqrt{c^2 d+e}}{c \sqrt{e} \sqrt{1-\frac{1}{c^2 x^2}} x}\right )}{2 e^{5/2} \sqrt{c^2 d+e}}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1-\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1+\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1-\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}+\sqrt{c^2 d+e}}\right )}{e^3}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1+\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}+\sqrt{c^2 d+e}}\right )}{e^3}+\frac{2 d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )}{e^3}-\frac{i b d \text{Li}_2\left (e^{2 i \csc ^{-1}(c x)}\right )}{e^3}-\frac{(i b d) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{i \sqrt{-d} x}{\frac{\sqrt{e}}{c}-\frac{\sqrt{c^2 d+e}}{c}}\right )}{x} \, dx,x,e^{i \csc ^{-1}(c x)}\right )}{e^3}-\frac{(i b d) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{i \sqrt{-d} x}{\frac{\sqrt{e}}{c}-\frac{\sqrt{c^2 d+e}}{c}}\right )}{x} \, dx,x,e^{i \csc ^{-1}(c x)}\right )}{e^3}-\frac{(i b d) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{i \sqrt{-d} x}{\frac{\sqrt{e}}{c}+\frac{\sqrt{c^2 d+e}}{c}}\right )}{x} \, dx,x,e^{i \csc ^{-1}(c x)}\right )}{e^3}-\frac{(i b d) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{i \sqrt{-d} x}{\frac{\sqrt{e}}{c}+\frac{\sqrt{c^2 d+e}}{c}}\right )}{x} \, dx,x,e^{i \csc ^{-1}(c x)}\right )}{e^3}\\ &=\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x}{2 c e^2}+\frac{d \left (a+b \csc ^{-1}(c x)\right )}{2 e^2 \left (e+\frac{d}{x^2}\right )}+\frac{x^2 \left (a+b \csc ^{-1}(c x)\right )}{2 e^2}-\frac{b d \tan ^{-1}\left (\frac{\sqrt{c^2 d+e}}{c \sqrt{e} \sqrt{1-\frac{1}{c^2 x^2}} x}\right )}{2 e^{5/2} \sqrt{c^2 d+e}}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1-\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1+\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1-\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}+\sqrt{c^2 d+e}}\right )}{e^3}-\frac{d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1+\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}+\sqrt{c^2 d+e}}\right )}{e^3}+\frac{2 d \left (a+b \csc ^{-1}(c x)\right ) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )}{e^3}+\frac{i b d \text{Li}_2\left (-\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}+\frac{i b d \text{Li}_2\left (\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}+\frac{i b d \text{Li}_2\left (-\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}+\sqrt{c^2 d+e}}\right )}{e^3}+\frac{i b d \text{Li}_2\left (\frac{i c \sqrt{-d} e^{i \csc ^{-1}(c x)}}{\sqrt{e}+\sqrt{c^2 d+e}}\right )}{e^3}-\frac{i b d \text{Li}_2\left (e^{2 i \csc ^{-1}(c x)}\right )}{e^3}\\ \end{align*}

Mathematica [B]  time = 5.61441, size = 1480, normalized size = 2.36 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^5*(a + b*ArcCsc[c*x]))/(d + e*x^2)^2,x]

[Out]

-(-2*a*e*x^2 + (2*a*d^2)/(d + e*x^2) + 4*a*d*Log[d + e*x^2] + b*(I*d*Pi^2 - (2*e*Sqrt[1 - 1/(c^2*x^2)]*x)/c -
(4*I)*d*Pi*ArcCsc[c*x] - 2*e*x^2*ArcCsc[c*x] + (d^(3/2)*ArcCsc[c*x])/(Sqrt[d] - I*Sqrt[e]*x) + (d^(3/2)*ArcCsc
[c*x])/(Sqrt[d] + I*Sqrt[e]*x) + (8*I)*d*ArcCsc[c*x]^2 - 2*d*ArcSin[1/(c*x)] - (16*I)*d*ArcSin[Sqrt[1 - (I*Sqr
t[e])/(c*Sqrt[d])]/Sqrt[2]]*ArcTan[(((-I)*c*Sqrt[d] + Sqrt[e])*Cot[(Pi + 2*ArcCsc[c*x])/4])/Sqrt[c^2*d + e]] -
(16*I)*d*ArcSin[Sqrt[1 + (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*ArcTan[((I*c*Sqrt[d] + Sqrt[e])*Cot[(Pi + 2*ArcCsc
[c*x])/4])/Sqrt[c^2*d + e]] - 2*d*Pi*Log[1 + (Sqrt[e] - Sqrt[c^2*d + e])/(c*Sqrt[d]*E^(I*ArcCsc[c*x]))] + 4*d*
ArcCsc[c*x]*Log[1 + (Sqrt[e] - Sqrt[c^2*d + e])/(c*Sqrt[d]*E^(I*ArcCsc[c*x]))] - 8*d*ArcSin[Sqrt[1 - (I*Sqrt[e
])/(c*Sqrt[d])]/Sqrt[2]]*Log[1 + (Sqrt[e] - Sqrt[c^2*d + e])/(c*Sqrt[d]*E^(I*ArcCsc[c*x]))] - 2*d*Pi*Log[1 + (
-Sqrt[e] + Sqrt[c^2*d + e])/(c*Sqrt[d]*E^(I*ArcCsc[c*x]))] + 4*d*ArcCsc[c*x]*Log[1 + (-Sqrt[e] + Sqrt[c^2*d +
e])/(c*Sqrt[d]*E^(I*ArcCsc[c*x]))] - 8*d*ArcSin[Sqrt[1 + (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*Log[1 + (-Sqrt[e] +
Sqrt[c^2*d + e])/(c*Sqrt[d]*E^(I*ArcCsc[c*x]))] - 2*d*Pi*Log[1 - (Sqrt[e] + Sqrt[c^2*d + e])/(c*Sqrt[d]*E^(I*
ArcCsc[c*x]))] + 4*d*ArcCsc[c*x]*Log[1 - (Sqrt[e] + Sqrt[c^2*d + e])/(c*Sqrt[d]*E^(I*ArcCsc[c*x]))] + 8*d*ArcS
in[Sqrt[1 + (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*Log[1 - (Sqrt[e] + Sqrt[c^2*d + e])/(c*Sqrt[d]*E^(I*ArcCsc[c*x])
)] - 2*d*Pi*Log[1 + (Sqrt[e] + Sqrt[c^2*d + e])/(c*Sqrt[d]*E^(I*ArcCsc[c*x]))] + 4*d*ArcCsc[c*x]*Log[1 + (Sqrt
[e] + Sqrt[c^2*d + e])/(c*Sqrt[d]*E^(I*ArcCsc[c*x]))] + 8*d*ArcSin[Sqrt[1 - (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*
Log[1 + (Sqrt[e] + Sqrt[c^2*d + e])/(c*Sqrt[d]*E^(I*ArcCsc[c*x]))] - 8*d*ArcCsc[c*x]*Log[1 - E^((2*I)*ArcCsc[c
*x])] + 2*d*Pi*Log[Sqrt[e] - (I*Sqrt[d])/x] + 2*d*Pi*Log[Sqrt[e] + (I*Sqrt[d])/x] + (d*Sqrt[e]*Log[(2*Sqrt[d]*
Sqrt[e]*(Sqrt[e] + c*((-I)*c*Sqrt[d] - Sqrt[-(c^2*d) - e]*Sqrt[1 - 1/(c^2*x^2)])*x))/(Sqrt[-(c^2*d) - e]*(Sqrt
[d] + I*Sqrt[e]*x))])/Sqrt[-(c^2*d) - e] + (d*Sqrt[e]*Log[(2*Sqrt[d]*Sqrt[e]*(-Sqrt[e] + c*((-I)*c*Sqrt[d] + S
qrt[-(c^2*d) - e]*Sqrt[1 - 1/(c^2*x^2)])*x))/(Sqrt[-(c^2*d) - e]*(Sqrt[d] - I*Sqrt[e]*x))])/Sqrt[-(c^2*d) - e]
+ (4*I)*d*PolyLog[2, (Sqrt[e] - Sqrt[c^2*d + e])/(c*Sqrt[d]*E^(I*ArcCsc[c*x]))] + (4*I)*d*PolyLog[2, (-Sqrt[e
] + Sqrt[c^2*d + e])/(c*Sqrt[d]*E^(I*ArcCsc[c*x]))] + (4*I)*d*PolyLog[2, -((Sqrt[e] + Sqrt[c^2*d + e])/(c*Sqrt
[d]*E^(I*ArcCsc[c*x])))] + (4*I)*d*PolyLog[2, (Sqrt[e] + Sqrt[c^2*d + e])/(c*Sqrt[d]*E^(I*ArcCsc[c*x]))] + (4*
I)*d*PolyLog[2, E^((2*I)*ArcCsc[c*x])]))/(4*e^3)

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Maple [C]  time = 0.734, size = 729, normalized size = 1.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*arccsc(c*x))/(e*x^2+d)^2,x)

[Out]

1/2*a*x^2/e^2-a/e^3*d*ln(c^2*e*x^2+c^2*d)-1/2*c^2*a/e^3*d^2/(c^2*e*x^2+c^2*d)+1/2*c^2*b/(c^2*e*x^2+c^2*d)/e*ar
ccsc(c*x)*x^4+c^2*b/(c^2*e*x^2+c^2*d)/e^2*arccsc(c*x)*d*x^2+1/2*c*b/(c^2*e*x^2+c^2*d)/e*((c^2*x^2-1)/c^2/x^2)^
(1/2)*x^3+1/2*c*b/(c^2*e*x^2+c^2*d)/e^2*((c^2*x^2-1)/c^2/x^2)^(1/2)*x*d-1/2*I*b/(c^2*e*x^2+c^2*d)/e*x^2-1/2*I*
b/(c^2*e*x^2+c^2*d)/e^2*d+1/2*I*b*(e*(c^2*d+e))^(1/2)/(c^2*d+e)/e^3*arctanh(1/4*(2*c^2*d*(I/c/x+(1-1/c^2/x^2)^
(1/2))^2-2*c^2*d-4*e)/(c^2*d*e+e^2)^(1/2))*d+1/2*I*b/e^3*d*sum((_R1^2*c^2*d-c^2*d-4*e)/(_R1^2*c^2*d-c^2*d-2*e)
*(I*arccsc(c*x)*ln((_R1-I/c/x-(1-1/c^2/x^2)^(1/2))/_R1)+dilog((_R1-I/c/x-(1-1/c^2/x^2)^(1/2))/_R1)),_R1=RootOf
(c^2*d*_Z^4+(-2*c^2*d-4*e)*_Z^2+c^2*d))+2*I*b/e^3*d*dilog(I/c/x+(1-1/c^2/x^2)^(1/2))+2*b/e^3*d*arccsc(c*x)*ln(
1+I/c/x+(1-1/c^2/x^2)^(1/2))-2*I*b/e^3*d*dilog(1+I/c/x+(1-1/c^2/x^2)^(1/2))+1/2*I*c^2*b/e^3*d^2*sum((_R1^2-1)/
(_R1^2*c^2*d-c^2*d-2*e)*(I*arccsc(c*x)*ln((_R1-I/c/x-(1-1/c^2/x^2)^(1/2))/_R1)+dilog((_R1-I/c/x-(1-1/c^2/x^2)^
(1/2))/_R1)),_R1=RootOf(c^2*d*_Z^4+(-2*c^2*d-4*e)*_Z^2+c^2*d))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arccsc(c*x))/(e*x^2+d)^2,x, algorithm="maxima")

[Out]

-1/2*a*(d^2/(e^4*x^2 + d*e^3) - x^2/e^2 + 2*d*log(e*x^2 + d)/e^3) + b*integrate(x^5*arctan2(1, sqrt(c*x + 1)*s
qrt(c*x - 1))/(e^2*x^4 + 2*d*e*x^2 + d^2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b x^{5} \operatorname{arccsc}\left (c x\right ) + a x^{5}}{e^{2} x^{4} + 2 \, d e x^{2} + d^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arccsc(c*x))/(e*x^2+d)^2,x, algorithm="fricas")

[Out]

integral((b*x^5*arccsc(c*x) + a*x^5)/(e^2*x^4 + 2*d*e*x^2 + d^2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*acsc(c*x))/(e*x**2+d)**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arccsc}\left (c x\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arccsc(c*x))/(e*x^2+d)^2,x, algorithm="giac")

[Out]

integrate((b*arccsc(c*x) + a)*x^5/(e*x^2 + d)^2, x)