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

Optimal. Leaf size=608 \[ \frac{i b d \text{PolyLog}\left (2,-\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}+\frac{i b d \text{PolyLog}\left (2,\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}+\frac{i b d \text{PolyLog}\left (2,-\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{e^3}+\frac{i b d \text{PolyLog}\left (2,\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{e^3}-\frac{i b d \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )}{e^3}-\frac{d \left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}-\frac{d \left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}-\frac{d \left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{e^3}-\frac{d \left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{e^3}+\frac{d \left (a+b \sec ^{-1}(c x)\right )}{2 e^2 \left (\frac{d}{x^2}+e\right )}+\frac{2 d \log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{e^3}+\frac{x^2 \left (a+b \sec ^{-1}(c x)\right )}{2 e^2}+\frac{b d \tan ^{-1}\left (\frac{\sqrt{c^2 d+e}}{c \sqrt{e} x \sqrt{1-\frac{1}{c^2 x^2}}}\right )}{2 e^{5/2} \sqrt{c^2 d+e}}-\frac{b x \sqrt{1-\frac{1}{c^2 x^2}}}{2 c e^2} \]

[Out]

-(b*Sqrt[1 - 1/(c^2*x^2)]*x)/(2*c*e^2) + (d*(a + b*ArcSec[c*x]))/(2*e^2*(e + d/x^2)) + (x^2*(a + b*ArcSec[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*ArcSec[c*x])*Log[1 - (c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e])])/e^3 - (d*(a + b*Arc
Sec[c*x])*Log[1 + (c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e])])/e^3 - (d*(a + b*ArcSec[c*x])*Lo
g[1 - (c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e])])/e^3 - (d*(a + b*ArcSec[c*x])*Log[1 + (c*Sqr
t[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e])])/e^3 + (2*d*(a + b*ArcSec[c*x])*Log[1 + E^((2*I)*ArcSec[
c*x])])/e^3 + (I*b*d*PolyLog[2, -((c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e]))])/e^3 + (I*b*d*P
olyLog[2, (c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e])])/e^3 + (I*b*d*PolyLog[2, -((c*Sqrt[-d]*E
^(I*ArcSec[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e]))])/e^3 + (I*b*d*PolyLog[2, (c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[
e] + Sqrt[c^2*d + e])])/e^3 - (I*b*d*PolyLog[2, -E^((2*I)*ArcSec[c*x])])/e^3

________________________________________________________________________________________

Rubi [A]  time = 1.30586, antiderivative size = 608, 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 = {5240, 4734, 4628, 264, 4626, 3719, 2190, 2279, 2391, 4730, 377, 205, 4742, 4520} \[ \frac{i b d \text{PolyLog}\left (2,-\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}+\frac{i b d \text{PolyLog}\left (2,\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}+\frac{i b d \text{PolyLog}\left (2,-\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{e^3}+\frac{i b d \text{PolyLog}\left (2,\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{e^3}-\frac{i b d \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )}{e^3}-\frac{d \left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}-\frac{d \left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{e^3}-\frac{d \left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{e^3}-\frac{d \left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{e^3}+\frac{d \left (a+b \sec ^{-1}(c x)\right )}{2 e^2 \left (\frac{d}{x^2}+e\right )}+\frac{2 d \log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{e^3}+\frac{x^2 \left (a+b \sec ^{-1}(c x)\right )}{2 e^2}+\frac{b d \tan ^{-1}\left (\frac{\sqrt{c^2 d+e}}{c \sqrt{e} x \sqrt{1-\frac{1}{c^2 x^2}}}\right )}{2 e^{5/2} \sqrt{c^2 d+e}}-\frac{b x \sqrt{1-\frac{1}{c^2 x^2}}}{2 c e^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*ArcSec[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*ArcSec[c*x]))/(2*e^2*(e + d/x^2)) + (x^2*(a + b*ArcSec[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*ArcSec[c*x])*Log[1 - (c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e])])/e^3 - (d*(a + b*Arc
Sec[c*x])*Log[1 + (c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e])])/e^3 - (d*(a + b*ArcSec[c*x])*Lo
g[1 - (c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e])])/e^3 - (d*(a + b*ArcSec[c*x])*Log[1 + (c*Sqr
t[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e])])/e^3 + (2*d*(a + b*ArcSec[c*x])*Log[1 + E^((2*I)*ArcSec[
c*x])])/e^3 + (I*b*d*PolyLog[2, -((c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e]))])/e^3 + (I*b*d*P
olyLog[2, (c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e])])/e^3 + (I*b*d*PolyLog[2, -((c*Sqrt[-d]*E
^(I*ArcSec[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e]))])/e^3 + (I*b*d*PolyLog[2, (c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[
e] + Sqrt[c^2*d + e])])/e^3 - (I*b*d*PolyLog[2, -E^((2*I)*ArcSec[c*x])])/e^3

Rule 5240

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> -Subst[Int[
((e + d*x^2)^p*(a + b*ArcCos[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 4734

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Int[
ExpandIntegrand[(a + b*ArcCos[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 4628

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcCo
s[c*x])^n)/(d*(m + 1)), x] + Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCos[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 4626

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

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
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 4730

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^(p + 1
)*(a + b*ArcCos[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 4742

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

Rule 4520

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

Mathematica [B]  time = 3.90373, size = 1255, normalized size = 2.06 \[ -\frac{\frac{2 a d^2}{e x^2+d}+4 a \log \left (e x^2+d\right ) d-2 a e x^2+b \left (-2 e \sec ^{-1}(c x) x^2+\frac{2 e \sqrt{1-\frac{1}{c^2 x^2}} x}{c}+\frac{d^{3/2} \sec ^{-1}(c x)}{\sqrt{d}-i \sqrt{e} x}+\frac{d^{3/2} \sec ^{-1}(c x)}{i \sqrt{e} x+\sqrt{d}}+2 d \sin ^{-1}\left (\frac{1}{c x}\right )+16 i d \sin ^{-1}\left (\frac{\sqrt{1-\frac{i \sqrt{e}}{c \sqrt{d}}}}{\sqrt{2}}\right ) \tan ^{-1}\left (\frac{\left (\sqrt{e}-i c \sqrt{d}\right ) \tan \left (\frac{1}{2} \sec ^{-1}(c x)\right )}{\sqrt{d c^2+e}}\right )+16 i d \sin ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{e}}{c \sqrt{d}}+1}}{\sqrt{2}}\right ) \tan ^{-1}\left (\frac{\left (i \sqrt{d} c+\sqrt{e}\right ) \tan \left (\frac{1}{2} \sec ^{-1}(c x)\right )}{\sqrt{d c^2+e}}\right )+4 d \sec ^{-1}(c x) \log \left (\frac{i e^{i \sec ^{-1}(c x)} \left (\sqrt{e}-\sqrt{d c^2+e}\right )}{c \sqrt{d}}+1\right )+8 d \sin ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{e}}{c \sqrt{d}}+1}}{\sqrt{2}}\right ) \log \left (\frac{i e^{i \sec ^{-1}(c x)} \left (\sqrt{e}-\sqrt{d c^2+e}\right )}{c \sqrt{d}}+1\right )+4 d \sec ^{-1}(c x) \log \left (\frac{i e^{i \sec ^{-1}(c x)} \left (\sqrt{d c^2+e}-\sqrt{e}\right )}{c \sqrt{d}}+1\right )+8 d \sin ^{-1}\left (\frac{\sqrt{1-\frac{i \sqrt{e}}{c \sqrt{d}}}}{\sqrt{2}}\right ) \log \left (\frac{i e^{i \sec ^{-1}(c x)} \left (\sqrt{d c^2+e}-\sqrt{e}\right )}{c \sqrt{d}}+1\right )+4 d \sec ^{-1}(c x) \log \left (1-\frac{i \left (\sqrt{e}+\sqrt{d c^2+e}\right ) e^{i \sec ^{-1}(c x)}}{c \sqrt{d}}\right )-8 d \sin ^{-1}\left (\frac{\sqrt{1-\frac{i \sqrt{e}}{c \sqrt{d}}}}{\sqrt{2}}\right ) \log \left (1-\frac{i \left (\sqrt{e}+\sqrt{d c^2+e}\right ) e^{i \sec ^{-1}(c x)}}{c \sqrt{d}}\right )+4 d \sec ^{-1}(c x) \log \left (\frac{i e^{i \sec ^{-1}(c x)} \left (\sqrt{e}+\sqrt{d c^2+e}\right )}{c \sqrt{d}}+1\right )-8 d \sin ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{e}}{c \sqrt{d}}+1}}{\sqrt{2}}\right ) \log \left (\frac{i e^{i \sec ^{-1}(c x)} \left (\sqrt{e}+\sqrt{d c^2+e}\right )}{c \sqrt{d}}+1\right )-8 d \sec ^{-1}(c x) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )-\frac{d \sqrt{e} \log \left (\frac{2 \sqrt{d} \sqrt{e} \left (c \left (i c \sqrt{d}-\sqrt{-d c^2-e} \sqrt{1-\frac{1}{c^2 x^2}}\right ) x+\sqrt{e}\right )}{\sqrt{-d c^2-e} \left (\sqrt{d}-i \sqrt{e} x\right )}\right )}{\sqrt{-d c^2-e}}-\frac{d \sqrt{e} \log \left (\frac{2 \sqrt{d} \sqrt{e} \left (c \left (i \sqrt{d} c+\sqrt{-d c^2-e} \sqrt{1-\frac{1}{c^2 x^2}}\right ) x-\sqrt{e}\right )}{\sqrt{-d c^2-e} \left (i \sqrt{e} x+\sqrt{d}\right )}\right )}{\sqrt{-d c^2-e}}-4 i d \text{PolyLog}\left (2,\frac{i \left (\sqrt{e}-\sqrt{d c^2+e}\right ) e^{i \sec ^{-1}(c x)}}{c \sqrt{d}}\right )-4 i d \text{PolyLog}\left (2,\frac{i \left (\sqrt{d c^2+e}-\sqrt{e}\right ) e^{i \sec ^{-1}(c x)}}{c \sqrt{d}}\right )-4 i d \text{PolyLog}\left (2,-\frac{i \left (\sqrt{e}+\sqrt{d c^2+e}\right ) e^{i \sec ^{-1}(c x)}}{c \sqrt{d}}\right )-4 i d \text{PolyLog}\left (2,\frac{i \left (\sqrt{e}+\sqrt{d c^2+e}\right ) e^{i \sec ^{-1}(c x)}}{c \sqrt{d}}\right )+4 i d \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )\right )}{4 e^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^5*(a + b*ArcSec[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*((2*e*Sqrt[1 - 1/(c^2*x^2)]*x)/c - 2*e*x^2*Arc
Sec[c*x] + (d^(3/2)*ArcSec[c*x])/(Sqrt[d] - I*Sqrt[e]*x) + (d^(3/2)*ArcSec[c*x])/(Sqrt[d] + I*Sqrt[e]*x) + 2*d
*ArcSin[1/(c*x)] + (16*I)*d*ArcSin[Sqrt[1 - (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*ArcTan[(((-I)*c*Sqrt[d] + Sqrt[e
])*Tan[ArcSec[c*x]/2])/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])*Tan[ArcSec[c*x]/2])/Sqrt[c^2*d + e]] + 4*d*ArcSec[c*x]*Log[1 + (I*(Sqrt[e] - Sqrt[c^2*d
+ e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + 8*d*ArcSin[Sqrt[1 + (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*Log[1 + (I*(Sqr
t[e] - Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + 4*d*ArcSec[c*x]*Log[1 + (I*(-Sqrt[e] + Sqrt[c^2*d +
e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + 8*d*ArcSin[Sqrt[1 - (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*Log[1 + (I*(-Sqrt[
e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + 4*d*ArcSec[c*x]*Log[1 - (I*(Sqrt[e] + Sqrt[c^2*d + e])
*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - 8*d*ArcSin[Sqrt[1 - (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*Log[1 - (I*(Sqrt[e] +
Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + 4*d*ArcSec[c*x]*Log[1 + (I*(Sqrt[e] + Sqrt[c^2*d + e])*E^(
I*ArcSec[c*x]))/(c*Sqrt[d])] - 8*d*ArcSin[Sqrt[1 + (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*Log[1 + (I*(Sqrt[e] + Sqr
t[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - 8*d*ArcSec[c*x]*Log[1 + E^((2*I)*ArcSec[c*x])] - (d*Sqrt[e]*Lo
g[(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] + 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] - (4*I)*d*PolyLog[2, (I*(Sqrt[e] - Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - (4*I)*d*PolyLog[
2, (I*(-Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - (4*I)*d*PolyLog[2, ((-I)*(Sqrt[e] + Sqrt[
c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - (4*I)*d*PolyLog[2, (I*(Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*
x]))/(c*Sqrt[d])] + (4*I)*d*PolyLog[2, -E^((2*I)*ArcSec[c*x])]))/(4*e^3)

________________________________________________________________________________________

Maple [C]  time = 0.695, size = 783, normalized size = 1.3 \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*arcsec(c*x))/(e*x^2+d)^2,x)

[Out]

1/2*a*x^2/e^2-1/2*c^2*a/e^3*d^2/(c^2*e*x^2+c^2*d)-a/e^3*d*ln(c^2*e*x^2+c^2*d)+1/2*c^2*b/(c^2*e*x^2+c^2*d)/e*ar
csec(c*x)*x^4+c^2*b/(c^2*e*x^2+c^2*d)/e^2*arcsec(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*(e*(c^2*d+e))^(1/2)/(c^2*d+e)/
e^3*arctanh(1/4*(2*c^2*d*(1/c/x+I*(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/(c^2*e*x^
2+c^2*d)/e^2*d-2*I*b/e^3*d*dilog(1-I*(1/c/x+I*(1-1/c^2/x^2)^(1/2)))-2*I*b/e^3*d*dilog(1+I*(1/c/x+I*(1-1/c^2/x^
2)^(1/2)))+2*b/e^3*d*arcsec(c*x)*ln(1+I*(1/c/x+I*(1-1/c^2/x^2)^(1/2)))+2*b/e^3*d*arcsec(c*x)*ln(1-I*(1/c/x+I*(
1-1/c^2/x^2)^(1/2)))-1/2*I*b/(c^2*e*x^2+c^2*d)/e*x^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*arcsec(c*x)*ln((_R1-1/c/x-I*(1-1/c^2/x^2)^(1/2))/_R1)+dilog((_R1-1/c/x-I*(1-1/c^2/x^2)^(1/2))/_R1)),_R1=Ro
otOf(c^2*d*_Z^4+(2*c^2*d+4*e)*_Z^2+c^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*arcsec(c*x)*ln((_R1-1/c/x-I*(1-1/c^2/x^2)^(1/2))/_R1)+dilog((_R1-1/c/x-I*(1-1/c^2/x^2)^(1/2))/_R1)),_R1=Root
Of(c^2*d*_Z^4+(2*c^2*d+4*e)*_Z^2+c^2*d))

________________________________________________________________________________________

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 (\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*arcsec(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*arctan(sqrt(c*x + 1)*sqrt(
c*x - 1))/(e^2*x^4 + 2*d*e*x^2 + d^2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b x^{5} \operatorname{arcsec}\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*arcsec(c*x))/(e*x^2+d)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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*asec(c*x))/(e*x**2+d)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arcsec}\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*arcsec(c*x))/(e*x^2+d)^2,x, algorithm="giac")

[Out]

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