3.104 \(\int \frac {x^5 (a+b \sec ^{-1}(c x))}{(d+e x^2)^3} \, dx\)

Optimal. Leaf size=707 \[ \frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 e^3}+\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 e^3}+\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {c^2 d+e}+\sqrt {e}}\right )}{2 e^3}+\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {c^2 d+e}+\sqrt {e}}\right )}{2 e^3}-\frac {a+b \sec ^{-1}(c x)}{2 e^2 \left (\frac {d}{x^2}+e\right )}-\frac {a+b \sec ^{-1}(c x)}{4 e \left (\frac {d}{x^2}+e\right )^2}-\frac {\log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{e^3}-\frac {b \left (c^2 d+2 e\right ) \tan ^{-1}\left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} x \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{8 e^{5/2} \left (c^2 d+e\right )^{3/2}}-\frac {b \tan ^{-1}\left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} x \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{2 e^{5/2} \sqrt {c^2 d+e}}-\frac {i b \text {Li}_2\left (-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {d c^2+e}}\right )}{2 e^3}-\frac {i b \text {Li}_2\left (\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {d c^2+e}}\right )}{2 e^3}-\frac {i b \text {Li}_2\left (-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {d c^2+e}}\right )}{2 e^3}-\frac {i b \text {Li}_2\left (\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {d c^2+e}}\right )}{2 e^3}-\frac {b c d \sqrt {1-\frac {1}{c^2 x^2}}}{8 e^2 x \left (c^2 d+e\right ) \left (\frac {d}{x^2}+e\right )}+\frac {i b \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )}{2 e^3} \]

[Out]

1/4*(-a-b*arcsec(c*x))/e/(e+d/x^2)^2+1/2*(-a-b*arcsec(c*x))/e^2/(e+d/x^2)-1/8*b*(c^2*d+2*e)*arctan((c^2*d+e)^(
1/2)/c/x/e^(1/2)/(1-1/c^2/x^2)^(1/2))/e^(5/2)/(c^2*d+e)^(3/2)-(a+b*arcsec(c*x))*ln(1+(1/c/x+I*(1-1/c^2/x^2)^(1
/2))^2)/e^3+1/2*(a+b*arcsec(c*x))*ln(1-c*(1/c/x+I*(1-1/c^2/x^2)^(1/2))*(-d)^(1/2)/(e^(1/2)-(c^2*d+e)^(1/2)))/e
^3+1/2*(a+b*arcsec(c*x))*ln(1+c*(1/c/x+I*(1-1/c^2/x^2)^(1/2))*(-d)^(1/2)/(e^(1/2)-(c^2*d+e)^(1/2)))/e^3+1/2*(a
+b*arcsec(c*x))*ln(1-c*(1/c/x+I*(1-1/c^2/x^2)^(1/2))*(-d)^(1/2)/(e^(1/2)+(c^2*d+e)^(1/2)))/e^3+1/2*(a+b*arcsec
(c*x))*ln(1+c*(1/c/x+I*(1-1/c^2/x^2)^(1/2))*(-d)^(1/2)/(e^(1/2)+(c^2*d+e)^(1/2)))/e^3+1/2*I*b*polylog(2,-(1/c/
x+I*(1-1/c^2/x^2)^(1/2))^2)/e^3-1/2*I*b*polylog(2,-c*(1/c/x+I*(1-1/c^2/x^2)^(1/2))*(-d)^(1/2)/(e^(1/2)-(c^2*d+
e)^(1/2)))/e^3-1/2*I*b*polylog(2,c*(1/c/x+I*(1-1/c^2/x^2)^(1/2))*(-d)^(1/2)/(e^(1/2)-(c^2*d+e)^(1/2)))/e^3-1/2
*I*b*polylog(2,-c*(1/c/x+I*(1-1/c^2/x^2)^(1/2))*(-d)^(1/2)/(e^(1/2)+(c^2*d+e)^(1/2)))/e^3-1/2*I*b*polylog(2,c*
(1/c/x+I*(1-1/c^2/x^2)^(1/2))*(-d)^(1/2)/(e^(1/2)+(c^2*d+e)^(1/2)))/e^3-1/2*b*arctan((c^2*d+e)^(1/2)/c/x/e^(1/
2)/(1-1/c^2/x^2)^(1/2))/e^(5/2)/(c^2*d+e)^(1/2)-1/8*b*c*d*(1-1/c^2/x^2)^(1/2)/e^2/(c^2*d+e)/(e+d/x^2)/x

________________________________________________________________________________________

Rubi [A]  time = 1.40, antiderivative size = 707, normalized size of antiderivative = 1.00, number of steps used = 33, number of rules used = 13, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {5240, 4734, 4626, 3719, 2190, 2279, 2391, 4730, 382, 377, 205, 4742, 4520} \[ -\frac {i b \text {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 e^3}-\frac {i b \text {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 e^3}-\frac {i b \text {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {c^2 d+e}+\sqrt {e}}\right )}{2 e^3}-\frac {i b \text {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {c^2 d+e}+\sqrt {e}}\right )}{2 e^3}+\frac {i b \text {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )}{2 e^3}+\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 e^3}+\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 e^3}+\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {c^2 d+e}+\sqrt {e}}\right )}{2 e^3}+\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {c^2 d+e}+\sqrt {e}}\right )}{2 e^3}-\frac {a+b \sec ^{-1}(c x)}{2 e^2 \left (\frac {d}{x^2}+e\right )}-\frac {a+b \sec ^{-1}(c x)}{4 e \left (\frac {d}{x^2}+e\right )^2}-\frac {\log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{e^3}-\frac {b c d \sqrt {1-\frac {1}{c^2 x^2}}}{8 e^2 x \left (c^2 d+e\right ) \left (\frac {d}{x^2}+e\right )}-\frac {b \left (c^2 d+2 e\right ) \tan ^{-1}\left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} x \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{8 e^{5/2} \left (c^2 d+e\right )^{3/2}}-\frac {b \tan ^{-1}\left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} x \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{2 e^{5/2} \sqrt {c^2 d+e}} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*ArcSec[c*x]))/(d + e*x^2)^3,x]

[Out]

-(b*c*d*Sqrt[1 - 1/(c^2*x^2)])/(8*e^2*(c^2*d + e)*(e + d/x^2)*x) - (a + b*ArcSec[c*x])/(4*e*(e + d/x^2)^2) - (
a + b*ArcSec[c*x])/(2*e^2*(e + d/x^2)) - (b*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]) - (b*(c^2*d + 2*e)*ArcTan[Sqrt[c^2*d + e]/(c*Sqrt[e]*Sqrt[1 - 1/(c^2*x^2)]*x)])/(8*e^(5
/2)*(c^2*d + e)^(3/2)) + ((a + b*ArcSec[c*x])*Log[1 - (c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e
])])/(2*e^3) + ((a + b*ArcSec[c*x])*Log[1 + (c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e])])/(2*e^
3) + ((a + b*ArcSec[c*x])*Log[1 - (c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e])])/(2*e^3) + ((a +
b*ArcSec[c*x])*Log[1 + (c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e])])/(2*e^3) - ((a + b*ArcSec[
c*x])*Log[1 + E^((2*I)*ArcSec[c*x])])/e^3 - ((I/2)*b*PolyLog[2, -((c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] - Sq
rt[c^2*d + e]))])/e^3 - ((I/2)*b*PolyLog[2, (c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e])])/e^3 -
((I/2)*b*PolyLog[2, -((c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e]))])/e^3 - ((I/2)*b*PolyLog[2,
(c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e])])/e^3 + ((I/2)*b*PolyLog[2, -E^((2*I)*ArcSec[c*x])
])/e^3

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 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 382

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

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 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 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]

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 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 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 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 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]

Rubi steps

\begin {align*} \int \frac {x^5 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx &=-\operatorname {Subst}\left (\int \frac {a+b \cos ^{-1}\left (\frac {x}{c}\right )}{x \left (e+d x^2\right )^3} \, dx,x,\frac {1}{x}\right )\\ &=-\operatorname {Subst}\left (\int \left (\frac {a+b \cos ^{-1}\left (\frac {x}{c}\right )}{e^3 x}-\frac {d x \left (a+b \cos ^{-1}\left (\frac {x}{c}\right )\right )}{e \left (e+d x^2\right )^3}-\frac {d x \left (a+b \cos ^{-1}\left (\frac {x}{c}\right )\right )}{e^2 \left (e+d x^2\right )^2}-\frac {d 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 {\operatorname {Subst}\left (\int \frac {a+b \cos ^{-1}\left (\frac {x}{c}\right )}{x} \, dx,x,\frac {1}{x}\right )}{e^3}+\frac {d \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 {d \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 \operatorname {Subst}\left (\int \frac {x \left (a+b \cos ^{-1}\left (\frac {x}{c}\right )\right )}{\left (e+d x^2\right )^3} \, dx,x,\frac {1}{x}\right )}{e}\\ &=-\frac {a+b \sec ^{-1}(c x)}{4 e \left (e+\frac {d}{x^2}\right )^2}-\frac {a+b \sec ^{-1}(c x)}{2 e^2 \left (e+\frac {d}{x^2}\right )}+\frac {\operatorname {Subst}\left (\int (a+b x) \tan (x) \, dx,x,\sec ^{-1}(c x)\right )}{e^3}+\frac {d \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}{\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 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{c^2}} \left (e+d x^2\right )^2} \, dx,x,\frac {1}{x}\right )}{4 c e}\\ &=-\frac {b c d \sqrt {1-\frac {1}{c^2 x^2}}}{8 e^2 \left (c^2 d+e\right ) \left (e+\frac {d}{x^2}\right ) x}-\frac {a+b \sec ^{-1}(c x)}{4 e \left (e+\frac {d}{x^2}\right )^2}-\frac {a+b \sec ^{-1}(c x)}{2 e^2 \left (e+\frac {d}{x^2}\right )}+\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b e^3}-\frac {(2 i) \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 {\sqrt {-d} \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 )}{2 e^3}+\frac {\sqrt {-d} \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 )}{2 e^3}-\frac {b \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 {\left (b \left (c^2 d+2 e\right )\right ) \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 )}{8 c e^2 \left (c^2 d+e\right )}\\ &=-\frac {b c d \sqrt {1-\frac {1}{c^2 x^2}}}{8 e^2 \left (c^2 d+e\right ) \left (e+\frac {d}{x^2}\right ) x}-\frac {a+b \sec ^{-1}(c x)}{4 e \left (e+\frac {d}{x^2}\right )^2}-\frac {a+b \sec ^{-1}(c x)}{2 e^2 \left (e+\frac {d}{x^2}\right )}+\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b e^3}-\frac {b \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 {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )}{e^3}+\frac {b \operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{e^3}+\frac {\sqrt {-d} \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 )}{2 e^3}-\frac {\sqrt {-d} \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 )}{2 e^3}-\frac {\left (b \left (c^2 d+2 e\right )\right ) \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 )}{8 c e^2 \left (c^2 d+e\right )}\\ &=-\frac {b c d \sqrt {1-\frac {1}{c^2 x^2}}}{8 e^2 \left (c^2 d+e\right ) \left (e+\frac {d}{x^2}\right ) x}-\frac {a+b \sec ^{-1}(c x)}{4 e \left (e+\frac {d}{x^2}\right )^2}-\frac {a+b \sec ^{-1}(c x)}{2 e^2 \left (e+\frac {d}{x^2}\right )}-\frac {b \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 {b \left (c^2 d+2 e\right ) \tan ^{-1}\left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} \sqrt {1-\frac {1}{c^2 x^2}} x}\right )}{8 e^{5/2} \left (c^2 d+e\right )^{3/2}}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )}{e^3}-\frac {(i b) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \sec ^{-1}(c x)}\right )}{2 e^3}-\frac {\left (i \sqrt {-d}\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 )}{2 e^3}-\frac {\left (i \sqrt {-d}\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 )}{2 e^3}+\frac {\left (i \sqrt {-d}\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 )}{2 e^3}+\frac {\left (i \sqrt {-d}\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 )}{2 e^3}\\ &=-\frac {b c d \sqrt {1-\frac {1}{c^2 x^2}}}{8 e^2 \left (c^2 d+e\right ) \left (e+\frac {d}{x^2}\right ) x}-\frac {a+b \sec ^{-1}(c x)}{4 e \left (e+\frac {d}{x^2}\right )^2}-\frac {a+b \sec ^{-1}(c x)}{2 e^2 \left (e+\frac {d}{x^2}\right )}-\frac {b \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 {b \left (c^2 d+2 e\right ) \tan ^{-1}\left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} \sqrt {1-\frac {1}{c^2 x^2}} x}\right )}{8 e^{5/2} \left (c^2 d+e\right )^{3/2}}+\frac {\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 )}{2 e^3}+\frac {\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 )}{2 e^3}+\frac {\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 )}{2 e^3}+\frac {\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 )}{2 e^3}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )}{e^3}+\frac {i b \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )}{2 e^3}-\frac {b \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 )}{2 e^3}-\frac {b \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 )}{2 e^3}-\frac {b \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 )}{2 e^3}-\frac {b \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 )}{2 e^3}\\ &=-\frac {b c d \sqrt {1-\frac {1}{c^2 x^2}}}{8 e^2 \left (c^2 d+e\right ) \left (e+\frac {d}{x^2}\right ) x}-\frac {a+b \sec ^{-1}(c x)}{4 e \left (e+\frac {d}{x^2}\right )^2}-\frac {a+b \sec ^{-1}(c x)}{2 e^2 \left (e+\frac {d}{x^2}\right )}-\frac {b \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 {b \left (c^2 d+2 e\right ) \tan ^{-1}\left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} \sqrt {1-\frac {1}{c^2 x^2}} x}\right )}{8 e^{5/2} \left (c^2 d+e\right )^{3/2}}+\frac {\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 )}{2 e^3}+\frac {\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 )}{2 e^3}+\frac {\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 )}{2 e^3}+\frac {\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 )}{2 e^3}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )}{e^3}+\frac {i b \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )}{2 e^3}+\frac {(i b) \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 )}{2 e^3}+\frac {(i b) \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 )}{2 e^3}+\frac {(i b) \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 )}{2 e^3}+\frac {(i b) \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 )}{2 e^3}\\ &=-\frac {b c d \sqrt {1-\frac {1}{c^2 x^2}}}{8 e^2 \left (c^2 d+e\right ) \left (e+\frac {d}{x^2}\right ) x}-\frac {a+b \sec ^{-1}(c x)}{4 e \left (e+\frac {d}{x^2}\right )^2}-\frac {a+b \sec ^{-1}(c x)}{2 e^2 \left (e+\frac {d}{x^2}\right )}-\frac {b \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 {b \left (c^2 d+2 e\right ) \tan ^{-1}\left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} \sqrt {1-\frac {1}{c^2 x^2}} x}\right )}{8 e^{5/2} \left (c^2 d+e\right )^{3/2}}+\frac {\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 )}{2 e^3}+\frac {\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 )}{2 e^3}+\frac {\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 )}{2 e^3}+\frac {\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 )}{2 e^3}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )}{e^3}-\frac {i b \text {Li}_2\left (-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 e^3}-\frac {i b \text {Li}_2\left (\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 e^3}-\frac {i b \text {Li}_2\left (-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 e^3}-\frac {i b \text {Li}_2\left (\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 e^3}+\frac {i b \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )}{2 e^3}\\ \end {align*}

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Mathematica [B]  time = 8.01, size = 1805, normalized size = 2.55 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^5*(a + b*ArcSec[c*x]))/(d + e*x^2)^3,x]

[Out]

-1/4*(a*d^2)/(e^3*(d + e*x^2)^2) + (a*d)/(e^3*(d + e*x^2)) + (a*Log[d + e*x^2])/(2*e^3) + b*((((-7*I)/16)*Sqrt
[d]*(-(ArcSec[c*x]/(I*Sqrt[d]*Sqrt[e] + e*x)) + (I*(ArcSin[1/(c*x)]/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]))/Sqrt[d]))/e^(5/2) + (((7*I)/16)*Sqrt[d]*(-(ArcSec[c*x]/((-I)*Sqrt[d]*Sqrt[e] + e*x)) - (
I*(ArcSin[1/(c*x)]/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]))/Sqrt[d]))/e^(5/2) - (d*(-(
ArcSec[c*x]/(Sqrt[e]*((-I)*Sqrt[d] + Sqrt[e]*x)^2)) + (ArcSin[1/(c*x)]/Sqrt[e] - I*((c*Sqrt[d]*Sqrt[e]*Sqrt[1
- 1/(c^2*x^2)]*x)/((c^2*d + e)*((-I)*Sqrt[d] + Sqrt[e]*x)) + ((2*c^2*d + e)*Log[(-4*d*Sqrt[e]*Sqrt[c^2*d + e]*
(I*Sqrt[e] + c*(c*Sqrt[d] - Sqrt[c^2*d + e]*Sqrt[1 - 1/(c^2*x^2)])*x))/((2*c^2*d + e)*((-I)*Sqrt[d] + Sqrt[e]*
x))])/(c^2*d + e)^(3/2)))/d))/(16*e^(5/2)) - (d*((I*c*Sqrt[e]*Sqrt[1 - 1/(c^2*x^2)]*x)/(Sqrt[d]*(c^2*d + e)*(I
*Sqrt[d] + Sqrt[e]*x)) - ArcSec[c*x]/(Sqrt[e]*(I*Sqrt[d] + Sqrt[e]*x)^2) + ArcSin[1/(c*x)]/(d*Sqrt[e]) - (I*(2
*c^2*d + e)*Log[(4*d*Sqrt[e]*Sqrt[c^2*d + e]*((-I)*Sqrt[e] + c*(c*Sqrt[d] + Sqrt[c^2*d + e]*Sqrt[1 - 1/(c^2*x^
2)])*x))/((2*c^2*d + e)*(I*Sqrt[d] + Sqrt[e]*x))])/(d*(c^2*d + e)^(3/2))))/(16*e^(5/2)) + ((I/4)*(8*ArcSin[Sqr
t[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]] -
(2*I)*ArcSec[c*x]*Log[1 + (I*(Sqrt[e] - Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - (4*I)*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])] -
(2*I)*ArcSec[c*x]*Log[1 + (I*(Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + (4*I)*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])] +
(2*I)*ArcSec[c*x]*Log[1 + E^((2*I)*ArcSec[c*x])] - 2*PolyLog[2, (I*(-Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c
*x]))/(c*Sqrt[d])] - 2*PolyLog[2, ((-I)*(Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + PolyLog[
2, -E^((2*I)*ArcSec[c*x])]))/e^3 + ((I/4)*(8*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]] - (2*I)*ArcSec[c*x]*Log[1 + (I*(-Sqrt[e] + Sqrt[c^2*d
+ e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - (4*I)*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])] - (2*I)*ArcSec[c*x]*Log[1 - (I*(Sqrt[e] + Sqrt[c^2*
d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + (4*I)*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])] + (2*I)*ArcSec[c*x]*Log[1 + E^((2*I)*ArcSec[c*x])]
- 2*PolyLog[2, ((-I)*(-Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - 2*PolyLog[2, (I*(Sqrt[e] +
Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + PolyLog[2, -E^((2*I)*ArcSec[c*x])]))/e^3)

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{5} \operatorname {arcsec}\left (c x\right ) + a x^{5}}{e^{3} x^{6} + 3 \, d e^{2} x^{4} + 3 \, d^{2} e x^{2} + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsec(c*x))/(e*x^2+d)^3,x, algorithm="fricas")

[Out]

integral((b*x^5*arcsec(c*x) + a*x^5)/(e^3*x^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 + d^3), x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsec(c*x))/(e*x^2+d)^3,x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 1.96, size = 1626, normalized size = 2.30 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*arcsec(c*x))/(e*x^2+d)^3,x)

[Out]

c^2*a/e^3*d/(c^2*e*x^2+c^2*d)-1/4*I*c^4*b/e/(c^2*e*x^2+c^2*d)^2/(c^2*d+e)*x^2*d+5/8*I*c^2*b*(e*(c^2*d+e))^(1/2
)/e^3/(c^2*d+e)^2*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
*c^4*b/e/(c^2*e*x^2+c^2*d)^2/(c^2*d+e)*arcsec(c*x)*d*x^2-1/8*c^5*b/e/(c^2*e*x^2+c^2*d)^2/(c^2*d+e)*((c^2*x^2-1
)/c^2/x^2)^(1/2)*x^3*d-1/8*c^5*b/e^2/(c^2*e*x^2+c^2*d)^2/(c^2*d+e)*((c^2*x^2-1)/c^2/x^2)^(1/2)*x*d^2-3/4*c^6*b
/e/(c^2*e*x^2+c^2*d)^2/(c^2*d+e)*arcsec(c*x)*d*x^4-1/2*c^6*b/e^2/(c^2*e*x^2+c^2*d)^2/(c^2*d+e)*arcsec(c*x)*d^2
*x^2-c^2*b/e^3/(c^2*d+e)*d*arcsec(c*x)*ln(1+I*(1/c/x+I*(1-1/c^2/x^2)^(1/2)))-c^2*b/e^3/(c^2*d+e)*d*arcsec(c*x)
*ln(1-I*(1/c/x+I*(1-1/c^2/x^2)^(1/2)))+I*c^2*b/e^3/(c^2*d+e)*d*dilog(1+I*(1/c/x+I*(1-1/c^2/x^2)^(1/2)))+I*c^2*
b/e^3/(c^2*d+e)*d*dilog(1-I*(1/c/x+I*(1-1/c^2/x^2)^(1/2)))-1/8*I*c^4*b/e^2/(c^2*e*x^2+c^2*d)^2/(c^2*d+e)*d^2-1
/4*I*c^2*b/e^2/(c^2*d+e)*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=RootOf(c^2*d*_Z^4+(2*c^2*d+4*e)*_Z^2+c^2*d))*d-1/4
*I*c^2*b/e^3/(c^2*d+e)*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=RootOf(c^2*d*_Z^4+(2*c^2*d+4*e)*_Z^2
+c^2*d))-1/4*I*c^4*b/e^3/(c^2*d+e)*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=RootOf(c^2*d*_Z^4+(2*c^2*d+4*e)*_Z^2
+c^2*d))+1/2*a/e^3*ln(c^2*e*x^2+c^2*d)-1/4*I*b/e^2/(c^2*d+e)*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=
RootOf(c^2*d*_Z^4+(2*c^2*d+4*e)*_Z^2+c^2*d))-1/4*c^4*a*d^2/e^3/(c^2*e*x^2+c^2*d)^2-3/4*c^4*b/(c^2*e*x^2+c^2*d)
^2/(c^2*d+e)*arcsec(c*x)*x^4+3/4*I*b*(e*(c^2*d+e))^(1/2)/e^2/(c^2*d+e)^2*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))-1/8*I*c^4*b/(c^2*e*x^2+c^2*d)^2/(c^2*d+e)*x^4+I*b/e^2/(c^2*d
+e)*dilog(1+I*(1/c/x+I*(1-1/c^2/x^2)^(1/2)))+I*b/e^2/(c^2*d+e)*dilog(1-I*(1/c/x+I*(1-1/c^2/x^2)^(1/2)))-b/e^2/
(c^2*d+e)*arcsec(c*x)*ln(1+I*(1/c/x+I*(1-1/c^2/x^2)^(1/2)))-b/e^2/(c^2*d+e)*arcsec(c*x)*ln(1-I*(1/c/x+I*(1-1/c
^2/x^2)^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, a {\left (\frac {4 \, d e x^{2} + 3 \, d^{2}}{e^{5} x^{4} + 2 \, d e^{4} x^{2} + d^{2} e^{3}} + \frac {2 \, \log \left (e x^{2} + d\right )}{e^{3}}\right )} + b \int \frac {x^{5} \arctan \left (\sqrt {c x + 1} \sqrt {c x - 1}\right )}{e^{3} x^{6} + 3 \, d e^{2} x^{4} + 3 \, d^{2} e x^{2} + d^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsec(c*x))/(e*x^2+d)^3,x, algorithm="maxima")

[Out]

1/4*a*((4*d*e*x^2 + 3*d^2)/(e^5*x^4 + 2*d*e^4*x^2 + d^2*e^3) + 2*log(e*x^2 + d)/e^3) + b*integrate(x^5*arctan(
sqrt(c*x + 1)*sqrt(c*x - 1))/(e^3*x^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 + d^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^5\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}{{\left (e\,x^2+d\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(a + b*acos(1/(c*x))))/(d + e*x^2)^3,x)

[Out]

int((x^5*(a + b*acos(1/(c*x))))/(d + e*x^2)^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*asec(c*x))/(e*x**2+d)**3,x)

[Out]

Timed out

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