3.1.99 \(\int \frac {a+b \sec ^{-1}(c x)}{x (d+e x^2)^2} \, dx\) [99]

Optimal. Leaf size=546 \[ -\frac {e \left (a+b \sec ^{-1}(c x)\right )}{2 d^2 \left (e+\frac {d}{x^2}\right )}+\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b d^2}-\frac {b \sqrt {e} \text {ArcTan}\left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} \sqrt {1-\frac {1}{c^2 x^2}} x}\right )}{2 d^2 \sqrt {c^2 d+e}}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d^2}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d^2}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d^2}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d^2}+\frac {i b \text {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d^2}+\frac {i b \text {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d^2}+\frac {i b \text {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d^2}+\frac {i b \text {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d^2} \]

[Out]

-1/2*e*(a+b*arcsec(c*x))/d^2/(e+d/x^2)+1/2*I*(a+b*arcsec(c*x))^2/b/d^2-1/2*(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)))/d^2-1/2*(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)))/d^2-1/2*(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)))/d^2-1/2*(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)))/d^2+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)))/d^2+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)))/d^2+
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)))/d^2+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)))/d^2-1/2*b*arctan((c^2*d+e)^(1/2)/c/x/e^
(1/2)/(1-1/c^2/x^2)^(1/2))*e^(1/2)/d^2/(c^2*d+e)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 1.05, antiderivative size = 546, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {5348, 4818, 4814, 385, 211, 4826, 4616, 2221, 2317, 2438} \begin {gather*} -\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 d^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 d^2}-\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 d^2}-\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 d^2}-\frac {e \left (a+b \sec ^{-1}(c x)\right )}{2 d^2 \left (\frac {d}{x^2}+e\right )}+\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b d^2}-\frac {b \sqrt {e} \text {ArcTan}\left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} x \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{2 d^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 d^2}+\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 d^2}+\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 d^2}+\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 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(e*(a + b*ArcSec[c*x]))/(d^2*(e + d/x^2)) + ((I/2)*(a + b*ArcSec[c*x])^2)/(b*d^2) - (b*Sqrt[e]*ArcTan[Sqr
t[c^2*d + e]/(c*Sqrt[e]*Sqrt[1 - 1/(c^2*x^2)]*x)])/(2*d^2*Sqrt[c^2*d + e]) - ((a + b*ArcSec[c*x])*Log[1 - (c*S
qrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e])])/(2*d^2) - ((a + b*ArcSec[c*x])*Log[1 + (c*Sqrt[-d]*E^
(I*ArcSec[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e])])/(2*d^2) - ((a + b*ArcSec[c*x])*Log[1 - (c*Sqrt[-d]*E^(I*ArcSec[
c*x]))/(Sqrt[e] + Sqrt[c^2*d + e])])/(2*d^2) - ((a + b*ArcSec[c*x])*Log[1 + (c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sq
rt[e] + Sqrt[c^2*d + e])])/(2*d^2) + ((I/2)*b*PolyLog[2, -((c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] - Sqrt[c^2*
d + e]))])/d^2 + ((I/2)*b*PolyLog[2, (c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] - Sqrt[c^2*d + e])])/d^2 + ((I/2)
*b*PolyLog[2, -((c*Sqrt[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e]))])/d^2 + ((I/2)*b*PolyLog[2, (c*Sqr
t[-d]*E^(I*ArcSec[c*x]))/(Sqrt[e] + Sqrt[c^2*d + e])])/d^2

Rule 211

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

Rule 385

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 4616

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 4814

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 4818

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 4826

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 5348

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

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Mathematica [F]
time = 34.53, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 5.16, size = 3095, normalized size = 5.67

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(3095\)
default	\(\text {Expression too large to display}\)	\(3095\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsec(c*x))/x/(e*x^2+d)^2,x,method=_RETURNVERBOSE)

[Out]

-1/4*I*b*(e*(c^2*d+e))^(1/2)/d^2/(c^2*d+e)*polylog(2,d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d+2*(e*(c^2*d
+e))^(1/2)-2*e))-I*b/c^2*arcsec(c*x)^2/d^3*(e*(c^2*d+e))^(1/2)-1/2*I*b/c^2*polylog(2,d*c^2*(1/c/x+I*(1-1/c^2/x
^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))/d^3*(e*(c^2*d+e))^(1/2)+2*I*b/c^2*arcsec(c*x)^2*e/d^3+2*I*b/c
^4*arcsec(c*x)^2*e^2/d^4+1/2*b*c^2/d/(c^2*d+e)*ln(1-d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+
e))^(1/2)-2*e))*arcsec(c*x)-2*b/c^2/d^3*e*ln(1-d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(
1/2)-2*e))*arcsec(c*x)+I*b/c^2*polylog(2,d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2
*e))*e/d^3+I*b/c^4*polylog(2,d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))*e^2/d^4
-2*b/c^4/d^4*e^2*ln(1-d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))*arcsec(c*x)-5/
4*I*b*polylog(2,d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))/d^2/(c^2*d+e)*e+1/8*
I*b*c^2*polylog(2,d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))/d/e/(c^2*d+e)*(e*(
c^2*d+e))^(1/2)+3*I*b/c^2*arcsec(c*x)^2/d^3/(c^2*d+e)*e*(e*(c^2*d+e))^(1/2)-3*b/c^2/d^3/(c^2*d+e)*ln(1-d*c^2*(
1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))*arcsec(c*x)*e*(e*(c^2*d+e))^(1/2)-2*b/c^4/d
^4*e^2/(c^2*d+e)*ln(1-d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))*arcsec(c*x)*(e
*(c^2*d+e))^(1/2)-1/4*b*c^2/d/e/(c^2*d+e)*ln(1-d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(
1/2)-2*e))*arcsec(c*x)*(e*(c^2*d+e))^(1/2)+3/4*I*b*polylog(2,d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(
e*(c^2*d+e))^(1/2)-2*e))/d^2/(c^2*d+e)*(e*(c^2*d+e))^(1/2)-1/2*I*b*c^2*arcsec(c*x)^2/d/(c^2*d+e)-1/4*I*b*c^2*p
olylog(2,d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))/d/(c^2*d+e)+3/2*I*b/c^2*pol
ylog(2,d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))/d^3/(c^2*d+e)*e*(e*(c^2*d+e))
^(1/2)-1/8*I*b*c^2*(e*(c^2*d+e))^(1/2)/d/e/(c^2*d+e)*polylog(2,d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d+2
*(e*(c^2*d+e))^(1/2)-2*e))+1/4*b*c^2*(e*(c^2*d+e))^(1/2)/d/e/(c^2*d+e)*arcsec(c*x)*ln(1-d*c^2*(1/c/x+I*(1-1/c^
2/x^2)^(1/2))^2/(-c^2*d+2*(e*(c^2*d+e))^(1/2)-2*e))-I*b/c^4*e^3*polylog(2,d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^
2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))/d^4/(c^2*d+e)+1/4*I*b*polylog(2,d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-
c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))/d^2+1/2*I*b*(e*(c^2*d+e))^(1/2)/d^2/(c^2*d+e)*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))-5/2*I*b*arcsec(c*x)^2/d^2/(c^2*d+e)*e+I*b*(e*(c^2*d+e
))^(1/2)/d^2/(c^2*d+e)*arcsec(c*x)^2+1/2*b*(e*(c^2*d+e))^(1/2)/d^2/(c^2*d+e)*arcsec(c*x)*ln(1-d*c^2*(1/c/x+I*(
1-1/c^2/x^2)^(1/2))^2/(-c^2*d+2*(e*(c^2*d+e))^(1/2)-2*e))-3/2*b/d^2/(c^2*d+e)*ln(1-d*c^2*(1/c/x+I*(1-1/c^2/x^2
)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))*arcsec(c*x)*(e*(c^2*d+e))^(1/2)+5/2*b/d^2/(c^2*d+e)*ln(1-d*c^2*
(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))*arcsec(c*x)*e+b/c^2/d^3*ln(1-d*c^2*(1/c/x+
I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))*arcsec(c*x)*(e*(c^2*d+e))^(1/2)+I*b/c^4*e^2*polyl
og(2,d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))/d^4/(c^2*d+e)*(e*(c^2*d+e))^(1/
2)+2*I*b/c^4*e^2*arcsec(c*x)^2/d^4/(c^2*d+e)*(e*(c^2*d+e))^(1/2)+1/2*a*c^2/d/(c^2*e*x^2+c^2*d)-1/2*b/d^2*ln(1-
d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))*arcsec(c*x)+I*b*arcsec(c*x)^2/d^2+1/
2*I*b/d^2*sum((_R1^2*c^2*d+2*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))+4*b/
c^2/d^3/(c^2*d+e)*ln(1-d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))*arcsec(c*x)*e
^2+2*b/c^4/d^4*e^3/(c^2*d+e)*ln(1-d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))*ar
csec(c*x)-2*I*b/c^2*polylog(2,d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))/d^3/(c
^2*d+e)*e^2-2*I*b/c^4*e^3*arcsec(c*x)^2/d^4/(c^2*d+e)-4*I*b/c^2*arcsec(c*x)^2/d^3/(c^2*d+e)*e^2-2*I*b/c^4*arcs
ec(c*x)^2*e/d^4*(e*(c^2*d+e))^(1/2)-I*b/c^4*polylog(2,d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2*(e*(c^2*
d+e))^(1/2)-2*e))*e/d^4*(e*(c^2*d+e))^(1/2)+2*b/c^4/d^4*e*ln(1-d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d-2
*(e*(c^2*d+e))^(1/2)-2*e))*arcsec(c*x)*(e*(c^2*d+e))^(1/2)-1/2*b*c^2*x^2*arcsec(c*x)*e/d^2/(c^2*e*x^2+c^2*d)+a
/d^2*ln(c*x)-1/2*a/d^2*ln(c^2*e*x^2+c^2*d)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Warning, integrat
ion of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(sageVARx)]s
ym2poly/r2sym(

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )}{x\,{\left (e\,x^2+d\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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