3.1.94 \(\int \frac {a+b \sec ^{-1}(c x)}{x (d+e x^2)} \, dx\) [94]

Optimal. Leaf size=459 \[ \frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b d}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d}+\frac {i b \text {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d}+\frac {i b \text {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d}+\frac {i b \text {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d}+\frac {i b \text {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.82, antiderivative size = 459, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5348, 4818, 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}-\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}-\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}-\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}+\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b d}+\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}+\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}+\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}+\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 )} \, dx &=-\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 )\\ &=-\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 )\\ &=-\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 \sqrt {-d}}+\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 \sqrt {-d}}\\ &=\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 \sqrt {-d}}-\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 \sqrt {-d}}\\ &=\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b d}-\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 \sqrt {-d}}-\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 \sqrt {-d}}+\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 \sqrt {-d}}+\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 \sqrt {-d}}\\ &=\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b d}-\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}-\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}-\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}-\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}+\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}+\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}+\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}+\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}\\ &=\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b d}-\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}-\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}-\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}-\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}-\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}-\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}-\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}-\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}\\ &=\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b d}-\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}-\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}-\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}-\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}+\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}+\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}+\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}+\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}\\ \end {align*}

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Mathematica [A]
time = 0.57, size = 402, normalized size = 0.88 \begin {gather*} \frac {4 a \log (x)-2 a \log \left (d+e x^2\right )+i b \left (2 \sec ^{-1}(c x)^2-4 \text {ArcSin}\left (\sqrt {1+\frac {e}{c^2 d}}\right ) \text {ArcTan}\left (\frac {c e \sqrt {1-\frac {1}{c^2 x^2}} x}{\sqrt {e \left (c^2 d+e\right )}}\right )+2 i \sec ^{-1}(c x) \log \left (1+\frac {\left (c^2 d+2 e-2 \sqrt {e \left (c^2 d+e\right )}\right ) e^{2 i \sec ^{-1}(c x)}}{c^2 d}\right )+2 i \text {ArcSin}\left (\sqrt {1+\frac {e}{c^2 d}}\right ) \log \left (1+\frac {\left (c^2 d+2 e-2 \sqrt {e \left (c^2 d+e\right )}\right ) e^{2 i \sec ^{-1}(c x)}}{c^2 d}\right )+2 i \sec ^{-1}(c x) \log \left (1+\frac {\left (c^2 d+2 \left (e+\sqrt {e \left (c^2 d+e\right )}\right )\right ) e^{2 i \sec ^{-1}(c x)}}{c^2 d}\right )-2 i \text {ArcSin}\left (\sqrt {1+\frac {e}{c^2 d}}\right ) \log \left (1+\frac {\left (c^2 d+2 \left (e+\sqrt {e \left (c^2 d+e\right )}\right )\right ) e^{2 i \sec ^{-1}(c x)}}{c^2 d}\right )+\text {PolyLog}\left (2,-\frac {\left (c^2 d+2 e-2 \sqrt {e \left (c^2 d+e\right )}\right ) e^{2 i \sec ^{-1}(c x)}}{c^2 d}\right )+\text {PolyLog}\left (2,-\frac {\left (c^2 d+2 \left (e+\sqrt {e \left (c^2 d+e\right )}\right )\right ) e^{2 i \sec ^{-1}(c x)}}{c^2 d}\right )\right )}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*Log[x] - 2*a*Log[d + e*x^2] + I*b*(2*ArcSec[c*x]^2 - 4*ArcSin[Sqrt[1 + e/(c^2*d)]]*ArcTan[(c*e*Sqrt[1 - 1
/(c^2*x^2)]*x)/Sqrt[e*(c^2*d + e)]] + (2*I)*ArcSec[c*x]*Log[1 + ((c^2*d + 2*e - 2*Sqrt[e*(c^2*d + e)])*E^((2*I
)*ArcSec[c*x]))/(c^2*d)] + (2*I)*ArcSin[Sqrt[1 + e/(c^2*d)]]*Log[1 + ((c^2*d + 2*e - 2*Sqrt[e*(c^2*d + e)])*E^
((2*I)*ArcSec[c*x]))/(c^2*d)] + (2*I)*ArcSec[c*x]*Log[1 + ((c^2*d + 2*(e + Sqrt[e*(c^2*d + e)]))*E^((2*I)*ArcS
ec[c*x]))/(c^2*d)] - (2*I)*ArcSin[Sqrt[1 + e/(c^2*d)]]*Log[1 + ((c^2*d + 2*(e + Sqrt[e*(c^2*d + e)]))*E^((2*I)
*ArcSec[c*x]))/(c^2*d)] + PolyLog[2, -(((c^2*d + 2*e - 2*Sqrt[e*(c^2*d + e)])*E^((2*I)*ArcSec[c*x]))/(c^2*d))]
+ PolyLog[2, -(((c^2*d + 2*(e + Sqrt[e*(c^2*d + e)]))*E^((2*I)*ArcSec[c*x]))/(c^2*d))]))/(4*d)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.92, size = 2933, normalized size = 6.39

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2933\)
default	\(\text {Expression too large to display}\)	\(2933\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*I*b*(e*(c^2*d+e))^(1/2)/(c^2*d+e)/d*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))+1/2*b*c^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)-1/8*I*b*c^2*(e*(c^2*d+e))^(1/2)/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))-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^3*(e*(c^2*d+e))^(1/2)-2*I*b/c^4*arcsec(c*x)^2*e/d^3*(e*(c^2*d+e))^(1/2)-2*I*b/c^2*e^2*poly
log(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))/(c^2*d+e)/d^2-I*b/c^4*e^3*poly
log(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))/(c^2*d+e)/d^3-1/4*b*c^2/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)+2*b/c^4*e^3/(c^2*d+e)/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)+2*b/c^4/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))*e*arc
sec(c*x)*(e*(c^2*d+e))^(1/2)+1/4*b*c^2*(e*(c^2*d+e))^(1/2)/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))+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))/e/(c^2*d+e)*(e*(c^2*d+e))^(1/2)-2*I*b/c^4*e^3*arcsec(c*x)^2/(c^2*d+e)/d^
3-4*I*b/c^2*arcsec(c*x)^2/(c^2*d+e)/d^2*e^2+4*b/c^2*e^2/(c^2*d+e)/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/c^2*arcsec(c*x)^2/d^2*(e*(c^2*d+e))^(1/2)-1/2*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^2*(e*(c^2*d+e))^(1/2)+2*I
*b/c^2*arcsec(c*x)^2/d^2*e-5/2*I*b*arcsec(c*x)^2/(c^2*d+e)/d*e+5/2*b*e/(c^2*d+e)/d*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)-3/2*b/(c^2*d+e)/d*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*(e*(c^2*d+e))^(1/2)/(
c^2*d+e)/d*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*
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^3+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^2*e-2*b/c^2/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))*e*arcsec(c*x)-2*b/c^4/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))*e^2*arcsec(c*x)+b/c^2/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)*(e*(c^2*d+e))^(1/2)+I*b*(e*(c^2*d+e))^(1/2)/(c^
2*d+e)/d*arcsec(c*x)^2+I*b*arcsec(c*x)^2/d-1/2*b/d*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)+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+1/2*I*b/d*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))-3*b/c^2*e/(c^2*d+e)/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)*(e*(c^2*d+e))^(1/2)-2*b/c^4*e^2/(c^2*d+e)/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*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))/(c^2*d+e)/d^3*(e*(c^2*d+e))^(1/2)+3*I*b/c^2*arcsec(c*x)^2/(c
^2*d+e)/d^2*(e*(c^2*d+e))^(1/2)*e+3/2*I*b/c^2*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))/(c^2*d+e)/d^2*(e*(c^2*d+e))^(1/2)+2*I*b/c^4*e^2*arcsec(c*x)^2/(c^2*d+e)/d^3*(e*(c^2*d+e)
)^(1/2)-1/2*I*b*c^2*arcsec(c*x)^2/(c^2*d+e)-1/4*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))/(c^2*d+e)-1/2*a/d*ln(c^2*e*x^2+c^2*d)+a/d*ln(c*x)+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))/(c^2*d+e)/d*(e*(c^2*d+e))^(1/2)-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))*e/(c^2*d+e)/d+2*I*b/c^4*arcsec(c*x
)^2*e^2/d^3

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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),x, algorithm="maxima")

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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),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 )} \,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)),x)

[Out]

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

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