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

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

[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

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Rubi [A]  time = 0.895521, antiderivative size = 459, normalized size of antiderivative = 1., 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 = {5240, 4734, 4742, 4520, 2190, 2279, 2391} \[ \frac{i b \text{PolyLog}\left (2,-\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{2 d}+\frac{i b \text{PolyLog}\left (2,\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{2 d}+\frac{i b \text{PolyLog}\left (2,-\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{2 d}+\frac{i b \text{PolyLog}\left (2,\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{2 d}-\frac{\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{2 d}-\frac{\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{e}-\sqrt{c^2 d+e}}\right )}{2 d}-\frac{\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{2 d}-\frac{\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac{c \sqrt{-d} e^{i \sec ^{-1}(c x)}}{\sqrt{c^2 d+e}+\sqrt{e}}\right )}{2 d}+\frac{i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b d} \]

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 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 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]

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]

Rubi steps

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

Mathematica [A]  time = 0.827598, size = 402, normalized size = 0.88 \[ \frac{i b \left (\text{PolyLog}\left (2,-\frac{\left (-2 \sqrt{e \left (c^2 d+e\right )}+c^2 d+2 e\right ) e^{2 i \sec ^{-1}(c x)}}{c^2 d}\right )+\text{PolyLog}\left (2,-\frac{\left (2 \left (\sqrt{e \left (c^2 d+e\right )}+e\right )+c^2 d\right ) e^{2 i \sec ^{-1}(c x)}}{c^2 d}\right )-4 \sin ^{-1}\left (\sqrt{\frac{e}{c^2 d}+1}\right ) \tan ^{-1}\left (\frac{c e x \sqrt{1-\frac{1}{c^2 x^2}}}{\sqrt{e \left (c^2 d+e\right )}}\right )+2 i \sec ^{-1}(c x) \log \left (1+\frac{\left (-2 \sqrt{e \left (c^2 d+e\right )}+c^2 d+2 e\right ) e^{2 i \sec ^{-1}(c x)}}{c^2 d}\right )+2 i \sec ^{-1}(c x) \log \left (1+\frac{\left (2 \left (\sqrt{e \left (c^2 d+e\right )}+e\right )+c^2 d\right ) e^{2 i \sec ^{-1}(c x)}}{c^2 d}\right )+2 i \sin ^{-1}\left (\sqrt{\frac{e}{c^2 d}+1}\right ) \log \left (1+\frac{\left (-2 \sqrt{e \left (c^2 d+e\right )}+c^2 d+2 e\right ) e^{2 i \sec ^{-1}(c x)}}{c^2 d}\right )-2 i \sin ^{-1}\left (\sqrt{\frac{e}{c^2 d}+1}\right ) \log \left (1+\frac{\left (2 \left (\sqrt{e \left (c^2 d+e\right )}+e\right )+c^2 d\right ) e^{2 i \sec ^{-1}(c x)}}{c^2 d}\right )+2 \sec ^{-1}(c x)^2\right )-2 a \log \left (d+e x^2\right )+4 a \log (x)}{4 d} \]

Warning: Unable to verify antiderivative.

[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]  time = 0.605, size = 2933, normalized size = 6.4 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \operatorname{arcsec}\left (c x\right ) + a}{e x^{3} + d x}, x\right ) \end{align*}

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)/(e*x^3 + d*x), x)

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

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

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]

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