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

Optimal. Leaf size=518 \[ -\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1+\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1+\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d}+\frac {\log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}+\frac {i b \text {Li}_2\left (-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )}{2 d}+\frac {i b \text {Li}_2\left (\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )}{2 d}+\frac {i b \text {Li}_2\left (-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )}{2 d}+\frac {i b \text {Li}_2\left (\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )}{2 d}-\frac {i b \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{2 d} \]

[Out]

(a+b*arcsin(c*x))*ln(1-(I*c*x+(-c^2*x^2+1)^(1/2))^2)/d-1/2*(a+b*arcsin(c*x))*ln(1-(I*c*x+(-c^2*x^2+1)^(1/2))*e
^(1/2)/(I*c*(-d)^(1/2)-(c^2*d+e)^(1/2)))/d-1/2*(a+b*arcsin(c*x))*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*
(-d)^(1/2)-(c^2*d+e)^(1/2)))/d-1/2*(a+b*arcsin(c*x))*ln(1-(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(-d)^(1/2)+(
c^2*d+e)^(1/2)))/d-1/2*(a+b*arcsin(c*x))*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(-d)^(1/2)+(c^2*d+e)^(1/
2)))/d-1/2*I*b*polylog(2,(I*c*x+(-c^2*x^2+1)^(1/2))^2)/d+1/2*I*b*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)
/(I*c*(-d)^(1/2)-(c^2*d+e)^(1/2)))/d+1/2*I*b*polylog(2,(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(-d)^(1/2)-(c^2
*d+e)^(1/2)))/d+1/2*I*b*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(-d)^(1/2)+(c^2*d+e)^(1/2)))/d+1/2*
I*b*polylog(2,(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(-d)^(1/2)+(c^2*d+e)^(1/2)))/d

________________________________________________________________________________________

Rubi [A]  time = 0.93, antiderivative size = 518, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {4733, 4625, 3717, 2190, 2279, 2391, 4741, 4521} \[ \frac {i b \text {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d}+\frac {i b \text {PolyLog}\left (2,\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d}+\frac {i b \text {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d}+\frac {i b \text {PolyLog}\left (2,\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d}-\frac {i b \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )}{2 d}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1+\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1+\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d}+\frac {\log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcSin[c*x])*Log[1 - (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e])])/(2*d) - ((a + b*A
rcSin[c*x])*Log[1 + (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e])])/(2*d) - ((a + b*ArcSin[c*x]
)*Log[1 - (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/(2*d) - ((a + b*ArcSin[c*x])*Log[1 +
(Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/(2*d) + ((a + b*ArcSin[c*x])*Log[1 - E^((2*I)*A
rcSin[c*x])])/d + ((I/2)*b*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e]))])/d + ((
I/2)*b*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e])])/d + ((I/2)*b*PolyLog[2, -((Sq
rt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e]))])/d + ((I/2)*b*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x]
))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/d - ((I/2)*b*PolyLog[2, E^((2*I)*ArcSin[c*x])])/d

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 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4521

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), 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)))/(I*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)))/(I*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] && NegQ[a^2 - b^2]

Rule 4625

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

Rule 4733

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Int[
ExpandIntegrand[(a + b*ArcSin[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 4741

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

Rubi steps

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

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Mathematica [A]  time = 0.76, size = 441, normalized size = 0.85 \[ -\frac {a \log \left (d+e x^2\right )}{2 d}+\frac {a \log (x)}{d}+\frac {b \left (i \text {Li}_2\left (\frac {\left (2 d c^2+e-2 \sqrt {c^2 d \left (d c^2+e\right )}\right ) e^{2 i \sin ^{-1}(c x)}}{e}\right )+i \text {Li}_2\left (\frac {\left (2 d c^2+e+2 \sqrt {c^2 d \left (d c^2+e\right )}\right ) e^{2 i \sin ^{-1}(c x)}}{e}\right )-4 i \sin ^{-1}\left (\sqrt {-\frac {c^2 d}{e}}\right ) \tan ^{-1}\left (\frac {c x \left (c^2 d+e\right )}{\sqrt {1-c^2 x^2} \sqrt {c^2 d \left (c^2 d+e\right )}}\right )-2 \sin ^{-1}\left (\sqrt {-\frac {c^2 d}{e}}\right ) \log \left (1-\frac {\left (-2 \sqrt {c^2 d \left (c^2 d+e\right )}+2 c^2 d+e\right ) e^{2 i \sin ^{-1}(c x)}}{e}\right )-2 \sin ^{-1}(c x) \log \left (1-\frac {\left (-2 \sqrt {c^2 d \left (c^2 d+e\right )}+2 c^2 d+e\right ) e^{2 i \sin ^{-1}(c x)}}{e}\right )+2 \sin ^{-1}\left (\sqrt {-\frac {c^2 d}{e}}\right ) \log \left (1-\frac {\left (2 \sqrt {c^2 d \left (c^2 d+e\right )}+2 c^2 d+e\right ) e^{2 i \sin ^{-1}(c x)}}{e}\right )-2 \sin ^{-1}(c x) \log \left (1-\frac {\left (2 \sqrt {c^2 d \left (c^2 d+e\right )}+2 c^2 d+e\right ) e^{2 i \sin ^{-1}(c x)}}{e}\right )-2 i \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )+4 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )\right )}{4 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arcsin \left (c x\right ) + a}{e x^{3} + d x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \arcsin \left (c x\right ) + a}{{\left (e x^{2} + d\right )} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.33, size = 355, normalized size = 0.69 \[ -\frac {a \ln \left (c^{2} e \,x^{2}+c^{2} d \right )}{2 d}+\frac {a \ln \left (c x \right )}{d}+\frac {i b \left (\munderset {\textit {\_R1} =\RootOf \left (e \,\textit {\_Z}^{4}+\left (-4 c^{2} d -2 e \right ) \textit {\_Z}^{2}+e \right )}{\sum }\frac {\left (\textit {\_R1}^{2} e -4 c^{2} d -e \right ) \left (i \arcsin \left (c x \right ) \ln \left (\frac {\textit {\_R1} -i c x -\sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )+\dilog \left (\frac {\textit {\_R1} -i c x -\sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )\right )}{\textit {\_R1}^{2} e -2 c^{2} d -e}\right )}{4 d}+\frac {b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}-\frac {i b \dilog \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {i b \dilog \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {i b \left (\munderset {\textit {\_R1} =\RootOf \left (e \,\textit {\_Z}^{4}+\left (-4 c^{2} d -2 e \right ) \textit {\_Z}^{2}+e \right )}{\sum }\frac {\left (\textit {\_R1}^{2}-1\right ) \left (i \arcsin \left (c x \right ) \ln \left (\frac {\textit {\_R1} -i c x -\sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )+\dilog \left (\frac {\textit {\_R1} -i c x -\sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )\right )}{\textit {\_R1}^{2} e -2 c^{2} d -e}\right ) e}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a/d*ln(c^2*e*x^2+c^2*d)+a/d*ln(c*x)+1/4*I*b*sum((_R1^2*e-4*c^2*d-e)/(_R1^2*e-2*c^2*d-e)*(I*arcsin(c*x)*ln
((_R1-I*c*x-(-c^2*x^2+1)^(1/2))/_R1)+dilog((_R1-I*c*x-(-c^2*x^2+1)^(1/2))/_R1)),_R1=RootOf(e*_Z^4+(-4*c^2*d-2*
e)*_Z^2+e))/d+b/d*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-I*b/d*dilog(1+I*c*x+(-c^2*x^2+1)^(1/2))+I*b*dilog
(I*c*x+(-c^2*x^2+1)^(1/2))/d+1/4*I*b*sum((_R1^2-1)/(_R1^2*e-2*c^2*d-e)*(I*arcsin(c*x)*ln((_R1-I*c*x-(-c^2*x^2+
1)^(1/2))/_R1)+dilog((_R1-I*c*x-(-c^2*x^2+1)^(1/2))/_R1)),_R1=RootOf(e*_Z^4+(-4*c^2*d-2*e)*_Z^2+e))*e/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(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(arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(e*x^3 + d*x),
x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asin}{\left (c x \right )}}{x \left (d + e x^{2}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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