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

Optimal. Leaf size=597 \[ -\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d^2}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1+\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d^2}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d^2}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1+\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d^2}+\frac {\log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}+\frac {a+b \sin ^{-1}(c x)}{2 d \left (d+e x^2\right )}-\frac {b c \tan ^{-1}\left (\frac {x \sqrt {c^2 d+e}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{2 d^{3/2} \sqrt {c^2 d+e}}+\frac {i b \text {Li}_2\left (-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )}{2 d^2}+\frac {i b \text {Li}_2\left (\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )}{2 d^2}+\frac {i b \text {Li}_2\left (-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )}{2 d^2}+\frac {i b \text {Li}_2\left (\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )}{2 d^2}-\frac {i b \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.01, antiderivative size = 597, normalized size of antiderivative = 1.00, number of steps used = 28, number of rules used = 11, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {4733, 4625, 3717, 2190, 2279, 2391, 4729, 377, 205, 4741, 4521} \[ \frac {i b \text {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d^2}+\frac {i b \text {PolyLog}\left (2,\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d^2}+\frac {i b \text {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d^2}+\frac {i b \text {PolyLog}\left (2,\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d^2}-\frac {i b \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d^2}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1+\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d^2}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d^2}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1+\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 d^2}+\frac {\log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}+\frac {a+b \sin ^{-1}(c x)}{2 d \left (d+e x^2\right )}-\frac {b c \tan ^{-1}\left (\frac {x \sqrt {c^2 d+e}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{2 d^{3/2} \sqrt {c^2 d+e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a + b*ArcSin[c*x])/(2*d*(d + e*x^2)) - (b*c*ArcTan[(Sqrt[c^2*d + e]*x)/(Sqrt[d]*Sqrt[1 - c^2*x^2])])/(2*d^(3/
2)*Sqrt[c^2*d + e]) - ((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^2) - ((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
^2) - ((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^2) - ((
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^2) + ((a + b*Ar
cSin[c*x])*Log[1 - E^((2*I)*ArcSin[c*x])])/d^2 + ((I/2)*b*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-
d] - Sqrt[c^2*d + e]))])/d^2 + ((I/2)*b*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e]
)])/d^2 + ((I/2)*b*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e]))])/d^2 + ((I/2)*b
*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/d^2 - ((I/2)*b*PolyLog[2, E^((2*I)*
ArcSin[c*x])])/d^2

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 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 4729

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^(p + 1
)*(a + b*ArcSin[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 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 )^2} \, dx &=\int \left (\frac {a+b \sin ^{-1}(c x)}{d^2 x}-\frac {e x \left (a+b \sin ^{-1}(c x)\right )}{d \left (d+e x^2\right )^2}-\frac {e x \left (a+b \sin ^{-1}(c x)\right )}{d^2 \left (d+e x^2\right )}\right ) \, dx\\ &=\frac {\int \frac {a+b \sin ^{-1}(c x)}{x} \, dx}{d^2}-\frac {e \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{d+e x^2} \, dx}{d^2}-\frac {e \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx}{d}\\ &=\frac {a+b \sin ^{-1}(c x)}{2 d \left (d+e x^2\right )}+\frac {\operatorname {Subst}\left (\int (a+b x) \cot (x) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}-\frac {(b c) \int \frac {1}{\sqrt {1-c^2 x^2} \left (d+e x^2\right )} \, dx}{2 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^2}\\ &=\frac {a+b \sin ^{-1}(c x)}{2 d \left (d+e x^2\right )}-\frac {i \left (a+b \sin ^{-1}(c x)\right )^2}{2 b d^2}-\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^2}-\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{d-\left (-c^2 d-e\right ) x^2} \, dx,x,\frac {x}{\sqrt {1-c^2 x^2}}\right )}{2 d}+\frac {\sqrt {e} \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {-d}-\sqrt {e} x} \, dx}{2 d^2}-\frac {\sqrt {e} \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {-d}+\sqrt {e} x} \, dx}{2 d^2}\\ &=\frac {a+b \sin ^{-1}(c x)}{2 d \left (d+e x^2\right )}-\frac {i \left (a+b \sin ^{-1}(c x)\right )^2}{2 b d^2}-\frac {b c \tan ^{-1}\left (\frac {\sqrt {c^2 d+e} x}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{2 d^{3/2} \sqrt {c^2 d+e}}+\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac {b \operatorname {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}+\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^2}-\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^2}\\ &=\frac {a+b \sin ^{-1}(c x)}{2 d \left (d+e x^2\right )}-\frac {b c \tan ^{-1}\left (\frac {\sqrt {c^2 d+e} x}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{2 d^{3/2} \sqrt {c^2 d+e}}+\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )}{d^2}+\frac {(i b) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}+\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^2}+\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^2}-\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^2}-\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^2}\\ &=\frac {a+b \sin ^{-1}(c x)}{2 d \left (d+e x^2\right )}-\frac {b c \tan ^{-1}\left (\frac {\sqrt {c^2 d+e} x}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{2 d^{3/2} \sqrt {c^2 d+e}}-\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^2}-\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^2}-\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^2}-\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^2}+\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac {i b \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}+\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^2}+\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^2}+\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^2}+\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^2}\\ &=\frac {a+b \sin ^{-1}(c x)}{2 d \left (d+e x^2\right )}-\frac {b c \tan ^{-1}\left (\frac {\sqrt {c^2 d+e} x}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{2 d^{3/2} \sqrt {c^2 d+e}}-\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^2}-\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^2}-\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^2}-\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^2}+\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac {i b \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}-\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^2}-\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^2}-\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^2}-\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^2}\\ &=\frac {a+b \sin ^{-1}(c x)}{2 d \left (d+e x^2\right )}-\frac {b c \tan ^{-1}\left (\frac {\sqrt {c^2 d+e} x}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{2 d^{3/2} \sqrt {c^2 d+e}}-\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^2}-\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^2}-\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^2}-\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^2}+\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )}{d^2}+\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^2}+\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^2}+\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^2}+\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^2}-\frac {i b \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}\\ \end {align*}

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Mathematica [F]  time = 3.92, size = 0, normalized size = 0.00 \[ \int \frac {a+b \sin ^{-1}(c x)}{x \left (d+e x^2\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arcsin \left (c x\right ) + a}{e^{2} x^{5} + 2 \, d e x^{3} + d^{2} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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maple [C]  time = 0.42, size = 491, normalized size = 0.82 \[ \frac {a \,c^{2}}{2 d \left (c^{2} e \,x^{2}+c^{2} d \right )}-\frac {a \ln \left (c^{2} e \,x^{2}+c^{2} d \right )}{2 d^{2}}+\frac {a \ln \left (c x \right )}{d^{2}}+\frac {b \,c^{2} \arcsin \left (c x \right )}{2 d \left (c^{2} e \,x^{2}+c^{2} d \right )}+\frac {i b \sqrt {c^{2} d \left (c^{2} d +e \right )}\, \arctanh \left (\frac {2 \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2} e -4 c^{2} d -2 e}{4 \sqrt {d^{2} c^{4}+c^{2} e d}}\right )}{2 d^{2} \left (c^{2} d +e \right )}+\frac {i b \left (\munderset {\textit {\_R1} =\RootOf \left (e \,\textit {\_Z}^{4}+\left (-4 c^{2} d -2 e \right ) \textit {\_Z}^{2}+e \right )}{\sum }\frac {\left (\textit {\_R1}^{2} e -4 c^{2} d -e \right ) \left (i \arcsin \left (c x \right ) \ln \left (\frac {\textit {\_R1} -i c x -\sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )+\dilog \left (\frac {\textit {\_R1} -i c x -\sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )\right )}{\textit {\_R1}^{2} e -2 c^{2} d -e}\right )}{4 d^{2}}+\frac {b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}-\frac {i b \dilog \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}+\frac {i b \dilog \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}+\frac {i b \left (\munderset {\textit {\_R1} =\RootOf \left (e \,\textit {\_Z}^{4}+\left (-4 c^{2} d -2 e \right ) \textit {\_Z}^{2}+e \right )}{\sum }\frac {\left (\textit {\_R1}^{2}-1\right ) \left (i \arcsin \left (c x \right ) \ln \left (\frac {\textit {\_R1} -i c x -\sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )+\dilog \left (\frac {\textit {\_R1} -i c x -\sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )\right )}{\textit {\_R1}^{2} e -2 c^{2} d -e}\right ) e}{4 d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*c^2/d/(c^2*e*x^2+c^2*d)-1/2*a/d^2*ln(c^2*e*x^2+c^2*d)+a/d^2*ln(c*x)+1/2*b*c^2*arcsin(c*x)/d/(c^2*e*x^2+c
^2*d)+1/2*I*b*(c^2*d*(c^2*d+e))^(1/2)/d^2/(c^2*d+e)*arctanh(1/4*(2*(I*c*x+(-c^2*x^2+1)^(1/2))^2*e-4*c^2*d-2*e)
/(c^4*d^2+c^2*d*e)^(1/2))+1/4*I*b/d^2*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))
+b/d^2*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-I*b/d^2*dilog(1+I*c*x+(-c^2*x^2+1)^(1/2))+I*b/d^2*dilog(I*c*
x+(-c^2*x^2+1)^(1/2))+1/4*I*b/d^2*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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*(1/(d*e*x^2 + d^2) - log(e*x^2 + d)/d^2 + 2*log(x)/d^2) + b*integrate(arctan2(c*x, sqrt(c*x + 1)*sqrt(-c
*x + 1))/(e^2*x^5 + 2*d*e*x^3 + d^2*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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