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

Optimal. Leaf size=541 \[ \frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1+\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1+\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {i b \text {Li}_2\left (-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {i b \text {Li}_2\left (\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {i b \text {Li}_2\left (-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {i b \text {Li}_2\left (\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )}{2 \sqrt {-d} \sqrt {e}} \]

[Out]

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)/e^(
1/2)-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)^(1/2
)/e^(1/2)+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)
^(1/2)/e^(1/2)-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)^(1/2)/e^(1/2)+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)^(1/2)/e^(1/2)-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)^(
1/2)/e^(1/2)+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)^(1/2
)/e^(1/2)-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)^(1/2)/e^
(1/2)

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Rubi [A]  time = 0.74, antiderivative size = 541, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4667, 4741, 4521, 2190, 2279, 2391} \[ \frac {i b \text {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {i b \text {PolyLog}\left (2,\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {i b \text {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {i b \text {PolyLog}\left (2,\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1+\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (1+\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*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*Sqrt[-d]*Sqrt[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*Sqrt[-d]*S
qrt[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*Sqrt[
-d]*Sqrt[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*
Sqrt[-d]*Sqrt[e]) + ((I/2)*b*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e]))])/(Sqr
t[-d]*Sqrt[e]) - ((I/2)*b*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e])])/(Sqrt[-d]*
Sqrt[e]) + ((I/2)*b*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e]))])/(Sqrt[-d]*Sqr
t[e]) - ((I/2)*b*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/(Sqrt[-d]*Sqrt[e])

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 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 4667

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a
+ b*ArcSin[c*x])^n, (d + e*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p]
&& (GtQ[p, 0] || IGtQ[n, 0])

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

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Mathematica [A]  time = 0.44, size = 490, normalized size = 0.91 \[ \frac {2 a \sqrt {-d} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )-i b \sqrt {d} \text {Li}_2\left (\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )+i b \sqrt {d} \text {Li}_2\left (\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {d c^2+e}-i c \sqrt {-d}}\right )+i b \sqrt {d} \text {Li}_2\left (-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )-i b \sqrt {d} \text {Li}_2\left (\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )-b \sqrt {d} \sin ^{-1}(c x) \log \left (1+\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )+b \sqrt {d} \sin ^{-1}(c x) \log \left (1+\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}-i c \sqrt {-d}}\right )+b \sqrt {d} \sin ^{-1}(c x) \log \left (1-\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )-b \sqrt {d} \sin ^{-1}(c x) \log \left (1+\frac {\sqrt {e} e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 \sqrt {-d^2} \sqrt {e}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*a*Sqrt[-d]*ArcTan[(Sqrt[e]*x)/Sqrt[d]] - b*Sqrt[d]*ArcSin[c*x]*Log[1 + (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqr
t[-d] - Sqrt[c^2*d + e])] + b*Sqrt[d]*ArcSin[c*x]*Log[1 + (Sqrt[e]*E^(I*ArcSin[c*x]))/((-I)*c*Sqrt[-d] + Sqrt[
c^2*d + e])] + b*Sqrt[d]*ArcSin[c*x]*Log[1 - (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])] - b
*Sqrt[d]*ArcSin[c*x]*Log[1 + (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])] - I*b*Sqrt[d]*PolyL
og[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e])] + I*b*Sqrt[d]*PolyLog[2, (Sqrt[e]*E^(I*Arc
Sin[c*x]))/((-I)*c*Sqrt[-d] + Sqrt[c^2*d + e])] + I*b*Sqrt[d]*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sq
rt[-d] + Sqrt[c^2*d + e]))] - I*b*Sqrt[d]*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d +
e])])/(2*Sqrt[-d^2]*Sqrt[e])

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arcsin \left (c x\right ) + a}{e x^{2} + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.09, size = 236, normalized size = 0.44 \[ \frac {a \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e}}+\frac {c b \left (\munderset {\textit {\_R1} =\RootOf \left (e \,\textit {\_Z}^{4}+\left (-4 c^{2} d -2 e \right ) \textit {\_Z}^{2}+e \right )}{\sum }\frac {i \arcsin \left (c x \right ) \ln \left (\frac {\textit {\_R1} -i c x -\sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )+\dilog \left (\frac {\textit {\_R1} -i c x -\sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )}{\textit {\_R1} \left (\textit {\_R1}^{2} e -2 c^{2} d -e \right )}\right )}{2}+\frac {c b \left (\munderset {\textit {\_R1} =\RootOf \left (e \,\textit {\_Z}^{4}+\left (-4 c^{2} d -2 e \right ) \textit {\_Z}^{2}+e \right )}{\sum }\frac {\textit {\_R1} \left (i \arcsin \left (c x \right ) \ln \left (\frac {\textit {\_R1} -i c x -\sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )+\dilog \left (\frac {\textit {\_R1} -i c x -\sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )\right )}{\textit {\_R1}^{2} e -2 c^{2} d -e}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))+1/2*c*b*sum(1/_R1/(_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))+1/2*c
*b*sum(_R1/(_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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))/(d + e*x^2),x)

[Out]

int((a + b*asin(c*x))/(d + e*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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