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

Optimal. Leaf size=214 \[ -\frac{1}{2} i b \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{-\sqrt{1-c^2}+i c}\right )-\frac{1}{2} i b \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{\sqrt{1-c^2}+i c}\right )+a \log (x)+\frac{1}{2} b \sin ^{-1}\left (c+d x^2\right ) \log \left (1-\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{-\sqrt{1-c^2}+i c}\right )+\frac{1}{2} b \sin ^{-1}\left (c+d x^2\right ) \log \left (1-\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{\sqrt{1-c^2}+i c}\right )-\frac{1}{4} i b \sin ^{-1}\left (c+d x^2\right )^2 \]

[Out]

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

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Rubi [A]  time = 0.375603, antiderivative size = 214, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {6742, 4805, 4741, 4521, 2190, 2279, 2391} \[ -\frac{1}{2} i b \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{-\sqrt{1-c^2}+i c}\right )-\frac{1}{2} i b \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{\sqrt{1-c^2}+i c}\right )+a \log (x)+\frac{1}{2} b \sin ^{-1}\left (c+d x^2\right ) \log \left (1-\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{-\sqrt{1-c^2}+i c}\right )+\frac{1}{2} b \sin ^{-1}\left (c+d x^2\right ) \log \left (1-\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{\sqrt{1-c^2}+i c}\right )-\frac{1}{4} i b \sin ^{-1}\left (c+d x^2\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x^2])/x,x]

[Out]

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

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

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]

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 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 \sin ^{-1}\left (c+d x^2\right )}{x} \, dx &=\int \left (\frac{a}{x}+\frac{b \sin ^{-1}\left (c+d x^2\right )}{x}\right ) \, dx\\ &=a \log (x)+b \int \frac{\sin ^{-1}\left (c+d x^2\right )}{x} \, dx\\ &=a \log (x)+\frac{1}{2} b \operatorname{Subst}\left (\int \frac{\sin ^{-1}(c+d x)}{x} \, dx,x,x^2\right )\\ &=a \log (x)+\frac{b \operatorname{Subst}\left (\int \frac{\sin ^{-1}(x)}{-\frac{c}{d}+\frac{x}{d}} \, dx,x,c+d x^2\right )}{2 d}\\ &=a \log (x)+\frac{b \operatorname{Subst}\left (\int \frac{x \cos (x)}{-\frac{c}{d}+\frac{\sin (x)}{d}} \, dx,x,\sin ^{-1}\left (c+d x^2\right )\right )}{2 d}\\ &=-\frac{1}{4} i b \sin ^{-1}\left (c+d x^2\right )^2+a \log (x)+\frac{(i b) \operatorname{Subst}\left (\int \frac{e^{i x} x}{-\frac{i c}{d}-\frac{\sqrt{1-c^2}}{d}+\frac{e^{i x}}{d}} \, dx,x,\sin ^{-1}\left (c+d x^2\right )\right )}{2 d}+\frac{(i b) \operatorname{Subst}\left (\int \frac{e^{i x} x}{-\frac{i c}{d}+\frac{\sqrt{1-c^2}}{d}+\frac{e^{i x}}{d}} \, dx,x,\sin ^{-1}\left (c+d x^2\right )\right )}{2 d}\\ &=-\frac{1}{4} i b \sin ^{-1}\left (c+d x^2\right )^2+\frac{1}{2} b \sin ^{-1}\left (c+d x^2\right ) \log \left (1-\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{i c-\sqrt{1-c^2}}\right )+\frac{1}{2} b \sin ^{-1}\left (c+d x^2\right ) \log \left (1-\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{i c+\sqrt{1-c^2}}\right )+a \log (x)-\frac{1}{2} b \operatorname{Subst}\left (\int \log \left (1+\frac{e^{i x}}{\left (-\frac{i c}{d}-\frac{\sqrt{1-c^2}}{d}\right ) d}\right ) \, dx,x,\sin ^{-1}\left (c+d x^2\right )\right )-\frac{1}{2} b \operatorname{Subst}\left (\int \log \left (1+\frac{e^{i x}}{\left (-\frac{i c}{d}+\frac{\sqrt{1-c^2}}{d}\right ) d}\right ) \, dx,x,\sin ^{-1}\left (c+d x^2\right )\right )\\ &=-\frac{1}{4} i b \sin ^{-1}\left (c+d x^2\right )^2+\frac{1}{2} b \sin ^{-1}\left (c+d x^2\right ) \log \left (1-\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{i c-\sqrt{1-c^2}}\right )+\frac{1}{2} b \sin ^{-1}\left (c+d x^2\right ) \log \left (1-\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{i c+\sqrt{1-c^2}}\right )+a \log (x)+\frac{1}{2} (i b) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{x}{\left (-\frac{i c}{d}-\frac{\sqrt{1-c^2}}{d}\right ) d}\right )}{x} \, dx,x,e^{i \sin ^{-1}\left (c+d x^2\right )}\right )+\frac{1}{2} (i b) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{x}{\left (-\frac{i c}{d}+\frac{\sqrt{1-c^2}}{d}\right ) d}\right )}{x} \, dx,x,e^{i \sin ^{-1}\left (c+d x^2\right )}\right )\\ &=-\frac{1}{4} i b \sin ^{-1}\left (c+d x^2\right )^2+\frac{1}{2} b \sin ^{-1}\left (c+d x^2\right ) \log \left (1-\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{i c-\sqrt{1-c^2}}\right )+\frac{1}{2} b \sin ^{-1}\left (c+d x^2\right ) \log \left (1-\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{i c+\sqrt{1-c^2}}\right )+a \log (x)-\frac{1}{2} i b \text{Li}_2\left (\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{i c-\sqrt{1-c^2}}\right )-\frac{1}{2} i b \text{Li}_2\left (\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{i c+\sqrt{1-c^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0203567, size = 230, normalized size = 1.07 \[ -\frac{1}{2} i b \text{PolyLog}\left (2,-\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{\sqrt{1-c^2}-i c}\right )-\frac{1}{2} i b \text{PolyLog}\left (2,\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{\sqrt{1-c^2}+i c}\right )+a \log (x)+\frac{1}{2} b \sin ^{-1}\left (c+d x^2\right ) \log \left (1+\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{d \left (-\frac{\sqrt{1-c^2}}{d}-\frac{i c}{d}\right )}\right )+\frac{1}{2} b \sin ^{-1}\left (c+d x^2\right ) \log \left (1+\frac{e^{i \sin ^{-1}\left (c+d x^2\right )}}{d \left (\frac{\sqrt{1-c^2}}{d}-\frac{i c}{d}\right )}\right )-\frac{1}{4} i b \sin ^{-1}\left (c+d x^2\right )^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.135, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\arcsin \left ( d{x}^{2}+c \right ) }{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \arcsin \left (d x^{2} + c\right ) + a}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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