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

Optimal. Leaf size=69 \[ -\frac{1}{4} i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}\left (c x^2\right )}\right )+a \log (x)-\frac{1}{4} i b \sin ^{-1}\left (c x^2\right )^2+\frac{1}{2} b \sin ^{-1}\left (c x^2\right ) \log \left (1-e^{2 i \sin ^{-1}\left (c x^2\right )}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0979946, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {6742, 4830, 3717, 2190, 2279, 2391} \[ -\frac{1}{4} i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}\left (c x^2\right )}\right )+a \log (x)-\frac{1}{4} i b \sin ^{-1}\left (c x^2\right )^2+\frac{1}{2} b \sin ^{-1}\left (c x^2\right ) \log \left (1-e^{2 i \sin ^{-1}\left (c x^2\right )}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x^2])/x,x]

[Out]

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

Rule 6742

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

Rule 4830

Int[ArcSin[(a_.)*(x_)^(p_)]^(n_.)/(x_), x_Symbol] :> Dist[1/p, Subst[Int[x^n*Cot[x], x], x, ArcSin[a*x^p]], x]
/; FreeQ[{a, p}, x] && IGtQ[n, 0]

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

Mathematica [A]  time = 0.0342929, size = 64, normalized size = 0.93 \[ a \log (x)+\frac{1}{2} b \left (\sin ^{-1}\left (c x^2\right ) \log \left (1-e^{2 i \sin ^{-1}\left (c x^2\right )}\right )-\frac{1}{2} i \left (\sin ^{-1}\left (c x^2\right )^2+\text{PolyLog}\left (2,e^{2 i \sin ^{-1}\left (c x^2\right )}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c*x^2])/x,x]

[Out]

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

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Maple [F]  time = 0.076, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\arcsin \left ( c{x}^{2} \right ) }{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arcsin(c*x^2) + 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 x^{2} \right )}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asin(c*x**2))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsin(c*x^2) + a)/x, x)