3.51 \(\int \frac{\cos ^{-1}(a x^2)}{x} \, dx\)

Optimal. Leaf size=62 \[ -\frac{1}{4} i \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}\left (a x^2\right )}\right )-\frac{1}{4} i \cos ^{-1}\left (a x^2\right )^2+\frac{1}{2} \cos ^{-1}\left (a x^2\right ) \log \left (1+e^{2 i \cos ^{-1}\left (a x^2\right )}\right ) \]

[Out]

(-I/4)*ArcCos[a*x^2]^2 + (ArcCos[a*x^2]*Log[1 + E^((2*I)*ArcCos[a*x^2])])/2 - (I/4)*PolyLog[2, -E^((2*I)*ArcCo
s[a*x^2])]

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Rubi [A]  time = 0.0584569, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4831, 3719, 2190, 2279, 2391} \[ -\frac{1}{4} i \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}\left (a x^2\right )}\right )-\frac{1}{4} i \cos ^{-1}\left (a x^2\right )^2+\frac{1}{2} \cos ^{-1}\left (a x^2\right ) \log \left (1+e^{2 i \cos ^{-1}\left (a x^2\right )}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a*x^2]/x,x]

[Out]

(-I/4)*ArcCos[a*x^2]^2 + (ArcCos[a*x^2]*Log[1 + E^((2*I)*ArcCos[a*x^2])])/2 - (I/4)*PolyLog[2, -E^((2*I)*ArcCo
s[a*x^2])]

Rule 4831

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

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
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{\cos ^{-1}\left (a x^2\right )}{x} \, dx &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int x \tan (x) \, dx,x,\cos ^{-1}\left (a x^2\right )\right )\right )\\ &=-\frac{1}{4} i \cos ^{-1}\left (a x^2\right )^2+i \operatorname{Subst}\left (\int \frac{e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\cos ^{-1}\left (a x^2\right )\right )\\ &=-\frac{1}{4} i \cos ^{-1}\left (a x^2\right )^2+\frac{1}{2} \cos ^{-1}\left (a x^2\right ) \log \left (1+e^{2 i \cos ^{-1}\left (a x^2\right )}\right )-\frac{1}{2} \operatorname{Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\cos ^{-1}\left (a x^2\right )\right )\\ &=-\frac{1}{4} i \cos ^{-1}\left (a x^2\right )^2+\frac{1}{2} \cos ^{-1}\left (a x^2\right ) \log \left (1+e^{2 i \cos ^{-1}\left (a x^2\right )}\right )+\frac{1}{4} i \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \cos ^{-1}\left (a x^2\right )}\right )\\ &=-\frac{1}{4} i \cos ^{-1}\left (a x^2\right )^2+\frac{1}{2} \cos ^{-1}\left (a x^2\right ) \log \left (1+e^{2 i \cos ^{-1}\left (a x^2\right )}\right )-\frac{1}{4} i \text{Li}_2\left (-e^{2 i \cos ^{-1}\left (a x^2\right )}\right )\\ \end{align*}

Mathematica [A]  time = 0.0309198, size = 56, normalized size = 0.9 \[ -\frac{1}{4} i \left (\text{PolyLog}\left (2,-e^{2 i \cos ^{-1}\left (a x^2\right )}\right )+\cos ^{-1}\left (a x^2\right ) \left (\cos ^{-1}\left (a x^2\right )+2 i \log \left (1+e^{2 i \cos ^{-1}\left (a x^2\right )}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a*x^2]/x,x]

[Out]

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

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Maple [F]  time = 0.066, size = 0, normalized size = 0. \begin{align*} \int{\frac{\arccos \left ( a{x}^{2} \right ) }{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(a*x^2)/x,x)

[Out]

int(arccos(a*x^2)/x,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x^2)/x,x, algorithm="maxima")

[Out]

integrate(arccos(a*x^2)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x^2)/x,x, algorithm="fricas")

[Out]

integral(arccos(a*x^2)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(a*x**2)/x,x)

[Out]

Integral(acos(a*x**2)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x^2)/x,x, algorithm="giac")

[Out]

integrate(arccos(a*x^2)/x, x)