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

Optimal. Leaf size=87 \[ (1+i) a e^{(1+i) \cos ^{-1}(a x)} \text{Hypergeometric2F1}\left (\frac{1}{2}-\frac{i}{2},1,\frac{3}{2}-\frac{i}{2},-e^{2 i \cos ^{-1}(a x)}\right )-(2+2 i) a e^{(1+i) \cos ^{-1}(a x)} \text{Hypergeometric2F1}\left (\frac{1}{2}-\frac{i}{2},2,\frac{3}{2}-\frac{i}{2},-e^{2 i \cos ^{-1}(a x)}\right ) \]

[Out]

(1 + I)*a*E^((1 + I)*ArcCos[a*x])*Hypergeometric2F1[1/2 - I/2, 1, 3/2 - I/2, -E^((2*I)*ArcCos[a*x])] - (2 + 2*
I)*a*E^((1 + I)*ArcCos[a*x])*Hypergeometric2F1[1/2 - I/2, 2, 3/2 - I/2, -E^((2*I)*ArcCos[a*x])]

________________________________________________________________________________________

Rubi [A]  time = 0.105852, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4837, 12, 4471, 2251} \[ (1+i) a e^{(1+i) \cos ^{-1}(a x)} \, _2F_1\left (\frac{1}{2}-\frac{i}{2},1;\frac{3}{2}-\frac{i}{2};-e^{2 i \cos ^{-1}(a x)}\right )-(2+2 i) a e^{(1+i) \cos ^{-1}(a x)} \, _2F_1\left (\frac{1}{2}-\frac{i}{2},2;\frac{3}{2}-\frac{i}{2};-e^{2 i \cos ^{-1}(a x)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + I)*a*E^((1 + I)*ArcCos[a*x])*Hypergeometric2F1[1/2 - I/2, 1, 3/2 - I/2, -E^((2*I)*ArcCos[a*x])] - (2 + 2*
I)*a*E^((1 + I)*ArcCos[a*x])*Hypergeometric2F1[1/2 - I/2, 2, 3/2 - I/2, -E^((2*I)*ArcCos[a*x])]

Rule 4837

Int[(u_.)*(f_)^(ArcCos[(a_.) + (b_.)*(x_)]^(n_.)*(c_.)), x_Symbol] :> -Dist[b^(-1), Subst[Int[(u /. x -> -(a/b
) + Cos[x]/b)*f^(c*x^n)*Sin[x], x], x, ArcCos[a + b*x]], x] /; FreeQ[{a, b, c, f}, x] && IGtQ[n, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 4471

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*(G_)[(d_.) + (e_.)*(x_)]^(m_.)*(H_)[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol]
:> Int[ExpandTrigToExp[F^(c*(a + b*x)), G[d + e*x]^m*H[d + e*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e}, x] &&
IGtQ[m, 0] && IGtQ[n, 0] && TrigQ[G] && TrigQ[H]

Rule 2251

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Simp
[(a^p*G^(h*(f + g*x))*Hypergeometric2F1[-p, (g*h*Log[G])/(d*e*Log[F]), (g*h*Log[G])/(d*e*Log[F]) + 1, Simplify
[-((b*F^(e*(c + d*x)))/a)]])/(g*h*Log[G]), x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x] && (ILtQ[p, 0] ||
GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{e^{\cos ^{-1}(a x)}}{x^2} \, dx &=-\frac{\operatorname{Subst}\left (\int a^2 e^x \sec (x) \tan (x) \, dx,x,\cos ^{-1}(a x)\right )}{a}\\ &=-\left (a \operatorname{Subst}\left (\int e^x \sec (x) \tan (x) \, dx,x,\cos ^{-1}(a x)\right )\right )\\ &=-\left (a \operatorname{Subst}\left (\int \left (\frac{4 i e^{(1+i) x}}{\left (1+e^{2 i x}\right )^2}-\frac{2 i e^{(1+i) x}}{1+e^{2 i x}}\right ) \, dx,x,\cos ^{-1}(a x)\right )\right )\\ &=(2 i a) \operatorname{Subst}\left (\int \frac{e^{(1+i) x}}{1+e^{2 i x}} \, dx,x,\cos ^{-1}(a x)\right )-(4 i a) \operatorname{Subst}\left (\int \frac{e^{(1+i) x}}{\left (1+e^{2 i x}\right )^2} \, dx,x,\cos ^{-1}(a x)\right )\\ &=(1+i) a e^{(1+i) \cos ^{-1}(a x)} \, _2F_1\left (\frac{1}{2}-\frac{i}{2},1;\frac{3}{2}-\frac{i}{2};-e^{2 i \cos ^{-1}(a x)}\right )-(2+2 i) a e^{(1+i) \cos ^{-1}(a x)} \, _2F_1\left (\frac{1}{2}-\frac{i}{2},2;\frac{3}{2}-\frac{i}{2};-e^{2 i \cos ^{-1}(a x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.0618076, size = 55, normalized size = 0.63 \[ -\frac{e^{\cos ^{-1}(a x)}}{x}+(1-i) a e^{(1+i) \cos ^{-1}(a x)} \text{Hypergeometric2F1}\left (\frac{1}{2}-\frac{i}{2},1,\frac{3}{2}-\frac{i}{2},-e^{2 i \cos ^{-1}(a x)}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(E^ArcCos[a*x]/x) + (1 - I)*a*E^((1 + I)*ArcCos[a*x])*Hypergeometric2F1[1/2 - I/2, 1, 3/2 - I/2, -E^((2*I)*Ar
cCos[a*x])]

________________________________________________________________________________________

Maple [F]  time = 0.007, size = 0, normalized size = 0. \begin{align*} \int{\frac{{{\rm e}^{\arccos \left ( ax \right ) }}}{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(e^(arccos(a*x))/x^2, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(e^(arccos(a*x))/x^2, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(acos(a*x))/x**2, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(e^(arccos(a*x))/x^2, x)