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

Optimal. Leaf size=87 \[ (1+i) a e^{(1+i) \cos ^{-1}(a x)} \, _2F_1\left (\frac {1}{2}-\frac {i}{2},1;\frac {3}{2}-\frac {i}{2};-e^{2 i \cos ^{-1}(a x)}\right )-(2+2 i) a e^{(1+i) \cos ^{-1}(a x)} \, _2F_1\left (\frac {1}{2}-\frac {i}{2},2;\frac {3}{2}-\frac {i}{2};-e^{2 i \cos ^{-1}(a x)}\right ) \]

[Out]

(1+I)*a*exp((1+I)*arccos(a*x))*hypergeom([1, 1/2-1/2*I],[3/2-1/2*I],-(a*x+I*(-a^2*x^2+1)^(1/2))^2)-(2+2*I)*a*e
xp((1+I)*arccos(a*x))*hypergeom([2, 1/2-1/2*I],[3/2-1/2*I],-(a*x+I*(-a^2*x^2+1)^(1/2))^2)

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Rubi [A]  time = 0.11, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 12

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

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])

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 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]

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*}

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Mathematica [A]  time = 0.07, size = 55, normalized size = 0.63 \[ -\frac {e^{\cos ^{-1}(a x)}}{x}+(1-i) a e^{(1+i) \cos ^{-1}(a x)} \, _2F_1\left (\frac {1}{2}-\frac {i}{2},1;\frac {3}{2}-\frac {i}{2};-e^{2 i \cos ^{-1}(a x)}\right ) \]

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])]

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e^{\left (\arccos \left (a x\right )\right )}}{x^{2}}, x\right ) \]

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)

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

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)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{\arccos \left (a x \right )}}{x^{2}}\, dx \]

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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{\left (\arccos \left (a x\right )\right )}}{x^{2}}\,{d x} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {e}}^{\mathrm {acos}\left (a\,x\right )}}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(exp(acos(a*x))/x^2, x)

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

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)

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