3.1.43 \(\int e^{\csc ^{-1}(a x)} \, dx\) [43]

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

[Out]

(-1+I)*exp((1+I)*arccsc(a*x))*hypergeom([1, 1/2-1/2*I],[3/2-1/2*I],(I/a/x+(1-1/a^2/x^2)^(1/2))^2)/a+(2-2*I)*ex
p((1+I)*arccsc(a*x))*hypergeom([2, 1/2-1/2*I],[3/2-1/2*I],(I/a/x+(1-1/a^2/x^2)^(1/2))^2)/a

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Rubi [A]
time = 0.07, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5375, 4559, 2283} \begin {gather*} \frac {(2-2 i) e^{(1+i) \csc ^{-1}(a x)} \, _2F_1\left (\frac {1}{2}-\frac {i}{2},2;\frac {3}{2}-\frac {i}{2};e^{2 i \csc ^{-1}(a x)}\right )}{a}-\frac {(1-i) e^{(1+i) \csc ^{-1}(a x)} \, _2F_1\left (\frac {1}{2}-\frac {i}{2},1;\frac {3}{2}-\frac {i}{2};e^{2 i \csc ^{-1}(a x)}\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^ArcCsc[a*x],x]

[Out]

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

Rule 2283

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

Rule 4559

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 5375

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

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 54, normalized size = 0.62 \begin {gather*} \frac {e^{\csc ^{-1}(a x)} \left (a x+(1+i) e^{i \csc ^{-1}(a x)} \, _2F_1\left (\frac {1}{2}-\frac {i}{2},1;\frac {3}{2}-\frac {i}{2};e^{2 i \csc ^{-1}(a x)}\right )\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^ArcCsc[a*x],x]

[Out]

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

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int {\mathrm e}^{\mathrm {arccsc}\left (a x \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(arccsc(a*x)),x)

[Out]

int(exp(arccsc(a*x)),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arccsc(a*x)),x, algorithm="maxima")

[Out]

integrate(e^(arccsc(a*x)), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arccsc(a*x)),x, algorithm="fricas")

[Out]

integral(e^(arccsc(a*x)), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int e^{\operatorname {acsc}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(acsc(a*x)),x)

[Out]

Integral(exp(acsc(a*x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arccsc(a*x)),x, algorithm="giac")

[Out]

integrate(e^(arccsc(a*x)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {e}}^{\mathrm {asin}\left (\frac {1}{a\,x}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(asin(1/(a*x))),x)

[Out]

int(exp(asin(1/(a*x))), x)

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