3.42 \(\int e^{\csc ^{-1}(a x)} x \, dx\)

Optimal. Leaf size=87 \[ \frac{\left (\frac{8}{5}+\frac{4 i}{5}\right ) e^{(1+2 i) \csc ^{-1}(a x)} \, _2F_1\left (1-\frac{i}{2},2;2-\frac{i}{2};e^{2 i \csc ^{-1}(a x)}\right )}{a^2}-\frac{\left (\frac{16}{5}+\frac{8 i}{5}\right ) e^{(1+2 i) \csc ^{-1}(a x)} \, _2F_1\left (1-\frac{i}{2},3;2-\frac{i}{2};e^{2 i \csc ^{-1}(a x)}\right )}{a^2} \]

[Out]

((8/5 + (4*I)/5)*E^((1 + 2*I)*ArcCsc[a*x])*Hypergeometric2F1[1 - I/2, 2, 2 - I/2, E^((2*I)*ArcCsc[a*x])])/a^2
- ((16/5 + (8*I)/5)*E^((1 + 2*I)*ArcCsc[a*x])*Hypergeometric2F1[1 - I/2, 3, 2 - I/2, E^((2*I)*ArcCsc[a*x])])/a
^2

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Rubi [A]  time = 0.106169, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5267, 12, 4471, 2251} \[ \frac{\left (\frac{8}{5}+\frac{4 i}{5}\right ) e^{(1+2 i) \csc ^{-1}(a x)} \, _2F_1\left (1-\frac{i}{2},2;2-\frac{i}{2};e^{2 i \csc ^{-1}(a x)}\right )}{a^2}-\frac{\left (\frac{16}{5}+\frac{8 i}{5}\right ) e^{(1+2 i) \csc ^{-1}(a x)} \, _2F_1\left (1-\frac{i}{2},3;2-\frac{i}{2};e^{2 i \csc ^{-1}(a x)}\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8/5 + (4*I)/5)*E^((1 + 2*I)*ArcCsc[a*x])*Hypergeometric2F1[1 - I/2, 2, 2 - I/2, E^((2*I)*ArcCsc[a*x])])/a^2
- ((16/5 + (8*I)/5)*E^((1 + 2*I)*ArcCsc[a*x])*Hypergeometric2F1[1 - I/2, 3, 2 - I/2, E^((2*I)*ArcCsc[a*x])])/a
^2

Rule 5267

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]

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 e^{\csc ^{-1}(a x)} x \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{e^x \cot (x) \csc ^2(x)}{a} \, dx,x,\csc ^{-1}(a x)\right )}{a}\\ &=-\frac{\operatorname{Subst}\left (\int e^x \cot (x) \csc ^2(x) \, dx,x,\csc ^{-1}(a x)\right )}{a^2}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-\frac{8 i e^{(1+2 i) x}}{\left (-1+e^{2 i x}\right )^3}-\frac{4 i e^{(1+2 i) x}}{\left (-1+e^{2 i x}\right )^2}\right ) \, dx,x,\csc ^{-1}(a x)\right )}{a^2}\\ &=\frac{(4 i) \operatorname{Subst}\left (\int \frac{e^{(1+2 i) x}}{\left (-1+e^{2 i x}\right )^2} \, dx,x,\csc ^{-1}(a x)\right )}{a^2}+\frac{(8 i) \operatorname{Subst}\left (\int \frac{e^{(1+2 i) x}}{\left (-1+e^{2 i x}\right )^3} \, dx,x,\csc ^{-1}(a x)\right )}{a^2}\\ &=\frac{\left (\frac{8}{5}+\frac{4 i}{5}\right ) e^{(1+2 i) \csc ^{-1}(a x)} \, _2F_1\left (1-\frac{i}{2},2;2-\frac{i}{2};e^{2 i \csc ^{-1}(a x)}\right )}{a^2}-\frac{\left (\frac{16}{5}+\frac{8 i}{5}\right ) e^{(1+2 i) \csc ^{-1}(a x)} \, _2F_1\left (1-\frac{i}{2},3;2-\frac{i}{2};e^{2 i \csc ^{-1}(a x)}\right )}{a^2}\\ \end{align*}

Mathematica [A]  time = 0.267743, size = 101, normalized size = 1.16 \[ \frac{\left (\frac{1}{5}+\frac{i}{10}\right ) e^{\csc ^{-1}(a x)} \left ((2-i) a x \left (\sqrt{1-\frac{1}{a^2 x^2}}+a x\right )+(1+2 i) \, _2F_1\left (-\frac{i}{2},1;1-\frac{i}{2};e^{2 i \csc ^{-1}(a x)}\right )+e^{2 i \csc ^{-1}(a x)} \, _2F_1\left (1,1-\frac{i}{2};2-\frac{i}{2};e^{2 i \csc ^{-1}(a x)}\right )\right )}{a^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1/5 + I/10)*E^ArcCsc[a*x]*((2 - I)*a*x*(Sqrt[1 - 1/(a^2*x^2)] + a*x) + (1 + 2*I)*Hypergeometric2F1[-I/2, 1,
1 - I/2, E^((2*I)*ArcCsc[a*x])] + E^((2*I)*ArcCsc[a*x])*Hypergeometric2F1[1, 1 - I/2, 2 - I/2, E^((2*I)*ArcCsc
[a*x])]))/a^2

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Maple [F]  time = 0.188, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{{\rm arccsc} \left (ax\right )}}x\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*e^(arccsc(a*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*e^(arccsc(a*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*exp(acsc(a*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*e^(arccsc(a*x)), x)