3.47 \(\int \frac{e^{\csc ^{-1}(a x)}}{x^4} \, dx\)

Optimal. Leaf size=84 \[ -\frac{1}{8} a^3 \sqrt{1-\frac{1}{a^2 x^2}} e^{\csc ^{-1}(a x)}-\frac{a^2 e^{\csc ^{-1}(a x)}}{8 x}+\frac{1}{40} a^3 e^{\csc ^{-1}(a x)} \cos \left (3 \csc ^{-1}(a x)\right )+\frac{3}{40} a^3 e^{\csc ^{-1}(a x)} \sin \left (3 \csc ^{-1}(a x)\right ) \]

[Out]

-(a^3*E^ArcCsc[a*x]*Sqrt[1 - 1/(a^2*x^2)])/8 - (a^2*E^ArcCsc[a*x])/(8*x) + (a^3*E^ArcCsc[a*x]*Cos[3*ArcCsc[a*x
]])/40 + (3*a^3*E^ArcCsc[a*x]*Sin[3*ArcCsc[a*x]])/40

________________________________________________________________________________________

Rubi [A]  time = 0.0671424, antiderivative size = 84, 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 = {5267, 12, 4469, 4433} \[ -\frac{1}{8} a^3 \sqrt{1-\frac{1}{a^2 x^2}} e^{\csc ^{-1}(a x)}-\frac{a^2 e^{\csc ^{-1}(a x)}}{8 x}+\frac{1}{40} a^3 e^{\csc ^{-1}(a x)} \cos \left (3 \csc ^{-1}(a x)\right )+\frac{3}{40} a^3 e^{\csc ^{-1}(a x)} \sin \left (3 \csc ^{-1}(a x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCsc[a*x]/x^4,x]

[Out]

-(a^3*E^ArcCsc[a*x]*Sqrt[1 - 1/(a^2*x^2)])/8 - (a^2*E^ArcCsc[a*x])/(8*x) + (a^3*E^ArcCsc[a*x]*Cos[3*ArcCsc[a*x
]])/40 + (3*a^3*E^ArcCsc[a*x]*Sin[3*ArcCsc[a*x]])/40

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 4469

Int[Cos[(f_.) + (g_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)]^(m_.), x_Symbol] :
> Int[ExpandTrigReduce[F^(c*(a + b*x)), Sin[d + e*x]^m*Cos[f + g*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e, f, g
}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 4433

Int[Cos[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*x))*C
os[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x] + Simp[(e*F^(c*(a + b*x))*Sin[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x]
/; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]

Rubi steps

\begin{align*} \int \frac{e^{\csc ^{-1}(a x)}}{x^4} \, dx &=-\frac{\operatorname{Subst}\left (\int a^4 e^x \cos (x) \sin ^2(x) \, dx,x,\csc ^{-1}(a x)\right )}{a}\\ &=-\left (a^3 \operatorname{Subst}\left (\int e^x \cos (x) \sin ^2(x) \, dx,x,\csc ^{-1}(a x)\right )\right )\\ &=-\left (a^3 \operatorname{Subst}\left (\int \left (\frac{1}{4} e^x \cos (x)-\frac{1}{4} e^x \cos (3 x)\right ) \, dx,x,\csc ^{-1}(a x)\right )\right )\\ &=-\left (\frac{1}{4} a^3 \operatorname{Subst}\left (\int e^x \cos (x) \, dx,x,\csc ^{-1}(a x)\right )\right )+\frac{1}{4} a^3 \operatorname{Subst}\left (\int e^x \cos (3 x) \, dx,x,\csc ^{-1}(a x)\right )\\ &=-\frac{1}{8} a^3 e^{\csc ^{-1}(a x)} \sqrt{1-\frac{1}{a^2 x^2}}-\frac{a^2 e^{\csc ^{-1}(a x)}}{8 x}+\frac{1}{40} a^3 e^{\csc ^{-1}(a x)} \cos \left (3 \csc ^{-1}(a x)\right )+\frac{3}{40} a^3 e^{\csc ^{-1}(a x)} \sin \left (3 \csc ^{-1}(a x)\right )\\ \end{align*}

Mathematica [A]  time = 0.132222, size = 54, normalized size = 0.64 \[ \frac{1}{40} a^3 e^{\csc ^{-1}(a x)} \left (-5 \sqrt{1-\frac{1}{a^2 x^2}}-\frac{5}{a x}+\cos \left (3 \csc ^{-1}(a x)\right )+3 \sin \left (3 \csc ^{-1}(a x)\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcCsc[a*x]/x^4,x]

[Out]

(a^3*E^ArcCsc[a*x]*(-5*Sqrt[1 - 1/(a^2*x^2)] - 5/(a*x) + Cos[3*ArcCsc[a*x]] + 3*Sin[3*ArcCsc[a*x]]))/40

________________________________________________________________________________________

Maple [F]  time = 0.181, size = 0, normalized size = 0. \begin{align*} \int{\frac{{{\rm e}^{{\rm arccsc} \left (ax\right )}}}{{x}^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(arccsc(a*x))/x^4,x)

[Out]

int(exp(arccsc(a*x))/x^4,x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (\operatorname{arccsc}\left (a x\right )\right )}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arccsc(a*x))/x^4,x, algorithm="maxima")

[Out]

integrate(e^(arccsc(a*x))/x^4, x)

________________________________________________________________________________________

Fricas [A]  time = 3.38539, size = 100, normalized size = 1.19 \begin{align*} \frac{{\left (a^{2} x^{2} -{\left (a^{2} x^{2} + 1\right )} \sqrt{a^{2} x^{2} - 1} - 3\right )} e^{\left (\operatorname{arccsc}\left (a x\right )\right )}}{10 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arccsc(a*x))/x^4,x, algorithm="fricas")

[Out]

1/10*(a^2*x^2 - (a^2*x^2 + 1)*sqrt(a^2*x^2 - 1) - 3)*e^(arccsc(a*x))/x^3

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\operatorname{acsc}{\left (a x \right )}}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(acsc(a*x))/x**4,x)

[Out]

Integral(exp(acsc(a*x))/x**4, x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (\operatorname{arccsc}\left (a x\right )\right )}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arccsc(a*x))/x^4,x, algorithm="giac")

[Out]

integrate(e^(arccsc(a*x))/x^4, x)