3.263 \(\int \frac{\sinh (\frac{\sqrt{1-a x}}{\sqrt{1+a x}})}{1-a^2 x^2} \, dx\)

Optimal. Leaf size=26 \[ -\frac{\text{Shi}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )}{a} \]

[Out]

-(SinhIntegral[Sqrt[1 - a*x]/Sqrt[1 + a*x]]/a)

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Rubi [A]  time = 0.0384661, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {6681, 3298} \[ -\frac{\text{Shi}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[Sqrt[1 - a*x]/Sqrt[1 + a*x]]/(1 - a^2*x^2),x]

[Out]

-(SinhIntegral[Sqrt[1 - a*x]/Sqrt[1 + a*x]]/a)

Rule 6681

Int[((a_.) + (b_.)*(F_)[((c_.)*Sqrt[(d_.) + (e_.)*(x_)])/Sqrt[(f_.) + (g_.)*(x_)]])^(n_.)/((A_.) + (C_.)*(x_)^
2), x_Symbol] :> Dist[(2*e*g)/(C*(e*f - d*g)), Subst[Int[(a + b*F[c*x])^n/x, x], x, Sqrt[d + e*x]/Sqrt[f + g*x
]], x] /; FreeQ[{a, b, c, d, e, f, g, A, C, F}, x] && EqQ[C*d*f - A*e*g, 0] && EqQ[e*f + d*g, 0] && IGtQ[n, 0]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rubi steps

\begin{align*} \int \frac{\sinh \left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )}{1-a^2 x^2} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\sinh (x)}{x} \, dx,x,\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )}{a}\\ &=-\frac{\text{Shi}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )}{a}\\ \end{align*}

Mathematica [A]  time = 0.0306777, size = 26, normalized size = 1. \[ -\frac{\text{Shi}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[Sqrt[1 - a*x]/Sqrt[1 + a*x]]/(1 - a^2*x^2),x]

[Out]

-(SinhIntegral[Sqrt[1 - a*x]/Sqrt[1 + a*x]]/a)

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Maple [F]  time = 0.024, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{-{a}^{2}{x}^{2}+1}\sinh \left ({\sqrt{-ax+1}{\frac{1}{\sqrt{ax+1}}}} \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh((-a*x+1)^(1/2)/(a*x+1)^(1/2))/(-a^2*x^2+1),x)

[Out]

int(sinh((-a*x+1)^(1/2)/(a*x+1)^(1/2))/(-a^2*x^2+1),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{\sinh \left (\frac{\sqrt{-a x + 1}}{\sqrt{a x + 1}}\right )}{a^{2} x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh((-a*x+1)^(1/2)/(a*x+1)^(1/2))/(-a^2*x^2+1),x, algorithm="maxima")

[Out]

-integrate(sinh(sqrt(-a*x + 1)/sqrt(a*x + 1))/(a^2*x^2 - 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sinh \left (\frac{\sqrt{-a x + 1}}{\sqrt{a x + 1}}\right )}{a^{2} x^{2} - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh((-a*x+1)^(1/2)/(a*x+1)^(1/2))/(-a^2*x^2+1),x, algorithm="fricas")

[Out]

integral(-sinh(sqrt(-a*x + 1)/sqrt(a*x + 1))/(a^2*x^2 - 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{\sinh{\left (\frac{\sqrt{- a x + 1}}{\sqrt{a x + 1}} \right )}}{a^{2} x^{2} - 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh((-a*x+1)**(1/2)/(a*x+1)**(1/2))/(-a**2*x**2+1),x)

[Out]

-Integral(sinh(sqrt(-a*x + 1)/sqrt(a*x + 1))/(a**2*x**2 - 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{\sinh \left (\frac{\sqrt{-a x + 1}}{\sqrt{a x + 1}}\right )}{a^{2} x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh((-a*x+1)^(1/2)/(a*x+1)^(1/2))/(-a^2*x^2+1),x, algorithm="giac")

[Out]

integrate(-sinh(sqrt(-a*x + 1)/sqrt(a*x + 1))/(a^2*x^2 - 1), x)