3.261 \(\int \frac{\sinh ^3(\frac{\sqrt{1-a x}}{\sqrt{1+a x}})}{1-a^2 x^2} \, dx\)

Optimal. Leaf size=58 \[ \frac{3 \text{Shi}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )}{4 a}-\frac{\text{Shi}\left (\frac{3 \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{4 a} \]

[Out]

(3*SinhIntegral[Sqrt[1 - a*x]/Sqrt[1 + a*x]])/(4*a) - SinhIntegral[(3*Sqrt[1 - a*x])/Sqrt[1 + a*x]]/(4*a)

________________________________________________________________________________________

Rubi [A]  time = 0.111941, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {6681, 3312, 3298} \[ \frac{3 \text{Shi}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )}{4 a}-\frac{\text{Shi}\left (\frac{3 \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{4 a} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[Sqrt[1 - a*x]/Sqrt[1 + a*x]]^3/(1 - a^2*x^2),x]

[Out]

(3*SinhIntegral[Sqrt[1 - a*x]/Sqrt[1 + a*x]])/(4*a) - SinhIntegral[(3*Sqrt[1 - a*x])/Sqrt[1 + a*x]]/(4*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 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

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 ^3\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )}{1-a^2 x^2} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\sinh ^3(x)}{x} \, dx,x,\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )}{a}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (\frac{3 i \sinh (x)}{4 x}-\frac{i \sinh (3 x)}{4 x}\right ) \, dx,x,\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )}{a}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\sinh (3 x)}{x} \, dx,x,\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )}{4 a}+\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (x)}{x} \, dx,x,\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )}{4 a}\\ &=\frac{3 \text{Shi}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right )}{4 a}-\frac{\text{Shi}\left (\frac{3 \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{4 a}\\ \end{align*}

Mathematica [A]  time = 0.0787227, size = 55, normalized size = 0.95 \[ \frac{3 \text{Shi}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )-\text{Shi}\left (\frac{3 \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{4 a} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[Sqrt[1 - a*x]/Sqrt[1 + a*x]]^3/(1 - a^2*x^2),x]

[Out]

(3*SinhIntegral[Sqrt[1 - a*x]/Sqrt[1 + a*x]] - SinhIntegral[(3*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/(4*a)

________________________________________________________________________________________

Maple [F]  time = 0.109, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{-{a}^{2}{x}^{2}+1} \left ( \sinh \left ({\sqrt{-ax+1}{\frac{1}{\sqrt{ax+1}}}} \right ) \right ) ^{3}}\, 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))^3/(-a^2*x^2+1),x)

[Out]

int(sinh((-a*x+1)^(1/2)/(a*x+1)^(1/2))^3/(-a^2*x^2+1),x)

________________________________________________________________________________________

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 )^{3}}{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))^3/(-a^2*x^2+1),x, algorithm="maxima")

[Out]

-integrate(sinh(sqrt(-a*x + 1)/sqrt(a*x + 1))^3/(a^2*x^2 - 1), x)

________________________________________________________________________________________

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 )^{3}}{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))^3/(-a^2*x^2+1),x, algorithm="fricas")

[Out]

integral(-sinh(sqrt(-a*x + 1)/sqrt(a*x + 1))^3/(a^2*x^2 - 1), x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**3/(-a**2*x**2+1),x)

[Out]

Timed out

________________________________________________________________________________________

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 )^{3}}{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))^3/(-a^2*x^2+1),x, algorithm="giac")

[Out]

integrate(-sinh(sqrt(-a*x + 1)/sqrt(a*x + 1))^3/(a^2*x^2 - 1), x)