3.263 \(\int \frac{(c-a^2 c x^2)^2}{\cosh ^{-1}(a x)} \, dx\)

Optimal. Leaf size=50 \[ \frac{5 c^2 \text{Shi}\left (\cosh ^{-1}(a x)\right )}{8 a}-\frac{5 c^2 \text{Shi}\left (3 \cosh ^{-1}(a x)\right )}{16 a}+\frac{c^2 \text{Shi}\left (5 \cosh ^{-1}(a x)\right )}{16 a} \]

[Out]

(5*c^2*SinhIntegral[ArcCosh[a*x]])/(8*a) - (5*c^2*SinhIntegral[3*ArcCosh[a*x]])/(16*a) + (c^2*SinhIntegral[5*A
rcCosh[a*x]])/(16*a)

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Rubi [A]  time = 0.113328, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {5700, 3312, 3298} \[ \frac{5 c^2 \text{Shi}\left (\cosh ^{-1}(a x)\right )}{8 a}-\frac{5 c^2 \text{Shi}\left (3 \cosh ^{-1}(a x)\right )}{16 a}+\frac{c^2 \text{Shi}\left (5 \cosh ^{-1}(a x)\right )}{16 a} \]

Antiderivative was successfully verified.

[In]

Int[(c - a^2*c*x^2)^2/ArcCosh[a*x],x]

[Out]

(5*c^2*SinhIntegral[ArcCosh[a*x]])/(8*a) - (5*c^2*SinhIntegral[3*ArcCosh[a*x]])/(16*a) + (c^2*SinhIntegral[5*A
rcCosh[a*x]])/(16*a)

Rule 5700

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(-d)^p/c, Subst[I
nt[(a + b*x)^n*Sinh[x]^(2*p + 1), x], x, ArcCosh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0
] && IGtQ[p, 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{\left (c-a^2 c x^2\right )^2}{\cosh ^{-1}(a x)} \, dx &=\frac{c^2 \operatorname{Subst}\left (\int \frac{\sinh ^5(x)}{x} \, dx,x,\cosh ^{-1}(a x)\right )}{a}\\ &=-\frac{\left (i c^2\right ) \operatorname{Subst}\left (\int \left (\frac{5 i \sinh (x)}{8 x}-\frac{5 i \sinh (3 x)}{16 x}+\frac{i \sinh (5 x)}{16 x}\right ) \, dx,x,\cosh ^{-1}(a x)\right )}{a}\\ &=\frac{c^2 \operatorname{Subst}\left (\int \frac{\sinh (5 x)}{x} \, dx,x,\cosh ^{-1}(a x)\right )}{16 a}-\frac{\left (5 c^2\right ) \operatorname{Subst}\left (\int \frac{\sinh (3 x)}{x} \, dx,x,\cosh ^{-1}(a x)\right )}{16 a}+\frac{\left (5 c^2\right ) \operatorname{Subst}\left (\int \frac{\sinh (x)}{x} \, dx,x,\cosh ^{-1}(a x)\right )}{8 a}\\ &=\frac{5 c^2 \text{Shi}\left (\cosh ^{-1}(a x)\right )}{8 a}-\frac{5 c^2 \text{Shi}\left (3 \cosh ^{-1}(a x)\right )}{16 a}+\frac{c^2 \text{Shi}\left (5 \cosh ^{-1}(a x)\right )}{16 a}\\ \end{align*}

Mathematica [A]  time = 0.164153, size = 34, normalized size = 0.68 \[ \frac{c^2 \left (10 \text{Shi}\left (\cosh ^{-1}(a x)\right )-5 \text{Shi}\left (3 \cosh ^{-1}(a x)\right )+\text{Shi}\left (5 \cosh ^{-1}(a x)\right )\right )}{16 a} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - a^2*c*x^2)^2/ArcCosh[a*x],x]

[Out]

(c^2*(10*SinhIntegral[ArcCosh[a*x]] - 5*SinhIntegral[3*ArcCosh[a*x]] + SinhIntegral[5*ArcCosh[a*x]]))/(16*a)

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Maple [A]  time = 0.032, size = 33, normalized size = 0.7 \begin{align*}{\frac{{c}^{2} \left ( 10\,{\it Shi} \left ({\rm arccosh} \left (ax\right ) \right ) -5\,{\it Shi} \left ( 3\,{\rm arccosh} \left (ax\right ) \right ) +{\it Shi} \left ( 5\,{\rm arccosh} \left (ax\right ) \right ) \right ) }{16\,a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*c*x^2+c)^2/arccosh(a*x),x)

[Out]

1/16/a*c^2*(10*Shi(arccosh(a*x))-5*Shi(3*arccosh(a*x))+Shi(5*arccosh(a*x)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a^2*c*x^2 - c)^2/arccosh(a*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{a^{4} c^{2} x^{4} - 2 \, a^{2} c^{2} x^{2} + c^{2}}{\operatorname{arcosh}\left (a x\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^4*c^2*x^4 - 2*a^2*c^2*x^2 + c^2)/arccosh(a*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} c^{2} \left (\int - \frac{2 a^{2} x^{2}}{\operatorname{acosh}{\left (a x \right )}}\, dx + \int \frac{a^{4} x^{4}}{\operatorname{acosh}{\left (a x \right )}}\, dx + \int \frac{1}{\operatorname{acosh}{\left (a x \right )}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*c*x**2+c)**2/acosh(a*x),x)

[Out]

c**2*(Integral(-2*a**2*x**2/acosh(a*x), x) + Integral(a**4*x**4/acosh(a*x), x) + Integral(1/acosh(a*x), x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a^2*c*x^2 - c)^2/arccosh(a*x), x)