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]

5/8*c^2*Shi(arccosh(a*x))/a-5/16*c^2*Shi(3*arccosh(a*x))/a+1/16*c^2*Shi(5*arccosh(a*x))/a

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Rubi [A]  time = 0.11, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.17, 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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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{4} c^{2} x^{4} - 2 \, a^{2} c^{2} x^{2} + c^{2}}{\operatorname {arcosh}\left (a x\right )}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a^{2} c x^{2} - c\right )}^{2}}{\operatorname {arcosh}\left (a x\right )}\,{d x} \]

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)

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maple [A]  time = 0.06, size = 33, normalized size = 0.66 \[ \frac {c^{2} \left (10 \Shi \left (\mathrm {arccosh}\left (a x \right )\right )-5 \Shi \left (3 \,\mathrm {arccosh}\left (a x \right )\right )+\Shi \left (5 \,\mathrm {arccosh}\left (a x \right )\right )\right )}{16 a} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a^{2} c x^{2} - c\right )}^{2}}{\operatorname {arcosh}\left (a x\right )}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (c-a^2\,c\,x^2\right )}^2}{\mathrm {acosh}\left (a\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - a^2*c*x^2)^2/acosh(a*x),x)

[Out]

int((c - a^2*c*x^2)^2/acosh(a*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ c^{2} \left (\int \left (- \frac {2 a^{2} x^{2}}{\operatorname {acosh}{\left (a x \right )}}\right )\, dx + \int \frac {a^{4} x^{4}}{\operatorname {acosh}{\left (a x \right )}}\, dx + \int \frac {1}{\operatorname {acosh}{\left (a x \right )}}\, dx\right ) \]

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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