3.141 \(\int \frac{a+b \cosh ^{-1}(c x)}{x^5} \, dx\)

Optimal. Leaf size=72 \[ -\frac{a+b \cosh ^{-1}(c x)}{4 x^4}+\frac{b c^3 \sqrt{c x-1} \sqrt{c x+1}}{6 x}+\frac{b c \sqrt{c x-1} \sqrt{c x+1}}{12 x^3} \]

[Out]

(b*c*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(12*x^3) + (b*c^3*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(6*x) - (a + b*ArcCosh[c*x]
)/(4*x^4)

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Rubi [A]  time = 0.0322867, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5662, 103, 12, 95} \[ -\frac{a+b \cosh ^{-1}(c x)}{4 x^4}+\frac{b c^3 \sqrt{c x-1} \sqrt{c x+1}}{6 x}+\frac{b c \sqrt{c x-1} \sqrt{c x+1}}{12 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCosh[c*x])/x^5,x]

[Out]

(b*c*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(12*x^3) + (b*c^3*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(6*x) - (a + b*ArcCosh[c*x]
)/(4*x^4)

Rule 5662

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcC
osh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCosh[c*x])^(n - 1))/(Sqr
t[-1 + c*x]*Sqrt[1 + c*x]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 95

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] /; FreeQ[{a, b, c, d,
e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && EqQ[a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1), 0
] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{a+b \cosh ^{-1}(c x)}{x^5} \, dx &=-\frac{a+b \cosh ^{-1}(c x)}{4 x^4}+\frac{1}{4} (b c) \int \frac{1}{x^4 \sqrt{-1+c x} \sqrt{1+c x}} \, dx\\ &=\frac{b c \sqrt{-1+c x} \sqrt{1+c x}}{12 x^3}-\frac{a+b \cosh ^{-1}(c x)}{4 x^4}+\frac{1}{12} (b c) \int \frac{2 c^2}{x^2 \sqrt{-1+c x} \sqrt{1+c x}} \, dx\\ &=\frac{b c \sqrt{-1+c x} \sqrt{1+c x}}{12 x^3}-\frac{a+b \cosh ^{-1}(c x)}{4 x^4}+\frac{1}{6} \left (b c^3\right ) \int \frac{1}{x^2 \sqrt{-1+c x} \sqrt{1+c x}} \, dx\\ &=\frac{b c \sqrt{-1+c x} \sqrt{1+c x}}{12 x^3}+\frac{b c^3 \sqrt{-1+c x} \sqrt{1+c x}}{6 x}-\frac{a+b \cosh ^{-1}(c x)}{4 x^4}\\ \end{align*}

Mathematica [A]  time = 0.0289037, size = 50, normalized size = 0.69 \[ \frac{-3 a+b c x \sqrt{c x-1} \sqrt{c x+1} \left (2 c^2 x^2+1\right )-3 b \cosh ^{-1}(c x)}{12 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCosh[c*x])/x^5,x]

[Out]

(-3*a + b*c*x*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(1 + 2*c^2*x^2) - 3*b*ArcCosh[c*x])/(12*x^4)

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Maple [A]  time = 0.004, size = 62, normalized size = 0.9 \begin{align*}{c}^{4} \left ( -{\frac{a}{4\,{c}^{4}{x}^{4}}}+b \left ( -{\frac{{\rm arccosh} \left (cx\right )}{4\,{c}^{4}{x}^{4}}}+{\frac{2\,{c}^{2}{x}^{2}+1}{12\,{c}^{3}{x}^{3}}\sqrt{cx-1}\sqrt{cx+1}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccosh(c*x))/x^5,x)

[Out]

c^4*(-1/4*a/c^4/x^4+b*(-1/4/c^4/x^4*arccosh(c*x)+1/12*(c*x-1)^(1/2)*(c*x+1)^(1/2)*(2*c^2*x^2+1)/c^3/x^3))

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Maxima [A]  time = 1.64523, size = 77, normalized size = 1.07 \begin{align*} \frac{1}{12} \,{\left ({\left (\frac{2 \, \sqrt{c^{2} x^{2} - 1} c^{2}}{x} + \frac{\sqrt{c^{2} x^{2} - 1}}{x^{3}}\right )} c - \frac{3 \, \operatorname{arcosh}\left (c x\right )}{x^{4}}\right )} b - \frac{a}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(c*x))/x^5,x, algorithm="maxima")

[Out]

1/12*((2*sqrt(c^2*x^2 - 1)*c^2/x + sqrt(c^2*x^2 - 1)/x^3)*c - 3*arccosh(c*x)/x^4)*b - 1/4*a/x^4

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Fricas [A]  time = 2.5122, size = 139, normalized size = 1.93 \begin{align*} \frac{3 \, a x^{4} - 3 \, b \log \left (c x + \sqrt{c^{2} x^{2} - 1}\right ) +{\left (2 \, b c^{3} x^{3} + b c x\right )} \sqrt{c^{2} x^{2} - 1} - 3 \, a}{12 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(c*x))/x^5,x, algorithm="fricas")

[Out]

1/12*(3*a*x^4 - 3*b*log(c*x + sqrt(c^2*x^2 - 1)) + (2*b*c^3*x^3 + b*c*x)*sqrt(c^2*x^2 - 1) - 3*a)/x^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{acosh}{\left (c x \right )}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acosh(c*x))/x**5,x)

[Out]

Integral((a + b*acosh(c*x))/x**5, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arcosh}\left (c x\right ) + a}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(c*x))/x^5,x, algorithm="giac")

[Out]

integrate((b*arccosh(c*x) + a)/x^5, x)