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]

1/4*(-a-b*arccosh(c*x))/x^4+1/12*b*c*(c*x-1)^(1/2)*(c*x+1)^(1/2)/x^3+1/6*b*c^3*(c*x-1)^(1/2)*(c*x+1)^(1/2)/x

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Rubi [A]  time = 0.03, antiderivative size = 72, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 0.51, size = 60, normalized size = 0.83 \[ \frac {3 \, a x^{4} - 3 \, b \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) + {\left (2 \, b c^{3} x^{3} + b c x\right )} \sqrt {c^{2} x^{2} - 1} - 3 \, a}{12 \, x^{4}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arcosh}\left (c x\right ) + a}{x^{5}}\,{d x} \]

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)

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maple [A]  time = 0.00, size = 62, normalized size = 0.86 \[ c^{4} \left (-\frac {a}{4 c^{4} x^{4}}+b \left (-\frac {\mathrm {arccosh}\left (c x \right )}{4 c^{4} x^{4}}+\frac {\sqrt {c x -1}\, \sqrt {c x +1}\, \left (2 c^{2} x^{2}+1\right )}{12 c^{3} x^{3}}\right )\right ) \]

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 = 0.68, size = 57, normalized size = 0.79 \[ \frac {1}{12} \, {\left ({\left (\frac {2 \, \sqrt {c^{2} x^{2} - 1} c^{2}}{x} + \frac {\sqrt {c^{2} x^{2} - 1}}{x^{3}}\right )} c - \frac {3 \, \operatorname {arcosh}\left (c x\right )}{x^{4}}\right )} b - \frac {a}{4 \, x^{4}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {acosh}\left (c\,x\right )}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(c*x))/x^5,x)

[Out]

int((a + b*acosh(c*x))/x^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {acosh}{\left (c x \right )}}{x^{5}}\, dx \]

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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