3.140 \(\int \frac{a+b \cosh ^{-1}(c x)}{x^4} \, dx\)

Optimal. Leaf size=71 \[ -\frac{a+b \cosh ^{-1}(c x)}{3 x^3}+\frac{1}{6} b c^3 \tan ^{-1}\left (\sqrt{c x-1} \sqrt{c x+1}\right )+\frac{b c \sqrt{c x-1} \sqrt{c x+1}}{6 x^2} \]

[Out]

(b*c*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(6*x^2) - (a + b*ArcCosh[c*x])/(3*x^3) + (b*c^3*ArcTan[Sqrt[-1 + c*x]*Sqrt[
1 + c*x]])/6

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Rubi [A]  time = 0.0334076, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {5662, 103, 12, 92, 205} \[ -\frac{a+b \cosh ^{-1}(c x)}{3 x^3}+\frac{1}{6} b c^3 \tan ^{-1}\left (\sqrt{c x-1} \sqrt{c x+1}\right )+\frac{b c \sqrt{c x-1} \sqrt{c x+1}}{6 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCosh[c*x])/x^4,x]

[Out]

(b*c*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(6*x^2) - (a + b*ArcCosh[c*x])/(3*x^3) + (b*c^3*ArcTan[Sqrt[-1 + c*x]*Sqrt[
1 + c*x]])/6

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 92

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.100809, size = 101, normalized size = 1.42 \[ -\frac{a}{3 x^3}+\frac{b c^3 \sqrt{c^2 x^2-1} \tan ^{-1}\left (\sqrt{c^2 x^2-1}\right )}{6 \sqrt{c x-1} \sqrt{c x+1}}+\frac{b c \sqrt{c x-1} \sqrt{c x+1}}{6 x^2}-\frac{b \cosh ^{-1}(c x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCosh[c*x])/x^4,x]

[Out]

-a/(3*x^3) + (b*c*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(6*x^2) - (b*ArcCosh[c*x])/(3*x^3) + (b*c^3*Sqrt[-1 + c^2*x^2]
*ArcTan[Sqrt[-1 + c^2*x^2]])/(6*Sqrt[-1 + c*x]*Sqrt[1 + c*x])

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Maple [A]  time = 0.006, size = 82, normalized size = 1.2 \begin{align*} -{\frac{a}{3\,{x}^{3}}}-{\frac{b{\rm arccosh} \left (cx\right )}{3\,{x}^{3}}}-{\frac{{c}^{3}b}{6}\sqrt{cx-1}\sqrt{cx+1}\arctan \left ({\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}}}+{\frac{bc}{6\,{x}^{2}}\sqrt{cx-1}\sqrt{cx+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccosh(c*x))/x^4,x)

[Out]

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

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Maxima [A]  time = 1.69955, size = 73, normalized size = 1.03 \begin{align*} -\frac{1}{6} \,{\left ({\left (c^{2} \arcsin \left (\frac{1}{\sqrt{c^{2}}{\left | x \right |}}\right ) - \frac{\sqrt{c^{2} x^{2} - 1}}{x^{2}}\right )} c + \frac{2 \, \operatorname{arcosh}\left (c x\right )}{x^{3}}\right )} b - \frac{a}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(c*x))/x^4,x, algorithm="maxima")

[Out]

-1/6*((c^2*arcsin(1/(sqrt(c^2)*abs(x))) - sqrt(c^2*x^2 - 1)/x^2)*c + 2*arccosh(c*x)/x^3)*b - 1/3*a/x^3

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Fricas [A]  time = 2.64239, size = 234, normalized size = 3.3 \begin{align*} \frac{2 \, b c^{3} x^{3} \arctan \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) + 2 \, b x^{3} \log \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) + \sqrt{c^{2} x^{2} - 1} b c x + 2 \,{\left (b x^{3} - b\right )} \log \left (c x + \sqrt{c^{2} x^{2} - 1}\right ) - 2 \, a}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(c*x))/x^4,x, algorithm="fricas")

[Out]

1/6*(2*b*c^3*x^3*arctan(-c*x + sqrt(c^2*x^2 - 1)) + 2*b*x^3*log(-c*x + sqrt(c^2*x^2 - 1)) + sqrt(c^2*x^2 - 1)*
b*c*x + 2*(b*x^3 - b)*log(c*x + sqrt(c^2*x^2 - 1)) - 2*a)/x^3

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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^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acosh(c*x))/x**4,x)

[Out]

Integral((a + b*acosh(c*x))/x**4, 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^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(c*x))/x^4,x, algorithm="giac")

[Out]

integrate((b*arccosh(c*x) + a)/x^4, x)