3.89 \(\int \frac {\cosh ^{-1}(a+b x)}{x^3} \, dx\)

Optimal. Leaf size=106 \[ -\frac {a b^2 \tan ^{-1}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {a+b x-1}}\right )}{\left (1-a^2\right )^{3/2}}+\frac {b \sqrt {a+b x-1} \sqrt {a+b x+1}}{2 \left (1-a^2\right ) x}-\frac {\cosh ^{-1}(a+b x)}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5866, 5802, 96, 93, 205} \[ -\frac {a b^2 \tan ^{-1}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {a+b x-1}}\right )}{\left (1-a^2\right )^{3/2}}+\frac {b \sqrt {a+b x-1} \sqrt {a+b x+1}}{2 \left (1-a^2\right ) x}-\frac {\cosh ^{-1}(a+b x)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcCosh[a + b*x]/x^3,x]

[Out]

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 96

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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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]

Rule 5802

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

Rule 5866

Int[((a_.) + ArcCosh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcCosh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rubi steps

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

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Mathematica [C]  time = 0.29, size = 136, normalized size = 1.28 \[ \frac {-\cosh ^{-1}(a+b x)+\frac {b x \left (-\sqrt {a+b x-1} \sqrt {a+b x+1}+\frac {i a b x \log \left (\frac {4 i \sqrt {1-a^2} \left (-i \sqrt {1-a^2} \sqrt {a+b x-1} \sqrt {a+b x+1}+a^2+a b x-1\right )}{a b^2 x}\right )}{\sqrt {1-a^2}}\right )}{a^2-1}}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCosh[a + b*x]/x^3,x]

[Out]

(-ArcCosh[a + b*x] + (b*x*(-(Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x]) + (I*a*b*x*Log[((4*I)*Sqrt[1 - a^2]*(-1 + a
^2 + a*b*x - I*Sqrt[1 - a^2]*Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x]))/(a*b^2*x)])/Sqrt[1 - a^2]))/(-1 + a^2))/(2
*x^2)

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fricas [B]  time = 0.68, size = 460, normalized size = 4.34 \[ \left [\frac {\sqrt {a^{2} - 1} a b^{2} x^{2} \log \left (\frac {a^{2} b x + a^{3} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} {\left (a^{2} + \sqrt {a^{2} - 1} a - 1\right )} + {\left (a b x + a^{2} - 1\right )} \sqrt {a^{2} - 1} - a}{x}\right ) - {\left (a^{2} - 1\right )} b^{2} x^{2} + {\left (a^{4} - 2 \, a^{2} + 1\right )} x^{2} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} {\left (a^{2} - 1\right )} b x - {\left (a^{4} - {\left (a^{4} - 2 \, a^{2} + 1\right )} x^{2} - 2 \, a^{2} + 1\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}{2 \, {\left (a^{4} - 2 \, a^{2} + 1\right )} x^{2}}, -\frac {2 \, \sqrt {-a^{2} + 1} a b^{2} x^{2} \arctan \left (-\frac {\sqrt {-a^{2} + 1} b x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} \sqrt {-a^{2} + 1}}{a^{2} - 1}\right ) + {\left (a^{2} - 1\right )} b^{2} x^{2} - {\left (a^{4} - 2 \, a^{2} + 1\right )} x^{2} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} {\left (a^{2} - 1\right )} b x + {\left (a^{4} - {\left (a^{4} - 2 \, a^{2} + 1\right )} x^{2} - 2 \, a^{2} + 1\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}{2 \, {\left (a^{4} - 2 \, a^{2} + 1\right )} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(b*x+a)/x^3,x, algorithm="fricas")

[Out]

[1/2*(sqrt(a^2 - 1)*a*b^2*x^2*log((a^2*b*x + a^3 + sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)*(a^2 + sqrt(a^2 - 1)*a -
1) + (a*b*x + a^2 - 1)*sqrt(a^2 - 1) - a)/x) - (a^2 - 1)*b^2*x^2 + (a^4 - 2*a^2 + 1)*x^2*log(-b*x - a + sqrt(b
^2*x^2 + 2*a*b*x + a^2 - 1)) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)*(a^2 - 1)*b*x - (a^4 - (a^4 - 2*a^2 + 1)*x^2
- 2*a^2 + 1)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)))/((a^4 - 2*a^2 + 1)*x^2), -1/2*(2*sqrt(-a^2 + 1)
*a*b^2*x^2*arctan(-(sqrt(-a^2 + 1)*b*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)*sqrt(-a^2 + 1))/(a^2 - 1)) + (a^2 -
1)*b^2*x^2 - (a^4 - 2*a^2 + 1)*x^2*log(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)) + sqrt(b^2*x^2 + 2*a*b*x
+ a^2 - 1)*(a^2 - 1)*b*x + (a^4 - (a^4 - 2*a^2 + 1)*x^2 - 2*a^2 + 1)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a
^2 - 1)))/((a^4 - 2*a^2 + 1)*x^2)]

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giac [A]  time = 0.53, size = 170, normalized size = 1.60 \[ -{\left (\frac {a b \arctan \left (-\frac {x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}}{\sqrt {-a^{2} + 1}}\right )}{{\left (a^{2} - 1\right )} \sqrt {-a^{2} + 1}} - \frac {{\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )} a b + a^{2} {\left | b \right |} - {\left | b \right |}}{{\left ({\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}^{2} - a^{2} + 1\right )} {\left (a^{2} - 1\right )}}\right )} b - \frac {\log \left (b x + a + \sqrt {{\left (b x + a\right )}^{2} - 1}\right )}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(b*x+a)/x^3,x, algorithm="giac")

[Out]

-(a*b*arctan(-(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))/sqrt(-a^2 + 1))/((a^2 - 1)*sqrt(-a^2 + 1)) - ((x*
abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))*a*b + a^2*abs(b) - abs(b))/(((x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x +
a^2 - 1))^2 - a^2 + 1)*(a^2 - 1)))*b - 1/2*log(b*x + a + sqrt((b*x + a)^2 - 1))/x^2

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maple [B]  time = 0.02, size = 181, normalized size = 1.71 \[ -\frac {\mathrm {arccosh}\left (b x +a \right )}{2 x^{2}}+\frac {b^{2} \sqrt {b x +a -1}\, \sqrt {b x +a +1}\, \ln \left (\frac {2 \sqrt {a^{2}-1}\, \sqrt {\left (b x +a \right )^{2}-1}+2 a \left (b x +a \right )-2}{b x}\right ) a}{2 \sqrt {a^{2}-1}\, \left (1+a \right ) \left (a -1\right ) \sqrt {\left (b x +a \right )^{2}-1}}-\frac {b \sqrt {b x +a -1}\, \sqrt {b x +a +1}\, a^{2}}{2 x \left (a^{2}-1\right ) \left (1+a \right ) \left (a -1\right )}+\frac {b \sqrt {b x +a -1}\, \sqrt {b x +a +1}}{2 x \left (a^{2}-1\right ) \left (1+a \right ) \left (a -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccosh(b*x+a)/x^3,x)

[Out]

-1/2*arccosh(b*x+a)/x^2+1/2*b^2*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)/(a^2-1)^(1/2)/(1+a)/(a-1)/((b*x+a)^2-1)^(1/2)*
ln(2*((a^2-1)^(1/2)*((b*x+a)^2-1)^(1/2)+a*(b*x+a)-1)/b/x)*a-1/2*b*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)/x/(a^2-1)/(1
+a)/(a-1)*a^2+1/2*b*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)/x/(a^2-1)/(1+a)/(a-1)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(b*x+a)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a-1>0)', see `assume?` for mor
e details)Is a-1 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acosh}\left (a+b\,x\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acosh(a + b*x)/x^3,x)

[Out]

int(acosh(a + b*x)/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acosh}{\left (a + b x \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acosh(b*x+a)/x**3,x)

[Out]

Integral(acosh(a + b*x)/x**3, x)

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