3.1.64 \(\int \frac {\sinh ^{-1}(a+b x)}{x^3} \, dx\) [64]

Optimal. Leaf size=92 \[ -\frac {b \sqrt {1+(a+b x)^2}}{2 \left (1+a^2\right ) x}-\frac {\sinh ^{-1}(a+b x)}{2 x^2}+\frac {a b^2 \tanh ^{-1}\left (\frac {1+a (a+b x)}{\sqrt {1+a^2} \sqrt {1+(a+b x)^2}}\right )}{2 \left (1+a^2\right )^{3/2}} \]

[Out]

-1/2*arcsinh(b*x+a)/x^2+1/2*a*b^2*arctanh((1+a*(b*x+a))/(a^2+1)^(1/2)/(1+(b*x+a)^2)^(1/2))/(a^2+1)^(3/2)-1/2*b
*(1+(b*x+a)^2)^(1/2)/(a^2+1)/x

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Rubi [A]
time = 0.07, antiderivative size = 92, 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 = {5859, 5828, 745, 739, 212} \begin {gather*} \frac {a b^2 \tanh ^{-1}\left (\frac {a (a+b x)+1}{\sqrt {a^2+1} \sqrt {(a+b x)^2+1}}\right )}{2 \left (a^2+1\right )^{3/2}}-\frac {b \sqrt {(a+b x)^2+1}}{2 \left (a^2+1\right ) x}-\frac {\sinh ^{-1}(a+b x)}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a + b*x]/x^3,x]

[Out]

-1/2*(b*Sqrt[1 + (a + b*x)^2])/((1 + a^2)*x) - ArcSinh[a + b*x]/(2*x^2) + (a*b^2*ArcTanh[(1 + a*(a + b*x))/(Sq
rt[1 + a^2]*Sqrt[1 + (a + b*x)^2])])/(2*(1 + a^2)^(3/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 739

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 745

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m + 1)*((a + c*x^2)^(p
+ 1)/((m + 1)*(c*d^2 + a*e^2))), x] + Dist[c*(d/(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x]
/; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 0]

Rule 5828

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*
((a + b*ArcSinh[c*x])^n/(e*(m + 1))), x] - Dist[b*c*(n/(e*(m + 1))), Int[(d + e*x)^(m + 1)*((a + b*ArcSinh[c*x
])^(n - 1)/Sqrt[1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5859

Int[((a_.) + ArcSinh[(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*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 110, normalized size = 1.20 \begin {gather*} -\frac {\sinh ^{-1}(a+b x)+\frac {b x \left (\sqrt {1+a^2} \sqrt {1+a^2+2 a b x+b^2 x^2}+a b x \log (x)-a b x \log \left (1+a^2+a b x+\sqrt {1+a^2} \sqrt {1+a^2+2 a b x+b^2 x^2}\right )\right )}{\left (1+a^2\right )^{3/2}}}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a + b*x]/x^3,x]

[Out]

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

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Maple [A]
time = 2.08, size = 112, normalized size = 1.22

method	result	size
derivativedivides	\(b^{2} \left (-\frac {\arcsinh \left (b x +a \right )}{2 b^{2} x^{2}}-\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 \left (a^{2}+1\right ) b x}+\frac {a \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b x}\right )}{2 \left (a^{2}+1\right )^{\frac {3}{2}}}\right )\)	\(112\)
default	\(b^{2} \left (-\frac {\arcsinh \left (b x +a \right )}{2 b^{2} x^{2}}-\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 \left (a^{2}+1\right ) b x}+\frac {a \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b x}\right )}{2 \left (a^{2}+1\right )^{\frac {3}{2}}}\right )\)	\(112\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(b*x+a)/x^3,x,method=_RETURNVERBOSE)

[Out]

b^2*(-1/2/b^2/x^2*arcsinh(b*x+a)-1/2/(a^2+1)/b/x*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+1/2*a/(a^2+1)^(3/2)*ln((2*a^2+2
+2*a*b*x+2*(a^2+1)^(1/2)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/b/x))

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Maxima [A]
time = 0.26, size = 146, normalized size = 1.59 \begin {gather*} \frac {1}{2} \, {\left (\frac {a b \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{{\left (a^{2} + 1\right )}^{\frac {3}{2}}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{{\left (a^{2} + 1\right )} x}\right )} b - \frac {\operatorname {arsinh}\left (b x + a\right )}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*b*arcsinh(2*a*b*x/(sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2)*abs(x)) + 2*a^2/(sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^
2)*abs(x)) + 2/(sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2)*abs(x)))/(a^2 + 1)^(3/2) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1
)/((a^2 + 1)*x))*b - 1/2*arcsinh(b*x + a)/x^2

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 236 vs. \(2 (78) = 156\).
time = 0.38, size = 236, normalized size = 2.57 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {asinh}{\left (a + b x \right )}}{x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a + b*x)/x**3, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (78) = 156\).
time = 0.46, size = 199, normalized size = 2.16 \begin {gather*} -\frac {1}{2} \, {\left (\frac {a b \log \left (\frac {{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 2 \, \sqrt {a^{2} + 1} \right |}}{{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 2 \, \sqrt {a^{2} + 1} \right |}}\right )}{{\left (a^{2} + 1\right )}^{\frac {3}{2}}} - \frac {2 \, {\left ({\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 |}\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a*b*log(abs(-2*x*abs(b) + 2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) - 2*sqrt(a^2 + 1))/abs(-2*x*abs(b) + 2*sqr
t(b^2*x^2 + 2*a*b*x + a^2 + 1) + 2*sqrt(a^2 + 1)))/(a^2 + 1)^(3/2) - 2*((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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {asinh}\left (a+b\,x\right )}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a + b*x)/x^3,x)

[Out]

int(asinh(a + b*x)/x^3, x)

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