3.63 \(\int \frac {\sinh ^{-1}(a+b x)}{x^2} \, dx\)

Optimal. Leaf size=57 \[ -\frac {b \tanh ^{-1}\left (\frac {a (a+b x)+1}{\sqrt {a^2+1} \sqrt {(a+b x)^2+1}}\right )}{\sqrt {a^2+1}}-\frac {\sinh ^{-1}(a+b x)}{x} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5865, 5801, 725, 206} \[ -\frac {b \tanh ^{-1}\left (\frac {a (a+b x)+1}{\sqrt {a^2+1} \sqrt {(a+b x)^2+1}}\right )}{\sqrt {a^2+1}}-\frac {\sinh ^{-1}(a+b x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a + b*x]/x^2,x]

[Out]

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

Rule 206

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

Rule 725

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 5801

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 5865

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

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Mathematica [A]  time = 0.04, size = 57, normalized size = 1.00 \[ -\frac {b \tanh ^{-1}\left (\frac {a^2+a b x+1}{\sqrt {a^2+1} \sqrt {(a+b x)^2+1}}\right )}{\sqrt {a^2+1}}-\frac {\sinh ^{-1}(a+b x)}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a + b*x]/x^2,x]

[Out]

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

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fricas [B]  time = 0.58, size = 167, normalized size = 2.93 \[ \frac {\sqrt {a^{2} + 1} b x \log \left (-\frac {a^{2} b x + a^{3} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a^{2} - \sqrt {a^{2} + 1} a + 1\right )} - {\left (a b x + a^{2} + 1\right )} \sqrt {a^{2} + 1} + a}{x}\right ) + {\left (a^{2} + 1\right )} x \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - {\left (a^{2} - {\left (a^{2} + 1\right )} x + 1\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{{\left (a^{2} + 1\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(b*x+a)/x^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.45, size = 110, normalized size = 1.93 \[ \frac {b \log \left (\frac {{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 2 \, \sqrt {a^{2} + 1} \right |}}{{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 2 \, \sqrt {a^{2} + 1} \right |}}\right )}{\sqrt {a^{2} + 1}} - \frac {\log \left (b x + a + \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(b*x+a)/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 71, normalized size = 1.25 \[ -\frac {\arcsinh \left (b x +a \right )}{x}-\frac {b \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b x}\right )}{\sqrt {a^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(b*x+a)/x^2,x)

[Out]

-arcsinh(b*x+a)/x-b/(a^2+1)^(1/2)*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 [B]  time = 0.32, size = 111, normalized size = 1.95 \[ -\frac {b \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{\sqrt {a^{2} + 1}} - \frac {\operatorname {arsinh}\left (b x + a\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(b*x+a)/x^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {asinh}\left (a+b\,x\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a + b*x)/x^2,x)

[Out]

int(asinh(a + b*x)/x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asinh}{\left (a + b x \right )}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(b*x+a)/x**2,x)

[Out]

Integral(asinh(a + b*x)/x**2, x)

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