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

Optimal. Leaf size=49 \[ \frac {c \tanh ^{-1}\left (\sqrt {\frac {1}{\left (\frac {a}{c}+\frac {b x}{c}\right )^2}+1}\right )}{b}+\frac {(a+b x) \text {csch}^{-1}\left (\frac {a}{c}+\frac {b x}{c}\right )}{b} \]

[Out]

(b*x+a)*arccsch(a/c+b*x/c)/b+c*arctanh((1+1/(a/c+b*x/c)^2)^(1/2))/b

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Rubi [A]  time = 0.03, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5892, 6314, 372, 266, 63, 207} \[ \frac {c \tanh ^{-1}\left (\sqrt {\frac {1}{\left (\frac {a}{c}+\frac {b x}{c}\right )^2}+1}\right )}{b}+\frac {(a+b x) \text {csch}^{-1}\left (\frac {a}{c}+\frac {b x}{c}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[c/(a + b*x)],x]

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 207

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 372

Int[(u_)^(m_.)*((a_) + (b_.)*(v_)^(n_))^(p_.), x_Symbol] :> Dist[u^m/(Coefficient[v, x, 1]*v^m), Subst[Int[x^m
*(a + b*x^n)^p, x], x, v], x] /; FreeQ[{a, b, m, n, p}, x] && LinearPairQ[u, v, x]

Rule 5892

Int[ArcSinh[(c_.)/((a_.) + (b_.)*(x_)^(n_.))]^(m_.)*(u_.), x_Symbol] :> Int[u*ArcCsch[a/c + (b*x^n)/c]^m, x] /
; FreeQ[{a, b, c, n, m}, x]

Rule 6314

Int[ArcCsch[(c_) + (d_.)*(x_)], x_Symbol] :> Simp[((c + d*x)*ArcCsch[c + d*x])/d, x] + Int[1/((c + d*x)*Sqrt[1
+ 1/(c + d*x)^2]), x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [B]  time = 0.13, size = 131, normalized size = 2.67 \[ \frac {(a+b x) \sqrt {\frac {a^2+2 a b x+b^2 x^2+c^2}{(a+b x)^2}} \left (c \tanh ^{-1}\left (\frac {a+b x}{\sqrt {a^2+2 a b x+b^2 x^2+c^2}}\right )+a \tanh ^{-1}\left (\frac {\sqrt {(a+b x)^2+c^2}}{c}\right )\right )}{b \sqrt {a^2+2 a b x+b^2 x^2+c^2}}+x \sinh ^{-1}\left (\frac {c}{a+b x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[c/(a + b*x)],x]

[Out]

x*ArcSinh[c/(a + b*x)] + ((a + b*x)*Sqrt[(a^2 + c^2 + 2*a*b*x + b^2*x^2)/(a + b*x)^2]*(c*ArcTanh[(a + b*x)/Sqr
t[a^2 + c^2 + 2*a*b*x + b^2*x^2]] + a*ArcTanh[Sqrt[c^2 + (a + b*x)^2]/c]))/(b*Sqrt[a^2 + c^2 + 2*a*b*x + b^2*x
^2])

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fricas [B]  time = 0.90, size = 242, normalized size = 4.94 \[ \frac {b x \log \left (\frac {{\left (b x + a\right )} \sqrt {\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + c^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + c}{b x + a}\right ) + a \log \left (-b x + {\left (b x + a\right )} \sqrt {\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + c^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - a + c\right ) - a \log \left (-b x + {\left (b x + a\right )} \sqrt {\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + c^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - a - c\right ) - c \log \left (-b x + {\left (b x + a\right )} \sqrt {\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + c^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - a\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(c/(b*x + a)), x)

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maple [A]  time = 0.04, size = 46, normalized size = 0.94 \[ -\frac {c \left (-\frac {\arcsinh \left (\frac {c}{b x +a}\right ) \left (b x +a \right )}{c}-\arctanh \left (\frac {1}{\sqrt {\frac {c^{2}}{\left (b x +a \right )^{2}}+1}}\right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(c/(b*x+a)),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {i \, c {\left (\log \left (\frac {i \, {\left (b^{2} x + a b\right )}}{b c} + 1\right ) - \log \left (-\frac {i \, {\left (b^{2} x + a b\right )}}{b c} + 1\right )\right )}}{2 \, b} + \frac {2 \, b x \log \left (c + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + c^{2}}\right ) + a \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + c^{2}\right ) - 2 \, {\left (b x + a\right )} \log \left (b x + a\right )}{2 \, b} + \int \frac {b^{2} c x^{2} + a b c x}{b^{2} c x^{2} + 2 \, a b c x + a^{2} c + c^{3} + {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + c^{2}\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*c*(log(I*(b^2*x + a*b)/(b*c) + 1) - log(-I*(b^2*x + a*b)/(b*c) + 1))/b + 1/2*(2*b*x*log(c + sqrt(b^2*x^
2 + 2*a*b*x + a^2 + c^2)) + a*log(b^2*x^2 + 2*a*b*x + a^2 + c^2) - 2*(b*x + a)*log(b*x + a))/b + integrate((b^
2*c*x^2 + a*b*c*x)/(b^2*c*x^2 + 2*a*b*c*x + a^2*c + c^3 + (b^2*x^2 + 2*a*b*x + a^2 + c^2)^(3/2)), x)

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mupad [B]  time = 1.09, size = 41, normalized size = 0.84 \[ \frac {c\,\mathrm {atanh}\left (\sqrt {\frac {c^2}{{\left (a+b\,x\right )}^2}+1}\right )}{b}+\frac {\mathrm {asinh}\left (\frac {c}{a+b\,x}\right )\,\left (a+b\,x\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(c/(a + b*x)),x)

[Out]

(c*atanh((c^2/(a + b*x)^2 + 1)^(1/2)))/b + (asinh(c/(a + b*x))*(a + b*x))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(c/(a + b*x)), x)

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