\(\int \frac {1}{\sqrt {1+a^2+2 a b x+b^2 x^2} \text {arcsinh}(a+b x)} \, dx\) [275]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [F]
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 30, antiderivative size = 11 \[ \int \frac {1}{\sqrt {1+a^2+2 a b x+b^2 x^2} \text {arcsinh}(a+b x)} \, dx=\frac {\log (\text {arcsinh}(a+b x))}{b} \]

[Out]

ln(arcsinh(b*x+a))/b

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {5860, 5782} \[ \int \frac {1}{\sqrt {1+a^2+2 a b x+b^2 x^2} \text {arcsinh}(a+b x)} \, dx=\frac {\log (\text {arcsinh}(a+b x))}{b} \]

[In]

Int[1/(Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*ArcSinh[a + b*x]),x]

[Out]

Log[ArcSinh[a + b*x]]/b

Rule 5782

Int[1/(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*Sqrt[(d_) + (e_.)*(x_)^2]), x_Symbol] :> Simp[(1/(b*c))*Simp[Sqrt[1
 + c^2*x^2]/Sqrt[d + e*x^2]]*Log[a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d]

Rule 5860

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(p_.), x_Symbol] :> D
ist[1/d, Subst[Int[(C/d^2 + (C/d^2)*x^2)^p*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A,
B, C, n, p}, x] && EqQ[B*(1 + c^2) - 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\sqrt {1+x^2} \text {arcsinh}(x)} \, dx,x,a+b x\right )}{b} \\ & = \frac {\log (\text {arcsinh}(a+b x))}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {1+a^2+2 a b x+b^2 x^2} \text {arcsinh}(a+b x)} \, dx=\frac {\log (\text {arcsinh}(a+b x))}{b} \]

[In]

Integrate[1/(Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*ArcSinh[a + b*x]),x]

[Out]

Log[ArcSinh[a + b*x]]/b

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.09

method result size
default \(\frac {\ln \left (\operatorname {arcsinh}\left (b x +a \right )\right )}{b}\) \(12\)

[In]

int(1/arcsinh(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

ln(arcsinh(b*x+a))/b

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 30 vs. \(2 (11) = 22\).

Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 2.73 \[ \int \frac {1}{\sqrt {1+a^2+2 a b x+b^2 x^2} \text {arcsinh}(a+b x)} \, dx=\frac {\log \left (\log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )\right )}{b} \]

[In]

integrate(1/arcsinh(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 22 vs. \(2 (8) = 16\).

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 2.00 \[ \int \frac {1}{\sqrt {1+a^2+2 a b x+b^2 x^2} \text {arcsinh}(a+b x)} \, dx=\begin {cases} \frac {\log {\left (\operatorname {asinh}{\left (a + b x \right )} \right )}}{b} & \text {for}\: b \neq 0 \\\frac {x}{\sqrt {a^{2} + 1} \operatorname {asinh}{\left (a \right )}} & \text {otherwise} \end {cases} \]

[In]

integrate(1/asinh(b*x+a)/(b**2*x**2+2*a*b*x+a**2+1)**(1/2),x)

[Out]

Piecewise((log(asinh(a + b*x))/b, Ne(b, 0)), (x/(sqrt(a**2 + 1)*asinh(a)), True))

Maxima [F]

\[ \int \frac {1}{\sqrt {1+a^2+2 a b x+b^2 x^2} \text {arcsinh}(a+b x)} \, dx=\int { \frac {1}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} \operatorname {arsinh}\left (b x + a\right )} \,d x } \]

[In]

integrate(1/arcsinh(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {1}{\sqrt {1+a^2+2 a b x+b^2 x^2} \text {arcsinh}(a+b x)} \, dx=\int { \frac {1}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} \operatorname {arsinh}\left (b x + a\right )} \,d x } \]

[In]

integrate(1/arcsinh(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 2.65 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {1+a^2+2 a b x+b^2 x^2} \text {arcsinh}(a+b x)} \, dx=\frac {\ln \left (\mathrm {asinh}\left (a+b\,x\right )\right )}{b} \]

[In]

int(1/(asinh(a + b*x)*(a^2 + b^2*x^2 + 2*a*b*x + 1)^(1/2)),x)

[Out]

log(asinh(a + b*x))/b