\(\int \frac {1}{\sqrt {1+a^2+2 a b x+b^2 x^2} \text {arcsinh}(a+b x)^2} \, dx\) [276]

Optimal result

Integrand size = 30, antiderivative size = 13 \[ \int \frac {1}{\sqrt {1+a^2+2 a b x+b^2 x^2} \text {arcsinh}(a+b x)^2} \, dx=-\frac {1}{b \text {arcsinh}(a+b x)} \]

[Out]

-1/b/arcsinh(b*x+a)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 13, 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, 5783} \[ \int \frac {1}{\sqrt {1+a^2+2 a b x+b^2 x^2} \text {arcsinh}(a+b x)^2} \, dx=-\frac {1}{b \text {arcsinh}(a+b x)} \]

[In]

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

[Out]

-(1/(b*ArcSinh[a + b*x]))

Rule 5783

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

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)^2} \, dx,x,a+b x\right )}{b} \\ & = -\frac {1}{b \text {arcsinh}(a+b x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {1+a^2+2 a b x+b^2 x^2} \text {arcsinh}(a+b x)^2} \, dx=-\frac {1}{b \text {arcsinh}(a+b x)} \]

[In]

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

[Out]

-(1/(b*ArcSinh[a + b*x]))

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08

method	result	size
default	\(-\frac {1}{b \,\operatorname {arcsinh}\left (b x +a \right )}\)	\(14\)

[In]

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

[Out]

-1/b/arcsinh(b*x+a)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 32 vs. \(2 (13) = 26\).

Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 2.46 \[ \int \frac {1}{\sqrt {1+a^2+2 a b x+b^2 x^2} \text {arcsinh}(a+b x)^2} \, dx=-\frac {1}{b \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (10) = 20\).

Time = 0.51 (sec) , antiderivative size = 26, normalized size of antiderivative = 2.00 \[ \int \frac {1}{\sqrt {1+a^2+2 a b x+b^2 x^2} \text {arcsinh}(a+b x)^2} \, dx=\begin {cases} - \frac {1}{b \operatorname {asinh}{\left (a + b x \right )}} & \text {for}\: b \neq 0 \\\frac {x}{\sqrt {a^{2} + 1} \operatorname {asinh}^{2}{\left (a \right )}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 150 vs. \(2 (13) = 26\).

Time = 0.33 (sec) , antiderivative size = 150, normalized size of antiderivative = 11.54 \[ \int \frac {1}{\sqrt {1+a^2+2 a b x+b^2 x^2} \text {arcsinh}(a+b x)^2} \, dx=-\frac {b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} b + b\right )} x + {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} + a}{{\left ({\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )} {\left (b^{2} x + a b\right )} + {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b + b\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} \]

[In]

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

[Out]

-(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b + b)*x + (b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + a)/(((b^2*x^2 + 2*a*b*
x + a^2 + 1)*(b^2*x + a*b) + (b^3*x^2 + 2*a*b^2*x + a^2*b + b)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*log(b*x + a
+ sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)))

Giac [F]

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

[In]

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

Mupad [B] (verification not implemented)

Time = 2.63 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {1+a^2+2 a b x+b^2 x^2} \text {arcsinh}(a+b x)^2} \, dx=-\frac {1}{b\,\mathrm {asinh}\left (a+b\,x\right )} \]

[In]

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

[Out]

-1/(b*asinh(a + b*x))