\(\int \frac {\text {arcsinh}(a+b x)}{(1+a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [280]

Optimal result

Integrand size = 28, antiderivative size = 46 \[ \int \frac {\text {arcsinh}(a+b x)}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {(a+b x) \text {arcsinh}(a+b x)}{b \sqrt {1+(a+b x)^2}}-\frac {\log \left (1+(a+b x)^2\right )}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {5860, 5787, 266} \[ \int \frac {\text {arcsinh}(a+b x)}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {(a+b x) \text {arcsinh}(a+b x)}{b \sqrt {(a+b x)^2+1}}-\frac {\log \left ((a+b x)^2+1\right )}{2 b} \]

[In]

Int[ArcSinh[a + b*x]/(1 + a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 5787

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[x*((a + b*ArcSinh
[c*x])^n/(d*Sqrt[d + e*x^2])), x] - Dist[b*c*(n/d)*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]], Int[x*((a + b*ArcS
inh[c*x])^(n - 1)/(1 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.35 \[ \int \frac {\text {arcsinh}(a+b x)}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {(a+b x) \text {arcsinh}(a+b x)}{b \sqrt {1+a^2+2 a b x+b^2 x^2}}-\frac {\log \left (1+a^2+2 a b x+b^2 x^2\right )}{2 b} \]

[In]

Integrate[ArcSinh[a + b*x]/(1 + a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(130\) vs. \(2(42)=84\).

Time = 0.61 (sec) , antiderivative size = 131, normalized size of antiderivative = 2.85

method	result	size
default	\(\frac {2 \,\operatorname {arcsinh}\left (b x +a \right )}{b}-\frac {\left (b^{2} x^{2}-\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b x +2 a b x -a \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a^{2}+1\right ) \operatorname {arcsinh}\left (b x +a \right )}{b \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}-\frac {\ln \left (1+\left (b x +a +\sqrt {1+\left (b x +a \right )^{2}}\right )^{2}\right )}{b}\)	\(131\)

[In]

int(arcsinh(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (42) = 84\).

Time = 0.26 (sec) , antiderivative size = 115, normalized size of antiderivative = 2.50 \[ \int \frac {\text {arcsinh}(a+b x)}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (b x + a\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + {\left (a^{2} + 1\right )} b\right )}} \]

[In]

integrate(arcsinh(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\text {arcsinh}(a+b x)}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\operatorname {asinh}{\left (a + b x \right )}}{\left (a^{2} + 2 a b x + b^{2} x^{2} + 1\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(asinh(b*x+a)/(b**2*x**2+2*a*b*x+a**2+1)**(3/2),x)

[Out]

Integral(asinh(a + b*x)/(a**2 + 2*a*b*x + b**2*x**2 + 1)**(3/2), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (42) = 84\).

Time = 0.34 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.59 \[ \int \frac {\text {arcsinh}(a+b x)}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-{\left (\frac {b^{2} x}{{\left (a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}} + \frac {a b}{{\left (a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}\right )} \operatorname {arsinh}\left (b x + a\right ) - \frac {\log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{2 \, b} \]

[In]

integrate(arcsinh(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(3/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.65 \[ \int \frac {\text {arcsinh}(a+b x)}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {{\left (x + \frac {a}{b}\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}} - \frac {\log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{2 \, b} \]

[In]

integrate(arcsinh(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(3/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {arcsinh}(a+b x)}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\mathrm {asinh}\left (a+b\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}^{3/2}} \,d x \]

[In]

int(asinh(a + b*x)/(a^2 + b^2*x^2 + 2*a*b*x + 1)^(3/2),x)

[Out]

int(asinh(a + b*x)/(a^2 + b^2*x^2 + 2*a*b*x + 1)^(3/2), x)