3.280 \(\int \frac{\sinh ^{-1}(a+b x)}{(1+a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=46 \[ \frac{(a+b x) \sinh ^{-1}(a+b x)}{b \sqrt{(a+b x)^2+1}}-\frac{\log \left ((a+b x)^2+1\right )}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0575618, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {5867, 5687, 260} \[ \frac{(a+b x) \sinh ^{-1}(a+b x)}{b \sqrt{(a+b x)^2+1}}-\frac{\log \left ((a+b x)^2+1\right )}{2 b} \]

Antiderivative was successfully verified.

[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 5867

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*x^2)/d^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]

Rule 5687

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*Sqrt[1 + c^2*x^2])/(d*Sqrt[d + e*x^2]), Int[(x*(a + b*ArcSinh[
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 260

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]

Rubi steps

\begin{align*} \int \frac{\sinh ^{-1}(a+b x)}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sinh ^{-1}(x)}{\left (1+x^2\right )^{3/2}} \, dx,x,a+b x\right )}{b}\\ &=\frac{(a+b x) \sinh ^{-1}(a+b x)}{b \sqrt{1+(a+b x)^2}}-\frac{\operatorname{Subst}\left (\int \frac{x}{1+x^2} \, dx,x,a+b x\right )}{b}\\ &=\frac{(a+b x) \sinh ^{-1}(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]  time = 0.0887991, size = 62, normalized size = 1.35 \[ \frac{(a+b x) \sinh ^{-1}(a+b x)}{b \sqrt{a^2+2 a b x+b^2 x^2+1}}-\frac{\log \left (a^2+2 a b x+b^2 x^2+1\right )}{2 b} \]

Antiderivative was successfully verified.

[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)

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Maple [B]  time = 0.094, size = 131, normalized size = 2.9 \begin{align*} 2\,{\frac{{\it Arcsinh} \left ( bx+a \right ) }{b}}-{\frac{{\it Arcsinh} \left ( bx+a \right ) }{b \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) } \left ({b}^{2}{x}^{2}-\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}xb+2\,xab-\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}a+{a}^{2}+1 \right ) }-{\frac{1}{b}\ln \left ( 1+ \left ( bx+a+\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.80635, size = 274, normalized size = 5.96 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.31941, size = 103, normalized size = 2.24 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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