3.264 \(\int \frac{\sqrt{1+a^2+2 a b x+b^2 x^2}}{\sinh ^{-1}(a+b x)^2} \, dx\)

Optimal. Leaf size=36 \[ \frac{\text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{b}-\frac{(a+b x)^2+1}{b \sinh ^{-1}(a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.115471, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5867, 5696, 5669, 5448, 12, 3298} \[ \frac{\text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{b}-\frac{(a+b x)^2+1}{b \sinh ^{-1}(a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 + (a + b*x)^2)/(b*ArcSinh[a + b*x])) + SinhIntegral[2*ArcSinh[a + b*x]]/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 5696

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(Sqrt[1 + c^2*x^2]
*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[(c*(2*p + 1)*d^IntPart[p]*(d + e*x^2)^Fr
acPart[p])/(b*(n + 1)*(1 + c^2*x^2)^FracPart[p]), Int[x*(1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n + 1),
x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && LtQ[n, -1]

Rule 5669

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*
Sinh[x]^m*Cosh[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rubi steps

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

Mathematica [A]  time = 0.0582283, size = 47, normalized size = 1.31 \[ -\frac{a^2-\sinh ^{-1}(a+b x) \text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )+2 a b x+b^2 x^2+1}{b \sinh ^{-1}(a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.049, size = 44, normalized size = 1.2 \begin{align*}{\frac{2\,{\it Shi} \left ( 2\,{\it Arcsinh} \left ( bx+a \right ) \right ){\it Arcsinh} \left ( bx+a \right ) -\cosh \left ( 2\,{\it Arcsinh} \left ( bx+a \right ) \right ) -1}{2\,b{\it Arcsinh} \left ( bx+a \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{2} +{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b + b\right )} x + a\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{{\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (b^{2} x + a b\right )} + b\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} + \int \frac{{\left (2 \, b^{2} x^{2} + 4 \, a b x + 2 \, a^{2} - 1\right )}{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac{3}{2}} + 2 \,{\left (2 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} + 2 \, a^{3} +{\left (6 \, a^{2} b + b\right )} x + a\right )}{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )} +{\left (2 \, b^{4} x^{4} + 8 \, a b^{3} x^{3} + 2 \, a^{4} + 3 \,{\left (4 \, a^{2} b^{2} + b^{2}\right )} x^{2} + 3 \, a^{2} + 2 \,{\left (4 \, a^{3} b + 3 \, a b\right )} x + 1\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{{\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + a^{4} + 2 \,{\left (3 \, a^{2} b^{2} + b^{2}\right )} x^{2} +{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} + 2 \, a^{2} + 4 \,{\left (a^{3} b + a b\right )} x + 2 \,{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b + b\right )} x + a\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 1\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((b^2*x^2 + 2*a*b*x + a^2 + 1)^2 + (b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b + b)*x + a)*sqrt(b^2*x^2 + 2*a*b*x
+ a^2 + 1))/((b^3*x^2 + 2*a*b^2*x + a^2*b + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(b^2*x + a*b) + b)*log(b*x + a
+ sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))) + integrate(((2*b^2*x^2 + 4*a*b*x + 2*a^2 - 1)*(b^2*x^2 + 2*a*b*x + a^2
+ 1)^(3/2) + 2*(2*b^3*x^3 + 6*a*b^2*x^2 + 2*a^3 + (6*a^2*b + b)*x + a)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + (2*b^4*
x^4 + 8*a*b^3*x^3 + 2*a^4 + 3*(4*a^2*b^2 + b^2)*x^2 + 3*a^2 + 2*(4*a^3*b + 3*a*b)*x + 1)*sqrt(b^2*x^2 + 2*a*b*
x + a^2 + 1))/((b^4*x^4 + 4*a*b^3*x^3 + a^4 + 2*(3*a^2*b^2 + b^2)*x^2 + (b^2*x^2 + 2*a*b*x + a^2 + 1)*(b^2*x^2
+ 2*a*b*x + a^2) + 2*a^2 + 4*(a^3*b + a*b)*x + 2*(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b + b)*x + a)*sqrt(b^2
*x^2 + 2*a*b*x + a^2 + 1) + 1)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{\operatorname{arsinh}\left (b x + a\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{\operatorname{asinh}^{2}{\left (a + b x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{\operatorname{arsinh}\left (b x + a\right )^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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