3.166 \(\int \frac{1}{(a+b \sinh ^{-1}(c+d x))^2} \, dx\)

Optimal. Leaf size=91 \[ -\frac{\sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{b^2 d}+\frac{\cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{b^2 d}-\frac{\sqrt{(c+d x)^2+1}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )} \]

[Out]

-(Sqrt[1 + (c + d*x)^2]/(b*d*(a + b*ArcSinh[c + d*x]))) - (CoshIntegral[(a + b*ArcSinh[c + d*x])/b]*Sinh[a/b])
/(b^2*d) + (Cosh[a/b]*SinhIntegral[(a + b*ArcSinh[c + d*x])/b])/(b^2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.174356, antiderivative size = 87, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5863, 5655, 5779, 3303, 3298, 3301} \[ -\frac{\sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{b^2 d}+\frac{\cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{b^2 d}-\frac{\sqrt{(c+d x)^2+1}}{b d \left (a+b \sinh ^{-1}(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c + d*x])^(-2),x]

[Out]

-(Sqrt[1 + (c + d*x)^2]/(b*d*(a + b*ArcSinh[c + d*x]))) - (CoshIntegral[a/b + ArcSinh[c + d*x]]*Sinh[a/b])/(b^
2*d) + (Cosh[a/b]*SinhIntegral[a/b + ArcSinh[c + d*x]])/(b^2*d)

Rule 5863

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcSinh[x])^n, x
], x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rule 5655

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^(n + 1
))/(b*c*(n + 1)), x] - Dist[c/(b*(n + 1)), Int[(x*(a + b*ArcSinh[c*x])^(n + 1))/Sqrt[1 + c^2*x^2], x], x] /; F
reeQ[{a, b, c}, x] && LtQ[n, -1]

Rule 5779

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Sinh[x]^m*Cosh[x]^(2*p + 1), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e,
n}, x] && EqQ[e, c^2*d] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

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]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rubi steps

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

Mathematica [A]  time = 0.271571, size = 179, normalized size = 1.97 \[ \frac{-\sinh \left (\frac{a}{b}\right ) \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )+\cosh \left (\frac{a}{b}\right ) \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )+b (-c) \sinh ^{-1}(c+d x) \log \left (a+b \sinh ^{-1}(c+d x)\right )+b c \sinh ^{-1}(c+d x) \log \left (d \left (a+b \sinh ^{-1}(c+d x)\right )\right )-a c \log \left (a+b \sinh ^{-1}(c+d x)\right )+a c \log \left (d \left (a+b \sinh ^{-1}(c+d x)\right )\right )-b \sqrt{c^2+2 c d x+d^2 x^2+1}}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c + d*x])^(-2),x]

[Out]

(-(b*Sqrt[1 + c^2 + 2*c*d*x + d^2*x^2]) - a*c*Log[a + b*ArcSinh[c + d*x]] - b*c*ArcSinh[c + d*x]*Log[a + b*Arc
Sinh[c + d*x]] + a*c*Log[d*(a + b*ArcSinh[c + d*x])] + b*c*ArcSinh[c + d*x]*Log[d*(a + b*ArcSinh[c + d*x])] -
(a + b*ArcSinh[c + d*x])*CoshIntegral[a/b + ArcSinh[c + d*x]]*Sinh[a/b] + (a + b*ArcSinh[c + d*x])*Cosh[a/b]*S
inhIntegral[a/b + ArcSinh[c + d*x]])/(b^2*d*(a + b*ArcSinh[c + d*x]))

________________________________________________________________________________________

Maple [A]  time = 0.038, size = 128, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ({\frac{1}{2\,b \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) } \left ( -\sqrt{1+ \left ( dx+c \right ) ^{2}}+dx+c \right ) }+{\frac{1}{2\,{b}^{2}}{{\rm e}^{{\frac{a}{b}}}}{\it Ei} \left ( 1,{\it Arcsinh} \left ( dx+c \right ) +{\frac{a}{b}} \right ) }-{\frac{1}{2\,b \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) } \left ( dx+c+\sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) }-{\frac{1}{2\,{b}^{2}}{{\rm e}^{-{\frac{a}{b}}}}{\it Ei} \left ( 1,-{\it Arcsinh} \left ( dx+c \right ) -{\frac{a}{b}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/2*(-(1+(d*x+c)^2)^(1/2)+d*x+c)/b/(a+b*arcsinh(d*x+c))+1/2/b^2*exp(a/b)*Ei(1,arcsinh(d*x+c)+a/b)-1/2/b*(
d*x+c+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))-1/2/b^2*exp(-a/b)*Ei(1,-arcsinh(d*x+c)-a/b))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asinh(c + d*x))**(-2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsinh(d*x + c) + a)^(-2), x)