3.7.26 \(\int \frac {1}{(a+b \sinh ^{-1}(c x))^2} \, dx\) [626]

Optimal. Leaf size=85 \[ -\frac {\sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {\text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right ) \sinh \left (\frac {a}{b}\right )}{b^2 c}+\frac {\cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b^2 c} \]

[Out]

cosh(a/b)*Shi((a+b*arcsinh(c*x))/b)/b^2/c-Chi((a+b*arcsinh(c*x))/b)*sinh(a/b)/b^2/c-(c^2*x^2+1)^(1/2)/b/c/(a+b
*arcsinh(c*x))

________________________________________________________________________________________

Rubi [A]
time = 0.12, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5773, 5819, 3384, 3379, 3382} \begin {gather*} -\frac {\sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b^2 c}+\frac {\cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b^2 c}-\frac {\sqrt {c^2 x^2+1}}{b c \left (a+b \sinh ^{-1}(c x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])^(-2),x]

[Out]

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

Rule 3379

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 3382

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]

Rule 3384

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 5773

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 5819

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.11, size = 71, normalized size = 0.84 \begin {gather*} \frac {-\frac {b \sqrt {1+c^2 x^2}}{a+b \sinh ^{-1}(c x)}-\text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right ) \sinh \left (\frac {a}{b}\right )+\cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{b^2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 3.17, size = 118, normalized size = 1.39

method	result	size
derivativedivides	\(\frac {\frac {-\sqrt {c^{2} x^{2}+1}+c x}{2 b \left (a +b \arcsinh \left (c x \right )\right )}+\frac {{\mathrm e}^{\frac {a}{b}} \expIntegral \left (1, \arcsinh \left (c x \right )+\frac {a}{b}\right )}{2 b^{2}}-\frac {c x +\sqrt {c^{2} x^{2}+1}}{2 b \left (a +b \arcsinh \left (c x \right )\right )}-\frac {{\mathrm e}^{-\frac {a}{b}} \expIntegral \left (1, -\arcsinh \left (c x \right )-\frac {a}{b}\right )}{2 b^{2}}}{c}\)	\(118\)
default	\(\frac {\frac {-\sqrt {c^{2} x^{2}+1}+c x}{2 b \left (a +b \arcsinh \left (c x \right )\right )}+\frac {{\mathrm e}^{\frac {a}{b}} \expIntegral \left (1, \arcsinh \left (c x \right )+\frac {a}{b}\right )}{2 b^{2}}-\frac {c x +\sqrt {c^{2} x^{2}+1}}{2 b \left (a +b \arcsinh \left (c x \right )\right )}-\frac {{\mathrm e}^{-\frac {a}{b}} \expIntegral \left (1, -\arcsinh \left (c x \right )-\frac {a}{b}\right )}{2 b^{2}}}{c}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asinh(c*x))**(-2), x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asinh(c*x))^2,x)

[Out]

int(1/(a + b*asinh(c*x))^2, x)

________________________________________________________________________________________