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

Optimal. Leaf size=58 \[ \frac {\cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{b d}-\frac {\sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{b d} \]

[Out]

Chi((a+b*arcsinh(d*x+c))/b)*cosh(a/b)/b/d-Shi((a+b*arcsinh(d*x+c))/b)*sinh(a/b)/b/d

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Rubi [A]  time = 0.09, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5863, 5657, 3303, 3298, 3301} \[ \frac {\cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{b d}-\frac {\sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{b d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c + d*x])^(-1),x]

[Out]

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

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]

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 5657

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 49, normalized size = 0.84 \[ \frac {\cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )-\sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )}{b d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c + d*x])^(-1),x]

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b \operatorname {arsinh}\left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{b \operatorname {arsinh}\left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 60, normalized size = 1.03 \[ \frac {-\frac {{\mathrm e}^{\frac {a}{b}} \Ei \left (1, \arcsinh \left (d x +c \right )+\frac {a}{b}\right )}{2 b}-\frac {{\mathrm e}^{-\frac {a}{b}} \Ei \left (1, -\arcsinh \left (d x +c \right )-\frac {a}{b}\right )}{2 b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{b \operatorname {arsinh}\left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{a+b\,\mathrm {asinh}\left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asinh(c + d*x)),x)

[Out]

int(1/(a + b*asinh(c + d*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{a + b \operatorname {asinh}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(a + b*asinh(c + d*x)), x)

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