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

Optimal. Leaf size=82 \[ \frac {\cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{2 b c}-\frac {\sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{2 b c}+\frac {\log \left (a+b \sinh ^{-1}(c x)\right )}{2 b c} \]

[Out]

1/2*Chi(2*(a+b*arcsinh(c*x))/b)*cosh(2*a/b)/b/c+1/2*ln(a+b*arcsinh(c*x))/b/c-1/2*Shi(2*(a+b*arcsinh(c*x))/b)*s
inh(2*a/b)/b/c

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Rubi [A]  time = 0.18, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {5699, 3312, 3303, 3298, 3301} \[ \frac {\cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c x)\right )}{2 b c}-\frac {\sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c x)\right )}{2 b c}+\frac {\log \left (a+b \sinh ^{-1}(c x)\right )}{2 b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 5699

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

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 63, normalized size = 0.77 \[ \frac {\cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-\sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+\log \left (a+b \sinh ^{-1}(c x)\right )}{2 b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c^{2} x^{2} + 1}}{b \operatorname {arsinh}\left (c x\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c^{2} x^{2} + 1}}{b \operatorname {arsinh}\left (c x\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.15, size = 79, normalized size = 0.96 \[ \frac {\ln \left (a +b \arcsinh \left (c x \right )\right )}{2 b c}-\frac {{\mathrm e}^{\frac {2 a}{b}} \Ei \left (1, 2 \arcsinh \left (c x \right )+\frac {2 a}{b}\right )}{4 c b}-\frac {{\mathrm e}^{-\frac {2 a}{b}} \Ei \left (1, -2 \arcsinh \left (c x \right )-\frac {2 a}{b}\right )}{4 c b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*ln(a+b*arcsinh(c*x))/b/c-1/4/c/b*exp(2*a/b)*Ei(1,2*arcsinh(c*x)+2*a/b)-1/4/c/b*exp(-2*a/b)*Ei(1,-2*arcsinh
(c*x)-2*a/b)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c^{2} x^{2} + 1}}{b \operatorname {arsinh}\left (c x\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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