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

Optimal. Leaf size=180 \[ -\frac{e \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{4 b c^3}+\frac{e \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{4 b c^3}+\frac{e \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{4 b c^3}-\frac{e \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{4 b c^3}+\frac{d \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{b c}-\frac{d \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{b c} \]

[Out]

(d*Cosh[a/b]*CoshIntegral[(a + b*ArcSinh[c*x])/b])/(b*c) - (e*Cosh[a/b]*CoshIntegral[(a + b*ArcSinh[c*x])/b])/
(4*b*c^3) + (e*Cosh[(3*a)/b]*CoshIntegral[(3*(a + b*ArcSinh[c*x]))/b])/(4*b*c^3) - (d*Sinh[a/b]*SinhIntegral[(
a + b*ArcSinh[c*x])/b])/(b*c) + (e*Sinh[a/b]*SinhIntegral[(a + b*ArcSinh[c*x])/b])/(4*b*c^3) - (e*Sinh[(3*a)/b
]*SinhIntegral[(3*(a + b*ArcSinh[c*x]))/b])/(4*b*c^3)

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Rubi [A]  time = 0.367419, antiderivative size = 176, normalized size of antiderivative = 0.98, number of steps used = 15, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {5706, 5657, 3303, 3298, 3301, 5669, 5448} \[ -\frac{e \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{4 b c^3}+\frac{e \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b c^3}+\frac{e \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{4 b c^3}-\frac{e \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b c^3}+\frac{d \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{b c}-\frac{d \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e*Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c*x]])/(4*b*c^3) + (e*Cosh[(3*a)/b]*CoshIntegral[(3*a)/b + 3*ArcSinh[
c*x]])/(4*b*c^3) + (d*Cosh[a/b]*CoshIntegral[(a + b*ArcSinh[c*x])/b])/(b*c) + (e*Sinh[a/b]*SinhIntegral[a/b +
ArcSinh[c*x]])/(4*b*c^3) - (e*Sinh[(3*a)/b]*SinhIntegral[(3*a)/b + 3*ArcSinh[c*x]])/(4*b*c^3) - (d*Sinh[a/b]*S
inhIntegral[(a + b*ArcSinh[c*x])/b])/(b*c)

Rule 5706

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a
+ b*ArcSinh[c*x])^n, (d + e*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && NeQ[e, c^2*d] && IntegerQ[p] &&
(p > 0 || IGtQ[n, 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 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]

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]

Rubi steps

\begin{align*} \int \frac{d+e x^2}{a+b \sinh ^{-1}(c x)} \, dx &=\int \left (\frac{d}{a+b \sinh ^{-1}(c x)}+\frac{e x^2}{a+b \sinh ^{-1}(c x)}\right ) \, dx\\ &=d \int \frac{1}{a+b \sinh ^{-1}(c x)} \, dx+e \int \frac{x^2}{a+b \sinh ^{-1}(c x)} \, dx\\ &=\frac{d \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}-\frac{x}{b}\right )}{x} \, dx,x,a+b \sinh ^{-1}(c x)\right )}{b c}+\frac{e \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh ^2(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{c^3}\\ &=\frac{e \operatorname{Subst}\left (\int \left (-\frac{\cosh (x)}{4 (a+b x)}+\frac{\cosh (3 x)}{4 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^3}+\frac{\left (d \cosh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{x}{b}\right )}{x} \, dx,x,a+b \sinh ^{-1}(c x)\right )}{b c}-\frac{\left (d \sinh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{x}{b}\right )}{x} \, dx,x,a+b \sinh ^{-1}(c x)\right )}{b c}\\ &=\frac{d \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{b c}-\frac{d \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{b c}-\frac{e \operatorname{Subst}\left (\int \frac{\cosh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^3}+\frac{e \operatorname{Subst}\left (\int \frac{\cosh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^3}\\ &=\frac{d \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{b c}-\frac{d \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{b c}-\frac{\left (e \cosh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^3}+\frac{\left (e \cosh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^3}+\frac{\left (e \sinh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^3}-\frac{\left (e \sinh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^3}\\ &=-\frac{e \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{4 b c^3}+\frac{e \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b c^3}+\frac{d \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{b c}+\frac{e \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{4 b c^3}-\frac{e \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b c^3}-\frac{d \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{b c}\\ \end{align*}

Mathematica [A]  time = 0.240057, size = 126, normalized size = 0.7 \[ \frac{\cosh \left (\frac{a}{b}\right ) \left (4 c^2 d-e\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )-4 c^2 d \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )+e \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+e \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )-e \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )}{4 b c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*c^2*d - e)*Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c*x]] + e*Cosh[(3*a)/b]*CoshIntegral[3*(a/b + ArcSinh[c*x]
)] - 4*c^2*d*Sinh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]] + e*Sinh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]] - e*Sin
h[(3*a)/b]*SinhIntegral[3*(a/b + ArcSinh[c*x])])/(4*b*c^3)

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Maple [A]  time = 0.086, size = 178, normalized size = 1. \begin{align*}{\frac{1}{c} \left ( -{\frac{e}{8\,{c}^{2}b}{{\rm e}^{-3\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-3\,{\it Arcsinh} \left ( cx \right ) -3\,{\frac{a}{b}} \right ) }-{\frac{e}{8\,{c}^{2}b}{{\rm e}^{3\,{\frac{a}{b}}}}{\it Ei} \left ( 1,3\,{\it Arcsinh} \left ( cx \right ) +3\,{\frac{a}{b}} \right ) }-{\frac{d}{2\,b}{{\rm e}^{{\frac{a}{b}}}}{\it Ei} \left ( 1,{\it Arcsinh} \left ( cx \right ) +{\frac{a}{b}} \right ) }+{\frac{e}{8\,{c}^{2}b}{{\rm e}^{{\frac{a}{b}}}}{\it Ei} \left ( 1,{\it Arcsinh} \left ( cx \right ) +{\frac{a}{b}} \right ) }-{\frac{d}{2\,b}{{\rm e}^{-{\frac{a}{b}}}}{\it Ei} \left ( 1,-{\it Arcsinh} \left ( cx \right ) -{\frac{a}{b}} \right ) }+{\frac{e}{8\,{c}^{2}b}{{\rm e}^{-{\frac{a}{b}}}}{\it Ei} \left ( 1,-{\it Arcsinh} \left ( cx \right ) -{\frac{a}{b}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(-1/8/c^2*e/b*exp(-3*a/b)*Ei(1,-3*arcsinh(c*x)-3*a/b)-1/8/c^2*e/b*exp(3*a/b)*Ei(1,3*arcsinh(c*x)+3*a/b)-1/
2/b*exp(a/b)*Ei(1,arcsinh(c*x)+a/b)*d+1/8/c^2/b*exp(a/b)*Ei(1,arcsinh(c*x)+a/b)*e-1/2/b*exp(-a/b)*Ei(1,-arcsin
h(c*x)-a/b)*d+1/8/c^2/b*exp(-a/b)*Ei(1,-arcsinh(c*x)-a/b)*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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