3.1.81 \(\int \frac {x^2}{\sinh ^{-1}(a+b x)} \, dx\) [81]

Optimal. Leaf size=60 \[ -\frac {\text {Chi}\left (\sinh ^{-1}(a+b x)\right )}{4 b^3}+\frac {a^2 \text {Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^3}+\frac {\text {Chi}\left (3 \sinh ^{-1}(a+b x)\right )}{4 b^3}-\frac {a \text {Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{b^3} \]

[Out]

-1/4*Chi(arcsinh(b*x+a))/b^3+a^2*Chi(arcsinh(b*x+a))/b^3+1/4*Chi(3*arcsinh(b*x+a))/b^3-a*Shi(2*arcsinh(b*x+a))
/b^3

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Rubi [A]
time = 0.38, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {5859, 5830, 6873, 12, 6874, 3382, 5556, 3379} \begin {gather*} \frac {a^2 \text {Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^3}-\frac {\text {Chi}\left (\sinh ^{-1}(a+b x)\right )}{4 b^3}+\frac {\text {Chi}\left (3 \sinh ^{-1}(a+b x)\right )}{4 b^3}-\frac {a \text {Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/ArcSinh[a + b*x],x]

[Out]

-1/4*CoshIntegral[ArcSinh[a + b*x]]/b^3 + (a^2*CoshIntegral[ArcSinh[a + b*x]])/b^3 + CoshIntegral[3*ArcSinh[a
+ b*x]]/(4*b^3) - (a*SinhIntegral[2*ArcSinh[a + b*x]])/b^3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 5556

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]

Rule 5830

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

Rule 5859

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

Rule 6873

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]
time = 0.09, size = 44, normalized size = 0.73 \begin {gather*} \frac {\left (-1+4 a^2\right ) \text {Chi}\left (\sinh ^{-1}(a+b x)\right )+\text {Chi}\left (3 \sinh ^{-1}(a+b x)\right )-4 a \text {Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{4 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/ArcSinh[a + b*x],x]

[Out]

((-1 + 4*a^2)*CoshIntegral[ArcSinh[a + b*x]] + CoshIntegral[3*ArcSinh[a + b*x]] - 4*a*SinhIntegral[2*ArcSinh[a
+ b*x]])/(4*b^3)

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Maple [A]
time = 3.25, size = 49, normalized size = 0.82

method	result	size
derivativedivides	\(\frac {a^{2} \hyperbolicCosineIntegral \left (\arcsinh \left (b x +a \right )\right )-a \hyperbolicSineIntegral \left (2 \arcsinh \left (b x +a \right )\right )-\frac {\hyperbolicCosineIntegral \left (\arcsinh \left (b x +a \right )\right )}{4}+\frac {\hyperbolicCosineIntegral \left (3 \arcsinh \left (b x +a \right )\right )}{4}}{b^{3}}\)	\(49\)
default	\(\frac {a^{2} \hyperbolicCosineIntegral \left (\arcsinh \left (b x +a \right )\right )-a \hyperbolicSineIntegral \left (2 \arcsinh \left (b x +a \right )\right )-\frac {\hyperbolicCosineIntegral \left (\arcsinh \left (b x +a \right )\right )}{4}+\frac {\hyperbolicCosineIntegral \left (3 \arcsinh \left (b x +a \right )\right )}{4}}{b^{3}}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^3*(a^2*Chi(arcsinh(b*x+a))-a*Shi(2*arcsinh(b*x+a))-1/4*Chi(arcsinh(b*x+a))+1/4*Chi(3*arcsinh(b*x+a)))

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

[Out]

integrate(x^2/arcsinh(b*x + a), x)

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

[Out]

integral(x^2/arcsinh(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/asinh(a + b*x), x)

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

[Out]

integrate(x^2/arcsinh(b*x + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^2}{\mathrm {asinh}\left (a+b\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/asinh(a + b*x),x)

[Out]

int(x^2/asinh(a + b*x), x)

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