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

Optimal. Leaf size=145 \[ -\frac{\sqrt{\frac{\pi }{2}} \sqrt{b} e^{\frac{2 a}{b}} \text{Erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 c^2}-\frac{\sqrt{\frac{\pi }{2}} \sqrt{b} e^{-\frac{2 a}{b}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 c^2}+\frac{\sqrt{a+b \sinh ^{-1}(c x)}}{4 c^2}+\frac{1}{2} x^2 \sqrt{a+b \sinh ^{-1}(c x)} \]

[Out]

Sqrt[a + b*ArcSinh[c*x]]/(4*c^2) + (x^2*Sqrt[a + b*ArcSinh[c*x]])/2 - (Sqrt[b]*E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqr
t[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(16*c^2) - (Sqrt[b]*Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]]
)/Sqrt[b]])/(16*c^2*E^((2*a)/b))

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Rubi [A]  time = 0.43199, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5663, 5779, 3312, 3307, 2180, 2204, 2205} \[ -\frac{\sqrt{\frac{\pi }{2}} \sqrt{b} e^{\frac{2 a}{b}} \text{Erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 c^2}-\frac{\sqrt{\frac{\pi }{2}} \sqrt{b} e^{-\frac{2 a}{b}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 c^2}+\frac{\sqrt{a+b \sinh ^{-1}(c x)}}{4 c^2}+\frac{1}{2} x^2 \sqrt{a+b \sinh ^{-1}(c x)} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[a + b*ArcSinh[c*x]],x]

[Out]

Sqrt[a + b*ArcSinh[c*x]]/(4*c^2) + (x^2*Sqrt[a + b*ArcSinh[c*x]])/2 - (Sqrt[b]*E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqr
t[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(16*c^2) - (Sqrt[b]*Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]]
)/Sqrt[b]])/(16*c^2*E^((2*a)/b))

Rule 5663

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^(m + 1)*(a + b*ArcSinh[c*x])^n)/
(m + 1), x] - Dist[(b*c*n)/(m + 1), Int[(x^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt[1 + c^2*x^2], x], x] /;
FreeQ[{a, b, c}, x] && IGtQ[m, 0] && GtQ[n, 0]

Rule 5779

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Sinh[x]^m*Cosh[x]^(2*p + 1), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e,
n}, x] && EqQ[e, c^2*d] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 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 3307

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/(E^(
I*k*Pi)*E^(I*(e + f*x))), x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d
, e, f, m}, x] && IntegerQ[2*k]

Rule 2180

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - (c*
f)/d) + (f*g*x^2)/d), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2205

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erf[(c + d*x)*Rt[-(b*Log[F]),
2]])/(2*d*Rt[-(b*Log[F]), 2]), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rubi steps

\begin{align*} \int x \sqrt{a+b \sinh ^{-1}(c x)} \, dx &=\frac{1}{2} x^2 \sqrt{a+b \sinh ^{-1}(c x)}-\frac{1}{4} (b c) \int \frac{x^2}{\sqrt{1+c^2 x^2} \sqrt{a+b \sinh ^{-1}(c x)}} \, dx\\ &=\frac{1}{2} x^2 \sqrt{a+b \sinh ^{-1}(c x)}-\frac{b \operatorname{Subst}\left (\int \frac{\sinh ^2(x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^2}\\ &=\frac{1}{2} x^2 \sqrt{a+b \sinh ^{-1}(c x)}+\frac{b \operatorname{Subst}\left (\int \left (\frac{1}{2 \sqrt{a+b x}}-\frac{\cosh (2 x)}{2 \sqrt{a+b x}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^2}\\ &=\frac{\sqrt{a+b \sinh ^{-1}(c x)}}{4 c^2}+\frac{1}{2} x^2 \sqrt{a+b \sinh ^{-1}(c x)}-\frac{b \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 c^2}\\ &=\frac{\sqrt{a+b \sinh ^{-1}(c x)}}{4 c^2}+\frac{1}{2} x^2 \sqrt{a+b \sinh ^{-1}(c x)}-\frac{b \operatorname{Subst}\left (\int \frac{e^{-2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^2}-\frac{b \operatorname{Subst}\left (\int \frac{e^{2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^2}\\ &=\frac{\sqrt{a+b \sinh ^{-1}(c x)}}{4 c^2}+\frac{1}{2} x^2 \sqrt{a+b \sinh ^{-1}(c x)}-\frac{\operatorname{Subst}\left (\int e^{\frac{2 a}{b}-\frac{2 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{8 c^2}-\frac{\operatorname{Subst}\left (\int e^{-\frac{2 a}{b}+\frac{2 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{8 c^2}\\ &=\frac{\sqrt{a+b \sinh ^{-1}(c x)}}{4 c^2}+\frac{1}{2} x^2 \sqrt{a+b \sinh ^{-1}(c x)}-\frac{\sqrt{b} e^{\frac{2 a}{b}} \sqrt{\frac{\pi }{2}} \text{erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 c^2}-\frac{\sqrt{b} e^{-\frac{2 a}{b}} \sqrt{\frac{\pi }{2}} \text{erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0950965, size = 127, normalized size = 0.88 \[ \frac{e^{-\frac{2 a}{b}} \sqrt{a+b \sinh ^{-1}(c x)} \left (\sqrt{\frac{a}{b}+\sinh ^{-1}(c x)} \text{Gamma}\left (\frac{3}{2},-\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+e^{\frac{4 a}{b}} \sqrt{-\frac{a+b \sinh ^{-1}(c x)}{b}} \text{Gamma}\left (\frac{3}{2},\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )\right )}{8 \sqrt{2} c^2 \sqrt{-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{b^2}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x*Sqrt[a + b*ArcSinh[c*x]],x]

[Out]

(Sqrt[a + b*ArcSinh[c*x]]*(Sqrt[a/b + ArcSinh[c*x]]*Gamma[3/2, (-2*(a + b*ArcSinh[c*x]))/b] + E^((4*a)/b)*Sqrt
[-((a + b*ArcSinh[c*x])/b)]*Gamma[3/2, (2*(a + b*ArcSinh[c*x]))/b]))/(8*Sqrt[2]*c^2*E^((2*a)/b)*Sqrt[-((a + b*
ArcSinh[c*x])^2/b^2)])

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Maple [F]  time = 0.059, size = 0, normalized size = 0. \begin{align*} \int x\sqrt{a+b{\it Arcsinh} \left ( cx \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*arcsinh(c*x) + a)*x, x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sqrt(a + b*asinh(c*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*arcsinh(c*x) + a)*x, x)