3.140 \(\int x (a+b \sinh ^{-1}(c x))^{3/2} \, dx\)

Optimal. Leaf size=179 \[ -\frac{3 \sqrt{\frac{\pi }{2}} b^{3/2} e^{\frac{2 a}{b}} \text{Erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{64 c^2}+\frac{3 \sqrt{\frac{\pi }{2}} b^{3/2} e^{-\frac{2 a}{b}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{64 c^2}-\frac{3 b x \sqrt{c^2 x^2+1} \sqrt{a+b \sinh ^{-1}(c x)}}{8 c}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac{1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2} \]

[Out]

(-3*b*x*Sqrt[1 + c^2*x^2]*Sqrt[a + b*ArcSinh[c*x]])/(8*c) + (a + b*ArcSinh[c*x])^(3/2)/(4*c^2) + (x^2*(a + b*A
rcSinh[c*x])^(3/2))/2 - (3*b^(3/2)*E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(64
*c^2) + (3*b^(3/2)*Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(64*c^2*E^((2*a)/b))

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Rubi [A]  time = 0.478203, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.714, Rules used = {5663, 5758, 5675, 5669, 5448, 12, 3308, 2180, 2204, 2205} \[ -\frac{3 \sqrt{\frac{\pi }{2}} b^{3/2} e^{\frac{2 a}{b}} \text{Erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{64 c^2}+\frac{3 \sqrt{\frac{\pi }{2}} b^{3/2} e^{-\frac{2 a}{b}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{64 c^2}-\frac{3 b x \sqrt{c^2 x^2+1} \sqrt{a+b \sinh ^{-1}(c x)}}{8 c}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac{1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcSinh[c*x])^(3/2),x]

[Out]

(-3*b*x*Sqrt[1 + c^2*x^2]*Sqrt[a + b*ArcSinh[c*x]])/(8*c) + (a + b*ArcSinh[c*x])^(3/2)/(4*c^2) + (x^2*(a + b*A
rcSinh[c*x])^(3/2))/2 - (3*b^(3/2)*E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(64
*c^2) + (3*b^(3/2)*Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(64*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 5758

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

Rule 5675

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSinh[c*x]
)^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && GtQ[d, 0] && NeQ[n, -1
]

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]

Rule 12

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

Rule 3308

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

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 \left (a+b \sinh ^{-1}(c x)\right )^{3/2} \, dx &=\frac{1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}-\frac{1}{4} (3 b c) \int \frac{x^2 \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{1+c^2 x^2}} \, dx\\ &=-\frac{3 b x \sqrt{1+c^2 x^2} \sqrt{a+b \sinh ^{-1}(c x)}}{8 c}+\frac{1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}+\frac{1}{16} \left (3 b^2\right ) \int \frac{x}{\sqrt{a+b \sinh ^{-1}(c x)}} \, dx+\frac{(3 b) \int \frac{\sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{1+c^2 x^2}} \, dx}{8 c}\\ &=-\frac{3 b x \sqrt{1+c^2 x^2} \sqrt{a+b \sinh ^{-1}(c x)}}{8 c}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac{1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh (x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^2}\\ &=-\frac{3 b x \sqrt{1+c^2 x^2} \sqrt{a+b \sinh ^{-1}(c x)}}{8 c}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac{1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{2 \sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^2}\\ &=-\frac{3 b x \sqrt{1+c^2 x^2} \sqrt{a+b \sinh ^{-1}(c x)}}{8 c}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac{1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^2}\\ &=-\frac{3 b x \sqrt{1+c^2 x^2} \sqrt{a+b \sinh ^{-1}(c x)}}{8 c}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac{1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}-\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{e^{-2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{64 c^2}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{64 c^2}\\ &=-\frac{3 b x \sqrt{1+c^2 x^2} \sqrt{a+b \sinh ^{-1}(c x)}}{8 c}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac{1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}-\frac{(3 b) \operatorname{Subst}\left (\int e^{\frac{2 a}{b}-\frac{2 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{32 c^2}+\frac{(3 b) \operatorname{Subst}\left (\int e^{-\frac{2 a}{b}+\frac{2 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{32 c^2}\\ &=-\frac{3 b x \sqrt{1+c^2 x^2} \sqrt{a+b \sinh ^{-1}(c x)}}{8 c}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{4 c^2}+\frac{1}{2} x^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}-\frac{3 b^{3/2} 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 )}{64 c^2}+\frac{3 b^{3/2} 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 )}{64 c^2}\\ \end{align*}

Mathematica [A]  time = 0.100346, size = 129, normalized size = 0.72 \[ \frac{b e^{-\frac{2 a}{b}} \sqrt{a+b \sinh ^{-1}(c x)} \left (e^{\frac{4 a}{b}} \sqrt{-\frac{a+b \sinh ^{-1}(c x)}{b}} \text{Gamma}\left (\frac{5}{2},\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-\sqrt{\frac{a}{b}+\sinh ^{-1}(c x)} \text{Gamma}\left (\frac{5}{2},-\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )\right )}{16 \sqrt{2} c^2 \sqrt{-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{b^2}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x*(a + b*ArcSinh[c*x])^(3/2),x]

[Out]

(b*Sqrt[a + b*ArcSinh[c*x]]*(-(Sqrt[a/b + ArcSinh[c*x]]*Gamma[5/2, (-2*(a + b*ArcSinh[c*x]))/b]) + E^((4*a)/b)
*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Gamma[5/2, (2*(a + b*ArcSinh[c*x]))/b]))/(16*Sqrt[2]*c^2*E^((2*a)/b)*Sqrt[-((
a + b*ArcSinh[c*x])^2/b^2)])

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Maple [F]  time = 0.052, size = 0, normalized size = 0. \begin{align*} \int x \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arcsinh(c*x))^(3/2),x)

[Out]

int(x*(a+b*arcsinh(c*x))^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{\frac{3}{2}} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsinh(c*x))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*arcsinh(c*x) + a)^(3/2)*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))^(3/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \left (a + b \operatorname{asinh}{\left (c x \right )}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*asinh(c*x))**(3/2),x)

[Out]

Integral(x*(a + b*asinh(c*x))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{\frac{3}{2}} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsinh(c*x))^(3/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)^(3/2)*x, x)