3.503 \(\int \frac{\sqrt{c+a^2 c x^2}}{\sinh ^{-1}(a x)^{3/2}} \, dx\)

Optimal. Leaf size=152 \[ -\frac{\sqrt{\frac{\pi }{2}} \sqrt{a^2 c x^2+c} \text{Erf}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{a \sqrt{a^2 x^2+1}}+\frac{\sqrt{\frac{\pi }{2}} \sqrt{a^2 c x^2+c} \text{Erfi}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{a \sqrt{a^2 x^2+1}}-\frac{2 \sqrt{a^2 x^2+1} \sqrt{a^2 c x^2+c}}{a \sqrt{\sinh ^{-1}(a x)}} \]

[Out]

(-2*Sqrt[1 + a^2*x^2]*Sqrt[c + a^2*c*x^2])/(a*Sqrt[ArcSinh[a*x]]) - (Sqrt[Pi/2]*Sqrt[c + a^2*c*x^2]*Erf[Sqrt[2
]*Sqrt[ArcSinh[a*x]]])/(a*Sqrt[1 + a^2*x^2]) + (Sqrt[Pi/2]*Sqrt[c + a^2*c*x^2]*Erfi[Sqrt[2]*Sqrt[ArcSinh[a*x]]
])/(a*Sqrt[1 + a^2*x^2])

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Rubi [A]  time = 0.124301, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {5696, 5669, 5448, 12, 3308, 2180, 2204, 2205} \[ -\frac{\sqrt{\frac{\pi }{2}} \sqrt{a^2 c x^2+c} \text{Erf}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{a \sqrt{a^2 x^2+1}}+\frac{\sqrt{\frac{\pi }{2}} \sqrt{a^2 c x^2+c} \text{Erfi}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{a \sqrt{a^2 x^2+1}}-\frac{2 \sqrt{a^2 x^2+1} \sqrt{a^2 c x^2+c}}{a \sqrt{\sinh ^{-1}(a x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + a^2*c*x^2]/ArcSinh[a*x]^(3/2),x]

[Out]

(-2*Sqrt[1 + a^2*x^2]*Sqrt[c + a^2*c*x^2])/(a*Sqrt[ArcSinh[a*x]]) - (Sqrt[Pi/2]*Sqrt[c + a^2*c*x^2]*Erf[Sqrt[2
]*Sqrt[ArcSinh[a*x]]])/(a*Sqrt[1 + a^2*x^2]) + (Sqrt[Pi/2]*Sqrt[c + a^2*c*x^2]*Erfi[Sqrt[2]*Sqrt[ArcSinh[a*x]]
])/(a*Sqrt[1 + a^2*x^2])

Rule 5696

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(Sqrt[1 + c^2*x^2]
*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[(c*(2*p + 1)*d^IntPart[p]*(d + e*x^2)^Fr
acPart[p])/(b*(n + 1)*(1 + c^2*x^2)^FracPart[p]), Int[x*(1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n + 1),
x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && LtQ[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 \frac{\sqrt{c+a^2 c x^2}}{\sinh ^{-1}(a x)^{3/2}} \, dx &=-\frac{2 \sqrt{1+a^2 x^2} \sqrt{c+a^2 c x^2}}{a \sqrt{\sinh ^{-1}(a x)}}+\frac{\left (4 a \sqrt{c+a^2 c x^2}\right ) \int \frac{x}{\sqrt{\sinh ^{-1}(a x)}} \, dx}{\sqrt{1+a^2 x^2}}\\ &=-\frac{2 \sqrt{1+a^2 x^2} \sqrt{c+a^2 c x^2}}{a \sqrt{\sinh ^{-1}(a x)}}+\frac{\left (4 \sqrt{c+a^2 c x^2}\right ) \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh (x)}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{a \sqrt{1+a^2 x^2}}\\ &=-\frac{2 \sqrt{1+a^2 x^2} \sqrt{c+a^2 c x^2}}{a \sqrt{\sinh ^{-1}(a x)}}+\frac{\left (4 \sqrt{c+a^2 c x^2}\right ) \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{2 \sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{a \sqrt{1+a^2 x^2}}\\ &=-\frac{2 \sqrt{1+a^2 x^2} \sqrt{c+a^2 c x^2}}{a \sqrt{\sinh ^{-1}(a x)}}+\frac{\left (2 \sqrt{c+a^2 c x^2}\right ) \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{a \sqrt{1+a^2 x^2}}\\ &=-\frac{2 \sqrt{1+a^2 x^2} \sqrt{c+a^2 c x^2}}{a \sqrt{\sinh ^{-1}(a x)}}-\frac{\sqrt{c+a^2 c x^2} \operatorname{Subst}\left (\int \frac{e^{-2 x}}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{a \sqrt{1+a^2 x^2}}+\frac{\sqrt{c+a^2 c x^2} \operatorname{Subst}\left (\int \frac{e^{2 x}}{\sqrt{x}} \, dx,x,\sinh ^{-1}(a x)\right )}{a \sqrt{1+a^2 x^2}}\\ &=-\frac{2 \sqrt{1+a^2 x^2} \sqrt{c+a^2 c x^2}}{a \sqrt{\sinh ^{-1}(a x)}}-\frac{\left (2 \sqrt{c+a^2 c x^2}\right ) \operatorname{Subst}\left (\int e^{-2 x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{a \sqrt{1+a^2 x^2}}+\frac{\left (2 \sqrt{c+a^2 c x^2}\right ) \operatorname{Subst}\left (\int e^{2 x^2} \, dx,x,\sqrt{\sinh ^{-1}(a x)}\right )}{a \sqrt{1+a^2 x^2}}\\ &=-\frac{2 \sqrt{1+a^2 x^2} \sqrt{c+a^2 c x^2}}{a \sqrt{\sinh ^{-1}(a x)}}-\frac{\sqrt{\frac{\pi }{2}} \sqrt{c+a^2 c x^2} \text{erf}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{a \sqrt{1+a^2 x^2}}+\frac{\sqrt{\frac{\pi }{2}} \sqrt{c+a^2 c x^2} \text{erfi}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )}{a \sqrt{1+a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.185339, size = 115, normalized size = 0.76 \[ -\frac{\sqrt{a^2 c x^2+c} \left (4 a^2 x^2+\sqrt{2 \pi } \sqrt{\sinh ^{-1}(a x)} \text{Erf}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )-\sqrt{2 \pi } \sqrt{\sinh ^{-1}(a x)} \text{Erfi}\left (\sqrt{2} \sqrt{\sinh ^{-1}(a x)}\right )+4\right )}{2 a \sqrt{a^2 x^2+1} \sqrt{\sinh ^{-1}(a x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[c + a^2*c*x^2]*(4 + 4*a^2*x^2 + Sqrt[2*Pi]*Sqrt[ArcSinh[a*x]]*Erf[Sqrt[2]*Sqrt[ArcSinh[a*x]]] - Sqrt[2*
Pi]*Sqrt[ArcSinh[a*x]]*Erfi[Sqrt[2]*Sqrt[ArcSinh[a*x]]]))/(2*a*Sqrt[1 + a^2*x^2]*Sqrt[ArcSinh[a*x]])

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Maple [F]  time = 0.229, size = 0, normalized size = 0. \begin{align*} \int{\sqrt{{a}^{2}c{x}^{2}+c} \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*c*x^2+c)^(1/2)/arcsinh(a*x)^(3/2),x)

[Out]

int((a^2*c*x^2+c)^(1/2)/arcsinh(a*x)^(3/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^(1/2)/arcsinh(a*x)^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a^2*c*x^2 + c)/arcsinh(a*x)^(3/2), 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((a^2*c*x^2+c)^(1/2)/arcsinh(a*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 \frac{\sqrt{c \left (a^{2} x^{2} + 1\right )}}{\operatorname{asinh}^{\frac{3}{2}}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**2*c*x**2+c)**(1/2)/asinh(a*x)**(3/2),x)

[Out]

Integral(sqrt(c*(a**2*x**2 + 1))/asinh(a*x)**(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^(1/2)/arcsinh(a*x)^(3/2),x, algorithm="giac")

[Out]

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