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

Optimal. Leaf size=135 \[ \frac{\sqrt{\frac{\pi }{2}} e^{\frac{2 a}{b}} \text{Erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{b^{3/2} c^2}+\frac{\sqrt{\frac{\pi }{2}} e^{-\frac{2 a}{b}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{b^{3/2} c^2}-\frac{2 x \sqrt{c^2 x^2+1}}{b c \sqrt{a+b \sinh ^{-1}(c x)}} \]

[Out]

(-2*x*Sqrt[1 + c^2*x^2])/(b*c*Sqrt[a + b*ArcSinh[c*x]]) + (E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[a + b*ArcS
inh[c*x]])/Sqrt[b]])/(b^(3/2)*c^2) + (Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(b^(3/2)*c^
2*E^((2*a)/b))

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Rubi [A]  time = 0.161703, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {5665, 3307, 2180, 2204, 2205} \[ \frac{\sqrt{\frac{\pi }{2}} e^{\frac{2 a}{b}} \text{Erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{b^{3/2} c^2}+\frac{\sqrt{\frac{\pi }{2}} e^{-\frac{2 a}{b}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{b^{3/2} c^2}-\frac{2 x \sqrt{c^2 x^2+1}}{b c \sqrt{a+b \sinh ^{-1}(c x)}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*ArcSinh[c*x])^(3/2),x]

[Out]

(-2*x*Sqrt[1 + c^2*x^2])/(b*c*Sqrt[a + b*ArcSinh[c*x]]) + (E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[a + b*ArcS
inh[c*x]])/Sqrt[b]])/(b^(3/2)*c^2) + (Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(b^(3/2)*c^
2*E^((2*a)/b))

Rule 5665

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 + c^2*x^2]*(a + b*ArcSi
nh[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n +
1), Sinh[x]^(m - 1)*(m + (m + 1)*Sinh[x]^2), x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0]
&& GeQ[n, -2] && LtQ[n, -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 \frac{x}{\left (a+b \sinh ^{-1}(c x)\right )^{3/2}} \, dx &=-\frac{2 x \sqrt{1+c^2 x^2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{2 \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^2}\\ &=-\frac{2 x \sqrt{1+c^2 x^2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{\operatorname{Subst}\left (\int \frac{e^{-2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^2}+\frac{\operatorname{Subst}\left (\int \frac{e^{2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^2}\\ &=-\frac{2 x \sqrt{1+c^2 x^2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{2 \operatorname{Subst}\left (\int e^{\frac{2 a}{b}-\frac{2 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{b^2 c^2}+\frac{2 \operatorname{Subst}\left (\int e^{-\frac{2 a}{b}+\frac{2 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{b^2 c^2}\\ &=-\frac{2 x \sqrt{1+c^2 x^2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{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 )}{b^{3/2} c^2}+\frac{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 )}{b^{3/2} c^2}\\ \end{align*}

Mathematica [A]  time = 0.108916, size = 134, normalized size = 0.99 \[ \frac{e^{-\frac{2 a}{b}} \left (\sqrt{2} \sqrt{-\frac{a+b \sinh ^{-1}(c x)}{b}} \text{Gamma}\left (\frac{1}{2},-\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-\sqrt{2} e^{\frac{4 a}{b}} \sqrt{\frac{a}{b}+\sinh ^{-1}(c x)} \text{Gamma}\left (\frac{1}{2},\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-2 e^{\frac{2 a}{b}} \sinh \left (2 \sinh ^{-1}(c x)\right )\right )}{2 b c^2 \sqrt{a+b \sinh ^{-1}(c x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[2]*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Gamma[1/2, (-2*(a + b*ArcSinh[c*x]))/b] - Sqrt[2]*E^((4*a)/b)*Sqrt[a/
b + ArcSinh[c*x]]*Gamma[1/2, (2*(a + b*ArcSinh[c*x]))/b] - 2*E^((2*a)/b)*Sinh[2*ArcSinh[c*x]])/(2*b*c^2*E^((2*
a)/b)*Sqrt[a + b*ArcSinh[c*x]])

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Maple [F]  time = 0.056, 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 \frac{x}{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{\frac{3}{2}}}\,{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(x/(b*arcsinh(c*x) + a)^(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(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 \frac{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 \frac{x}{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{\frac{3}{2}}}\,{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(x/(b*arcsinh(c*x) + a)^(3/2), x)