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

Optimal. Leaf size=102 \[ \frac {\sqrt {\pi } \sqrt {b} e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c}-\frac {\sqrt {\pi } \sqrt {b} e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c}+x \sqrt {a+b \sinh ^{-1}(c x)} \]

[Out]

1/4*exp(a/b)*erf((a+b*arcsinh(c*x))^(1/2)/b^(1/2))*b^(1/2)*Pi^(1/2)/c-1/4*erfi((a+b*arcsinh(c*x))^(1/2)/b^(1/2
))*b^(1/2)*Pi^(1/2)/c/exp(a/b)+x*(a+b*arcsinh(c*x))^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5653, 5779, 3308, 2180, 2204, 2205} \[ \frac {\sqrt {\pi } \sqrt {b} e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c}-\frac {\sqrt {\pi } \sqrt {b} e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{4 c}+x \sqrt {a+b \sinh ^{-1}(c x)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*ArcSinh[c*x]],x]

[Out]

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

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]

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 5653

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

Rubi steps

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

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Mathematica [A]  time = 0.24, size = 101, normalized size = 0.99 \[ \frac {e^{-\frac {a}{b}} \sqrt {a+b \sinh ^{-1}(c x)} \left (\frac {\Gamma \left (\frac {3}{2},-\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{\sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}}}-\frac {e^{\frac {2 a}{b}} \Gamma \left (\frac {3}{2},\frac {a}{b}+\sinh ^{-1}(c x)\right )}{\sqrt {\frac {a}{b}+\sinh ^{-1}(c x)}}\right )}{2 c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[a + b*ArcSinh[c*x]],x]

[Out]

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

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \operatorname {arsinh}\left (c x\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int \sqrt {a +b \arcsinh \left (c x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \operatorname {arsinh}\left (c x\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {a+b\,\mathrm {asinh}\left (c\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))^(1/2),x)

[Out]

int((a + b*asinh(c*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \operatorname {asinh}{\left (c x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*asinh(c*x)), x)

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