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

Optimal. Leaf size=92 \[ \frac {\sqrt {\pi } e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{2 \sqrt {b} d}+\frac {\sqrt {\pi } e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{2 \sqrt {b} d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5863, 5657, 3307, 2180, 2205, 2204} \[ \frac {\sqrt {\pi } e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{2 \sqrt {b} d}+\frac {\sqrt {\pi } e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{2 \sqrt {b} d} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + b*ArcSinh[c + d*x]],x]

[Out]

(E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(2*Sqrt[b]*d) + (Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh
[c + d*x]]/Sqrt[b]])/(2*Sqrt[b]*d*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 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 5657

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

Rule 5863

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcSinh[x])^n, x
], x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 111, normalized size = 1.21 \[ \frac {e^{-\frac {a}{b}} \left (\sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \Gamma \left (\frac {1}{2},-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )-e^{\frac {2 a}{b}} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \Gamma \left (\frac {1}{2},\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )}{2 d \sqrt {a+b \sinh ^{-1}(c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/Sqrt[a + b*ArcSinh[c + d*x]],x]

[Out]

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

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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(1/(a+b*arcsinh(d*x+c))^(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 \[ \int \frac {1}{\sqrt {b \operatorname {arsinh}\left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(b*arcsinh(d*x + c) + a), x)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a +b \arcsinh \left (d x +c \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \operatorname {arsinh}\left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(b*arcsinh(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asinh(c + d*x))^(1/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(a + b*asinh(c + d*x)), x)

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