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

Optimal. Leaf size=156 \[ \frac {\sqrt {\frac {\pi }{2}} \sqrt {a^2 c x^2+c} \text {erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{4 a \sqrt {a^2 x^2+1}}+\frac {\sqrt {\frac {\pi }{2}} \sqrt {a^2 c x^2+c} \text {erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{4 a \sqrt {a^2 x^2+1}}+\frac {\sqrt {a^2 c x^2+c} \sqrt {\sinh ^{-1}(a x)}}{a \sqrt {a^2 x^2+1}} \]

[Out]

1/8*erf(2^(1/2)*arcsinh(a*x)^(1/2))*2^(1/2)*Pi^(1/2)*(a^2*c*x^2+c)^(1/2)/a/(a^2*x^2+1)^(1/2)+1/8*erfi(2^(1/2)*
arcsinh(a*x)^(1/2))*2^(1/2)*Pi^(1/2)*(a^2*c*x^2+c)^(1/2)/a/(a^2*x^2+1)^(1/2)+(a^2*c*x^2+c)^(1/2)*arcsinh(a*x)^
(1/2)/a/(a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {5702, 5699, 3312, 3307, 2180, 2204, 2205} \[ \frac {\sqrt {\frac {\pi }{2}} \sqrt {a^2 c x^2+c} \text {Erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{4 a \sqrt {a^2 x^2+1}}+\frac {\sqrt {\frac {\pi }{2}} \sqrt {a^2 c x^2+c} \text {Erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{4 a \sqrt {a^2 x^2+1}}+\frac {\sqrt {a^2 c x^2+c} \sqrt {\sinh ^{-1}(a x)}}{a \sqrt {a^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 5699

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c, Subst[Int[
(a + b*x)^n*Cosh[x]^(2*p + 1), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IG
tQ[2*p, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 5702

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

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 101, normalized size = 0.65 \[ \frac {\sqrt {c \left (a^2 x^2+1\right )} \left (8 \sinh ^{-1}(a x)-\sqrt {2} \sqrt {\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},2 \sinh ^{-1}(a x)\right )+\sqrt {2} \sqrt {-\sinh ^{-1}(a x)} \Gamma \left (\frac {1}{2},-2 \sinh ^{-1}(a x)\right )\right )}{8 a \sqrt {a^2 x^2+1} \sqrt {\sinh ^{-1}(a x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[c*(1 + a^2*x^2)]*(8*ArcSinh[a*x] + Sqrt[2]*Sqrt[-ArcSinh[a*x]]*Gamma[1/2, -2*ArcSinh[a*x]] - Sqrt[2]*Sqr
t[ArcSinh[a*x]]*Gamma[1/2, 2*ArcSinh[a*x]]))/(8*a*Sqrt[1 + a^2*x^2]*Sqrt[ArcSinh[a*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((a^2*c*x^2+c)^(1/2)/arcsinh(a*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 \[ \int \frac {\sqrt {a^{2} c x^{2} + c}}{\sqrt {\operatorname {arsinh}\left (a x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a^{2} c \,x^{2}+c}}{\sqrt {\arcsinh \left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a^{2} c x^{2} + c}}{\sqrt {\operatorname {arsinh}\left (a x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {c\,a^2\,x^2+c}}{\sqrt {\mathrm {asinh}\left (a\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c \left (a^{2} x^{2} + 1\right )}}{\sqrt {\operatorname {asinh}{\left (a x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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