3.5.97 \(\int \frac {\sqrt {c+a^2 c x^2}}{\sqrt {\sinh ^{-1}(a x)}} \, dx\) [497]

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

[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 = 0.09, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5791, 3393, 3388, 2211, 2235, 2236} \begin {gather*} \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}} \end {gather*}

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 2211

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] &&  !TrueQ[$UseGamma]

Rule 2235

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 2236

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 3388

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 3393

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 5791

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(1/(b*c))*Simp[(d
+ e*x^2)^p/(1 + c^2*x^2)^p], Subst[Int[x^n*Cosh[-a/b + x/b]^(2*p + 1), x], x, a + b*ArcSinh[c*x]], x] /; Free
Q[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IGtQ[2*p, 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} \text {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} \text {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} \text {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} \text {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} \text {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} \text {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} \text {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 = 0.07, size = 101, normalized size = 0.65 \begin {gather*} \frac {\sqrt {c \left (1+a^2 x^2\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 {1+a^2 x^2} \sqrt {\sinh ^{-1}(a x)}} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [F]
time = 180.00, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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