3.6.3 \(\int \frac {\sqrt {c+a^2 c x^2}}{\sinh ^{-1}(a x)^{3/2}} \, dx\) [503]

Optimal. Leaf size=152 \[ -\frac {2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}{a \sqrt {\sinh ^{-1}(a x)}}-\frac {\sqrt {\frac {\pi }{2}} \sqrt {c+a^2 c x^2} \text {Erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{a \sqrt {1+a^2 x^2}}+\frac {\sqrt {\frac {\pi }{2}} \sqrt {c+a^2 c x^2} \text {Erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{a \sqrt {1+a^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {5790, 5780, 5556, 12, 3389, 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 )}{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 )}{a \sqrt {a^2 x^2+1}}-\frac {2 \sqrt {a^2 x^2+1} \sqrt {a^2 c x^2+c}}{a \sqrt {\sinh ^{-1}(a x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + a^2*c*x^2]/ArcSinh[a*x]^(3/2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 3389

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 5556

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 5780

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/(b*c^(m + 1)), Subst[Int[x^n*Sinh
[-a/b + x/b]^m*Cosh[-a/b + x/b], x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 5790

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

Rubi steps

\begin {align*} \int \frac {\sqrt {c+a^2 c x^2}}{\sinh ^{-1}(a x)^{3/2}} \, dx &=-\frac {2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}{a \sqrt {\sinh ^{-1}(a x)}}+\frac {\left (4 a \sqrt {c+a^2 c x^2}\right ) \int \frac {x}{\sqrt {\sinh ^{-1}(a x)}} \, dx}{\sqrt {1+a^2 x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}{a \sqrt {\sinh ^{-1}(a x)}}+\frac {\left (4 \sqrt {c+a^2 c x^2}\right ) \text {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{a \sqrt {1+a^2 x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}{a \sqrt {\sinh ^{-1}(a x)}}+\frac {\left (4 \sqrt {c+a^2 c x^2}\right ) \text {Subst}\left (\int \frac {\sinh (2 x)}{2 \sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{a \sqrt {1+a^2 x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}{a \sqrt {\sinh ^{-1}(a x)}}+\frac {\left (2 \sqrt {c+a^2 c x^2}\right ) \text {Subst}\left (\int \frac {\sinh (2 x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{a \sqrt {1+a^2 x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}{a \sqrt {\sinh ^{-1}(a x)}}-\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 )}{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 )}{a \sqrt {1+a^2 x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}{a \sqrt {\sinh ^{-1}(a x)}}-\frac {\left (2 \sqrt {c+a^2 c x^2}\right ) \text {Subst}\left (\int e^{-2 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{a \sqrt {1+a^2 x^2}}+\frac {\left (2 \sqrt {c+a^2 c x^2}\right ) \text {Subst}\left (\int e^{2 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{a \sqrt {1+a^2 x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}{a \sqrt {\sinh ^{-1}(a x)}}-\frac {\sqrt {\frac {\pi }{2}} \sqrt {c+a^2 c x^2} \text {erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{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 )}{a \sqrt {1+a^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 115, normalized size = 0.76 \begin {gather*} -\frac {\sqrt {c+a^2 c x^2} \left (4+4 a^2 x^2+\sqrt {2 \pi } \sqrt {\sinh ^{-1}(a x)} \text {Erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )-\sqrt {2 \pi } \sqrt {\sinh ^{-1}(a x)} \text {Erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )\right )}{2 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]/ArcSinh[a*x]^(3/2),x]

[Out]

-1/2*(Sqrt[c + a^2*c*x^2]*(4 + 4*a^2*x^2 + Sqrt[2*Pi]*Sqrt[ArcSinh[a*x]]*Erf[Sqrt[2]*Sqrt[ArcSinh[a*x]]] - Sqr
t[2*Pi]*Sqrt[ArcSinh[a*x]]*Erfi[Sqrt[2]*Sqrt[ArcSinh[a*x]]]))/(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 \[\int \frac {\sqrt {a^{2} c \,x^{2}+c}}{\arcsinh \left (a x \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*c*x^2+c)^(1/2)/arcsinh(a*x)^(3/2),x)

[Out]

int((a^2*c*x^2+c)^(1/2)/arcsinh(a*x)^(3/2),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)^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a^2*c*x^2 + c)/arcsinh(a*x)^(3/2), 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)^(3/2),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 )}}{\operatorname {asinh}^{\frac {3}{2}}{\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)**(3/2),x)

[Out]

Integral(sqrt(c*(a**2*x**2 + 1))/asinh(a*x)**(3/2), 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)^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt(a^2*c*x^2 + c)/arcsinh(a*x)^(3/2), 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}}{{\mathrm {asinh}\left (a\,x\right )}^{3/2}} \,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)^(3/2),x)

[Out]

int((c + a^2*c*x^2)^(1/2)/asinh(a*x)^(3/2), x)

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