3.508 \(\int \frac {\sqrt {c+a^2 c x^2}}{\sinh ^{-1}(a x)^{5/2}} \, dx\)

Optimal. Leaf size=182 \[ \frac {2 \sqrt {2 \pi } \sqrt {a^2 c x^2+c} \text {erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{3 a \sqrt {a^2 x^2+1}}+\frac {2 \sqrt {2 \pi } \sqrt {a^2 c x^2+c} \text {erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{3 a \sqrt {a^2 x^2+1}}-\frac {8 x \sqrt {a^2 c x^2+c}}{3 \sqrt {\sinh ^{-1}(a x)}}-\frac {2 \sqrt {a^2 x^2+1} \sqrt {a^2 c x^2+c}}{3 a \sinh ^{-1}(a x)^{3/2}} \]

[Out]

2/3*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)+2/3*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)-2/3*(a^2*x^2+1)^(1/2)*(a^2*c*x^2+
c)^(1/2)/a/arcsinh(a*x)^(3/2)-8/3*x*(a^2*c*x^2+c)^(1/2)/arcsinh(a*x)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5696, 5665, 3307, 2180, 2204, 2205} \[ \frac {2 \sqrt {2 \pi } \sqrt {a^2 c x^2+c} \text {Erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{3 a \sqrt {a^2 x^2+1}}+\frac {2 \sqrt {2 \pi } \sqrt {a^2 c x^2+c} \text {Erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{3 a \sqrt {a^2 x^2+1}}-\frac {8 x \sqrt {a^2 c x^2+c}}{3 \sqrt {\sinh ^{-1}(a x)}}-\frac {2 \sqrt {a^2 x^2+1} \sqrt {a^2 c x^2+c}}{3 a \sinh ^{-1}(a x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + a^2*c*x^2]/ArcSinh[a*x]^(5/2),x]

[Out]

(-2*Sqrt[1 + a^2*x^2]*Sqrt[c + a^2*c*x^2])/(3*a*ArcSinh[a*x]^(3/2)) - (8*x*Sqrt[c + a^2*c*x^2])/(3*Sqrt[ArcSin
h[a*x]]) + (2*Sqrt[2*Pi]*Sqrt[c + a^2*c*x^2]*Erf[Sqrt[2]*Sqrt[ArcSinh[a*x]]])/(3*a*Sqrt[1 + a^2*x^2]) + (2*Sqr
t[2*Pi]*Sqrt[c + a^2*c*x^2]*Erfi[Sqrt[2]*Sqrt[ArcSinh[a*x]]])/(3*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 5665

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 + c^2*x^2]*(a + b*ArcSi
nh[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n +
1), Sinh[x]^(m - 1)*(m + (m + 1)*Sinh[x]^2), x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0]
&& GeQ[n, -2] && LtQ[n, -1]

Rule 5696

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> 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)*d^IntPart[p]*(d + e*x^2)^Fr
acPart[p])/(b*(n + 1)*(1 + c^2*x^2)^FracPart[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)^{5/2}} \, dx &=-\frac {2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}+\frac {\left (4 a \sqrt {c+a^2 c x^2}\right ) \int \frac {x}{\sinh ^{-1}(a x)^{3/2}} \, dx}{3 \sqrt {1+a^2 x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac {8 x \sqrt {c+a^2 c x^2}}{3 \sqrt {\sinh ^{-1}(a x)}}+\frac {\left (8 \sqrt {c+a^2 c x^2}\right ) \operatorname {Subst}\left (\int \frac {\cosh (2 x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a \sqrt {1+a^2 x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac {8 x \sqrt {c+a^2 c x^2}}{3 \sqrt {\sinh ^{-1}(a x)}}+\frac {\left (4 \sqrt {c+a^2 c x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a \sqrt {1+a^2 x^2}}+\frac {\left (4 \sqrt {c+a^2 c x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a \sqrt {1+a^2 x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac {8 x \sqrt {c+a^2 c x^2}}{3 \sqrt {\sinh ^{-1}(a x)}}+\frac {\left (8 \sqrt {c+a^2 c x^2}\right ) \operatorname {Subst}\left (\int e^{-2 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{3 a \sqrt {1+a^2 x^2}}+\frac {\left (8 \sqrt {c+a^2 c x^2}\right ) \operatorname {Subst}\left (\int e^{2 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{3 a \sqrt {1+a^2 x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}{3 a \sinh ^{-1}(a x)^{3/2}}-\frac {8 x \sqrt {c+a^2 c x^2}}{3 \sqrt {\sinh ^{-1}(a x)}}+\frac {2 \sqrt {2 \pi } \sqrt {c+a^2 c x^2} \text {erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{3 a \sqrt {1+a^2 x^2}}+\frac {2 \sqrt {2 \pi } \sqrt {c+a^2 c x^2} \text {erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{3 a \sqrt {1+a^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 122, normalized size = 0.67 \[ -\frac {2 \sqrt {a^2 c x^2+c} \left (a^2 x^2+4 a x \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)+\sqrt {2} \left (-\sinh ^{-1}(a x)\right )^{3/2} \Gamma \left (\frac {1}{2},-2 \sinh ^{-1}(a x)\right )+\sqrt {2} \sinh ^{-1}(a x)^{3/2} \Gamma \left (\frac {1}{2},2 \sinh ^{-1}(a x)\right )+1\right )}{3 a \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[c + a^2*c*x^2]/ArcSinh[a*x]^(5/2),x]

[Out]

(-2*Sqrt[c + a^2*c*x^2]*(1 + a^2*x^2 + 4*a*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x] + Sqrt[2]*(-ArcSinh[a*x])^(3/2)*Ga
mma[1/2, -2*ArcSinh[a*x]] + Sqrt[2]*ArcSinh[a*x]^(3/2)*Gamma[1/2, 2*ArcSinh[a*x]]))/(3*a*Sqrt[1 + a^2*x^2]*Arc
Sinh[a*x]^(3/2))

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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)^(5/2),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 \[ \int \frac {\sqrt {a^{2} c x^{2} + c}}{\operatorname {arsinh}\left (a x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^(1/2)/arcsinh(a*x)^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(a^2*c*x^2 + c)/arcsinh(a*x)^(5/2), x)

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maple [F]  time = 0.35, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a^{2} c \,x^{2}+c}}{\arcsinh \left (a x \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*c*x^2+c)^(1/2)/arcsinh(a*x)^(5/2),x)

[Out]

int((a^2*c*x^2+c)^(1/2)/arcsinh(a*x)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a^{2} c x^{2} + c}}{\operatorname {arsinh}\left (a x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^(1/2)/arcsinh(a*x)^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a^2*c*x^2 + c)/arcsinh(a*x)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {c\,a^2\,x^2+c}}{{\mathrm {asinh}\left (a\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + a^2*c*x^2)^(1/2)/asinh(a*x)^(5/2),x)

[Out]

int((c + a^2*c*x^2)^(1/2)/asinh(a*x)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c \left (a^{2} x^{2} + 1\right )}}{\operatorname {asinh}^{\frac {5}{2}}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**2*c*x**2+c)**(1/2)/asinh(a*x)**(5/2),x)

[Out]

Integral(sqrt(c*(a**2*x**2 + 1))/asinh(a*x)**(5/2), x)

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