3.337 \(\int \frac {\sinh ^{-1}(a x)^3}{\sqrt {c+a^2 c x^2}} \, dx\)

Optimal. Leaf size=40 \[ \frac {\sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^4}{4 a \sqrt {a^2 c x^2+c}} \]

[Out]

1/4*arcsinh(a*x)^4*(a^2*x^2+1)^(1/2)/a/(a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {5677, 5675} \[ \frac {\sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^4}{4 a \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^4)/(4*a*Sqrt[c + a^2*c*x^2])

Rule 5675

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

Rule 5677

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

Rubi steps

\begin {align*} \int \frac {\sinh ^{-1}(a x)^3}{\sqrt {c+a^2 c x^2}} \, dx &=\frac {\sqrt {1+a^2 x^2} \int \frac {\sinh ^{-1}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx}{\sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^4}{4 a \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 40, normalized size = 1.00 \[ \frac {\sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^4}{4 a \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a*x]^3/Sqrt[c + a^2*c*x^2],x]

[Out]

(Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^4)/(4*a*Sqrt[c + a^2*c*x^2])

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arsinh}\left (a x\right )^{3}}{\sqrt {a^{2} c x^{2} + c}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsinh(a*x)^3/sqrt(a^2*c*x^2 + c), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arsinh}\left (a x\right )^{3}}{\sqrt {a^{2} c x^{2} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(a*x)^3/sqrt(a^2*c*x^2 + c), x)

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maple [A]  time = 0.04, size = 39, normalized size = 0.98 \[ \frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \arcsinh \left (a x \right )^{4}}{4 \sqrt {a^{2} x^{2}+1}\, a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(c*(a^2*x^2+1))^(1/2)/(a^2*x^2+1)^(1/2)/a/c*arcsinh(a*x)^4

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maxima [A]  time = 0.37, size = 14, normalized size = 0.35 \[ \frac {\operatorname {arsinh}\left (a x\right )^{4}}{4 \, a \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*arcsinh(a*x)^4/(a*sqrt(c))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\mathrm {asinh}\left (a\,x\right )}^3}{\sqrt {c\,a^2\,x^2+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asinh}^{3}{\left (a x \right )}}{\sqrt {c \left (a^{2} x^{2} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a*x)**3/sqrt(c*(a**2*x**2 + 1)), x)

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