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

Optimal. Leaf size=15 \[ \frac{\sinh ^{-1}(a+b x)^4}{4 b} \]

[Out]

ArcSinh[a + b*x]^4/(4*b)

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Rubi [A]  time = 0.0690976, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {5867, 5675} \[ \frac{\sinh ^{-1}(a+b x)^4}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a + b*x]^3/Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

ArcSinh[a + b*x]^4/(4*b)

Rule 5867

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(p_.), x_Symbol] :> D
ist[1/d, Subst[Int[(C/d^2 + (C*x^2)/d^2)^p*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A,
B, C, n, p}, x] && EqQ[B*(1 + c^2) - 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]

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
]

Rubi steps

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

Mathematica [A]  time = 0.0186737, size = 15, normalized size = 1. \[ \frac{\sinh ^{-1}(a+b x)^4}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a + b*x]^3/Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

ArcSinh[a + b*x]^4/(4*b)

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Maple [A]  time = 0.049, size = 14, normalized size = 0.9 \begin{align*}{\frac{ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{4}}{4\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*arcsinh(b*x+a)^4/b

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.56151, size = 78, normalized size = 5.2 \begin{align*} \frac{\log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{4}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^4/b

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Sympy [A]  time = 1.40799, size = 26, normalized size = 1.73 \begin{align*} \begin{cases} \frac{\operatorname{asinh}^{4}{\left (a + b x \right )}}{4 b} & \text{for}\: b \neq 0 \\\frac{x \operatorname{asinh}^{3}{\left (a \right )}}{\sqrt{a^{2} + 1}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((asinh(a + b*x)**4/(4*b), Ne(b, 0)), (x*asinh(a)**3/sqrt(a**2 + 1), True))

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Giac [B]  time = 1.71263, size = 43, normalized size = 2.87 \begin{align*} \frac{\log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{4}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^4/b