3.292 \(\int x^3 \cosh ^{-1}(a+b x^4) \, dx\)

Optimal. Leaf size=54 \[ \frac{\left (a+b x^4\right ) \cosh ^{-1}\left (a+b x^4\right )}{4 b}-\frac{\sqrt{a+b x^4-1} \sqrt{a+b x^4+1}}{4 b} \]

[Out]

-(Sqrt[-1 + a + b*x^4]*Sqrt[1 + a + b*x^4])/(4*b) + ((a + b*x^4)*ArcCosh[a + b*x^4])/(4*b)

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Rubi [A]  time = 0.0521281, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6715, 5864, 5654, 74} \[ \frac{\left (a+b x^4\right ) \cosh ^{-1}\left (a+b x^4\right )}{4 b}-\frac{\sqrt{a+b x^4-1} \sqrt{a+b x^4+1}}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcCosh[a + b*x^4],x]

[Out]

-(Sqrt[-1 + a + b*x^4]*Sqrt[1 + a + b*x^4])/(4*b) + ((a + b*x^4)*ArcCosh[a + b*x^4])/(4*b)

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 5864

Int[((a_.) + ArcCosh[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcCosh[x])^n, x
], x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rule 5654

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcCosh[c*x])^n, x] - Dist[b*c*n, In
t[(x*(a + b*ArcCosh[c*x])^(n - 1))/(Sqrt[-1 + c*x]*Sqrt[1 + c*x]), x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 74

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

Rubi steps

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

Mathematica [A]  time = 0.0332246, size = 50, normalized size = 0.93 \[ \frac{\left (a+b x^4\right ) \cosh ^{-1}\left (a+b x^4\right )-\sqrt{a+b x^4-1} \sqrt{a+b x^4+1}}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcCosh[a + b*x^4],x]

[Out]

(-(Sqrt[-1 + a + b*x^4]*Sqrt[1 + a + b*x^4]) + (a + b*x^4)*ArcCosh[a + b*x^4])/(4*b)

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Maple [A]  time = 0.001, size = 45, normalized size = 0.8 \begin{align*}{\frac{1}{4\,b} \left ( \left ( b{x}^{4}+a \right ){\rm arccosh} \left (b{x}^{4}+a\right )-\sqrt{b{x}^{4}+a-1}\sqrt{b{x}^{4}+a+1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arccosh(b*x^4+a),x)

[Out]

1/4/b*((b*x^4+a)*arccosh(b*x^4+a)-(b*x^4+a-1)^(1/2)*(b*x^4+a+1)^(1/2))

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Maxima [A]  time = 0.972995, size = 50, normalized size = 0.93 \begin{align*} \frac{{\left (b x^{4} + a\right )} \operatorname{arcosh}\left (b x^{4} + a\right ) - \sqrt{{\left (b x^{4} + a\right )}^{2} - 1}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccosh(b*x^4+a),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.08293, size = 151, normalized size = 2.8 \begin{align*} \frac{{\left (b x^{4} + a\right )} \log \left (b x^{4} + a + \sqrt{b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}\right ) - \sqrt{b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccosh(b*x^4+a),x, algorithm="fricas")

[Out]

1/4*((b*x^4 + a)*log(b*x^4 + a + sqrt(b^2*x^8 + 2*a*b*x^4 + a^2 - 1)) - sqrt(b^2*x^8 + 2*a*b*x^4 + a^2 - 1))/b

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Sympy [A]  time = 1.21483, size = 61, normalized size = 1.13 \begin{align*} \begin{cases} \frac{a \operatorname{acosh}{\left (a + b x^{4} \right )}}{4 b} + \frac{x^{4} \operatorname{acosh}{\left (a + b x^{4} \right )}}{4} - \frac{\sqrt{a^{2} + 2 a b x^{4} + b^{2} x^{8} - 1}}{4 b} & \text{for}\: b \neq 0 \\\frac{x^{4} \operatorname{acosh}{\left (a \right )}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*acosh(b*x**4+a),x)

[Out]

Piecewise((a*acosh(a + b*x**4)/(4*b) + x**4*acosh(a + b*x**4)/4 - sqrt(a**2 + 2*a*b*x**4 + b**2*x**8 - 1)/(4*b
), Ne(b, 0)), (x**4*acosh(a)/4, True))

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Giac [B]  time = 1.17469, size = 143, normalized size = 2.65 \begin{align*} \frac{1}{4} \, x^{4} \log \left (b x^{4} + a + \sqrt{{\left (b x^{4} + a\right )}^{2} - 1}\right ) - \frac{1}{4} \, b{\left (\frac{a \log \left ({\left | -a b -{\left (x^{4}{\left | b \right |} - \sqrt{b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}\right )}{\left | b \right |} \right |}\right )}{b{\left | b \right |}} + \frac{\sqrt{b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}}{b^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccosh(b*x^4+a),x, algorithm="giac")

[Out]

1/4*x^4*log(b*x^4 + a + sqrt((b*x^4 + a)^2 - 1)) - 1/4*b*(a*log(abs(-a*b - (x^4*abs(b) - sqrt(b^2*x^8 + 2*a*b*
x^4 + a^2 - 1))*abs(b)))/(b*abs(b)) + sqrt(b^2*x^8 + 2*a*b*x^4 + a^2 - 1)/b^2)