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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 54, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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 = 0.04, 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)

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 66, normalized size = 1.22 \[ \frac {{\left (b x^{4} + a\right )} \log \left (b x^{4} + a + \sqrt {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}\right ) - \sqrt {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}}{4 \, b} \]

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

________________________________________________________________________________________

giac [B]  time = 0.19, size = 106, normalized size = 1.96 \[ \frac {1}{4} \, x^{4} \log \left (b x^{4} + a + \sqrt {{\left (b x^{4} + a\right )}^{2} - 1}\right ) - \frac {1}{4} \, b {\left (\frac {a \log \left ({\left | -a b - {\left (x^{4} {\left | b \right |} - \sqrt {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}\right )} {\left | b \right |} \right |}\right )}{b {\left | b \right |}} + \frac {\sqrt {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}}{b^{2}}\right )} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.00, size = 45, normalized size = 0.83 \[ \frac {\left (b \,x^{4}+a \right ) \mathrm {arccosh}\left (b \,x^{4}+a \right )-\sqrt {b \,x^{4}+a -1}\, \sqrt {b \,x^{4}+a +1}}{4 b} \]

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))

________________________________________________________________________________________

maxima [A]  time = 0.30, size = 37, normalized size = 0.69 \[ \frac {{\left (b x^{4} + a\right )} \operatorname {arcosh}\left (b x^{4} + a\right ) - \sqrt {{\left (b x^{4} + a\right )}^{2} - 1}}{4 \, b} \]

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

________________________________________________________________________________________

mupad [B]  time = 4.68, size = 295, normalized size = 5.46 \[ \frac {x^4\,\mathrm {acosh}\left (b\,x^4+a\right )}{4}-\frac {\frac {4\,a\,\left (\sqrt {a-1}-\sqrt {b\,x^4+a-1}\right )}{b\,\left (\sqrt {a+1}-\sqrt {b\,x^4+a+1}\right )}+\frac {4\,a\,{\left (\sqrt {a-1}-\sqrt {b\,x^4+a-1}\right )}^3}{b\,{\left (\sqrt {a+1}-\sqrt {b\,x^4+a+1}\right )}^3}-\frac {8\,{\left (\sqrt {a-1}-\sqrt {b\,x^4+a-1}\right )}^2\,\sqrt {a-1}\,\sqrt {a+1}}{b\,{\left (\sqrt {a+1}-\sqrt {b\,x^4+a+1}\right )}^2}}{4\,\left (\frac {{\left (\sqrt {a-1}-\sqrt {b\,x^4+a-1}\right )}^4}{{\left (\sqrt {a+1}-\sqrt {b\,x^4+a+1}\right )}^4}-\frac {2\,{\left (\sqrt {a-1}-\sqrt {b\,x^4+a-1}\right )}^2}{{\left (\sqrt {a+1}-\sqrt {b\,x^4+a+1}\right )}^2}+1\right )}+\frac {a\,\mathrm {atanh}\left (\frac {\sqrt {a-1}-\sqrt {b\,x^4+a-1}}{\sqrt {a+1}-\sqrt {b\,x^4+a+1}}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acosh(a + b*x^4),x)

[Out]

(x^4*acosh(a + b*x^4))/4 - ((4*a*((a - 1)^(1/2) - (a + b*x^4 - 1)^(1/2)))/(b*((a + 1)^(1/2) - (a + b*x^4 + 1)^
(1/2))) + (4*a*((a - 1)^(1/2) - (a + b*x^4 - 1)^(1/2))^3)/(b*((a + 1)^(1/2) - (a + b*x^4 + 1)^(1/2))^3) - (8*(
(a - 1)^(1/2) - (a + b*x^4 - 1)^(1/2))^2*(a - 1)^(1/2)*(a + 1)^(1/2))/(b*((a + 1)^(1/2) - (a + b*x^4 + 1)^(1/2
))^2))/(4*(((a - 1)^(1/2) - (a + b*x^4 - 1)^(1/2))^4/((a + 1)^(1/2) - (a + b*x^4 + 1)^(1/2))^4 - (2*((a - 1)^(
1/2) - (a + b*x^4 - 1)^(1/2))^2)/((a + 1)^(1/2) - (a + b*x^4 + 1)^(1/2))^2 + 1)) + (a*atanh(((a - 1)^(1/2) - (
a + b*x^4 - 1)^(1/2))/((a + 1)^(1/2) - (a + b*x^4 + 1)^(1/2))))/b

________________________________________________________________________________________

sympy [A]  time = 0.79, size = 61, normalized size = 1.13 \[ \begin {cases} \frac {a \operatorname {acosh}{\left (a + b x^{4} \right )}}{4 b} + \frac {x^{4} \operatorname {acosh}{\left (a + b x^{4} \right )}}{4} - \frac {\sqrt {a^{2} + 2 a b x^{4} + b^{2} x^{8} - 1}}{4 b} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {acosh}{\relax (a )}}{4} & \text {otherwise} \end {cases} \]

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))

________________________________________________________________________________________