\(\int x^3 \text {arccosh}(a x) \, dx\) [2]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [F(-2)]
   Mupad [F(-1)]

Optimal result

Integrand size = 8, antiderivative size = 77 \[ \int x^3 \text {arccosh}(a x) \, dx=-\frac {3 x \sqrt {-1+a x} \sqrt {1+a x}}{32 a^3}-\frac {x^3 \sqrt {-1+a x} \sqrt {1+a x}}{16 a}-\frac {3 \text {arccosh}(a x)}{32 a^4}+\frac {1}{4} x^4 \text {arccosh}(a x) \]

[Out]

-3/32*arccosh(a*x)/a^4+1/4*x^4*arccosh(a*x)-3/32*x*(a*x-1)^(1/2)*(a*x+1)^(1/2)/a^3-1/16*x^3*(a*x-1)^(1/2)*(a*x
+1)^(1/2)/a

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5883, 102, 12, 92, 54} \[ \int x^3 \text {arccosh}(a x) \, dx=-\frac {3 \text {arccosh}(a x)}{32 a^4}-\frac {3 x \sqrt {a x-1} \sqrt {a x+1}}{32 a^3}+\frac {1}{4} x^4 \text {arccosh}(a x)-\frac {x^3 \sqrt {a x-1} \sqrt {a x+1}}{16 a} \]

[In]

Int[x^3*ArcCosh[a*x],x]

[Out]

(-3*x*Sqrt[-1 + a*x]*Sqrt[1 + a*x])/(32*a^3) - (x^3*Sqrt[-1 + a*x]*Sqrt[1 + a*x])/(16*a) - (3*ArcCosh[a*x])/(3
2*a^4) + (x^4*ArcCosh[a*x])/4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 54

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ArcCosh[b*(x/a)]/b, x] /; FreeQ[{a,
 b, c, d}, x] && EqQ[a + c, 0] && EqQ[b - d, 0] && GtQ[a, 0]

Rule 92

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a + b*x
)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
 f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 102

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

Rule 5883

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcC
osh[c*x])^n/(d*(m + 1))), x] - Dist[b*c*(n/(d*(m + 1))), Int[(d*x)^(m + 1)*((a + b*ArcCosh[c*x])^(n - 1)/(Sqrt
[1 + c*x]*Sqrt[-1 + c*x])), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} x^4 \text {arccosh}(a x)-\frac {1}{4} a \int \frac {x^4}{\sqrt {-1+a x} \sqrt {1+a x}} \, dx \\ & = -\frac {x^3 \sqrt {-1+a x} \sqrt {1+a x}}{16 a}+\frac {1}{4} x^4 \text {arccosh}(a x)-\frac {\int \frac {3 x^2}{\sqrt {-1+a x} \sqrt {1+a x}} \, dx}{16 a} \\ & = -\frac {x^3 \sqrt {-1+a x} \sqrt {1+a x}}{16 a}+\frac {1}{4} x^4 \text {arccosh}(a x)-\frac {3 \int \frac {x^2}{\sqrt {-1+a x} \sqrt {1+a x}} \, dx}{16 a} \\ & = -\frac {3 x \sqrt {-1+a x} \sqrt {1+a x}}{32 a^3}-\frac {x^3 \sqrt {-1+a x} \sqrt {1+a x}}{16 a}+\frac {1}{4} x^4 \text {arccosh}(a x)-\frac {3 \int \frac {1}{\sqrt {-1+a x} \sqrt {1+a x}} \, dx}{32 a^3} \\ & = -\frac {3 x \sqrt {-1+a x} \sqrt {1+a x}}{32 a^3}-\frac {x^3 \sqrt {-1+a x} \sqrt {1+a x}}{16 a}-\frac {3 \text {arccosh}(a x)}{32 a^4}+\frac {1}{4} x^4 \text {arccosh}(a x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.92 \[ \int x^3 \text {arccosh}(a x) \, dx=-\frac {a x \sqrt {-1+a x} \sqrt {1+a x} \left (3+2 a^2 x^2\right )-8 a^4 x^4 \text {arccosh}(a x)+6 \text {arctanh}\left (\sqrt {\frac {-1+a x}{1+a x}}\right )}{32 a^4} \]

[In]

Integrate[x^3*ArcCosh[a*x],x]

[Out]

-1/32*(a*x*Sqrt[-1 + a*x]*Sqrt[1 + a*x]*(3 + 2*a^2*x^2) - 8*a^4*x^4*ArcCosh[a*x] + 6*ArcTanh[Sqrt[(-1 + a*x)/(
1 + a*x)]])/a^4

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.27

method result size
derivativedivides \(\frac {\frac {a^{4} x^{4} \operatorname {arccosh}\left (a x \right )}{4}-\frac {\sqrt {a x -1}\, \sqrt {a x +1}\, \left (2 a^{3} x^{3} \sqrt {a^{2} x^{2}-1}+3 a x \sqrt {a^{2} x^{2}-1}+3 \ln \left (a x +\sqrt {a^{2} x^{2}-1}\right )\right )}{32 \sqrt {a^{2} x^{2}-1}}}{a^{4}}\) \(98\)
default \(\frac {\frac {a^{4} x^{4} \operatorname {arccosh}\left (a x \right )}{4}-\frac {\sqrt {a x -1}\, \sqrt {a x +1}\, \left (2 a^{3} x^{3} \sqrt {a^{2} x^{2}-1}+3 a x \sqrt {a^{2} x^{2}-1}+3 \ln \left (a x +\sqrt {a^{2} x^{2}-1}\right )\right )}{32 \sqrt {a^{2} x^{2}-1}}}{a^{4}}\) \(98\)
parts \(\frac {x^{4} \operatorname {arccosh}\left (a x \right )}{4}-\frac {\sqrt {a x -1}\, \sqrt {a x +1}\, \left (2 \,\operatorname {csgn}\left (a \right ) a^{3} x^{3} \sqrt {a^{2} x^{2}-1}+3 x \sqrt {a^{2} x^{2}-1}\, \operatorname {csgn}\left (a \right ) a +3 \ln \left (\left (\operatorname {csgn}\left (a \right ) \sqrt {a^{2} x^{2}-1}+a x \right ) \operatorname {csgn}\left (a \right )\right )\right ) \operatorname {csgn}\left (a \right )}{32 a^{4} \sqrt {a^{2} x^{2}-1}}\) \(106\)

[In]

int(x^3*arccosh(a*x),x,method=_RETURNVERBOSE)

[Out]

1/a^4*(1/4*a^4*x^4*arccosh(a*x)-1/32*(a*x-1)^(1/2)*(a*x+1)^(1/2)*(2*a^3*x^3*(a^2*x^2-1)^(1/2)+3*a*x*(a^2*x^2-1
)^(1/2)+3*ln(a*x+(a^2*x^2-1)^(1/2)))/(a^2*x^2-1)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.77 \[ \int x^3 \text {arccosh}(a x) \, dx=\frac {{\left (8 \, a^{4} x^{4} - 3\right )} \log \left (a x + \sqrt {a^{2} x^{2} - 1}\right ) - {\left (2 \, a^{3} x^{3} + 3 \, a x\right )} \sqrt {a^{2} x^{2} - 1}}{32 \, a^{4}} \]

[In]

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

[Out]

1/32*((8*a^4*x^4 - 3)*log(a*x + sqrt(a^2*x^2 - 1)) - (2*a^3*x^3 + 3*a*x)*sqrt(a^2*x^2 - 1))/a^4

Sympy [F]

\[ \int x^3 \text {arccosh}(a x) \, dx=\int x^{3} \operatorname {acosh}{\left (a x \right )}\, dx \]

[In]

integrate(x**3*acosh(a*x),x)

[Out]

Integral(x**3*acosh(a*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00 \[ \int x^3 \text {arccosh}(a x) \, dx=\frac {1}{4} \, x^{4} \operatorname {arcosh}\left (a x\right ) - \frac {1}{32} \, {\left (\frac {2 \, \sqrt {a^{2} x^{2} - 1} x^{3}}{a^{2}} + \frac {3 \, \sqrt {a^{2} x^{2} - 1} x}{a^{4}} + \frac {3 \, \log \left (2 \, a^{2} x + 2 \, \sqrt {a^{2} x^{2} - 1} a\right )}{a^{5}}\right )} a \]

[In]

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

[Out]

1/4*x^4*arccosh(a*x) - 1/32*(2*sqrt(a^2*x^2 - 1)*x^3/a^2 + 3*sqrt(a^2*x^2 - 1)*x/a^4 + 3*log(2*a^2*x + 2*sqrt(
a^2*x^2 - 1)*a)/a^5)*a

Giac [F(-2)]

Exception generated. \[ \int x^3 \text {arccosh}(a x) \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int x^3 \text {arccosh}(a x) \, dx=\int x^3\,\mathrm {acosh}\left (a\,x\right ) \,d x \]

[In]

int(x^3*acosh(a*x),x)

[Out]

int(x^3*acosh(a*x), x)