3.2.33 \(\int x^3 (a+b \cosh ^{-1}(c x)) \, dx\) [133]

Optimal. Leaf size=84 \[ -\frac {3 b x \sqrt {-1+c x} \sqrt {1+c x}}{32 c^3}-\frac {b x^3 \sqrt {-1+c x} \sqrt {1+c x}}{16 c}-\frac {3 b \cosh ^{-1}(c x)}{32 c^4}+\frac {1}{4} x^4 \left (a+b \cosh ^{-1}(c x)\right ) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5883, 102, 12, 92, 54} \begin {gather*} \frac {1}{4} x^4 \left (a+b \cosh ^{-1}(c x)\right )-\frac {3 b \cosh ^{-1}(c x)}{32 c^4}-\frac {3 b x \sqrt {c x-1} \sqrt {c x+1}}{32 c^3}-\frac {b x^3 \sqrt {c x-1} \sqrt {c x+1}}{16 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcCosh[c*x]),x]

[Out]

(-3*b*x*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(32*c^3) - (b*x^3*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(16*c) - (3*b*ArcCosh[c*
x])/(32*c^4) + (x^4*(a + b*ArcCosh[c*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*} \int x^3 \left (a+b \cosh ^{-1}(c x)\right ) \, dx &=\frac {1}{4} x^4 \left (a+b \cosh ^{-1}(c x)\right )-\frac {1}{4} (b c) \int \frac {x^4}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx\\ &=-\frac {b x^3 \sqrt {-1+c x} \sqrt {1+c x}}{16 c}+\frac {1}{4} x^4 \left (a+b \cosh ^{-1}(c x)\right )-\frac {b \int \frac {3 x^2}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{16 c}\\ &=-\frac {b x^3 \sqrt {-1+c x} \sqrt {1+c x}}{16 c}+\frac {1}{4} x^4 \left (a+b \cosh ^{-1}(c x)\right )-\frac {(3 b) \int \frac {x^2}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{16 c}\\ &=-\frac {3 b x \sqrt {-1+c x} \sqrt {1+c x}}{32 c^3}-\frac {b x^3 \sqrt {-1+c x} \sqrt {1+c x}}{16 c}+\frac {1}{4} x^4 \left (a+b \cosh ^{-1}(c x)\right )-\frac {(3 b) \int \frac {1}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{32 c^3}\\ &=-\frac {3 b x \sqrt {-1+c x} \sqrt {1+c x}}{32 c^3}-\frac {b x^3 \sqrt {-1+c x} \sqrt {1+c x}}{16 c}-\frac {3 b \cosh ^{-1}(c x)}{32 c^4}+\frac {1}{4} x^4 \left (a+b \cosh ^{-1}(c x)\right )\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 105, normalized size = 1.25 \begin {gather*} \frac {a x^4}{4}-\frac {3 b x \sqrt {-1+c x} \sqrt {1+c x}}{32 c^3}-\frac {b x^3 \sqrt {-1+c x} \sqrt {1+c x}}{16 c}+\frac {1}{4} b x^4 \cosh ^{-1}(c x)-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {-1+c x}}{\sqrt {1+c x}}\right )}{16 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcCosh[c*x]),x]

[Out]

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

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Maple [A]
time = 1.97, size = 114, normalized size = 1.36

method	result	size
derivativedivides	\(\frac {\frac {c^{4} x^{4} a}{4}+\frac {b \,c^{4} x^{4} \mathrm {arccosh}\left (c x \right )}{4}-\frac {b \sqrt {c x -1}\, \sqrt {c x +1}\, c^{3} x^{3}}{16}-\frac {3 b \sqrt {c x -1}\, \sqrt {c x +1}\, c x}{32}-\frac {3 b \sqrt {c x -1}\, \sqrt {c x +1}\, \ln \left (c x +\sqrt {c^{2} x^{2}-1}\right )}{32 \sqrt {c^{2} x^{2}-1}}}{c^{4}}\)	\(114\)
default	\(\frac {\frac {c^{4} x^{4} a}{4}+\frac {b \,c^{4} x^{4} \mathrm {arccosh}\left (c x \right )}{4}-\frac {b \sqrt {c x -1}\, \sqrt {c x +1}\, c^{3} x^{3}}{16}-\frac {3 b \sqrt {c x -1}\, \sqrt {c x +1}\, c x}{32}-\frac {3 b \sqrt {c x -1}\, \sqrt {c x +1}\, \ln \left (c x +\sqrt {c^{2} x^{2}-1}\right )}{32 \sqrt {c^{2} x^{2}-1}}}{c^{4}}\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 87, normalized size = 1.04 \begin {gather*} \frac {1}{4} \, a x^{4} + \frac {1}{32} \, {\left (8 \, x^{4} \operatorname {arcosh}\left (c x\right ) - {\left (\frac {2 \, \sqrt {c^{2} x^{2} - 1} x^{3}}{c^{2}} + \frac {3 \, \sqrt {c^{2} x^{2} - 1} x}{c^{4}} + \frac {3 \, \log \left (2 \, c^{2} x + 2 \, \sqrt {c^{2} x^{2} - 1} c\right )}{c^{5}}\right )} c\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*x^4 + 1/32*(8*x^4*arccosh(c*x) - (2*sqrt(c^2*x^2 - 1)*x^3/c^2 + 3*sqrt(c^2*x^2 - 1)*x/c^4 + 3*log(2*c^2*
x + 2*sqrt(c^2*x^2 - 1)*c)/c^5)*c)*b

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Fricas [A]
time = 0.35, size = 73, normalized size = 0.87 \begin {gather*} \frac {8 \, a c^{4} x^{4} + {\left (8 \, b c^{4} x^{4} - 3 \, b\right )} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) - {\left (2 \, b c^{3} x^{3} + 3 \, b c x\right )} \sqrt {c^{2} x^{2} - 1}}{32 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 0.22, size = 87, normalized size = 1.04 \begin {gather*} \begin {cases} \frac {a x^{4}}{4} + \frac {b x^{4} \operatorname {acosh}{\left (c x \right )}}{4} - \frac {b x^{3} \sqrt {c^{2} x^{2} - 1}}{16 c} - \frac {3 b x \sqrt {c^{2} x^{2} - 1}}{32 c^{3}} - \frac {3 b \operatorname {acosh}{\left (c x \right )}}{32 c^{4}} & \text {for}\: c \neq 0 \\\frac {x^{4} \left (a + \frac {i \pi b}{2}\right )}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*x**4/4 + b*x**4*acosh(c*x)/4 - b*x**3*sqrt(c**2*x**2 - 1)/(16*c) - 3*b*x*sqrt(c**2*x**2 - 1)/(32*
c**3) - 3*b*acosh(c*x)/(32*c**4), Ne(c, 0)), (x**4*(a + I*pi*b/2)/4, True))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccosh(c*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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*acosh(c*x)),x)

[Out]

int(x^3*(a + b*acosh(c*x)), x)

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