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

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

[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

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Rubi [A]  time = 0.0390041, antiderivative size = 84, normalized size of antiderivative = 1., 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 = {5662, 100, 12, 90, 52} \[ \frac{1}{4} x^4 \left (a+b \cosh ^{-1}(c x)\right )-\frac{3 b x \sqrt{c x-1} \sqrt{c x+1}}{32 c^3}-\frac{3 b \cosh ^{-1}(c x)}{32 c^4}-\frac{b x^3 \sqrt{c x-1} \sqrt{c x+1}}{16 c} \]

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 5662

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))/(Sqr
t[-1 + c*x]*Sqrt[1 + c*x]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 100

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 12

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

Rule 90

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 52

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]

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*}

Mathematica [A]  time = 0.037223, size = 105, normalized size = 1.25 \[ \frac{a x^4}{4}-\frac{3 b x \sqrt{c x-1} \sqrt{c x+1}}{32 c^3}-\frac{3 b \tanh ^{-1}\left (\frac{\sqrt{c x-1}}{\sqrt{c x+1}}\right )}{16 c^4}-\frac{b x^3 \sqrt{c x-1} \sqrt{c x+1}}{16 c}+\frac{1}{4} b x^4 \cosh ^{-1}(c x) \]

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 = 0.012, size = 109, normalized size = 1.3 \begin{align*}{\frac{{x}^{4}a}{4}}+{\frac{b{x}^{4}{\rm arccosh} \left (cx\right )}{4}}-{\frac{b{x}^{3}}{16\,c}\sqrt{cx-1}\sqrt{cx+1}}-{\frac{3\,bx}{32\,{c}^{3}}\sqrt{cx-1}\sqrt{cx+1}}-{\frac{3\,b}{32\,{c}^{4}}\sqrt{cx-1}\sqrt{cx+1}\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}-1} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*a+1/4*b*x^4*arccosh(c*x)-1/16*b*x^3*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c-3/32*b*x*(c*x-1)^(1/2)*(c*x+1)^(1/2)
/c^3-3/32/c^4*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 = 1.1697, size = 130, normalized size = 1.55 \begin{align*} \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} \sqrt{c^{2}}\right )}{\sqrt{c^{2}} c^{4}}\right )} c\right )} b \end{align*}

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)*sqrt(c^2))/(sqrt(c^2)*c^4))*c)*b

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Fricas [A]  time = 2.45928, size = 161, normalized size = 1.92 \begin{align*} \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{align*}

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 [A]  time = 1.45855, size = 87, normalized size = 1.04 \begin{align*} \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{align*}

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 [A]  time = 1.46898, size = 123, normalized size = 1.46 \begin{align*} \frac{1}{4} \, a x^{4} + \frac{1}{32} \,{\left (8 \, x^{4} \log \left (c x + \sqrt{c^{2} x^{2} - 1}\right ) -{\left (\sqrt{c^{2} x^{2} - 1} x{\left (\frac{2 \, x^{2}}{c^{2}} + \frac{3}{c^{4}}\right )} - \frac{3 \, \log \left ({\left | -x{\left | c \right |} + \sqrt{c^{2} x^{2} - 1} \right |}\right )}{c^{4}{\left | c \right |}}\right )} c\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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