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

Optimal. Leaf size=152 \[ \frac{\sqrt{a+b x-1} \sqrt{a+b x+1} \left (4 a \left (19 a^2+16\right )-\left (26 a^2+9\right ) (a+b x)\right )}{96 b^4}-\frac{\left (8 a^4+24 a^2+3\right ) \cosh ^{-1}(a+b x)}{32 b^4}+\frac{7 a x^2 \sqrt{a+b x-1} \sqrt{a+b x+1}}{48 b^2}-\frac{x^3 \sqrt{a+b x-1} \sqrt{a+b x+1}}{16 b}+\frac{1}{4} x^4 \cosh ^{-1}(a+b x) \]

[Out]

(7*a*x^2*Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x])/(48*b^2) - (x^3*Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x])/(16*b) +
(Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x]*(4*a*(16 + 19*a^2) - (9 + 26*a^2)*(a + b*x)))/(96*b^4) - ((3 + 24*a^2 +
8*a^4)*ArcCosh[a + b*x])/(32*b^4) + (x^4*ArcCosh[a + b*x])/4

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Rubi [A]  time = 0.186638, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {5866, 5802, 100, 153, 147, 52} \[ \frac{\sqrt{a+b x-1} \sqrt{a+b x+1} \left (4 a \left (19 a^2+16\right )-\left (26 a^2+9\right ) (a+b x)\right )}{96 b^4}-\frac{\left (8 a^4+24 a^2+3\right ) \cosh ^{-1}(a+b x)}{32 b^4}+\frac{7 a x^2 \sqrt{a+b x-1} \sqrt{a+b x+1}}{48 b^2}-\frac{x^3 \sqrt{a+b x-1} \sqrt{a+b x+1}}{16 b}+\frac{1}{4} x^4 \cosh ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcCosh[a + b*x],x]

[Out]

(7*a*x^2*Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x])/(48*b^2) - (x^3*Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x])/(16*b) +
(Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x]*(4*a*(16 + 19*a^2) - (9 + 26*a^2)*(a + b*x)))/(96*b^4) - ((3 + 24*a^2 +
8*a^4)*ArcCosh[a + b*x])/(32*b^4) + (x^4*ArcCosh[a + b*x])/4

Rule 5866

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

Rule 5802

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)
*(a + b*ArcCosh[c*x])^n)/(e*(m + 1)), x] - Dist[(b*c*n)/(e*(m + 1)), Int[((d + e*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, e, 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 153

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

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 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 \cosh ^{-1}(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a}{b}+\frac{x}{b}\right )^3 \cosh ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac{1}{4} x^4 \cosh ^{-1}(a+b x)-\frac{1}{4} \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{b}+\frac{x}{b}\right )^4}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,a+b x\right )\\ &=-\frac{x^3 \sqrt{-1+a+b x} \sqrt{1+a+b x}}{16 b}+\frac{1}{4} x^4 \cosh ^{-1}(a+b x)-\frac{1}{16} \operatorname{Subst}\left (\int \frac{\left (\frac{3+4 a^2}{b^2}-\frac{7 a x}{b^2}\right ) \left (-\frac{a}{b}+\frac{x}{b}\right )^2}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,a+b x\right )\\ &=\frac{7 a x^2 \sqrt{-1+a+b x} \sqrt{1+a+b x}}{48 b^2}-\frac{x^3 \sqrt{-1+a+b x} \sqrt{1+a+b x}}{16 b}+\frac{1}{4} x^4 \cosh ^{-1}(a+b x)-\frac{1}{48} \operatorname{Subst}\left (\int \frac{\left (-\frac{a \left (23+12 a^2\right )}{b^3}+\frac{\left (9+26 a^2\right ) x}{b^3}\right ) \left (-\frac{a}{b}+\frac{x}{b}\right )}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,a+b x\right )\\ &=\frac{7 a x^2 \sqrt{-1+a+b x} \sqrt{1+a+b x}}{48 b^2}-\frac{x^3 \sqrt{-1+a+b x} \sqrt{1+a+b x}}{16 b}+\frac{\sqrt{-1+a+b x} \sqrt{1+a+b x} \left (4 a \left (16+19 a^2\right )-\left (9+26 a^2\right ) (a+b x)\right )}{96 b^4}+\frac{1}{4} x^4 \cosh ^{-1}(a+b x)-\frac{\left (3+24 a^2+8 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,a+b x\right )}{32 b^4}\\ &=\frac{7 a x^2 \sqrt{-1+a+b x} \sqrt{1+a+b x}}{48 b^2}-\frac{x^3 \sqrt{-1+a+b x} \sqrt{1+a+b x}}{16 b}+\frac{\sqrt{-1+a+b x} \sqrt{1+a+b x} \left (4 a \left (16+19 a^2\right )-\left (9+26 a^2\right ) (a+b x)\right )}{96 b^4}-\frac{\left (3+24 a^2+8 a^4\right ) \cosh ^{-1}(a+b x)}{32 b^4}+\frac{1}{4} x^4 \cosh ^{-1}(a+b x)\\ \end{align*}

Mathematica [A]  time = 0.17599, size = 121, normalized size = 0.8 \[ \frac{\sqrt{a+b x-1} \sqrt{a+b x+1} \left (-26 a^2 b x+50 a^3+14 a b^2 x^2+55 a-6 b^3 x^3-9 b x\right )-3 \left (8 a^4+24 a^2+3\right ) \log \left (\sqrt{a+b x-1} \sqrt{a+b x+1}+a+b x\right )+24 b^4 x^4 \cosh ^{-1}(a+b x)}{96 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcCosh[a + b*x],x]

[Out]

(Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x]*(55*a + 50*a^3 - 9*b*x - 26*a^2*b*x + 14*a*b^2*x^2 - 6*b^3*x^3) + 24*b^4
*x^4*ArcCosh[a + b*x] - 3*(3 + 24*a^2 + 8*a^4)*Log[a + b*x + Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x]])/(96*b^4)

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Maple [B]  time = 0.033, size = 308, normalized size = 2. \begin{align*}{\frac{{x}^{4}{\rm arccosh} \left (bx+a\right )}{4}}-{\frac{{x}^{3}}{16\,b}\sqrt{bx+a-1}\sqrt{bx+a+1}}+{\frac{7\,a{x}^{2}}{48\,{b}^{2}}\sqrt{bx+a-1}\sqrt{bx+a+1}}-{\frac{{a}^{4}}{4\,{b}^{4}}\sqrt{bx+a-1}\sqrt{bx+a+1}\ln \left ( bx+a+\sqrt{ \left ( bx+a \right ) ^{2}-1} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}-1}}}}-{\frac{13\,x{a}^{2}}{48\,{b}^{3}}\sqrt{bx+a-1}\sqrt{bx+a+1}}+{\frac{25\,{a}^{3}}{48\,{b}^{4}}\sqrt{bx+a-1}\sqrt{bx+a+1}}-{\frac{3\,{a}^{2}}{4\,{b}^{4}}\sqrt{bx+a-1}\sqrt{bx+a+1}\ln \left ( bx+a+\sqrt{ \left ( bx+a \right ) ^{2}-1} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}-1}}}}-{\frac{3\,x}{32\,{b}^{3}}\sqrt{bx+a-1}\sqrt{bx+a+1}}+{\frac{55\,a}{96\,{b}^{4}}\sqrt{bx+a-1}\sqrt{bx+a+1}}-{\frac{3}{32\,{b}^{4}}\sqrt{bx+a-1}\sqrt{bx+a+1}\ln \left ( bx+a+\sqrt{ \left ( bx+a \right ) ^{2}-1} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*arccosh(b*x+a)-1/16*x^3*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)/b+7/48*a*x^2*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)/b
^2-1/4/b^4*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)/((b*x+a)^2-1)^(1/2)*a^4*ln(b*x+a+((b*x+a)^2-1)^(1/2))-13/48/b^3*(b*
x+a-1)^(1/2)*(b*x+a+1)^(1/2)*x*a^2+25/48/b^4*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)*a^3-3/4/b^4*(b*x+a-1)^(1/2)*(b*x+
a+1)^(1/2)/((b*x+a)^2-1)^(1/2)*a^2*ln(b*x+a+((b*x+a)^2-1)^(1/2))-3/32/b^3*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)*x+55
/96/b^4*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)*a-3/32/b^4*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)/((b*x+a)^2-1)^(1/2)*ln(b*x+
a+((b*x+a)^2-1)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.27272, size = 262, normalized size = 1.72 \begin{align*} \frac{3 \,{\left (8 \, b^{4} x^{4} - 8 \, a^{4} - 24 \, a^{2} - 3\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) -{\left (6 \, b^{3} x^{3} - 14 \, a b^{2} x^{2} - 50 \, a^{3} +{\left (26 \, a^{2} + 9\right )} b x - 55 \, a\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}}{96 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(8*b^4*x^4 - 8*a^4 - 24*a^2 - 3)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)) - (6*b^3*x^3 - 14*a*
b^2*x^2 - 50*a^3 + (26*a^2 + 9)*b*x - 55*a)*sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))/b^4

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Sympy [A]  time = 1.93184, size = 255, normalized size = 1.68 \begin{align*} \begin{cases} - \frac{a^{4} \operatorname{acosh}{\left (a + b x \right )}}{4 b^{4}} + \frac{25 a^{3} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} - 1}}{48 b^{4}} - \frac{13 a^{2} x \sqrt{a^{2} + 2 a b x + b^{2} x^{2} - 1}}{48 b^{3}} - \frac{3 a^{2} \operatorname{acosh}{\left (a + b x \right )}}{4 b^{4}} + \frac{7 a x^{2} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} - 1}}{48 b^{2}} + \frac{55 a \sqrt{a^{2} + 2 a b x + b^{2} x^{2} - 1}}{96 b^{4}} + \frac{x^{4} \operatorname{acosh}{\left (a + b x \right )}}{4} - \frac{x^{3} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} - 1}}{16 b} - \frac{3 x \sqrt{a^{2} + 2 a b x + b^{2} x^{2} - 1}}{32 b^{3}} - \frac{3 \operatorname{acosh}{\left (a + b x \right )}}{32 b^{4}} & \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+a),x)

[Out]

Piecewise((-a**4*acosh(a + b*x)/(4*b**4) + 25*a**3*sqrt(a**2 + 2*a*b*x + b**2*x**2 - 1)/(48*b**4) - 13*a**2*x*
sqrt(a**2 + 2*a*b*x + b**2*x**2 - 1)/(48*b**3) - 3*a**2*acosh(a + b*x)/(4*b**4) + 7*a*x**2*sqrt(a**2 + 2*a*b*x
+ b**2*x**2 - 1)/(48*b**2) + 55*a*sqrt(a**2 + 2*a*b*x + b**2*x**2 - 1)/(96*b**4) + x**4*acosh(a + b*x)/4 - x*
*3*sqrt(a**2 + 2*a*b*x + b**2*x**2 - 1)/(16*b) - 3*x*sqrt(a**2 + 2*a*b*x + b**2*x**2 - 1)/(32*b**3) - 3*acosh(
a + b*x)/(32*b**4), Ne(b, 0)), (x**4*acosh(a)/4, True))

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Giac [A]  time = 1.20281, size = 220, normalized size = 1.45 \begin{align*} \frac{1}{4} \, x^{4} \log \left (b x + a + \sqrt{{\left (b x + a\right )}^{2} - 1}\right ) - \frac{1}{96} \,{\left (\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{\left ({\left (2 \, x{\left (\frac{3 \, x}{b^{2}} - \frac{7 \, a}{b^{3}}\right )} + \frac{26 \, a^{2} b^{3} + 9 \, b^{3}}{b^{7}}\right )} x - \frac{5 \,{\left (10 \, a^{3} b^{2} + 11 \, a b^{2}\right )}}{b^{7}}\right )} - \frac{3 \,{\left (8 \, a^{4} + 24 \, a^{2} + 3\right )} \log \left ({\left | -a b -{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}{\left | b \right |} \right |}\right )}{b^{4}{\left | b \right |}}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*log(b*x + a + sqrt((b*x + a)^2 - 1)) - 1/96*(sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)*((2*x*(3*x/b^2 - 7*a/b^
3) + (26*a^2*b^3 + 9*b^3)/b^7)*x - 5*(10*a^3*b^2 + 11*a*b^2)/b^7) - 3*(8*a^4 + 24*a^2 + 3)*log(abs(-a*b - (x*a
bs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))*abs(b)))/(b^4*abs(b)))*b