3.1 \(\int (d+e x)^3 \cosh ^{-1}(c x) \, dx\)

Optimal. Leaf size=183 \[ -\frac{\sqrt{c x-1} \sqrt{c x+1} \left (e x \left (26 c^2 d^2+9 e^2\right )+4 d \left (19 c^2 d^2+16 e^2\right )\right )}{96 c^3}-\frac{\left (24 c^2 d^2 e^2+8 c^4 d^4+3 e^4\right ) \cosh ^{-1}(c x)}{32 c^4 e}-\frac{\sqrt{c x-1} \sqrt{c x+1} (d+e x)^3}{16 c}-\frac{7 d \sqrt{c x-1} \sqrt{c x+1} (d+e x)^2}{48 c}+\frac{\cosh ^{-1}(c x) (d+e x)^4}{4 e} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.153374, antiderivative size = 183, 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 = {5802, 100, 153, 147, 52} \[ -\frac{\sqrt{c x-1} \sqrt{c x+1} \left (e x \left (26 c^2 d^2+9 e^2\right )+4 d \left (19 c^2 d^2+16 e^2\right )\right )}{96 c^3}-\frac{\left (24 c^2 d^2 e^2+8 c^4 d^4+3 e^4\right ) \cosh ^{-1}(c x)}{32 c^4 e}-\frac{\sqrt{c x-1} \sqrt{c x+1} (d+e x)^3}{16 c}-\frac{7 d \sqrt{c x-1} \sqrt{c x+1} (d+e x)^2}{48 c}+\frac{\cosh ^{-1}(c x) (d+e x)^4}{4 e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3*ArcCosh[c*x],x]

[Out]

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

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 (d+e x)^3 \cosh ^{-1}(c x) \, dx &=\frac{(d+e x)^4 \cosh ^{-1}(c x)}{4 e}-\frac{c \int \frac{(d+e x)^4}{\sqrt{-1+c x} \sqrt{1+c x}} \, dx}{4 e}\\ &=-\frac{\sqrt{-1+c x} \sqrt{1+c x} (d+e x)^3}{16 c}+\frac{(d+e x)^4 \cosh ^{-1}(c x)}{4 e}-\frac{\int \frac{(d+e x)^2 \left (4 c^2 d^2+3 e^2+7 c^2 d e x\right )}{\sqrt{-1+c x} \sqrt{1+c x}} \, dx}{16 c e}\\ &=-\frac{7 d \sqrt{-1+c x} \sqrt{1+c x} (d+e x)^2}{48 c}-\frac{\sqrt{-1+c x} \sqrt{1+c x} (d+e x)^3}{16 c}+\frac{(d+e x)^4 \cosh ^{-1}(c x)}{4 e}-\frac{\int \frac{(d+e x) \left (c^2 d \left (12 c^2 d^2+23 e^2\right )+c^2 e \left (26 c^2 d^2+9 e^2\right ) x\right )}{\sqrt{-1+c x} \sqrt{1+c x}} \, dx}{48 c^3 e}\\ &=-\frac{7 d \sqrt{-1+c x} \sqrt{1+c x} (d+e x)^2}{48 c}-\frac{\sqrt{-1+c x} \sqrt{1+c x} (d+e x)^3}{16 c}-\frac{\sqrt{-1+c x} \sqrt{1+c x} \left (4 d \left (19 c^2 d^2+16 e^2\right )+e \left (26 c^2 d^2+9 e^2\right ) x\right )}{96 c^3}+\frac{(d+e x)^4 \cosh ^{-1}(c x)}{4 e}-\frac{\left (8 c^4 d^4+24 c^2 d^2 e^2+3 e^4\right ) \int \frac{1}{\sqrt{-1+c x} \sqrt{1+c x}} \, dx}{32 c^3 e}\\ &=-\frac{7 d \sqrt{-1+c x} \sqrt{1+c x} (d+e x)^2}{48 c}-\frac{\sqrt{-1+c x} \sqrt{1+c x} (d+e x)^3}{16 c}-\frac{\sqrt{-1+c x} \sqrt{1+c x} \left (4 d \left (19 c^2 d^2+16 e^2\right )+e \left (26 c^2 d^2+9 e^2\right ) x\right )}{96 c^3}-\frac{\left (8 c^4 d^4+24 c^2 d^2 e^2+3 e^4\right ) \cosh ^{-1}(c x)}{32 c^4 e}+\frac{(d+e x)^4 \cosh ^{-1}(c x)}{4 e}\\ \end{align*}

Mathematica [A]  time = 0.248598, size = 153, normalized size = 0.84 \[ -\frac{c \sqrt{c x-1} \sqrt{c x+1} \left (c^2 \left (72 d^2 e x+96 d^3+32 d e^2 x^2+6 e^3 x^3\right )+e^2 (64 d+9 e x)\right )-24 c^4 x \cosh ^{-1}(c x) \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )+9 \left (8 c^2 d^2 e+e^3\right ) \log \left (c x+\sqrt{c x-1} \sqrt{c x+1}\right )}{96 c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3*ArcCosh[c*x],x]

[Out]

-(c*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(e^2*(64*d + 9*e*x) + c^2*(96*d^3 + 72*d^2*e*x + 32*d*e^2*x^2 + 6*e^3*x^3)) -
24*c^4*x*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3)*ArcCosh[c*x] + 9*(8*c^2*d^2*e + e^3)*Log[c*x + Sqrt[-1 +
c*x]*Sqrt[1 + c*x]])/(96*c^4)

________________________________________________________________________________________

Maple [B]  time = 0.032, size = 351, normalized size = 1.9 \begin{align*}{\frac{{e}^{3}{\rm arccosh} \left (cx\right ){x}^{4}}{4}}+{e}^{2}{\rm arccosh} \left (cx\right ){x}^{3}d+{\frac{3\,e{\rm arccosh} \left (cx\right ){x}^{2}{d}^{2}}{2}}+{\rm arccosh} \left (cx\right )x{d}^{3}+{\frac{{\rm arccosh} \left (cx\right ){d}^{4}}{4\,e}}-{\frac{{e}^{3}{x}^{3}}{16\,c}\sqrt{cx-1}\sqrt{cx+1}}-{\frac{{e}^{2}{x}^{2}d}{3\,c}\sqrt{cx-1}\sqrt{cx+1}}-{\frac{{d}^{4}}{4\,e}\sqrt{cx-1}\sqrt{cx+1}\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}-1} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}}}-{\frac{3\,e{d}^{2}x}{4\,c}\sqrt{cx-1}\sqrt{cx+1}}-{\frac{{d}^{3}}{c}\sqrt{cx-1}\sqrt{cx+1}}-{\frac{3\,e{d}^{2}}{4\,{c}^{2}}\sqrt{cx-1}\sqrt{cx+1}\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}-1} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}}}-{\frac{3\,{e}^{3}x}{32\,{c}^{3}}\sqrt{cx-1}\sqrt{cx+1}}-{\frac{2\,d{e}^{2}}{3\,{c}^{3}}\sqrt{cx-1}\sqrt{cx+1}}-{\frac{3\,{e}^{3}}{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((e*x+d)^3*arccosh(c*x),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.19868, size = 333, normalized size = 1.82 \begin{align*} -\frac{1}{96} \,{\left (\frac{6 \, \sqrt{c^{2} x^{2} - 1} e^{3} x^{3}}{c^{2}} + \frac{32 \, \sqrt{c^{2} x^{2} - 1} d e^{2} x^{2}}{c^{2}} + \frac{72 \, \sqrt{c^{2} x^{2} - 1} d^{2} e x}{c^{2}} + \frac{96 \, \sqrt{c^{2} x^{2} - 1} d^{3}}{c^{2}} + \frac{72 \, d^{2} e \log \left (2 \, c^{2} x + 2 \, \sqrt{c^{2} x^{2} - 1} \sqrt{c^{2}}\right )}{\sqrt{c^{2}} c^{2}} + \frac{9 \, \sqrt{c^{2} x^{2} - 1} e^{3} x}{c^{4}} + \frac{64 \, \sqrt{c^{2} x^{2} - 1} d e^{2}}{c^{4}} + \frac{9 \, e^{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 + \frac{1}{4} \,{\left (e^{3} x^{4} + 4 \, d e^{2} x^{3} + 6 \, d^{2} e x^{2} + 4 \, d^{3} x\right )} \operatorname{arcosh}\left (c x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*arccosh(c*x),x, algorithm="maxima")

[Out]

-1/96*(6*sqrt(c^2*x^2 - 1)*e^3*x^3/c^2 + 32*sqrt(c^2*x^2 - 1)*d*e^2*x^2/c^2 + 72*sqrt(c^2*x^2 - 1)*d^2*e*x/c^2
+ 96*sqrt(c^2*x^2 - 1)*d^3/c^2 + 72*d^2*e*log(2*c^2*x + 2*sqrt(c^2*x^2 - 1)*sqrt(c^2))/(sqrt(c^2)*c^2) + 9*sq
rt(c^2*x^2 - 1)*e^3*x/c^4 + 64*sqrt(c^2*x^2 - 1)*d*e^2/c^4 + 9*e^3*log(2*c^2*x + 2*sqrt(c^2*x^2 - 1)*sqrt(c^2)
)/(sqrt(c^2)*c^4))*c + 1/4*(e^3*x^4 + 4*d*e^2*x^3 + 6*d^2*e*x^2 + 4*d^3*x)*arccosh(c*x)

________________________________________________________________________________________

Fricas [A]  time = 1.92111, size = 329, normalized size = 1.8 \begin{align*} \frac{3 \,{\left (8 \, c^{4} e^{3} x^{4} + 32 \, c^{4} d e^{2} x^{3} + 48 \, c^{4} d^{2} e x^{2} + 32 \, c^{4} d^{3} x - 24 \, c^{2} d^{2} e - 3 \, e^{3}\right )} \log \left (c x + \sqrt{c^{2} x^{2} - 1}\right ) -{\left (6 \, c^{3} e^{3} x^{3} + 32 \, c^{3} d e^{2} x^{2} + 96 \, c^{3} d^{3} + 64 \, c d e^{2} + 9 \,{\left (8 \, c^{3} d^{2} e + c e^{3}\right )} x\right )} \sqrt{c^{2} x^{2} - 1}}{96 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*arccosh(c*x),x, algorithm="fricas")

[Out]

1/96*(3*(8*c^4*e^3*x^4 + 32*c^4*d*e^2*x^3 + 48*c^4*d^2*e*x^2 + 32*c^4*d^3*x - 24*c^2*d^2*e - 3*e^3)*log(c*x +
sqrt(c^2*x^2 - 1)) - (6*c^3*e^3*x^3 + 32*c^3*d*e^2*x^2 + 96*c^3*d^3 + 64*c*d*e^2 + 9*(8*c^3*d^2*e + c*e^3)*x)*
sqrt(c^2*x^2 - 1))/c^4

________________________________________________________________________________________

Sympy [A]  time = 1.79107, size = 258, normalized size = 1.41 \begin{align*} \begin{cases} d^{3} x \operatorname{acosh}{\left (c x \right )} + \frac{3 d^{2} e x^{2} \operatorname{acosh}{\left (c x \right )}}{2} + d e^{2} x^{3} \operatorname{acosh}{\left (c x \right )} + \frac{e^{3} x^{4} \operatorname{acosh}{\left (c x \right )}}{4} - \frac{d^{3} \sqrt{c^{2} x^{2} - 1}}{c} - \frac{3 d^{2} e x \sqrt{c^{2} x^{2} - 1}}{4 c} - \frac{d e^{2} x^{2} \sqrt{c^{2} x^{2} - 1}}{3 c} - \frac{e^{3} x^{3} \sqrt{c^{2} x^{2} - 1}}{16 c} - \frac{3 d^{2} e \operatorname{acosh}{\left (c x \right )}}{4 c^{2}} - \frac{2 d e^{2} \sqrt{c^{2} x^{2} - 1}}{3 c^{3}} - \frac{3 e^{3} x \sqrt{c^{2} x^{2} - 1}}{32 c^{3}} - \frac{3 e^{3} \operatorname{acosh}{\left (c x \right )}}{32 c^{4}} & \text{for}\: c \neq 0 \\\frac{i \pi \left (d^{3} x + \frac{3 d^{2} e x^{2}}{2} + d e^{2} x^{3} + \frac{e^{3} x^{4}}{4}\right )}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*acosh(c*x),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.20013, size = 225, normalized size = 1.23 \begin{align*} \frac{1}{4} \,{\left (x e + d\right )}^{4} e^{\left (-1\right )} \log \left (c x + \sqrt{c^{2} x^{2} - 1}\right ) - \frac{1}{96} \,{\left (\sqrt{c^{2} x^{2} - 1}{\left ({\left (2 \, x{\left (\frac{3 \, x e^{4}}{c} + \frac{16 \, d e^{3}}{c}\right )} + \frac{9 \,{\left (8 \, c^{5} d^{2} e^{2} + c^{3} e^{4}\right )}}{c^{6}}\right )} x + \frac{32 \,{\left (3 \, c^{5} d^{3} e + 2 \, c^{3} d e^{3}\right )}}{c^{6}}\right )} - \frac{3 \,{\left (8 \, c^{4} d^{4} + 24 \, c^{2} d^{2} e^{2} + 3 \, e^{4}\right )} \log \left ({\left | -x{\left | c \right |} + \sqrt{c^{2} x^{2} - 1} \right |}\right )}{c^{3}{\left | c \right |}}\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*arccosh(c*x),x, algorithm="giac")

[Out]

1/4*(x*e + d)^4*e^(-1)*log(c*x + sqrt(c^2*x^2 - 1)) - 1/96*(sqrt(c^2*x^2 - 1)*((2*x*(3*x*e^4/c + 16*d*e^3/c) +
9*(8*c^5*d^2*e^2 + c^3*e^4)/c^6)*x + 32*(3*c^5*d^3*e + 2*c^3*d*e^3)/c^6) - 3*(8*c^4*d^4 + 24*c^2*d^2*e^2 + 3*
e^4)*log(abs(-x*abs(c) + sqrt(c^2*x^2 - 1)))/(c^3*abs(c)))*e^(-1)