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

Optimal. Leaf size=97 \[ -\frac {1}{4} \left (\frac {e}{c^2}+\frac {2 d^2}{e}\right ) \cosh ^{-1}(c x)-\frac {\sqrt {c x-1} \sqrt {c x+1} (d+e x)}{4 c}+\frac {\cosh ^{-1}(c x) (d+e x)^2}{2 e}-\frac {3 d \sqrt {c x-1} \sqrt {c x+1}}{4 c} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5802, 90, 80, 52} \[ -\frac {1}{4} \left (\frac {e}{c^2}+\frac {2 d^2}{e}\right ) \cosh ^{-1}(c x)-\frac {\sqrt {c x-1} \sqrt {c x+1} (d+e x)}{4 c}+\frac {\cosh ^{-1}(c x) (d+e x)^2}{2 e}-\frac {3 d \sqrt {c x-1} \sqrt {c x+1}}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*ArcCosh[c*x],x]

[Out]

(-3*d*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(4*c) - (Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(d + e*x))/(4*c) - (((2*d^2)/e + e/c
^2)*ArcCosh[c*x])/4 + ((d + e*x)^2*ArcCosh[c*x])/(2*e)

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]

Rule 80

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

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 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]

Rubi steps

\begin {align*} \int (d+e x) \cosh ^{-1}(c x) \, dx &=\frac {(d+e x)^2 \cosh ^{-1}(c x)}{2 e}-\frac {c \int \frac {(d+e x)^2}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{2 e}\\ &=-\frac {\sqrt {-1+c x} \sqrt {1+c x} (d+e x)}{4 c}+\frac {(d+e x)^2 \cosh ^{-1}(c x)}{2 e}-\frac {\int \frac {2 c^2 d^2+e^2+3 c^2 d e x}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{4 c e}\\ &=-\frac {3 d \sqrt {-1+c x} \sqrt {1+c x}}{4 c}-\frac {\sqrt {-1+c x} \sqrt {1+c x} (d+e x)}{4 c}+\frac {(d+e x)^2 \cosh ^{-1}(c x)}{2 e}-\frac {1}{4} \left (\frac {2 c d^2}{e}+\frac {e}{c}\right ) \int \frac {1}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx\\ &=-\frac {3 d \sqrt {-1+c x} \sqrt {1+c x}}{4 c}-\frac {\sqrt {-1+c x} \sqrt {1+c x} (d+e x)}{4 c}-\frac {1}{4} \left (\frac {2 d^2}{e}+\frac {e}{c^2}\right ) \cosh ^{-1}(c x)+\frac {(d+e x)^2 \cosh ^{-1}(c x)}{2 e}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 73, normalized size = 0.75 \[ -\frac {-2 c^2 x \cosh ^{-1}(c x) (2 d+e x)+c \sqrt {c x-1} \sqrt {c x+1} (4 d+e x)+2 e \tanh ^{-1}\left (\sqrt {\frac {c x-1}{c x+1}}\right )}{4 c^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d + e*x)*ArcCosh[c*x],x]

[Out]

-1/4*(c*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(4*d + e*x) - 2*c^2*x*(2*d + e*x)*ArcCosh[c*x] + 2*e*ArcTanh[Sqrt[(-1 + c
*x)/(1 + c*x)]])/c^2

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fricas [A]  time = 0.49, size = 65, normalized size = 0.67 \[ \frac {{\left (2 \, c^{2} e x^{2} + 4 \, c^{2} d x - e\right )} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) - \sqrt {c^{2} x^{2} - 1} {\left (c e x + 4 \, c d\right )}}{4 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*c^2*e*x^2 + 4*c^2*d*x - e)*log(c*x + sqrt(c^2*x^2 - 1)) - sqrt(c^2*x^2 - 1)*(c*e*x + 4*c*d))/c^2

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giac [A]  time = 0.29, size = 87, normalized size = 0.90 \[ \frac {1}{2} \, {\left (x^{2} e + 2 \, d x\right )} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) - \frac {1}{4} \, \sqrt {c^{2} x^{2} - 1} {\left (\frac {x e}{c} + \frac {4 \, d}{c}\right )} + \frac {e \log \left ({\left | -x {\left | c \right |} + \sqrt {c^{2} x^{2} - 1} \right |}\right )}{4 \, c {\left | c \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x^2*e + 2*d*x)*log(c*x + sqrt(c^2*x^2 - 1)) - 1/4*sqrt(c^2*x^2 - 1)*(x*e/c + 4*d/c) + 1/4*e*log(abs(-x*ab
s(c) + sqrt(c^2*x^2 - 1)))/(c*abs(c))

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maple [A]  time = 0.01, size = 107, normalized size = 1.10 \[ \frac {\mathrm {arccosh}\left (c x \right ) x^{2} e}{2}+\mathrm {arccosh}\left (c x \right ) x d -\frac {\sqrt {c x -1}\, \sqrt {c x +1}\, e x}{4 c}-\frac {d \sqrt {c x -1}\, \sqrt {c x +1}}{c}-\frac {\sqrt {c x -1}\, \sqrt {c x +1}\, e \ln \left (c x +\sqrt {c^{2} x^{2}-1}\right )}{4 c^{2} \sqrt {c^{2} x^{2}-1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*arccosh(c*x),x)

[Out]

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

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maxima [A]  time = 0.46, size = 82, normalized size = 0.85 \[ -\frac {1}{4} \, c {\left (\frac {\sqrt {c^{2} x^{2} - 1} e x}{c^{2}} + \frac {4 \, \sqrt {c^{2} x^{2} - 1} d}{c^{2}} + \frac {e \log \left (2 \, c^{2} x + 2 \, \sqrt {c^{2} x^{2} - 1} c\right )}{c^{3}}\right )} + \frac {1}{2} \, {\left (e x^{2} + 2 \, d x\right )} \operatorname {arcosh}\left (c x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*c*(sqrt(c^2*x^2 - 1)*e*x/c^2 + 4*sqrt(c^2*x^2 - 1)*d/c^2 + e*log(2*c^2*x + 2*sqrt(c^2*x^2 - 1)*c)/c^3) +
1/2*(e*x^2 + 2*d*x)*arccosh(c*x)

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mupad [B]  time = 0.71, size = 68, normalized size = 0.70 \[ d\,x\,\mathrm {acosh}\left (c\,x\right )+e\,x\,\mathrm {acosh}\left (c\,x\right )\,\left (\frac {x}{2}-\frac {1}{4\,c^2\,x}\right )-\frac {d\,\sqrt {c\,x-1}\,\sqrt {c\,x+1}}{c}-\frac {e\,x\,\sqrt {c\,x-1}\,\sqrt {c\,x+1}}{4\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acosh(c*x)*(d + e*x),x)

[Out]

d*x*acosh(c*x) + e*x*acosh(c*x)*(x/2 - 1/(4*c^2*x)) - (d*(c*x - 1)^(1/2)*(c*x + 1)^(1/2))/c - (e*x*(c*x - 1)^(
1/2)*(c*x + 1)^(1/2))/(4*c)

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sympy [A]  time = 0.26, size = 80, normalized size = 0.82 \[ \begin {cases} d x \operatorname {acosh}{\left (c x \right )} + \frac {e x^{2} \operatorname {acosh}{\left (c x \right )}}{2} - \frac {d \sqrt {c^{2} x^{2} - 1}}{c} - \frac {e x \sqrt {c^{2} x^{2} - 1}}{4 c} - \frac {e \operatorname {acosh}{\left (c x \right )}}{4 c^{2}} & \text {for}\: c \neq 0 \\\frac {i \pi \left (d x + \frac {e x^{2}}{2}\right )}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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