3.100 \(\int \frac{a+b \cosh ^{-1}(c+d x)}{(c e+d e x)^3} \, dx\)

Optimal. Leaf size=66 \[ \frac{b \sqrt{c+d x-1} \sqrt{c+d x+1}}{2 d e^3 (c+d x)}-\frac{a+b \cosh ^{-1}(c+d x)}{2 d e^3 (c+d x)^2} \]

[Out]

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

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Rubi [A]  time = 0.0542246, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {5866, 12, 5662, 95} \[ \frac{b \sqrt{c+d x-1} \sqrt{c+d x+1}}{2 d e^3 (c+d x)}-\frac{a+b \cosh ^{-1}(c+d x)}{2 d e^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCosh[c + d*x])/(c*e + d*e*x)^3,x]

[Out]

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

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 12

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

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 95

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))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] /; FreeQ[{a, b, c, d,
e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && EqQ[a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1), 0
] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0561951, size = 55, normalized size = 0.83 \[ -\frac{a-b \sqrt{c+d x-1} (c+d x) \sqrt{c+d x+1}+b \cosh ^{-1}(c+d x)}{2 d e^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCosh[c + d*x])/(c*e + d*e*x)^3,x]

[Out]

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

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Maple [A]  time = 0.011, size = 65, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( -{\frac{a}{2\,{e}^{3} \left ( dx+c \right ) ^{2}}}+{\frac{b}{{e}^{3}} \left ( -{\frac{{\rm arccosh} \left (dx+c\right )}{2\, \left ( dx+c \right ) ^{2}}}+{\frac{1}{2\,dx+2\,c}\sqrt{dx+c-1}\sqrt{dx+c+1}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.76058, size = 159, normalized size = 2.41 \begin{align*} \frac{1}{2} \, b{\left (\frac{\sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} - 1} d}{d^{3} e^{3} x + c d^{2} e^{3}} - \frac{\operatorname{arcosh}\left (d x + c\right )}{d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}}\right )} - \frac{a}{2 \,{\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.41292, size = 257, normalized size = 3.89 \begin{align*} \frac{a d^{2} x^{2} + 2 \, a c d x - b c^{2} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right ) +{\left (b c^{2} d x + b c^{3}\right )} \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}}{2 \,{\left (c^{2} d^{3} e^{3} x^{2} + 2 \, c^{3} d^{2} e^{3} x + c^{4} d e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac{b \operatorname{acosh}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx}{e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3), x) + Integral(b*acosh(c + d*x)/(c**3 + 3*c**2*d*x
+ 3*c*d**2*x**2 + d**3*x**3), x))/e**3

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arcosh}\left (d x + c\right ) + a}{{\left (d e x + c e\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arccosh(d*x + c) + a)/(d*e*x + c*e)^3, x)