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

Optimal. Leaf size=132 \[ -\frac{c \sqrt{c x-1} \sqrt{c x+1}}{2 \left (c^2 d^2-e^2\right ) (d+e x)}+\frac{c^3 d \tanh ^{-1}\left (\frac{\sqrt{c x+1} \sqrt{c d+e}}{\sqrt{c x-1} \sqrt{c d-e}}\right )}{e (c d-e)^{3/2} (c d+e)^{3/2}}-\frac{\cosh ^{-1}(c x)}{2 e (d+e x)^2} \]

[Out]

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

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Rubi [A]  time = 0.134899, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5802, 96, 93, 208} \[ -\frac{c \sqrt{c x-1} \sqrt{c x+1}}{2 \left (c^2 d^2-e^2\right ) (d+e x)}+\frac{c^3 d \tanh ^{-1}\left (\frac{\sqrt{c x+1} \sqrt{c d+e}}{\sqrt{c x-1} \sqrt{c d-e}}\right )}{e (c d-e)^{3/2} (c d+e)^{3/2}}-\frac{\cosh ^{-1}(c x)}{2 e (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcCosh[c*x]/(d + e*x)^3,x]

[Out]

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

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 96

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] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.185377, size = 190, normalized size = 1.44 \[ \frac{c (d+e x) \left (-e \sqrt{c x-1} \sqrt{c x+1} \sqrt{c^2 d^2-e^2}-c^2 d (d+e x) \log \left (-\sqrt{c x-1} \sqrt{c x+1} \sqrt{c^2 d^2-e^2}+c^2 d x+e\right )+c^2 d (d+e x) \log (d+e x)\right )-\left (c^2 d^2-e^2\right )^{3/2} \cosh ^{-1}(c x)}{2 e (c d-e) (c d+e) \sqrt{c^2 d^2-e^2} (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCosh[c*x]/(d + e*x)^3,x]

[Out]

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

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Maple [B]  time = 0.025, size = 338, normalized size = 2.6 \begin{align*} -{\frac{{c}^{2}{\rm arccosh} \left (cx\right )}{2\, \left ( cxe+cd \right ) ^{2}e}}-{\frac{{c}^{4}xd}{2\,e \left ( cd+e \right ) \left ( cd-e \right ) \left ( cxe+cd \right ) }\sqrt{cx-1}\sqrt{cx+1}\ln \left ( -2\,{\frac{1}{cxe+cd} \left ({c}^{2}dx-\sqrt{{c}^{2}{x}^{2}-1}\sqrt{{\frac{{c}^{2}{d}^{2}-{e}^{2}}{{e}^{2}}}}e+e \right ) } \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}}{\frac{1}{\sqrt{{\frac{{c}^{2}{d}^{2}-{e}^{2}}{{e}^{2}}}}}}}-{\frac{{d}^{2}{c}^{4}}{2\,{e}^{2} \left ( cd+e \right ) \left ( cd-e \right ) \left ( cxe+cd \right ) }\sqrt{cx-1}\sqrt{cx+1}\ln \left ( -2\,{\frac{1}{cxe+cd} \left ({c}^{2}dx-\sqrt{{c}^{2}{x}^{2}-1}\sqrt{{\frac{{c}^{2}{d}^{2}-{e}^{2}}{{e}^{2}}}}e+e \right ) } \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}}{\frac{1}{\sqrt{{\frac{{c}^{2}{d}^{2}-{e}^{2}}{{e}^{2}}}}}}}-{\frac{{c}^{2}}{ \left ( 2\,cd+2\,e \right ) \left ( cd-e \right ) \left ( cxe+cd \right ) }\sqrt{cx-1}\sqrt{cx+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.959, size = 2040, normalized size = 15.45 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(c^4*d^6 - c^2*d^4*e^2 + (c^4*d^4*e^2 - c^2*d^2*e^4)*x^2 + (c^3*d^3*e^2*x^2 + 2*c^3*d^4*e*x + c^3*d^5)*s
qrt(c^2*d^2 - e^2)*log((c^3*d^2*x + c*d*e - sqrt(c^2*d^2 - e^2)*(c^2*d*x + e) + (c^2*d^2 - sqrt(c^2*d^2 - e^2)
*c*d - e^2)*sqrt(c^2*x^2 - 1))/(e*x + d)) + 2*(c^4*d^5*e - c^2*d^3*e^3)*x - ((c^4*d^4*e^2 - 2*c^2*d^2*e^4 + e^
6)*x^2 + 2*(c^4*d^5*e - 2*c^2*d^3*e^3 + d*e^5)*x)*log(c*x + sqrt(c^2*x^2 - 1)) - (c^4*d^6 - 2*c^2*d^4*e^2 + d^
2*e^4 + (c^4*d^4*e^2 - 2*c^2*d^2*e^4 + e^6)*x^2 + 2*(c^4*d^5*e - 2*c^2*d^3*e^3 + d*e^5)*x)*log(-c*x + sqrt(c^2
*x^2 - 1)) + (c^3*d^5*e - c*d^3*e^3 + (c^3*d^4*e^2 - c*d^2*e^4)*x)*sqrt(c^2*x^2 - 1))/(c^4*d^8*e - 2*c^2*d^6*e
^3 + d^4*e^5 + (c^4*d^6*e^3 - 2*c^2*d^4*e^5 + d^2*e^7)*x^2 + 2*(c^4*d^7*e^2 - 2*c^2*d^5*e^4 + d^3*e^6)*x), -1/
2*(c^4*d^6 - c^2*d^4*e^2 + (c^4*d^4*e^2 - c^2*d^2*e^4)*x^2 + 2*(c^3*d^3*e^2*x^2 + 2*c^3*d^4*e*x + c^3*d^5)*sqr
t(-c^2*d^2 + e^2)*arctan(-(sqrt(-c^2*d^2 + e^2)*sqrt(c^2*x^2 - 1)*e - sqrt(-c^2*d^2 + e^2)*(c*e*x + c*d))/(c^2
*d^2 - e^2)) + 2*(c^4*d^5*e - c^2*d^3*e^3)*x - ((c^4*d^4*e^2 - 2*c^2*d^2*e^4 + e^6)*x^2 + 2*(c^4*d^5*e - 2*c^2
*d^3*e^3 + d*e^5)*x)*log(c*x + sqrt(c^2*x^2 - 1)) - (c^4*d^6 - 2*c^2*d^4*e^2 + d^2*e^4 + (c^4*d^4*e^2 - 2*c^2*
d^2*e^4 + e^6)*x^2 + 2*(c^4*d^5*e - 2*c^2*d^3*e^3 + d*e^5)*x)*log(-c*x + sqrt(c^2*x^2 - 1)) + (c^3*d^5*e - c*d
^3*e^3 + (c^3*d^4*e^2 - c*d^2*e^4)*x)*sqrt(c^2*x^2 - 1))/(c^4*d^8*e - 2*c^2*d^6*e^3 + d^4*e^5 + (c^4*d^6*e^3 -
2*c^2*d^4*e^5 + d^2*e^7)*x^2 + 2*(c^4*d^7*e^2 - 2*c^2*d^5*e^4 + d^3*e^6)*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acosh(c*x)/(d + e*x)**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError