3.467 \(\int \frac{(d+e x^2) (a+b \cosh ^{-1}(c x))}{x^2} \, dx\)

Optimal. Leaf size=75 \[ -\frac{d \left (a+b \cosh ^{-1}(c x)\right )}{x}+e x \left (a+b \cosh ^{-1}(c x)\right )+b c d \tan ^{-1}\left (\sqrt{c x-1} \sqrt{c x+1}\right )-\frac{b e \sqrt{c x-1} \sqrt{c x+1}}{c} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0991049, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {5786, 460, 92, 205} \[ -\frac{d \left (a+b \cosh ^{-1}(c x)\right )}{x}+e x \left (a+b \cosh ^{-1}(c x)\right )+b c d \tan ^{-1}\left (\sqrt{c x-1} \sqrt{c x+1}\right )-\frac{b e \sqrt{c x-1} \sqrt{c x+1}}{c} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)*(a + b*ArcCosh[c*x]))/x^2,x]

[Out]

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

Rule 5786

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(d*(f*x)^(
m + 1)*(a + b*ArcCosh[c*x]))/(f*(m + 1)), x] + (-Dist[(b*c)/(f*(m + 1)*(m + 3)), Int[((f*x)^(m + 1)*(d*(m + 3)
+ e*(m + 1)*x^2))/(Sqrt[1 + c*x]*Sqrt[-1 + c*x]), x], x] + Simp[(e*(f*x)^(m + 3)*(a + b*ArcCosh[c*x]))/(f^3*(
m + 3)), x]) /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[c^2*d + e, 0] && NeQ[m, -1] && NeQ[m, -3]

Rule 460

Int[((e_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.)*(x_)^(non2_.))^(p_.)*((c_) + (d_.)
*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m + 1)*(a1 + b1*x^(n/2))^(p + 1)*(a2 + b2*x^(n/2))^(p + 1))/(b1*b2*e*
(m + n*(p + 1) + 1)), x] - Dist[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/(b1*b2*(m + n*(p + 1) + 1)), I
nt[(e*x)^m*(a1 + b1*x^(n/2))^p*(a2 + b2*x^(n/2))^p, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, e, m, n, p}, x] &&
EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && NeQ[m + n*(p + 1) + 1, 0]

Rule 92

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 205

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

Rubi steps

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

Mathematica [A]  time = 0.131937, size = 105, normalized size = 1.4 \[ -\frac{a d}{x}+a e x+\frac{b c d \sqrt{c^2 x^2-1} \tan ^{-1}\left (\sqrt{c^2 x^2-1}\right )}{\sqrt{c x-1} \sqrt{c x+1}}-\frac{b d \cosh ^{-1}(c x)}{x}-\frac{b e \sqrt{c x-1} \sqrt{c x+1}}{c}+b e x \cosh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)*(a + b*ArcCosh[c*x]))/x^2,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.017, size = 95, normalized size = 1.3 \begin{align*} axe-{\frac{ad}{x}}+b{\rm arccosh} \left (cx\right )xe-{\frac{bd{\rm arccosh} \left (cx\right )}{x}}-{cbd\sqrt{cx-1}\sqrt{cx+1}\arctan \left ({\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}}}-{\frac{be}{c}\sqrt{cx-1}\sqrt{cx+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)*(a+b*arccosh(c*x))/x^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.71594, size = 88, normalized size = 1.17 \begin{align*} -{\left (c \arcsin \left (\frac{1}{\sqrt{c^{2}}{\left | x \right |}}\right ) + \frac{\operatorname{arcosh}\left (c x\right )}{x}\right )} b d + a e x + \frac{{\left (c x \operatorname{arcosh}\left (c x\right ) - \sqrt{c^{2} x^{2} - 1}\right )} b e}{c} - \frac{a d}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arccosh(c*x))/x^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.61705, size = 298, normalized size = 3.97 \begin{align*} \frac{2 \, b c^{2} d x \arctan \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) + a c e x^{2} - \sqrt{c^{2} x^{2} - 1} b e x - a c d +{\left (b c d - b c e\right )} x \log \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) +{\left (b c e x^{2} - b c d +{\left (b c d - b c e\right )} x\right )} \log \left (c x + \sqrt{c^{2} x^{2} - 1}\right )}{c x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arccosh(c*x))/x^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arccosh(c*x))/x^2,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)*(b*arccosh(c*x) + a)/x^2, x)