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

Optimal. Leaf size=160 \[ -\frac{d^2 \left (a+b \cosh ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac{b e \left (1-c^2 x^2\right ) \left (6 c^2 d+e\right )}{3 c^3 \sqrt{c x-1} \sqrt{c x+1}}-\frac{b e^2 \left (1-c^2 x^2\right )^2}{9 c^3 \sqrt{c x-1} \sqrt{c x+1}}+b c d^2 \tan ^{-1}\left (\sqrt{c x-1} \sqrt{c x+1}\right ) \]

[Out]

(b*e*(6*c^2*d + e)*(1 - c^2*x^2))/(3*c^3*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) - (b*e^2*(1 - c^2*x^2)^2)/(9*c^3*Sqrt[-
1 + c*x]*Sqrt[1 + c*x]) - (d^2*(a + b*ArcCosh[c*x]))/x + 2*d*e*x*(a + b*ArcCosh[c*x]) + (e^2*x^3*(a + b*ArcCos
h[c*x]))/3 + b*c*d^2*ArcTan[Sqrt[-1 + c*x]*Sqrt[1 + c*x]]

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Rubi [A]  time = 0.302226, antiderivative size = 185, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {270, 5790, 520, 1251, 897, 1153, 205} \[ -\frac{d^2 \left (a+b \cosh ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac{b c d^2 \sqrt{c^2 x^2-1} \tan ^{-1}\left (\sqrt{c^2 x^2-1}\right )}{\sqrt{c x-1} \sqrt{c x+1}}+\frac{b e \left (1-c^2 x^2\right ) \left (6 c^2 d+e\right )}{3 c^3 \sqrt{c x-1} \sqrt{c x+1}}-\frac{b e^2 \left (1-c^2 x^2\right )^2}{9 c^3 \sqrt{c x-1} \sqrt{c x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e*(6*c^2*d + e)*(1 - c^2*x^2))/(3*c^3*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) - (b*e^2*(1 - c^2*x^2)^2)/(9*c^3*Sqrt[-
1 + c*x]*Sqrt[1 + c*x]) - (d^2*(a + b*ArcCosh[c*x]))/x + 2*d*e*x*(a + b*ArcCosh[c*x]) + (e^2*x^3*(a + b*ArcCos
h[c*x]))/3 + (b*c*d^2*Sqrt[-1 + c^2*x^2]*ArcTan[Sqrt[-1 + c^2*x^2]])/(Sqrt[-1 + c*x]*Sqrt[1 + c*x])

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 5790

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcCosh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(Sqrt[
1 + c*x]*Sqrt[-1 + c*x]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p] &
& (GtQ[p, 0] || (IGtQ[(m - 1)/2, 0] && LeQ[m + p, 0]))

Rule 520

Int[(u_.)*((c_) + (d_.)*(x_)^(n_.) + (e_.)*(x_)^(n2_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b
2_.)*(x_)^(non2_.))^(p_.), x_Symbol] :> Dist[((a1 + b1*x^(n/2))^FracPart[p]*(a2 + b2*x^(n/2))^FracPart[p])/(a1
*a2 + b1*b2*x^n)^FracPart[p], Int[u*(a1*a2 + b1*b2*x^n)^p*(c + d*x^n + e*x^(2*n))^q, x], x] /; FreeQ[{a1, b1,
a2, b2, c, d, e, n, p, q}, x] && EqQ[non2, n/2] && EqQ[n2, 2*n] && EqQ[a2*b1 + a1*b2, 0]

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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 )^2 \left (a+b \cosh ^{-1}(c x)\right )}{x^2} \, dx &=-\frac{d^2 \left (a+b \cosh ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )-(b c) \int \frac{-d^2+2 d e x^2+\frac{e^2 x^4}{3}}{x \sqrt{-1+c x} \sqrt{1+c x}} \, dx\\ &=-\frac{d^2 \left (a+b \cosh ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )-\frac{\left (b c \sqrt{-1+c^2 x^2}\right ) \int \frac{-d^2+2 d e x^2+\frac{e^2 x^4}{3}}{x \sqrt{-1+c^2 x^2}} \, dx}{\sqrt{-1+c x} \sqrt{1+c x}}\\ &=-\frac{d^2 \left (a+b \cosh ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )-\frac{\left (b c \sqrt{-1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{-d^2+2 d e x+\frac{e^2 x^2}{3}}{x \sqrt{-1+c^2 x}} \, dx,x,x^2\right )}{2 \sqrt{-1+c x} \sqrt{1+c x}}\\ &=-\frac{d^2 \left (a+b \cosh ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )-\frac{\left (b \sqrt{-1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\frac{-c^4 d^2+2 c^2 d e+\frac{e^2}{3}}{c^4}-\frac{\left (-2 c^2 d e-\frac{2 e^2}{3}\right ) x^2}{c^4}+\frac{e^2 x^4}{3 c^4}}{\frac{1}{c^2}+\frac{x^2}{c^2}} \, dx,x,\sqrt{-1+c^2 x^2}\right )}{c \sqrt{-1+c x} \sqrt{1+c x}}\\ &=-\frac{d^2 \left (a+b \cosh ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )-\frac{\left (b \sqrt{-1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{3} e \left (6 d+\frac{e}{c^2}\right )+\frac{e^2 x^2}{3 c^2}-\frac{d^2}{\frac{1}{c^2}+\frac{x^2}{c^2}}\right ) \, dx,x,\sqrt{-1+c^2 x^2}\right )}{c \sqrt{-1+c x} \sqrt{1+c x}}\\ &=\frac{b e \left (6 c^2 d+e\right ) \left (1-c^2 x^2\right )}{3 c^3 \sqrt{-1+c x} \sqrt{1+c x}}-\frac{b e^2 \left (1-c^2 x^2\right )^2}{9 c^3 \sqrt{-1+c x} \sqrt{1+c x}}-\frac{d^2 \left (a+b \cosh ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac{\left (b d^2 \sqrt{-1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{c^2}+\frac{x^2}{c^2}} \, dx,x,\sqrt{-1+c^2 x^2}\right )}{c \sqrt{-1+c x} \sqrt{1+c x}}\\ &=\frac{b e \left (6 c^2 d+e\right ) \left (1-c^2 x^2\right )}{3 c^3 \sqrt{-1+c x} \sqrt{1+c x}}-\frac{b e^2 \left (1-c^2 x^2\right )^2}{9 c^3 \sqrt{-1+c x} \sqrt{1+c x}}-\frac{d^2 \left (a+b \cosh ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac{b c d^2 \sqrt{-1+c^2 x^2} \tan ^{-1}\left (\sqrt{-1+c^2 x^2}\right )}{\sqrt{-1+c x} \sqrt{1+c x}}\\ \end{align*}

Mathematica [A]  time = 0.2309, size = 128, normalized size = 0.8 \[ \frac{1}{3} \left (-\frac{3 a d^2}{x}+6 a d e x+a e^2 x^3-\frac{b e \sqrt{c x-1} \sqrt{c x+1} \left (c^2 \left (18 d+e x^2\right )+2 e\right )}{3 c^3}+\frac{b \cosh ^{-1}(c x) \left (-3 d^2+6 d e x^2+e^2 x^4\right )}{x}-3 b c d^2 \tan ^{-1}\left (\frac{1}{\sqrt{c x-1} \sqrt{c x+1}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*a*d^2)/x + 6*a*d*e*x + a*e^2*x^3 - (b*e*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(2*e + c^2*(18*d + e*x^2)))/(3*c^3)
+ (b*(-3*d^2 + 6*d*e*x^2 + e^2*x^4)*ArcCosh[c*x])/x - 3*b*c*d^2*ArcTan[1/(Sqrt[-1 + c*x]*Sqrt[1 + c*x])])/3

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Maple [A]  time = 0.018, size = 177, normalized size = 1.1 \begin{align*}{\frac{a{x}^{3}{e}^{2}}{3}}+2\,axde-{\frac{{d}^{2}a}{x}}+{\frac{b{\rm arccosh} \left (cx\right ){x}^{3}{e}^{2}}{3}}+2\,b{\rm arccosh} \left (cx\right )xde-{\frac{b{d}^{2}{\rm arccosh} \left (cx\right )}{x}}-{c{d}^{2}b\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{b{x}^{2}{e}^{2}}{9\,c}\sqrt{cx-1}\sqrt{cx+1}}-2\,{\frac{\sqrt{cx+1}\sqrt{cx-1}bde}{c}}-{\frac{2\,b{e}^{2}}{9\,{c}^{3}}\sqrt{cx-1}\sqrt{cx+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*x^3*e^2+2*a*x*d*e-d^2*a/x+1/3*b*arccosh(c*x)*x^3*e^2+2*b*arccosh(c*x)*x*d*e-d^2*b*arccosh(c*x)/x-c*d^2*b
*(c*x-1)^(1/2)*(c*x+1)^(1/2)/(c^2*x^2-1)^(1/2)*arctan(1/(c^2*x^2-1)^(1/2))-1/9*b/c*(c*x-1)^(1/2)*(c*x+1)^(1/2)
*x^2*e^2-2*b/c*(c*x-1)^(1/2)*(c*x+1)^(1/2)*d*e-2/9*b/c^3*(c*x-1)^(1/2)*(c*x+1)^(1/2)*e^2

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Maxima [A]  time = 1.72476, size = 184, normalized size = 1.15 \begin{align*} \frac{1}{3} \, a e^{2} x^{3} -{\left (c \arcsin \left (\frac{1}{\sqrt{c^{2}}{\left | x \right |}}\right ) + \frac{\operatorname{arcosh}\left (c x\right )}{x}\right )} b d^{2} + \frac{1}{9} \,{\left (3 \, x^{3} \operatorname{arcosh}\left (c x\right ) - c{\left (\frac{\sqrt{c^{2} x^{2} - 1} x^{2}}{c^{2}} + \frac{2 \, \sqrt{c^{2} x^{2} - 1}}{c^{4}}\right )}\right )} b e^{2} + 2 \, a d e x + \frac{2 \,{\left (c x \operatorname{arcosh}\left (c x\right ) - \sqrt{c^{2} x^{2} - 1}\right )} b d e}{c} - \frac{a d^{2}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*e^2*x^3 - (c*arcsin(1/(sqrt(c^2)*abs(x))) + arccosh(c*x)/x)*b*d^2 + 1/9*(3*x^3*arccosh(c*x) - c*(sqrt(c^
2*x^2 - 1)*x^2/c^2 + 2*sqrt(c^2*x^2 - 1)/c^4))*b*e^2 + 2*a*d*e*x + 2*(c*x*arccosh(c*x) - sqrt(c^2*x^2 - 1))*b*
d*e/c - a*d^2/x

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Fricas [A]  time = 2.96235, size = 512, normalized size = 3.2 \begin{align*} \frac{3 \, a c^{3} e^{2} x^{4} + 18 \, b c^{4} d^{2} x \arctan \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) + 18 \, a c^{3} d e x^{2} - 9 \, a c^{3} d^{2} + 3 \,{\left (3 \, b c^{3} d^{2} - 6 \, b c^{3} d e - b c^{3} e^{2}\right )} x \log \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) + 3 \,{\left (b c^{3} e^{2} x^{4} + 6 \, b c^{3} d e x^{2} - 3 \, b c^{3} d^{2} +{\left (3 \, b c^{3} d^{2} - 6 \, b c^{3} d e - b c^{3} e^{2}\right )} x\right )} \log \left (c x + \sqrt{c^{2} x^{2} - 1}\right ) -{\left (b c^{2} e^{2} x^{3} + 2 \,{\left (9 \, b c^{2} d e + b e^{2}\right )} x\right )} \sqrt{c^{2} x^{2} - 1}}{9 \, c^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(3*a*c^3*e^2*x^4 + 18*b*c^4*d^2*x*arctan(-c*x + sqrt(c^2*x^2 - 1)) + 18*a*c^3*d*e*x^2 - 9*a*c^3*d^2 + 3*(3
*b*c^3*d^2 - 6*b*c^3*d*e - b*c^3*e^2)*x*log(-c*x + sqrt(c^2*x^2 - 1)) + 3*(b*c^3*e^2*x^4 + 6*b*c^3*d*e*x^2 - 3
*b*c^3*d^2 + (3*b*c^3*d^2 - 6*b*c^3*d*e - b*c^3*e^2)*x)*log(c*x + sqrt(c^2*x^2 - 1)) - (b*c^2*e^2*x^3 + 2*(9*b
*c^2*d*e + b*e^2)*x)*sqrt(c^2*x^2 - 1))/(c^3*x)

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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 )^{2}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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