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]

-d^2*(a+b*arccosh(c*x))/x+2*d*e*x*(a+b*arccosh(c*x))+1/3*e^2*x^3*(a+b*arccosh(c*x))+b*c*d^2*arctan((c*x-1)^(1/
2)*(c*x+1)^(1/2))+1/3*b*e*(6*c^2*d+e)*(-c^2*x^2+1)/c^3/(c*x-1)^(1/2)/(c*x+1)^(1/2)-1/9*b*e^2*(-c^2*x^2+1)^2/c^
3/(c*x-1)^(1/2)/(c*x+1)^(1/2)

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Rubi [A]  time = 0.30, 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 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]

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 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 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 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 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]))

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*}

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Mathematica [A]  time = 0.26, size = 128, normalized size = 0.80 \[ \frac {1}{3} \left (-\frac {3 a d^2}{x}+6 a d e x+a e^2 x^3-\frac {b e \sqrt {c x-1} \sqrt {c x+1} \left (c^2 \left (18 d+e x^2\right )+2 e\right )}{3 c^3}+\frac {b \cosh ^{-1}(c x) \left (-3 d^2+6 d e x^2+e^2 x^4\right )}{x}-3 b c d^2 \tan ^{-1}\left (\frac {1}{\sqrt {c x-1} \sqrt {c x+1}}\right )\right ) \]

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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fricas [A]  time = 0.86, size = 236, normalized size = 1.48 \[ \frac {3 \, a c^{3} e^{2} x^{4} + 18 \, b c^{4} d^{2} x \arctan \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) + 18 \, a c^{3} d e x^{2} - 9 \, a c^{3} d^{2} + 3 \, {\left (3 \, b c^{3} d^{2} - 6 \, b c^{3} d e - b c^{3} e^{2}\right )} x \log \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) + 3 \, {\left (b c^{3} e^{2} x^{4} + 6 \, b c^{3} d e x^{2} - 3 \, b c^{3} d^{2} + {\left (3 \, b c^{3} d^{2} - 6 \, b c^{3} d e - b c^{3} e^{2}\right )} x\right )} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) - {\left (b c^{2} e^{2} x^{3} + 2 \, {\left (9 \, b c^{2} d e + b e^{2}\right )} x\right )} \sqrt {c^{2} x^{2} - 1}}{9 \, c^{3} x} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arcosh}\left (c x\right ) + a\right )}}{x^{2}}\,{d x} \]

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)

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maple [A]  time = 0.02, size = 177, normalized size = 1.11 \[ \frac {a \,x^{3} e^{2}}{3}+2 a d e x -\frac {d^{2} a}{x}+\frac {b \,\mathrm {arccosh}\left (c x \right ) x^{3} e^{2}}{3}+2 b \,\mathrm {arccosh}\left (c x \right ) d e x -\frac {d^{2} b \,\mathrm {arccosh}\left (c x \right )}{x}-\frac {c \,d^{2} b \sqrt {c x -1}\, \sqrt {c x +1}\, \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )}{\sqrt {c^{2} x^{2}-1}}-\frac {b \sqrt {c x -1}\, \sqrt {c x +1}\, x^{2} e^{2}}{9 c}-\frac {2 b \sqrt {c x -1}\, \sqrt {c x +1}\, d e}{c}-\frac {2 b \sqrt {c x -1}\, \sqrt {c x +1}\, e^{2}}{9 c^{3}} \]

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*d*e*x-d^2*a/x+1/3*b*arccosh(c*x)*x^3*e^2+2*b*arccosh(c*x)*d*e*x-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 = 0.77, size = 134, normalized size = 0.84 \[ \frac {1}{3} \, a e^{2} x^{3} - {\left (c \arcsin \left (\frac {1}{c {\left | x \right |}}\right ) + \frac {\operatorname {arcosh}\left (c x\right )}{x}\right )} b d^{2} + \frac {1}{9} \, {\left (3 \, x^{3} \operatorname {arcosh}\left (c x\right ) - c {\left (\frac {\sqrt {c^{2} x^{2} - 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {c^{2} x^{2} - 1}}{c^{4}}\right )}\right )} b e^{2} + 2 \, a d e x + \frac {2 \, {\left (c x \operatorname {arcosh}\left (c x\right ) - \sqrt {c^{2} x^{2} - 1}\right )} b d e}{c} - \frac {a d^{2}}{x} \]

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/(c*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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^2}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {acosh}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}}{x^{2}}\, dx \]

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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