3.469 \(\int \frac {(d+e x^2) (a+b \cosh ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=94 \[ -\frac {d \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac {e \left (a+b \cosh ^{-1}(c x)\right )}{x}+\frac {1}{6} b c \left (c^2 d+6 e\right ) \tan ^{-1}\left (\sqrt {c x-1} \sqrt {c x+1}\right )+\frac {b c d \sqrt {c x-1} \sqrt {c x+1}}{6 x^2} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {5786, 454, 92, 205} \[ -\frac {d \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac {e \left (a+b \cosh ^{-1}(c x)\right )}{x}+\frac {1}{6} b c \left (c^2 d+6 e\right ) \tan ^{-1}\left (\sqrt {c x-1} \sqrt {c x+1}\right )+\frac {b c d \sqrt {c x-1} \sqrt {c x+1}}{6 x^2} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)*(a + b*ArcCosh[c*x]))/x^4,x]

[Out]

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

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]

Rule 454

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.27, size = 128, normalized size = 1.36 \[ \frac {\frac {-2 a \sqrt {c x-1} \sqrt {c x+1} \left (d+3 e x^2\right )+b c x^3 \sqrt {c^2 x^2-1} \left (c^2 d+6 e\right ) \tan ^{-1}\left (\sqrt {c^2 x^2-1}\right )+b c d x \left (c^2 x^2-1\right )}{\sqrt {c x-1} \sqrt {c x+1}}-2 b \cosh ^{-1}(c x) \left (d+3 e x^2\right )}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)*(a + b*ArcCosh[c*x]))/x^4,x]

[Out]

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

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fricas [A]  time = 0.88, size = 139, normalized size = 1.48 \[ \frac {2 \, {\left (b c^{3} d + 6 \, b c e\right )} x^{3} \arctan \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) + 2 \, {\left (b d + 3 \, b e\right )} x^{3} \log \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) + \sqrt {c^{2} x^{2} - 1} b c d x - 6 \, a e x^{2} - 2 \, a d - 2 \, {\left (3 \, b e x^{2} - {\left (b d + 3 \, b e\right )} x^{3} + b d\right )} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right )}{6 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arccosh(c*x))/x^4,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )} {\left (b \operatorname {arcosh}\left (c x\right ) + a\right )}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arccosh(c*x))/x^4,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)*(b*arccosh(c*x) + a)/x^4, x)

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maple [A]  time = 0.02, size = 146, normalized size = 1.55 \[ -\frac {a e}{x}-\frac {d a}{3 x^{3}}-\frac {b \,\mathrm {arccosh}\left (c x \right ) e}{x}-\frac {d b \,\mathrm {arccosh}\left (c x \right )}{3 x^{3}}-\frac {c^{3} d b \sqrt {c x -1}\, \sqrt {c x +1}\, \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )}{6 \sqrt {c^{2} x^{2}-1}}+\frac {b c d \sqrt {c x -1}\, \sqrt {c x +1}}{6 x^{2}}-\frac {c b \sqrt {c x -1}\, \sqrt {c x +1}\, \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right ) e}{\sqrt {c^{2} x^{2}-1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)*(a+b*arccosh(c*x))/x^4,x)

[Out]

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

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maxima [A]  time = 0.47, size = 85, normalized size = 0.90 \[ -\frac {1}{6} \, {\left ({\left (c^{2} \arcsin \left (\frac {1}{c {\left | x \right |}}\right ) - \frac {\sqrt {c^{2} x^{2} - 1}}{x^{2}}\right )} c + \frac {2 \, \operatorname {arcosh}\left (c x\right )}{x^{3}}\right )} b d - {\left (c \arcsin \left (\frac {1}{c {\left | x \right |}}\right ) + \frac {\operatorname {arcosh}\left (c x\right )}{x}\right )} b e - \frac {a e}{x} - \frac {a d}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arccosh(c*x))/x^4,x, algorithm="maxima")

[Out]

-1/6*((c^2*arcsin(1/(c*abs(x))) - sqrt(c^2*x^2 - 1)/x^2)*c + 2*arccosh(c*x)/x^3)*b*d - (c*arcsin(1/(c*abs(x)))
+ arccosh(c*x)/x)*b*e - a*e/x - 1/3*a*d/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,\left (e\,x^2+d\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*acosh(c*x))*(d + e*x^2))/x^4,x)

[Out]

int(((a + b*acosh(c*x))*(d + e*x^2))/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {acosh}{\left (c x \right )}\right ) \left (d + e x^{2}\right )}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)*(a+b*acosh(c*x))/x**4,x)

[Out]

Integral((a + b*acosh(c*x))*(d + e*x**2)/x**4, x)

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