\(\int \frac {(d+e x^2) (a+b \text {arccosh}(c x))}{x^4} \, dx\) [469]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 19, antiderivative size = 94 \[ \int \frac {\left (d+e x^2\right ) (a+b \text {arccosh}(c x))}{x^4} \, dx=\frac {b c d \sqrt {-1+c x} \sqrt {1+c x}}{6 x^2}-\frac {d (a+b \text {arccosh}(c x))}{3 x^3}-\frac {e (a+b \text {arccosh}(c x))}{x}+\frac {1}{6} b c \left (c^2 d+6 e\right ) \arctan \left (\sqrt {-1+c x} \sqrt {1+c x}\right ) \]

[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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {5956, 465, 94, 211} \[ \int \frac {\left (d+e x^2\right ) (a+b \text {arccosh}(c x))}{x^4} \, dx=-\frac {d (a+b \text {arccosh}(c x))}{3 x^3}-\frac {e (a+b \text {arccosh}(c x))}{x}+\frac {1}{6} b c \arctan \left (\sqrt {c x-1} \sqrt {c x+1}\right ) \left (c^2 d+6 e\right )+\frac {b c d \sqrt {c x-1} \sqrt {c x+1}}{6 x^2} \]

[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 94

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 211

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

Rule 465

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 5956

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*} \text {integral}& = -\frac {d (a+b \text {arccosh}(c x))}{3 x^3}-\frac {e (a+b \text {arccosh}(c x))}{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 (a+b \text {arccosh}(c x))}{3 x^3}-\frac {e (a+b \text {arccosh}(c x))}{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 (a+b \text {arccosh}(c x))}{3 x^3}-\frac {e (a+b \text {arccosh}(c x))}{x}+\frac {1}{6} \left (b c^2 \left (c^2 d+6 e\right )\right ) \text {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 (a+b \text {arccosh}(c x))}{3 x^3}-\frac {e (a+b \text {arccosh}(c x))}{x}+\frac {1}{6} b c \left (c^2 d+6 e\right ) \arctan \left (\sqrt {-1+c x} \sqrt {1+c x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.36 \[ \int \frac {\left (d+e x^2\right ) (a+b \text {arccosh}(c x))}{x^4} \, dx=\frac {-2 b \left (d+3 e x^2\right ) \text {arccosh}(c x)+\frac {b c d x \left (-1+c^2 x^2\right )-2 a \sqrt {-1+c x} \sqrt {1+c x} \left (d+3 e x^2\right )+b c \left (c^2 d+6 e\right ) x^3 \sqrt {-1+c^2 x^2} \arctan \left (\sqrt {-1+c^2 x^2}\right )}{\sqrt {-1+c x} \sqrt {1+c x}}}{6 x^3} \]

[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)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.50

method result size
parts \(a \left (-\frac {e}{x}-\frac {d}{3 x^{3}}\right )+b \,c^{3} \left (-\frac {\operatorname {arccosh}\left (c x \right ) e}{c^{3} x}-\frac {\operatorname {arccosh}\left (c x \right ) d}{3 c^{3} x^{3}}-\frac {\sqrt {c x -1}\, \sqrt {c x +1}\, \left (\arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right ) c^{4} d \,x^{2}+6 \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right ) e \,c^{2} x^{2}-\sqrt {c^{2} x^{2}-1}\, c^{2} d \right )}{6 c^{4} \sqrt {c^{2} x^{2}-1}\, x^{2}}\right )\) \(141\)
derivativedivides \(c^{3} \left (\frac {a \left (-\frac {d}{3 c \,x^{3}}-\frac {e}{c x}\right )}{c^{2}}+\frac {b \left (-\frac {\operatorname {arccosh}\left (c x \right ) d}{3 c \,x^{3}}-\frac {\operatorname {arccosh}\left (c x \right ) e}{c x}-\frac {\sqrt {c x -1}\, \sqrt {c x +1}\, \left (\arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right ) c^{4} d \,x^{2}+6 \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right ) e \,c^{2} x^{2}-\sqrt {c^{2} x^{2}-1}\, c^{2} d \right )}{6 \sqrt {c^{2} x^{2}-1}\, c^{2} x^{2}}\right )}{c^{2}}\right )\) \(154\)
default \(c^{3} \left (\frac {a \left (-\frac {d}{3 c \,x^{3}}-\frac {e}{c x}\right )}{c^{2}}+\frac {b \left (-\frac {\operatorname {arccosh}\left (c x \right ) d}{3 c \,x^{3}}-\frac {\operatorname {arccosh}\left (c x \right ) e}{c x}-\frac {\sqrt {c x -1}\, \sqrt {c x +1}\, \left (\arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right ) c^{4} d \,x^{2}+6 \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right ) e \,c^{2} x^{2}-\sqrt {c^{2} x^{2}-1}\, c^{2} d \right )}{6 \sqrt {c^{2} x^{2}-1}\, c^{2} x^{2}}\right )}{c^{2}}\right )\) \(154\)

[In]

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

[Out]

a*(-e/x-1/3*d/x^3)+b*c^3*(-1/c^3*arccosh(c*x)*e/x-1/3*arccosh(c*x)*d/c^3/x^3-1/6/c^4*(c*x-1)^(1/2)*(c*x+1)^(1/
2)*(arctan(1/(c^2*x^2-1)^(1/2))*c^4*d*x^2+6*arctan(1/(c^2*x^2-1)^(1/2))*e*c^2*x^2-(c^2*x^2-1)^(1/2)*c^2*d)/(c^
2*x^2-1)^(1/2)/x^2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.48 \[ \int \frac {\left (d+e x^2\right ) (a+b \text {arccosh}(c x))}{x^4} \, dx=\frac {2 \, {\left (b c^{3} d + 6 \, b c e\right )} x^{3} \arctan \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) + 2 \, {\left (b d + 3 \, b e\right )} x^{3} \log \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) + \sqrt {c^{2} x^{2} - 1} b c d x - 6 \, a e x^{2} - 2 \, a d - 2 \, {\left (3 \, b e x^{2} - {\left (b d + 3 \, b e\right )} x^{3} + b d\right )} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right )}{6 \, x^{3}} \]

[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

Sympy [F]

\[ \int \frac {\left (d+e x^2\right ) (a+b \text {arccosh}(c x))}{x^4} \, dx=\int \frac {\left (a + b \operatorname {acosh}{\left (c x \right )}\right ) \left (d + e x^{2}\right )}{x^{4}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.90 \[ \int \frac {\left (d+e x^2\right ) (a+b \text {arccosh}(c x))}{x^4} \, dx=-\frac {1}{6} \, {\left ({\left (c^{2} \arcsin \left (\frac {1}{c {\left | x \right |}}\right ) - \frac {\sqrt {c^{2} x^{2} - 1}}{x^{2}}\right )} c + \frac {2 \, \operatorname {arcosh}\left (c x\right )}{x^{3}}\right )} b d - {\left (c \arcsin \left (\frac {1}{c {\left | x \right |}}\right ) + \frac {\operatorname {arcosh}\left (c x\right )}{x}\right )} b e - \frac {a e}{x} - \frac {a d}{3 \, x^{3}} \]

[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

Giac [F]

\[ \int \frac {\left (d+e x^2\right ) (a+b \text {arccosh}(c x))}{x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \operatorname {arcosh}\left (c x\right ) + a\right )}}{x^{4}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right ) (a+b \text {arccosh}(c x))}{x^4} \, dx=\int \frac {\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,\left (e\,x^2+d\right )}{x^4} \,d x \]

[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)