3.487 \(\int \frac {(d+e x^2)^3 (a+b \cosh ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=260 \[ -\frac {d^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \cosh ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{3} e^3 x^3 \left (a+b \cosh ^{-1}(c x)\right )-\frac {b c d^3 \left (1-c^2 x^2\right )}{6 x^2 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b c d^2 \sqrt {c^2 x^2-1} \left (c^2 d+18 e\right ) \tan ^{-1}\left (\sqrt {c^2 x^2-1}\right )}{6 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b e^2 \left (1-c^2 x^2\right ) \left (9 c^2 d+e\right )}{3 c^3 \sqrt {c x-1} \sqrt {c x+1}}-\frac {b e^3 \left (1-c^2 x^2\right )^2}{9 c^3 \sqrt {c x-1} \sqrt {c x+1}} \]

[Out]

-1/3*d^3*(a+b*arccosh(c*x))/x^3-3*d^2*e*(a+b*arccosh(c*x))/x+3*d*e^2*x*(a+b*arccosh(c*x))+1/3*e^3*x^3*(a+b*arc
cosh(c*x))+1/3*b*e^2*(9*c^2*d+e)*(-c^2*x^2+1)/c^3/(c*x-1)^(1/2)/(c*x+1)^(1/2)-1/6*b*c*d^3*(-c^2*x^2+1)/x^2/(c*
x-1)^(1/2)/(c*x+1)^(1/2)-1/9*b*e^3*(-c^2*x^2+1)^2/c^3/(c*x-1)^(1/2)/(c*x+1)^(1/2)+1/6*b*c*d^2*(c^2*d+18*e)*arc
tan((c^2*x^2-1)^(1/2))*(c^2*x^2-1)^(1/2)/(c*x-1)^(1/2)/(c*x+1)^(1/2)

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Rubi [A]  time = 0.46, antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {270, 5790, 12, 1610, 1799, 1621, 897, 1153, 205} \[ -\frac {3 d^2 e \left (a+b \cosh ^{-1}(c x)\right )}{x}-\frac {d^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}+3 d e^2 x \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{3} e^3 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac {b c d^2 \sqrt {c^2 x^2-1} \left (c^2 d+18 e\right ) \tan ^{-1}\left (\sqrt {c^2 x^2-1}\right )}{6 \sqrt {c x-1} \sqrt {c x+1}}-\frac {b c d^3 \left (1-c^2 x^2\right )}{6 x^2 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b e^2 \left (1-c^2 x^2\right ) \left (9 c^2 d+e\right )}{3 c^3 \sqrt {c x-1} \sqrt {c x+1}}-\frac {b e^3 \left (1-c^2 x^2\right )^2}{9 c^3 \sqrt {c x-1} \sqrt {c x+1}} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^3*(a + b*ArcCosh[c*x]))/x^4,x]

[Out]

(b*e^2*(9*c^2*d + e)*(1 - c^2*x^2))/(3*c^3*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) - (b*c*d^3*(1 - c^2*x^2))/(6*x^2*Sqrt
[-1 + c*x]*Sqrt[1 + c*x]) - (b*e^3*(1 - c^2*x^2)^2)/(9*c^3*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) - (d^3*(a + b*ArcCosh
[c*x]))/(3*x^3) - (3*d^2*e*(a + b*ArcCosh[c*x]))/x + 3*d*e^2*x*(a + b*ArcCosh[c*x]) + (e^3*x^3*(a + b*ArcCosh[
c*x]))/3 + (b*c*d^2*(c^2*d + 18*e)*Sqrt[-1 + c^2*x^2]*ArcTan[Sqrt[-1 + c^2*x^2]])/(6*Sqrt[-1 + c*x]*Sqrt[1 + c
*x])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 1610

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Dist[((
a + b*x)^FracPart[m]*(c + d*x)^FracPart[m])/(a*c + b*d*x^2)^FracPart[m], Int[Px*(a*c + b*d*x^2)^m*(e + f*x)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[Px, x] && EqQ[b*c + a*d, 0] && EqQ[m, n] &&  !Intege
rQ[m]

Rule 1621

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> With[{Qx = PolynomialQuotient[Px,
a + b*x, x], R = PolynomialRemainder[Px, a + b*x, x]}, Simp[(R*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/((m + 1)*
(b*c - a*d)), x] + Dist[1/((m + 1)*(b*c - a*d)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*ExpandToSum[(m + 1)*(b*c -
a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; FreeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && ILtQ[m, -1] && GtQ[Expo
n[Px, x], 2]

Rule 1799

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*SubstFor[x^2,
Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && 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 )^3 \left (a+b \cosh ^{-1}(c x)\right )}{x^4} \, dx &=-\frac {d^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \cosh ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{3} e^3 x^3 \left (a+b \cosh ^{-1}(c x)\right )-(b c) \int \frac {-d^3-9 d^2 e x^2+9 d e^2 x^4+e^3 x^6}{3 x^3 \sqrt {-1+c x} \sqrt {1+c x}} \, dx\\ &=-\frac {d^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \cosh ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{3} e^3 x^3 \left (a+b \cosh ^{-1}(c x)\right )-\frac {1}{3} (b c) \int \frac {-d^3-9 d^2 e x^2+9 d e^2 x^4+e^3 x^6}{x^3 \sqrt {-1+c x} \sqrt {1+c x}} \, dx\\ &=-\frac {d^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \cosh ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{3} e^3 x^3 \left (a+b \cosh ^{-1}(c x)\right )-\frac {\left (b c \sqrt {-1+c^2 x^2}\right ) \int \frac {-d^3-9 d^2 e x^2+9 d e^2 x^4+e^3 x^6}{x^3 \sqrt {-1+c^2 x^2}} \, dx}{3 \sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {d^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \cosh ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{3} e^3 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^3-9 d^2 e x+9 d e^2 x^2+e^3 x^3}{x^2 \sqrt {-1+c^2 x}} \, dx,x,x^2\right )}{6 \sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {b c d^3 \left (1-c^2 x^2\right )}{6 x^2 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {d^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \cosh ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{3} e^3 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 {-\frac {1}{2} d^2 \left (c^2 d+18 e\right )+9 d e^2 x+e^3 x^2}{x \sqrt {-1+c^2 x}} \, dx,x,x^2\right )}{6 \sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {b c d^3 \left (1-c^2 x^2\right )}{6 x^2 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {d^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \cosh ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{3} e^3 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 {9 c^2 d e^2+e^3-\frac {1}{2} c^4 d^2 \left (c^2 d+18 e\right )}{c^4}-\frac {\left (-9 c^2 d e^2-2 e^3\right ) x^2}{c^4}+\frac {e^3 x^4}{c^4}}{\frac {1}{c^2}+\frac {x^2}{c^2}} \, dx,x,\sqrt {-1+c^2 x^2}\right )}{3 c \sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {b c d^3 \left (1-c^2 x^2\right )}{6 x^2 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {d^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \cosh ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{3} e^3 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 (e^2 \left (9 d+\frac {e}{c^2}\right )+\frac {e^3 x^2}{c^2}+\frac {-c^2 d^3-18 d^2 e}{2 \left (\frac {1}{c^2}+\frac {x^2}{c^2}\right )}\right ) \, dx,x,\sqrt {-1+c^2 x^2}\right )}{3 c \sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {b e^2 \left (9 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 c d^3 \left (1-c^2 x^2\right )}{6 x^2 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {b e^3 \left (1-c^2 x^2\right )^2}{9 c^3 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {d^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \cosh ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{3} e^3 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac {\left (b d^2 \left (c^2 d+18 e\right ) \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 )}{6 c \sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {b e^2 \left (9 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 c d^3 \left (1-c^2 x^2\right )}{6 x^2 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {b e^3 \left (1-c^2 x^2\right )^2}{9 c^3 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {d^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \cosh ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{3} e^3 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac {b c d^2 \left (c^2 d+18 e\right ) \sqrt {-1+c^2 x^2} \tan ^{-1}\left (\sqrt {-1+c^2 x^2}\right )}{6 \sqrt {-1+c x} \sqrt {1+c x}}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 184, normalized size = 0.71 \[ \frac {1}{6} \left (-\frac {2 a d^3}{x^3}-\frac {18 a d^2 e}{x}+18 a d e^2 x+2 a e^3 x^3-b c d^2 \left (c^2 d+18 e\right ) \tan ^{-1}\left (\frac {1}{\sqrt {c x-1} \sqrt {c x+1}}\right )-\frac {b \sqrt {c x-1} \sqrt {c x+1} \left (-3 c^4 d^3+2 c^2 e^2 x^2 \left (27 d+e x^2\right )+4 e^3 x^2\right )}{3 c^3 x^2}+\frac {2 b \cosh ^{-1}(c x) \left (-d^3-9 d^2 e x^2+9 d e^2 x^4+e^3 x^6\right )}{x^3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^3*(a + b*ArcCosh[c*x]))/x^4,x]

[Out]

((-2*a*d^3)/x^3 - (18*a*d^2*e)/x + 18*a*d*e^2*x + 2*a*e^3*x^3 - (b*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(-3*c^4*d^3 +
4*e^3*x^2 + 2*c^2*e^2*x^2*(27*d + e*x^2)))/(3*c^3*x^2) + (2*b*(-d^3 - 9*d^2*e*x^2 + 9*d*e^2*x^4 + e^3*x^6)*Arc
Cosh[c*x])/x^3 - b*c*d^2*(c^2*d + 18*e)*ArcTan[1/(Sqrt[-1 + c*x]*Sqrt[1 + c*x])])/6

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fricas [A]  time = 0.81, size = 322, normalized size = 1.24 \[ \frac {6 \, a c^{3} e^{3} x^{6} + 54 \, a c^{3} d e^{2} x^{4} - 54 \, a c^{3} d^{2} e x^{2} - 6 \, a c^{3} d^{3} + 6 \, {\left (b c^{6} d^{3} + 18 \, b c^{4} d^{2} e\right )} x^{3} \arctan \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) + 6 \, {\left (b c^{3} d^{3} + 9 \, b c^{3} d^{2} e - 9 \, b c^{3} d e^{2} - b c^{3} e^{3}\right )} x^{3} \log \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) + 6 \, {\left (b c^{3} e^{3} x^{6} + 9 \, b c^{3} d e^{2} x^{4} - 9 \, b c^{3} d^{2} e x^{2} - b c^{3} d^{3} + {\left (b c^{3} d^{3} + 9 \, b c^{3} d^{2} e - 9 \, b c^{3} d e^{2} - b c^{3} e^{3}\right )} x^{3}\right )} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) - {\left (2 \, b c^{2} e^{3} x^{5} - 3 \, b c^{4} d^{3} x + 2 \, {\left (27 \, b c^{2} d e^{2} + 2 \, b e^{3}\right )} x^{3}\right )} \sqrt {c^{2} x^{2} - 1}}{18 \, c^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*arccosh(c*x))/x^4,x, algorithm="fricas")

[Out]

1/18*(6*a*c^3*e^3*x^6 + 54*a*c^3*d*e^2*x^4 - 54*a*c^3*d^2*e*x^2 - 6*a*c^3*d^3 + 6*(b*c^6*d^3 + 18*b*c^4*d^2*e)
*x^3*arctan(-c*x + sqrt(c^2*x^2 - 1)) + 6*(b*c^3*d^3 + 9*b*c^3*d^2*e - 9*b*c^3*d*e^2 - b*c^3*e^3)*x^3*log(-c*x
+ sqrt(c^2*x^2 - 1)) + 6*(b*c^3*e^3*x^6 + 9*b*c^3*d*e^2*x^4 - 9*b*c^3*d^2*e*x^2 - b*c^3*d^3 + (b*c^3*d^3 + 9*
b*c^3*d^2*e - 9*b*c^3*d*e^2 - b*c^3*e^3)*x^3)*log(c*x + sqrt(c^2*x^2 - 1)) - (2*b*c^2*e^3*x^5 - 3*b*c^4*d^3*x
+ 2*(27*b*c^2*d*e^2 + 2*b*e^3)*x^3)*sqrt(c^2*x^2 - 1))/(c^3*x^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )}^{3} {\left (b \operatorname {arcosh}\left (c x\right ) + a\right )}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*arccosh(c*x))/x^4,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^3*(b*arccosh(c*x) + a)/x^4, x)

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maple [A]  time = 0.02, size = 278, normalized size = 1.07 \[ \frac {a \,x^{3} e^{3}}{3}+3 a x d \,e^{2}-\frac {3 a \,d^{2} e}{x}-\frac {a \,d^{3}}{3 x^{3}}+\frac {b \,\mathrm {arccosh}\left (c x \right ) x^{3} e^{3}}{3}+3 b \,\mathrm {arccosh}\left (c x \right ) x d \,e^{2}-\frac {3 b \,\mathrm {arccosh}\left (c x \right ) d^{2} e}{x}-\frac {b \,\mathrm {arccosh}\left (c x \right ) d^{3}}{3 x^{3}}-\frac {c^{3} b \sqrt {c x -1}\, \sqrt {c x +1}\, d^{3} \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )}{6 \sqrt {c^{2} x^{2}-1}}+\frac {b c \,d^{3} \sqrt {c x -1}\, \sqrt {c x +1}}{6 x^{2}}-\frac {3 c b \sqrt {c x -1}\, \sqrt {c x +1}\, \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right ) d^{2} e}{\sqrt {c^{2} x^{2}-1}}-\frac {b \sqrt {c x -1}\, \sqrt {c x +1}\, x^{2} e^{3}}{9 c}-\frac {3 b \sqrt {c x -1}\, \sqrt {c x +1}\, d \,e^{2}}{c}-\frac {2 b \sqrt {c x -1}\, \sqrt {c x +1}\, e^{3}}{9 c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^3*(a+b*arccosh(c*x))/x^4,x)

[Out]

1/3*a*x^3*e^3+3*a*x*d*e^2-3*a*d^2*e/x-1/3*a*d^3/x^3+1/3*b*arccosh(c*x)*x^3*e^3+3*b*arccosh(c*x)*x*d*e^2-3*b*ar
ccosh(c*x)*d^2*e/x-1/3*b*arccosh(c*x)*d^3/x^3-1/6*c^3*b*(c*x-1)^(1/2)*(c*x+1)^(1/2)/(c^2*x^2-1)^(1/2)*d^3*arct
an(1/(c^2*x^2-1)^(1/2))+1/6*b*c*d^3*(c*x-1)^(1/2)*(c*x+1)^(1/2)/x^2-3*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))*d^2*e-1/9/c*b*(c*x-1)^(1/2)*(c*x+1)^(1/2)*x^2*e^3-3/c*b*(c*x-1)^(1/2)*(c
*x+1)^(1/2)*d*e^2-2/9/c^3*b*(c*x-1)^(1/2)*(c*x+1)^(1/2)*e^3

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maxima [A]  time = 0.57, size = 197, normalized size = 0.76 \[ \frac {1}{3} \, a e^{3} x^{3} - \frac {1}{6} \, {\left ({\left (c^{2} \arcsin \left (\frac {1}{c {\left | x \right |}}\right ) - \frac {\sqrt {c^{2} x^{2} - 1}}{x^{2}}\right )} c + \frac {2 \, \operatorname {arcosh}\left (c x\right )}{x^{3}}\right )} b d^{3} - 3 \, {\left (c \arcsin \left (\frac {1}{c {\left | x \right |}}\right ) + \frac {\operatorname {arcosh}\left (c x\right )}{x}\right )} b d^{2} e + \frac {1}{9} \, {\left (3 \, x^{3} \operatorname {arcosh}\left (c x\right ) - c {\left (\frac {\sqrt {c^{2} x^{2} - 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {c^{2} x^{2} - 1}}{c^{4}}\right )}\right )} b e^{3} + 3 \, a d e^{2} x + \frac {3 \, {\left (c x \operatorname {arcosh}\left (c x\right ) - \sqrt {c^{2} x^{2} - 1}\right )} b d e^{2}}{c} - \frac {3 \, a d^{2} e}{x} - \frac {a d^{3}}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*arccosh(c*x))/x^4,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^3}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*acosh(c*x))*(d + e*x^2)^3)/x^4,x)

[Out]

int(((a + b*acosh(c*x))*(d + e*x^2)^3)/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {acosh}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{3}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**3*(a+b*acosh(c*x))/x**4,x)

[Out]

Integral((a + b*acosh(c*x))*(d + e*x**2)**3/x**4, x)

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