\(\int x^2 (d-c^2 d x^2) (a+b \text {arccosh}(c x)) \, dx\) [3]

Optimal result

Integrand size = 23, antiderivative size = 121 \[ \int x^2 \left (d-c^2 d x^2\right ) (a+b \text {arccosh}(c x)) \, dx=-\frac {26 b d \sqrt {-1+c x} \sqrt {1+c x}}{225 c^3}-\frac {13 b d x^2 \sqrt {-1+c x} \sqrt {1+c x}}{225 c}+\frac {1}{25} b c d x^4 \sqrt {-1+c x} \sqrt {1+c x}+\frac {1}{3} d x^3 (a+b \text {arccosh}(c x))-\frac {1}{5} c^2 d x^5 (a+b \text {arccosh}(c x)) \]

[Out]

1/3*d*x^3*(a+b*arccosh(c*x))-1/5*c^2*d*x^5*(a+b*arccosh(c*x))-26/225*b*d*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c^3-13/22
5*b*d*x^2*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c+1/25*b*c*d*x^4*(c*x-1)^(1/2)*(c*x+1)^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {14, 5921, 12, 471, 102, 75} \[ \int x^2 \left (d-c^2 d x^2\right ) (a+b \text {arccosh}(c x)) \, dx=-\frac {1}{5} c^2 d x^5 (a+b \text {arccosh}(c x))+\frac {1}{3} d x^3 (a+b \text {arccosh}(c x))-\frac {26 b d \sqrt {c x-1} \sqrt {c x+1}}{225 c^3}+\frac {1}{25} b c d x^4 \sqrt {c x-1} \sqrt {c x+1}-\frac {13 b d x^2 \sqrt {c x-1} \sqrt {c x+1}}{225 c} \]

[In]

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

[Out]

(-26*b*d*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(225*c^3) - (13*b*d*x^2*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(225*c) + (b*c*d*
x^4*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/25 + (d*x^3*(a + b*ArcCosh[c*x]))/3 - (c^2*d*x^5*(a + b*ArcCosh[c*x]))/5

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 75

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 102

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(m + n + p + 1))), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 471

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

Rule 5921

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] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} d x^3 (a+b \text {arccosh}(c x))-\frac {1}{5} c^2 d x^5 (a+b \text {arccosh}(c x))-(b c) \int \frac {d x^3 \left (5-3 c^2 x^2\right )}{15 \sqrt {-1+c x} \sqrt {1+c x}} \, dx \\ & = \frac {1}{3} d x^3 (a+b \text {arccosh}(c x))-\frac {1}{5} c^2 d x^5 (a+b \text {arccosh}(c x))-\frac {1}{15} (b c d) \int \frac {x^3 \left (5-3 c^2 x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx \\ & = \frac {1}{25} b c d x^4 \sqrt {-1+c x} \sqrt {1+c x}+\frac {1}{3} d x^3 (a+b \text {arccosh}(c x))-\frac {1}{5} c^2 d x^5 (a+b \text {arccosh}(c x))-\frac {1}{75} (13 b c d) \int \frac {x^3}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx \\ & = -\frac {13 b d x^2 \sqrt {-1+c x} \sqrt {1+c x}}{225 c}+\frac {1}{25} b c d x^4 \sqrt {-1+c x} \sqrt {1+c x}+\frac {1}{3} d x^3 (a+b \text {arccosh}(c x))-\frac {1}{5} c^2 d x^5 (a+b \text {arccosh}(c x))-\frac {(13 b d) \int \frac {2 x}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{225 c} \\ & = -\frac {13 b d x^2 \sqrt {-1+c x} \sqrt {1+c x}}{225 c}+\frac {1}{25} b c d x^4 \sqrt {-1+c x} \sqrt {1+c x}+\frac {1}{3} d x^3 (a+b \text {arccosh}(c x))-\frac {1}{5} c^2 d x^5 (a+b \text {arccosh}(c x))-\frac {(26 b d) \int \frac {x}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{225 c} \\ & = -\frac {26 b d \sqrt {-1+c x} \sqrt {1+c x}}{225 c^3}-\frac {13 b d x^2 \sqrt {-1+c x} \sqrt {1+c x}}{225 c}+\frac {1}{25} b c d x^4 \sqrt {-1+c x} \sqrt {1+c x}+\frac {1}{3} d x^3 (a+b \text {arccosh}(c x))-\frac {1}{5} c^2 d x^5 (a+b \text {arccosh}(c x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.74 \[ \int x^2 \left (d-c^2 d x^2\right ) (a+b \text {arccosh}(c x)) \, dx=-\frac {d \left (15 a c^3 x^3 \left (-5+3 c^2 x^2\right )+b \sqrt {-1+c x} \sqrt {1+c x} \left (26+13 c^2 x^2-9 c^4 x^4\right )+15 b c^3 x^3 \left (-5+3 c^2 x^2\right ) \text {arccosh}(c x)\right )}{225 c^3} \]

[In]

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

[Out]

-1/225*(d*(15*a*c^3*x^3*(-5 + 3*c^2*x^2) + b*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(26 + 13*c^2*x^2 - 9*c^4*x^4) + 15*b
*c^3*x^3*(-5 + 3*c^2*x^2)*ArcCosh[c*x]))/c^3

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.71

method	result	size
parts	\(-d a \left (\frac {1}{5} c^{2} x^{5}-\frac {1}{3} x^{3}\right )-\frac {d b \left (\frac {\operatorname {arccosh}\left (c x \right ) c^{5} x^{5}}{5}-\frac {c^{3} x^{3} \operatorname {arccosh}\left (c x \right )}{3}-\frac {\sqrt {c x -1}\, \sqrt {c x +1}\, \left (9 c^{4} x^{4}-13 c^{2} x^{2}-26\right )}{225}\right )}{c^{3}}\)	\(86\)
derivativedivides	\(\frac {-d a \left (\frac {1}{5} c^{5} x^{5}-\frac {1}{3} c^{3} x^{3}\right )-d b \left (\frac {\operatorname {arccosh}\left (c x \right ) c^{5} x^{5}}{5}-\frac {c^{3} x^{3} \operatorname {arccosh}\left (c x \right )}{3}-\frac {\sqrt {c x -1}\, \sqrt {c x +1}\, \left (9 c^{4} x^{4}-13 c^{2} x^{2}-26\right )}{225}\right )}{c^{3}}\)	\(90\)
default	\(\frac {-d a \left (\frac {1}{5} c^{5} x^{5}-\frac {1}{3} c^{3} x^{3}\right )-d b \left (\frac {\operatorname {arccosh}\left (c x \right ) c^{5} x^{5}}{5}-\frac {c^{3} x^{3} \operatorname {arccosh}\left (c x \right )}{3}-\frac {\sqrt {c x -1}\, \sqrt {c x +1}\, \left (9 c^{4} x^{4}-13 c^{2} x^{2}-26\right )}{225}\right )}{c^{3}}\)	\(90\)

[In]

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

[Out]

-d*a*(1/5*c^2*x^5-1/3*x^3)-d*b/c^3*(1/5*arccosh(c*x)*c^5*x^5-1/3*c^3*x^3*arccosh(c*x)-1/225*(c*x-1)^(1/2)*(c*x
+1)^(1/2)*(9*c^4*x^4-13*c^2*x^2-26))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.85 \[ \int x^2 \left (d-c^2 d x^2\right ) (a+b \text {arccosh}(c x)) \, dx=-\frac {45 \, a c^{5} d x^{5} - 75 \, a c^{3} d x^{3} + 15 \, {\left (3 \, b c^{5} d x^{5} - 5 \, b c^{3} d x^{3}\right )} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) - {\left (9 \, b c^{4} d x^{4} - 13 \, b c^{2} d x^{2} - 26 \, b d\right )} \sqrt {c^{2} x^{2} - 1}}{225 \, c^{3}} \]

[In]

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

[Out]

-1/225*(45*a*c^5*d*x^5 - 75*a*c^3*d*x^3 + 15*(3*b*c^5*d*x^5 - 5*b*c^3*d*x^3)*log(c*x + sqrt(c^2*x^2 - 1)) - (9
*b*c^4*d*x^4 - 13*b*c^2*d*x^2 - 26*b*d)*sqrt(c^2*x^2 - 1))/c^3

Sympy [F]

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

[In]

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

[Out]

-d*(Integral(-a*x**2, x) + Integral(a*c**2*x**4, x) + Integral(-b*x**2*acosh(c*x), x) + Integral(b*c**2*x**4*a
cosh(c*x), x))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.20 \[ \int x^2 \left (d-c^2 d x^2\right ) (a+b \text {arccosh}(c x)) \, dx=-\frac {1}{5} \, a c^{2} d x^{5} - \frac {1}{75} \, {\left (15 \, x^{5} \operatorname {arcosh}\left (c x\right ) - {\left (\frac {3 \, \sqrt {c^{2} x^{2} - 1} x^{4}}{c^{2}} + \frac {4 \, \sqrt {c^{2} x^{2} - 1} x^{2}}{c^{4}} + \frac {8 \, \sqrt {c^{2} x^{2} - 1}}{c^{6}}\right )} c\right )} b c^{2} d + \frac {1}{3} \, a d x^{3} + \frac {1}{9} \, {\left (3 \, x^{3} \operatorname {arcosh}\left (c x\right ) - c {\left (\frac {\sqrt {c^{2} x^{2} - 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {c^{2} x^{2} - 1}}{c^{4}}\right )}\right )} b d \]

[In]

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

[Out]

-1/5*a*c^2*d*x^5 - 1/75*(15*x^5*arccosh(c*x) - (3*sqrt(c^2*x^2 - 1)*x^4/c^2 + 4*sqrt(c^2*x^2 - 1)*x^2/c^4 + 8*
sqrt(c^2*x^2 - 1)/c^6)*c)*b*c^2*d + 1/3*a*d*x^3 + 1/9*(3*x^3*arccosh(c*x) - c*(sqrt(c^2*x^2 - 1)*x^2/c^2 + 2*s
qrt(c^2*x^2 - 1)/c^4))*b*d

Giac [F(-2)]

Exception generated. \[ \int x^2 \left (d-c^2 d x^2\right ) (a+b \text {arccosh}(c x)) \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> an error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

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

[In]

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

[Out]

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