3.3 \(\int x^2 (d-c^2 d x^2) (a+b \cosh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=121 \[ -\frac{1}{5} c^2 d x^5 \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{3} d x^3 \left (a+b \cosh ^{-1}(c x)\right )-\frac{26 b d \sqrt{c x-1} \sqrt{c x+1}}{225 c^3}+\frac{1}{25} b c d x^4 \sqrt{c x-1} \sqrt{c x+1}-\frac{13 b d x^2 \sqrt{c x-1} \sqrt{c x+1}}{225 c} \]

[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

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Rubi [A]  time = 0.132875, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {14, 5731, 12, 460, 100, 74} \[ -\frac{1}{5} c^2 d x^5 \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{3} d x^3 \left (a+b \cosh ^{-1}(c x)\right )-\frac{26 b d \sqrt{c x-1} \sqrt{c x+1}}{225 c^3}+\frac{1}{25} b c d x^4 \sqrt{c x-1} \sqrt{c x+1}-\frac{13 b d x^2 \sqrt{c x-1} \sqrt{c x+1}}{225 c} \]

Antiderivative was successfully verified.

[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 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 5731

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]

Rule 12

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

Rule 460

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 100

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 74

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]

Rubi steps

\begin{align*} \int x^2 \left (d-c^2 d x^2\right ) \left (a+b \cosh ^{-1}(c x)\right ) \, dx &=\frac{1}{3} d x^3 \left (a+b \cosh ^{-1}(c x)\right )-\frac{1}{5} c^2 d x^5 \left (a+b \cosh ^{-1}(c x)\right )-(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 \left (a+b \cosh ^{-1}(c x)\right )-\frac{1}{5} c^2 d x^5 \left (a+b \cosh ^{-1}(c x)\right )-\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 \left (a+b \cosh ^{-1}(c x)\right )-\frac{1}{5} c^2 d x^5 \left (a+b \cosh ^{-1}(c x)\right )-\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 \left (a+b \cosh ^{-1}(c x)\right )-\frac{1}{5} c^2 d x^5 \left (a+b \cosh ^{-1}(c x)\right )-\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 \left (a+b \cosh ^{-1}(c x)\right )-\frac{1}{5} c^2 d x^5 \left (a+b \cosh ^{-1}(c x)\right )-\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 \left (a+b \cosh ^{-1}(c x)\right )-\frac{1}{5} c^2 d x^5 \left (a+b \cosh ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.107724, size = 89, normalized size = 0.74 \[ -\frac{d \left (15 a c^3 x^3 \left (3 c^2 x^2-5\right )+b \sqrt{c x-1} \sqrt{c x+1} \left (-9 c^4 x^4+13 c^2 x^2+26\right )+15 b c^3 x^3 \left (3 c^2 x^2-5\right ) \cosh ^{-1}(c x)\right )}{225 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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]))/(225*c^3)

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Maple [A]  time = 0.01, size = 90, normalized size = 0.7 \begin{align*}{\frac{1}{{c}^{3}} \left ( -da \left ({\frac{{c}^{5}{x}^{5}}{5}}-{\frac{{c}^{3}{x}^{3}}{3}} \right ) -db \left ({\frac{{\rm arccosh} \left (cx\right ){c}^{5}{x}^{5}}{5}}-{\frac{{c}^{3}{x}^{3}{\rm arccosh} \left (cx\right )}{3}}-{\frac{9\,{c}^{4}{x}^{4}-13\,{c}^{2}{x}^{2}-26}{225}\sqrt{cx-1}\sqrt{cx+1}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^3*(-d*a*(1/5*c^5*x^5-1/3*c^3*x^3)-d*b*(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)))

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Maxima [A]  time = 1.08639, size = 196, normalized size = 1.62 \begin{align*} -\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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.82329, size = 235, normalized size = 1.94 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 3.12227, size = 133, normalized size = 1.1 \begin{align*} \begin{cases} - \frac{a c^{2} d x^{5}}{5} + \frac{a d x^{3}}{3} - \frac{b c^{2} d x^{5} \operatorname{acosh}{\left (c x \right )}}{5} + \frac{b c d x^{4} \sqrt{c^{2} x^{2} - 1}}{25} + \frac{b d x^{3} \operatorname{acosh}{\left (c x \right )}}{3} - \frac{13 b d x^{2} \sqrt{c^{2} x^{2} - 1}}{225 c} - \frac{26 b d \sqrt{c^{2} x^{2} - 1}}{225 c^{3}} & \text{for}\: c \neq 0 \\\frac{d x^{3} \left (a + \frac{i \pi b}{2}\right )}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a*c**2*d*x**5/5 + a*d*x**3/3 - b*c**2*d*x**5*acosh(c*x)/5 + b*c*d*x**4*sqrt(c**2*x**2 - 1)/25 + b*
d*x**3*acosh(c*x)/3 - 13*b*d*x**2*sqrt(c**2*x**2 - 1)/(225*c) - 26*b*d*sqrt(c**2*x**2 - 1)/(225*c**3), Ne(c, 0
)), (d*x**3*(a + I*pi*b/2)/3, True))

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Giac [A]  time = 1.34541, size = 200, normalized size = 1.65 \begin{align*} -\frac{1}{5} \, a c^{2} d x^{5} - \frac{1}{75} \,{\left (15 \, x^{5} \log \left (c x + \sqrt{c^{2} x^{2} - 1}\right ) - \frac{3 \,{\left (c^{2} x^{2} - 1\right )}^{\frac{5}{2}} + 10 \,{\left (c^{2} x^{2} - 1\right )}^{\frac{3}{2}} + 15 \, \sqrt{c^{2} x^{2} - 1}}{c^{5}}\right )} b c^{2} d + \frac{1}{3} \, a d x^{3} + \frac{1}{9} \,{\left (3 \, x^{3} \log \left (c x + \sqrt{c^{2} x^{2} - 1}\right ) - \frac{{\left (c^{2} x^{2} - 1\right )}^{\frac{3}{2}} + 3 \, \sqrt{c^{2} x^{2} - 1}}{c^{3}}\right )} b d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*a*c^2*d*x^5 - 1/75*(15*x^5*log(c*x + sqrt(c^2*x^2 - 1)) - (3*(c^2*x^2 - 1)^(5/2) + 10*(c^2*x^2 - 1)^(3/2)
+ 15*sqrt(c^2*x^2 - 1))/c^5)*b*c^2*d + 1/3*a*d*x^3 + 1/9*(3*x^3*log(c*x + sqrt(c^2*x^2 - 1)) - ((c^2*x^2 - 1)
^(3/2) + 3*sqrt(c^2*x^2 - 1))/c^3)*b*d