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

Optimal. Leaf size=138 \[ \frac {1}{3} d x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \cosh ^{-1}(c x)\right )-\frac {2 b \sqrt {c x-1} \sqrt {c x+1} \left (25 c^2 d+12 e\right )}{225 c^5}-\frac {b x^2 \sqrt {c x-1} \sqrt {c x+1} \left (25 c^2 d+12 e\right )}{225 c^3}-\frac {b e x^4 \sqrt {c x-1} \sqrt {c x+1}}{25 c} \]

[Out]

1/3*d*x^3*(a+b*arccosh(c*x))+1/5*e*x^5*(a+b*arccosh(c*x))-2/225*b*(25*c^2*d+12*e)*(c*x-1)^(1/2)*(c*x+1)^(1/2)/
c^5-1/225*b*(25*c^2*d+12*e)*x^2*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c^3-1/25*b*e*x^4*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c

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Rubi [A]  time = 0.12, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {5786, 460, 100, 12, 74} \[ \frac {1}{3} d x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \cosh ^{-1}(c x)\right )-\frac {b x^2 \sqrt {c x-1} \sqrt {c x+1} \left (25 c^2 d+12 e\right )}{225 c^3}-\frac {2 b \sqrt {c x-1} \sqrt {c x+1} \left (25 c^2 d+12 e\right )}{225 c^5}-\frac {b e x^4 \sqrt {c x-1} \sqrt {c x+1}}{25 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*(25*c^2*d + 12*e)*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(225*c^5) - (b*(25*c^2*d + 12*e)*x^2*Sqrt[-1 + c*x]*Sqrt
[1 + c*x])/(225*c^3) - (b*e*x^4*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(25*c) + (d*x^3*(a + b*ArcCosh[c*x]))/3 + (e*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 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]

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 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 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 x^2 \left (d+e 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} e x^5 \left (a+b \cosh ^{-1}(c x)\right )-\frac {1}{15} (b c) \int \frac {x^3 \left (5 d+3 e x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx\\ &=-\frac {b e x^4 \sqrt {-1+c x} \sqrt {1+c x}}{25 c}+\frac {1}{3} d x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \cosh ^{-1}(c x)\right )-\frac {1}{75} \left (b c \left (25 d+\frac {12 e}{c^2}\right )\right ) \int \frac {x^3}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx\\ &=-\frac {b \left (25 c^2 d+12 e\right ) x^2 \sqrt {-1+c x} \sqrt {1+c x}}{225 c^3}-\frac {b e x^4 \sqrt {-1+c x} \sqrt {1+c x}}{25 c}+\frac {1}{3} d x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \cosh ^{-1}(c x)\right )-\frac {\left (b \left (25 c^2 d+12 e\right )\right ) \int \frac {2 x}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{225 c^3}\\ &=-\frac {b \left (25 c^2 d+12 e\right ) x^2 \sqrt {-1+c x} \sqrt {1+c x}}{225 c^3}-\frac {b e x^4 \sqrt {-1+c x} \sqrt {1+c x}}{25 c}+\frac {1}{3} d x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \cosh ^{-1}(c x)\right )-\frac {\left (2 b \left (25 c^2 d+12 e\right )\right ) \int \frac {x}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{225 c^3}\\ &=-\frac {2 b \left (25 c^2 d+12 e\right ) \sqrt {-1+c x} \sqrt {1+c x}}{225 c^5}-\frac {b \left (25 c^2 d+12 e\right ) x^2 \sqrt {-1+c x} \sqrt {1+c x}}{225 c^3}-\frac {b e x^4 \sqrt {-1+c x} \sqrt {1+c x}}{25 c}+\frac {1}{3} d x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \cosh ^{-1}(c x)\right )\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 101, normalized size = 0.73 \[ \frac {1}{225} \left (15 a x^3 \left (5 d+3 e x^2\right )-\frac {b \sqrt {c x-1} \sqrt {c x+1} \left (c^4 \left (25 d x^2+9 e x^4\right )+2 c^2 \left (25 d+6 e x^2\right )+24 e\right )}{c^5}+15 b x^3 \cosh ^{-1}(c x) \left (5 d+3 e x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*a*x^3*(5*d + 3*e*x^2) - (b*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(24*e + 2*c^2*(25*d + 6*e*x^2) + c^4*(25*d*x^2 + 9
*e*x^4)))/c^5 + 15*b*x^3*(5*d + 3*e*x^2)*ArcCosh[c*x])/225

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fricas [A]  time = 0.60, size = 119, normalized size = 0.86 \[ \frac {45 \, a c^{5} e x^{5} + 75 \, a c^{5} d x^{3} + 15 \, {\left (3 \, b c^{5} e x^{5} + 5 \, b c^{5} d x^{3}\right )} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) - {\left (9 \, b c^{4} e x^{4} + 50 \, b c^{2} d + {\left (25 \, b c^{4} d + 12 \, b c^{2} e\right )} x^{2} + 24 \, b e\right )} \sqrt {c^{2} x^{2} - 1}}{225 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/225*(45*a*c^5*e*x^5 + 75*a*c^5*d*x^3 + 15*(3*b*c^5*e*x^5 + 5*b*c^5*d*x^3)*log(c*x + sqrt(c^2*x^2 - 1)) - (9*
b*c^4*e*x^4 + 50*b*c^2*d + (25*b*c^4*d + 12*b*c^2*e)*x^2 + 24*b*e)*sqrt(c^2*x^2 - 1))/c^5

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*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

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maple [A]  time = 0.01, size = 115, normalized size = 0.83 \[ \frac {\frac {a \left (\frac {1}{5} c^{5} x^{5} e +\frac {1}{3} c^{5} x^{3} d \right )}{c^{2}}+\frac {b \left (\frac {\mathrm {arccosh}\left (c x \right ) c^{5} x^{5} e}{5}+\frac {\mathrm {arccosh}\left (c x \right ) c^{5} x^{3} d}{3}-\frac {\sqrt {c x -1}\, \sqrt {c x +1}\, \left (9 c^{4} e \,x^{4}+25 c^{4} d \,x^{2}+12 c^{2} x^{2} e +50 c^{2} d +24 e \right )}{225}\right )}{c^{2}}}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^3*(a/c^2*(1/5*c^5*x^5*e+1/3*c^5*x^3*d)+b/c^2*(1/5*arccosh(c*x)*c^5*x^5*e+1/3*arccosh(c*x)*c^5*x^3*d-1/225*
(c*x-1)^(1/2)*(c*x+1)^(1/2)*(9*c^4*e*x^4+25*c^4*d*x^2+12*c^2*e*x^2+50*c^2*d+24*e)))

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maxima [A]  time = 0.38, size = 139, normalized size = 1.01 \[ \frac {1}{5} \, a e x^{5} + \frac {1}{3} \, a d x^{3} + \frac {1}{9} \, {\left (3 \, x^{3} \operatorname {arcosh}\left (c x\right ) - c {\left (\frac {\sqrt {c^{2} x^{2} - 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {c^{2} x^{2} - 1}}{c^{4}}\right )}\right )} b d + \frac {1}{75} \, {\left (15 \, x^{5} \operatorname {arcosh}\left (c x\right ) - {\left (\frac {3 \, \sqrt {c^{2} x^{2} - 1} x^{4}}{c^{2}} + \frac {4 \, \sqrt {c^{2} x^{2} - 1} x^{2}}{c^{4}} + \frac {8 \, \sqrt {c^{2} x^{2} - 1}}{c^{6}}\right )} c\right )} b e \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*e*x^5 + 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*sqrt(c^2*x^2 - 1)/c^4))
*b*d + 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*e

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.15, size = 178, normalized size = 1.29 \[ \begin {cases} \frac {a d x^{3}}{3} + \frac {a e x^{5}}{5} + \frac {b d x^{3} \operatorname {acosh}{\left (c x \right )}}{3} + \frac {b e x^{5} \operatorname {acosh}{\left (c x \right )}}{5} - \frac {b d x^{2} \sqrt {c^{2} x^{2} - 1}}{9 c} - \frac {b e x^{4} \sqrt {c^{2} x^{2} - 1}}{25 c} - \frac {2 b d \sqrt {c^{2} x^{2} - 1}}{9 c^{3}} - \frac {4 b e x^{2} \sqrt {c^{2} x^{2} - 1}}{75 c^{3}} - \frac {8 b e \sqrt {c^{2} x^{2} - 1}}{75 c^{5}} & \text {for}\: c \neq 0 \\\left (a + \frac {i \pi b}{2}\right ) \left (\frac {d x^{3}}{3} + \frac {e x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*d*x**3/3 + a*e*x**5/5 + b*d*x**3*acosh(c*x)/3 + b*e*x**5*acosh(c*x)/5 - b*d*x**2*sqrt(c**2*x**2 -
1)/(9*c) - b*e*x**4*sqrt(c**2*x**2 - 1)/(25*c) - 2*b*d*sqrt(c**2*x**2 - 1)/(9*c**3) - 4*b*e*x**2*sqrt(c**2*x*
*2 - 1)/(75*c**3) - 8*b*e*sqrt(c**2*x**2 - 1)/(75*c**5), Ne(c, 0)), ((a + I*pi*b/2)*(d*x**3/3 + e*x**5/5), Tru
e))

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