3.66 \(\int x^3 \sqrt {d-c^2 d x^2} (a+b \cosh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=195 \[ \frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \cosh ^{-1}(c x)\right )}{5 c^4 d^2}-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \cosh ^{-1}(c x)\right )}{3 c^4 d}-\frac {b c x^5 \sqrt {d-c^2 d x^2}}{25 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b x^3 \sqrt {d-c^2 d x^2}}{45 c \sqrt {c x-1} \sqrt {c x+1}}+\frac {2 b x \sqrt {d-c^2 d x^2}}{15 c^3 \sqrt {c x-1} \sqrt {c x+1}} \]

[Out]

-1/3*(-c^2*d*x^2+d)^(3/2)*(a+b*arccosh(c*x))/c^4/d+1/5*(-c^2*d*x^2+d)^(5/2)*(a+b*arccosh(c*x))/c^4/d^2+2/15*b*
x*(-c^2*d*x^2+d)^(1/2)/c^3/(c*x-1)^(1/2)/(c*x+1)^(1/2)+1/45*b*x^3*(-c^2*d*x^2+d)^(1/2)/c/(c*x-1)^(1/2)/(c*x+1)
^(1/2)-1/25*b*c*x^5*(-c^2*d*x^2+d)^(1/2)/(c*x-1)^(1/2)/(c*x+1)^(1/2)

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Rubi [A]  time = 0.32, antiderivative size = 214, normalized size of antiderivative = 1.10, number of steps used = 4, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {5798, 100, 12, 74, 5733} \[ -\frac {x^2 (1-c x) (c x+1) \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{5 c^2}-\frac {2 (1-c x) (c x+1) \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{15 c^4}-\frac {b c x^5 \sqrt {d-c^2 d x^2}}{25 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b x^3 \sqrt {d-c^2 d x^2}}{45 c \sqrt {c x-1} \sqrt {c x+1}}+\frac {2 b x \sqrt {d-c^2 d x^2}}{15 c^3 \sqrt {c x-1} \sqrt {c x+1}} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[d - c^2*d*x^2]*(a + b*ArcCosh[c*x]),x]

[Out]

(2*b*x*Sqrt[d - c^2*d*x^2])/(15*c^3*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) + (b*x^3*Sqrt[d - c^2*d*x^2])/(45*c*Sqrt[-1
+ c*x]*Sqrt[1 + c*x]) - (b*c*x^5*Sqrt[d - c^2*d*x^2])/(25*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) - (2*(1 - c*x)*(1 + c*
x)*Sqrt[d - c^2*d*x^2]*(a + b*ArcCosh[c*x]))/(15*c^4) - (x^2*(1 - c*x)*(1 + c*x)*Sqrt[d - c^2*d*x^2]*(a + b*Ar
cCosh[c*x]))/(5*c^2)

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 5733

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d1_) + (e1_.)*(x_))^(p_)*((d2_) + (e2_.)*(x_))^(p_), x_Sym
bol] :> With[{u = IntHide[x^m*(1 + c*x)^p*(-1 + c*x)^p, x]}, Dist[(-(d1*d2))^p*(a + b*ArcCosh[c*x]), u, x] - D
ist[b*c*(-(d1*d2))^p, Int[SimplifyIntegrand[u/(Sqrt[1 + c*x]*Sqrt[-1 + c*x]), x], x], x]] /; FreeQ[{a, b, c, d
1, e1, d2, e2}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2, 0] && IntegerQ[p - 1/2] && (IGtQ[(m + 1)/2, 0] || IL
tQ[(m + 2*p + 3)/2, 0]) && NeQ[p, -2^(-1)] && GtQ[d1, 0] && LtQ[d2, 0]

Rule 5798

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

Rubi steps

\begin {align*} \int x^3 \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right ) \, dx &=\frac {\sqrt {d-c^2 d x^2} \int x^3 \sqrt {-1+c x} \sqrt {1+c x} \left (a+b \cosh ^{-1}(c x)\right ) \, dx}{\sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {2 (1-c x) (1+c x) \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{15 c^4}-\frac {x^2 (1-c x) (1+c x) \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{5 c^2}-\frac {\left (b c \sqrt {d-c^2 d x^2}\right ) \int \frac {-2-c^2 x^2+3 c^4 x^4}{15 c^4} \, dx}{\sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {2 (1-c x) (1+c x) \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{15 c^4}-\frac {x^2 (1-c x) (1+c x) \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{5 c^2}-\frac {\left (b \sqrt {d-c^2 d x^2}\right ) \int \left (-2-c^2 x^2+3 c^4 x^4\right ) \, dx}{15 c^3 \sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {2 b x \sqrt {d-c^2 d x^2}}{15 c^3 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b x^3 \sqrt {d-c^2 d x^2}}{45 c \sqrt {-1+c x} \sqrt {1+c x}}-\frac {b c x^5 \sqrt {d-c^2 d x^2}}{25 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {2 (1-c x) (1+c x) \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{15 c^4}-\frac {x^2 (1-c x) (1+c x) \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{5 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 128, normalized size = 0.66 \[ \frac {\sqrt {d-c^2 d x^2} \left (3 c^2 x^2 (c x-1)^{3/2} (c x+1)^{3/2} \left (a+b \cosh ^{-1}(c x)\right )+2 (c x-1)^{3/2} (c x+1)^{3/2} \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{15} b c x \left (-9 c^4 x^4+5 c^2 x^2+30\right )\right )}{15 c^4 \sqrt {c x-1} \sqrt {c x+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sqrt[d - c^2*d*x^2]*(a + b*ArcCosh[c*x]),x]

[Out]

(Sqrt[d - c^2*d*x^2]*((b*c*x*(30 + 5*c^2*x^2 - 9*c^4*x^4))/15 + 2*(-1 + c*x)^(3/2)*(1 + c*x)^(3/2)*(a + b*ArcC
osh[c*x]) + 3*c^2*x^2*(-1 + c*x)^(3/2)*(1 + c*x)^(3/2)*(a + b*ArcCosh[c*x])))/(15*c^4*Sqrt[-1 + c*x]*Sqrt[1 +
c*x])

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fricas [A]  time = 0.51, size = 176, normalized size = 0.90 \[ \frac {15 \, {\left (3 \, b c^{6} x^{6} - 4 \, b c^{4} x^{4} - b c^{2} x^{2} + 2 \, b\right )} \sqrt {-c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) - {\left (9 \, b c^{5} x^{5} - 5 \, b c^{3} x^{3} - 30 \, b c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} + 15 \, {\left (3 \, a c^{6} x^{6} - 4 \, a c^{4} x^{4} - a c^{2} x^{2} + 2 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{225 \, {\left (c^{6} x^{2} - c^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccosh(c*x))*(-c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

1/225*(15*(3*b*c^6*x^6 - 4*b*c^4*x^4 - b*c^2*x^2 + 2*b)*sqrt(-c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 - 1)) - (9
*b*c^5*x^5 - 5*b*c^3*x^3 - 30*b*c*x)*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1) + 15*(3*a*c^6*x^6 - 4*a*c^4*x^4 -
a*c^2*x^2 + 2*a)*sqrt(-c^2*d*x^2 + d))/(c^6*x^2 - c^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccosh(c*x))*(-c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.54, size = 640, normalized size = 3.28 \[ a \left (-\frac {x^{2} \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{5 c^{2} d}-\frac {2 \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{15 d \,c^{4}}\right )+b \left (\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (16 c^{6} x^{6}-28 c^{4} x^{4}+16 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{5} c^{5}+13 c^{2} x^{2}-20 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{3} c^{3}+5 \sqrt {c x +1}\, \sqrt {c x -1}\, x c -1\right ) \left (-1+5 \,\mathrm {arccosh}\left (c x \right )\right )}{800 \left (c x +1\right ) c^{4} \left (c x -1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (4 c^{4} x^{4}-5 c^{2} x^{2}+4 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{3} c^{3}-3 \sqrt {c x +1}\, \sqrt {c x -1}\, x c +1\right ) \left (-1+3 \,\mathrm {arccosh}\left (c x \right )\right )}{288 \left (c x +1\right ) c^{4} \left (c x -1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (\sqrt {c x +1}\, \sqrt {c x -1}\, x c +c^{2} x^{2}-1\right ) \left (-1+\mathrm {arccosh}\left (c x \right )\right )}{16 \left (c x +1\right ) c^{4} \left (c x -1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-\sqrt {c x +1}\, \sqrt {c x -1}\, x c +c^{2} x^{2}-1\right ) \left (1+\mathrm {arccosh}\left (c x \right )\right )}{16 \left (c x +1\right ) c^{4} \left (c x -1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-4 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{3} c^{3}+4 c^{4} x^{4}+3 \sqrt {c x +1}\, \sqrt {c x -1}\, x c -5 c^{2} x^{2}+1\right ) \left (1+3 \,\mathrm {arccosh}\left (c x \right )\right )}{288 \left (c x +1\right ) c^{4} \left (c x -1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-16 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{5} c^{5}+16 c^{6} x^{6}+20 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{3} c^{3}-28 c^{4} x^{4}-5 \sqrt {c x +1}\, \sqrt {c x -1}\, x c +13 c^{2} x^{2}-1\right ) \left (1+5 \,\mathrm {arccosh}\left (c x \right )\right )}{800 \left (c x +1\right ) c^{4} \left (c x -1\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arccosh(c*x))*(-c^2*d*x^2+d)^(1/2),x)

[Out]

a*(-1/5*x^2*(-c^2*d*x^2+d)^(3/2)/c^2/d-2/15/d/c^4*(-c^2*d*x^2+d)^(3/2))+b*(1/800*(-d*(c^2*x^2-1))^(1/2)*(16*c^
6*x^6-28*c^4*x^4+16*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^5*c^5+13*c^2*x^2-20*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^3*c^3+5*(c
*x+1)^(1/2)*(c*x-1)^(1/2)*x*c-1)*(-1+5*arccosh(c*x))/(c*x+1)/c^4/(c*x-1)+1/288*(-d*(c^2*x^2-1))^(1/2)*(4*c^4*x
^4-5*c^2*x^2+4*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^3*c^3-3*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x*c+1)*(-1+3*arccosh(c*x))/(c
*x+1)/c^4/(c*x-1)-1/16*(-d*(c^2*x^2-1))^(1/2)*((c*x+1)^(1/2)*(c*x-1)^(1/2)*x*c+c^2*x^2-1)*(-1+arccosh(c*x))/(c
*x+1)/c^4/(c*x-1)-1/16*(-d*(c^2*x^2-1))^(1/2)*(-(c*x+1)^(1/2)*(c*x-1)^(1/2)*x*c+c^2*x^2-1)*(1+arccosh(c*x))/(c
*x+1)/c^4/(c*x-1)+1/288*(-d*(c^2*x^2-1))^(1/2)*(-4*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^3*c^3+4*c^4*x^4+3*(c*x+1)^(1/
2)*(c*x-1)^(1/2)*x*c-5*c^2*x^2+1)*(1+3*arccosh(c*x))/(c*x+1)/c^4/(c*x-1)+1/800*(-d*(c^2*x^2-1))^(1/2)*(-16*(c*
x+1)^(1/2)*(c*x-1)^(1/2)*x^5*c^5+16*c^6*x^6+20*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^3*c^3-28*c^4*x^4-5*(c*x+1)^(1/2)*
(c*x-1)^(1/2)*x*c+13*c^2*x^2-1)*(1+5*arccosh(c*x))/(c*x+1)/c^4/(c*x-1))

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maxima [A]  time = 0.51, size = 144, normalized size = 0.74 \[ -\frac {1}{15} \, b {\left (\frac {3 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x^{2}}{c^{2} d} + \frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{c^{4} d}\right )} \operatorname {arcosh}\left (c x\right ) - \frac {1}{15} \, a {\left (\frac {3 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x^{2}}{c^{2} d} + \frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{c^{4} d}\right )} - \frac {{\left (9 \, c^{4} \sqrt {-d} x^{5} - 5 \, c^{2} \sqrt {-d} x^{3} - 30 \, \sqrt {-d} x\right )} b}{225 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccosh(c*x))*(-c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

-1/15*b*(3*(-c^2*d*x^2 + d)^(3/2)*x^2/(c^2*d) + 2*(-c^2*d*x^2 + d)^(3/2)/(c^4*d))*arccosh(c*x) - 1/15*a*(3*(-c
^2*d*x^2 + d)^(3/2)*x^2/(c^2*d) + 2*(-c^2*d*x^2 + d)^(3/2)/(c^4*d)) - 1/225*(9*c^4*sqrt(-d)*x^5 - 5*c^2*sqrt(-
d)*x^3 - 30*sqrt(-d)*x)*b/c^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,\sqrt {d-c^2\,d\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*acosh(c*x))*(d - c^2*d*x^2)^(1/2),x)

[Out]

int(x^3*(a + b*acosh(c*x))*(d - c^2*d*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )} \left (a + b \operatorname {acosh}{\left (c x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*acosh(c*x))*(-c**2*d*x**2+d)**(1/2),x)

[Out]

Integral(x**3*sqrt(-d*(c*x - 1)*(c*x + 1))*(a + b*acosh(c*x)), x)

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