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

Optimal. Leaf size=136 \[ \frac {x^4 \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {b \sqrt {c x-1}}{4 c^4 d^3 \sqrt {c x+1}}+\frac {b}{4 c^4 d^3 \sqrt {c x-1} \sqrt {c x+1}}-\frac {b \cosh ^{-1}(c x)}{4 c^4 d^3}+\frac {b x^3}{12 c d^3 (c x-1)^{3/2} (c x+1)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {5722, 98, 21, 89, 12, 78, 52} \[ \frac {x^4 \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {b \sqrt {c x-1}}{4 c^4 d^3 \sqrt {c x+1}}+\frac {b}{4 c^4 d^3 \sqrt {c x-1} \sqrt {c x+1}}-\frac {b \cosh ^{-1}(c x)}{4 c^4 d^3}+\frac {b x^3}{12 c d^3 (c x-1)^{3/2} (c x+1)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x^3)/(12*c*d^3*(-1 + c*x)^(3/2)*(1 + c*x)^(3/2)) + b/(4*c^4*d^3*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) + (b*Sqrt[-1
+ c*x])/(4*c^4*d^3*Sqrt[1 + c*x]) - (b*ArcCosh[c*x])/(4*c^4*d^3) + (x^4*(a + b*ArcCosh[c*x]))/(4*d^3*(1 - c^2*
x^2)^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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 52

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ArcCosh[(b*x)/a]/b, x] /; FreeQ[{a,
b, c, d}, x] && EqQ[a + c, 0] && EqQ[b - d, 0] && GtQ[a, 0]

Rule 78

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

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 98

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

Rule 5722

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcCosh[c*x])^n)/(d*f*(m + 1)), x] + Dist[(b*c*n*(-d)^p)/(f*(m + 1)
), Int[(f*x)^(m + 1)*(1 + c*x)^(p + 1/2)*(-1 + c*x)^(p + 1/2)*(a + b*ArcCosh[c*x])^(n - 1), x], x] /; FreeQ[{a
, b, c, d, e, f, m, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1] && IntegerQ[
p]

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 83, normalized size = 0.61 \[ \frac {a \left (6 c^2 x^2-3\right )+b c x \sqrt {c x-1} \sqrt {c x+1} \left (4 c^2 x^2-3\right )+3 b \left (2 c^2 x^2-1\right ) \cosh ^{-1}(c x)}{12 c^4 d^3 \left (c^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*x*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(-3 + 4*c^2*x^2) + a*(-3 + 6*c^2*x^2) + 3*b*(-1 + 2*c^2*x^2)*ArcCosh[c*x])
/(12*c^4*d^3*(-1 + c^2*x^2)^2)

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fricas [A]  time = 0.72, size = 101, normalized size = 0.74 \[ \frac {3 \, a c^{4} x^{4} + 3 \, {\left (2 \, b c^{2} x^{2} - b\right )} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) + {\left (4 \, b c^{3} x^{3} - 3 \, b c x\right )} \sqrt {c^{2} x^{2} - 1}}{12 \, {\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}} \]

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)^3,x, algorithm="fricas")

[Out]

1/12*(3*a*c^4*x^4 + 3*(2*b*c^2*x^2 - b)*log(c*x + sqrt(c^2*x^2 - 1)) + (4*b*c^3*x^3 - 3*b*c*x)*sqrt(c^2*x^2 -
1))/(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4*d^3)

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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)^3,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 [A]  time = 0.02, size = 136, normalized size = 1.00 \[ \frac {-\frac {a \left (-\frac {1}{16 \left (c x -1\right )^{2}}-\frac {3}{16 \left (c x -1\right )}-\frac {1}{16 \left (c x +1\right )^{2}}+\frac {3}{16 \left (c x +1\right )}\right )}{d^{3}}-\frac {b \left (-\frac {\mathrm {arccosh}\left (c x \right )}{16 \left (c x -1\right )^{2}}-\frac {3 \,\mathrm {arccosh}\left (c x \right )}{16 \left (c x -1\right )}-\frac {\mathrm {arccosh}\left (c x \right )}{16 \left (c x +1\right )^{2}}+\frac {3 \,\mathrm {arccosh}\left (c x \right )}{16 \left (c x +1\right )}-\frac {c x \left (4 c^{2} x^{2}-3\right )}{12 \left (c x +1\right )^{\frac {3}{2}} \left (c x -1\right )^{\frac {3}{2}}}\right )}{d^{3}}}{c^{4}} \]

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)^3,x)

[Out]

1/c^4*(-a/d^3*(-1/16/(c*x-1)^2-3/16/(c*x-1)-1/16/(c*x+1)^2+3/16/(c*x+1))-b/d^3*(-1/16*arccosh(c*x)/(c*x-1)^2-3
/16*arccosh(c*x)/(c*x-1)-1/16*arccosh(c*x)/(c*x+1)^2+3/16*arccosh(c*x)/(c*x+1)-1/12*c*x*(4*c^2*x^2-3)/(c*x+1)^
(3/2)/(c*x-1)^(3/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{16} \, b {\left (\frac {4 \, c^{2} x^{2} + 4 \, {\left (2 \, c^{2} x^{2} - 1\right )} \log \left (c x + \sqrt {c x + 1} \sqrt {c x - 1}\right ) - 3}{c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}} + 16 \, \int \frac {2 \, c^{2} x^{2} - 1}{4 \, {\left (c^{10} d^{3} x^{7} - 3 \, c^{8} d^{3} x^{5} + 3 \, c^{6} d^{3} x^{3} - c^{4} d^{3} x + {\left (c^{9} d^{3} x^{6} - 3 \, c^{7} d^{3} x^{4} + 3 \, c^{5} d^{3} x^{2} - c^{3} d^{3}\right )} e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (c x - 1\right )\right )}\right )}}\,{d x}\right )} + \frac {{\left (2 \, c^{2} x^{2} - 1\right )} a}{4 \, {\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}} \]

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)^3,x, algorithm="maxima")

[Out]

1/16*b*((4*c^2*x^2 + 4*(2*c^2*x^2 - 1)*log(c*x + sqrt(c*x + 1)*sqrt(c*x - 1)) - 3)/(c^8*d^3*x^4 - 2*c^6*d^3*x^
2 + c^4*d^3) + 16*integrate(1/4*(2*c^2*x^2 - 1)/(c^10*d^3*x^7 - 3*c^8*d^3*x^5 + 3*c^6*d^3*x^3 - c^4*d^3*x + (c
^9*d^3*x^6 - 3*c^7*d^3*x^4 + 3*c^5*d^3*x^2 - c^3*d^3)*e^(1/2*log(c*x + 1) + 1/2*log(c*x - 1))), x)) + 1/4*(2*c
^2*x^2 - 1)*a/(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4*d^3)

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

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)^3,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {a x^{3}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx + \int \frac {b x^{3} \operatorname {acosh}{\left (c x \right )}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx}{d^{3}} \]

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)**3,x)

[Out]

-(Integral(a*x**3/(c**6*x**6 - 3*c**4*x**4 + 3*c**2*x**2 - 1), x) + Integral(b*x**3*acosh(c*x)/(c**6*x**6 - 3*
c**4*x**4 + 3*c**2*x**2 - 1), x))/d**3

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