3.2.6 \(\int \frac {x^3 (a+b \cosh ^{-1}(c x))}{\sqrt {d-c^2 d x^2}} \, dx\) [106]

Optimal. Leaf size=156 \[ -\frac {2 b x \sqrt {-1+c x} \sqrt {1+c x}}{3 c^3 \sqrt {d-c^2 d x^2}}-\frac {b x^3 \sqrt {-1+c x} \sqrt {1+c x}}{9 c \sqrt {d-c^2 d x^2}}-\frac {2 \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{3 c^4 d}-\frac {x^2 \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{3 c^2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.12, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {5938, 5914, 8, 30} \begin {gather*} -\frac {x^2 \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{3 c^2 d}-\frac {2 \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{3 c^4 d}-\frac {b x^3 \sqrt {c x-1} \sqrt {c x+1}}{9 c \sqrt {d-c^2 d x^2}}-\frac {2 b x \sqrt {c x-1} \sqrt {c x+1}}{3 c^3 \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 5914

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

Rule 5938

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

Rubi steps

\begin {align*} \int \frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}} \, dx &=\frac {\left (\sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{\sqrt {d-c^2 d x^2}}\\ &=-\frac {x^2 (1-c x) (1+c x) \left (a+b \cosh ^{-1}(c x)\right )}{3 c^2 \sqrt {d-c^2 d x^2}}+\frac {\left (2 \sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {x \left (a+b \cosh ^{-1}(c x)\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{3 c^2 \sqrt {d-c^2 d x^2}}-\frac {\left (b \sqrt {-1+c x} \sqrt {1+c x}\right ) \int x^2 \, dx}{3 c \sqrt {d-c^2 d x^2}}\\ &=-\frac {b x^3 \sqrt {-1+c x} \sqrt {1+c x}}{9 c \sqrt {d-c^2 d x^2}}-\frac {2 (1-c x) (1+c x) \left (a+b \cosh ^{-1}(c x)\right )}{3 c^4 \sqrt {d-c^2 d x^2}}-\frac {x^2 (1-c x) (1+c x) \left (a+b \cosh ^{-1}(c x)\right )}{3 c^2 \sqrt {d-c^2 d x^2}}-\frac {\left (2 b \sqrt {-1+c x} \sqrt {1+c x}\right ) \int 1 \, dx}{3 c^3 \sqrt {d-c^2 d x^2}}\\ &=-\frac {2 b x \sqrt {-1+c x} \sqrt {1+c x}}{3 c^3 \sqrt {d-c^2 d x^2}}-\frac {b x^3 \sqrt {-1+c x} \sqrt {1+c x}}{9 c \sqrt {d-c^2 d x^2}}-\frac {2 (1-c x) (1+c x) \left (a+b \cosh ^{-1}(c x)\right )}{3 c^4 \sqrt {d-c^2 d x^2}}-\frac {x^2 (1-c x) (1+c x) \left (a+b \cosh ^{-1}(c x)\right )}{3 c^2 \sqrt {d-c^2 d x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.13, size = 113, normalized size = 0.72 \begin {gather*} \frac {\sqrt {d-c^2 d x^2} \left (b c x \sqrt {-1+c x} \sqrt {1+c x} \left (6+c^2 x^2\right )-3 a \left (-2+c^2 x^2+c^4 x^4\right )-3 b \left (-2+c^2 x^2+c^4 x^4\right ) \cosh ^{-1}(c x)\right )}{9 c^4 d (-1+c x) (1+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d - c^2*d*x^2]*(b*c*x*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(6 + c^2*x^2) - 3*a*(-2 + c^2*x^2 + c^4*x^4) - 3*b*(-
2 + c^2*x^2 + c^4*x^4)*ArcCosh[c*x]))/(9*c^4*d*(-1 + c*x)*(1 + c*x))

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(381\) vs. \(2(132)=264\).
time = 4.16, size = 382, normalized size = 2.45

method	result	size
default	\(a \left (-\frac {x^{2} \sqrt {-c^{2} d \,x^{2}+d}}{3 c^{2} d}-\frac {2 \sqrt {-c^{2} d \,x^{2}+d}}{3 d \,c^{4}}\right )+b \left (-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (4 c^{4} x^{4}-5 c^{2} x^{2}+4 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{3} c^{3}-3 \sqrt {c x +1}\, \sqrt {c x -1}\, x c +1\right ) \left (-1+3 \,\mathrm {arccosh}\left (c x \right )\right )}{72 c^{4} d \left (c^{2} x^{2}-1\right )}-\frac {3 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (\sqrt {c x +1}\, \sqrt {c x -1}\, x c +c^{2} x^{2}-1\right ) \left (-1+\mathrm {arccosh}\left (c x \right )\right )}{8 c^{4} d \left (c^{2} x^{2}-1\right )}-\frac {3 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-\sqrt {c x +1}\, \sqrt {c x -1}\, x c +c^{2} x^{2}-1\right ) \left (1+\mathrm {arccosh}\left (c x \right )\right )}{8 c^{4} d \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-4 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{3} c^{3}+4 c^{4} x^{4}+3 \sqrt {c x +1}\, \sqrt {c x -1}\, x c -5 c^{2} x^{2}+1\right ) \left (1+3 \,\mathrm {arccosh}\left (c x \right )\right )}{72 c^{4} d \left (c^{2} x^{2}-1\right )}\right )\)	\(382\)

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,method=_RETURNVERBOSE)

[Out]

a*(-1/3*x^2/c^2/d*(-c^2*d*x^2+d)^(1/2)-2/3/d/c^4*(-c^2*d*x^2+d)^(1/2))+b*(-1/72*(-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
^4/d/(c^2*x^2-1)-3/8*(-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^4/
d/(c^2*x^2-1)-3/8*(-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^4/d/(
c^2*x^2-1)-1/72*(-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^4/d/(c^2*x^2-1))

________________________________________________________________________________________

Maxima [A]
time = 0.48, size = 131, normalized size = 0.84 \begin {gather*} -\frac {1}{3} \, b {\left (\frac {\sqrt {-c^{2} d x^{2} + d} x^{2}}{c^{2} d} + \frac {2 \, \sqrt {-c^{2} d x^{2} + d}}{c^{4} d}\right )} \operatorname {arcosh}\left (c x\right ) - \frac {1}{3} \, a {\left (\frac {\sqrt {-c^{2} d x^{2} + d} x^{2}}{c^{2} d} + \frac {2 \, \sqrt {-c^{2} d x^{2} + d}}{c^{4} d}\right )} + \frac {{\left (c^{2} \sqrt {-d} x^{3} + 6 \, \sqrt {-d} x\right )} b}{9 \, c^{3} d} \end {gather*}

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]

-1/3*b*(sqrt(-c^2*d*x^2 + d)*x^2/(c^2*d) + 2*sqrt(-c^2*d*x^2 + d)/(c^4*d))*arccosh(c*x) - 1/3*a*(sqrt(-c^2*d*x
^2 + d)*x^2/(c^2*d) + 2*sqrt(-c^2*d*x^2 + d)/(c^4*d)) + 1/9*(c^2*sqrt(-d)*x^3 + 6*sqrt(-d)*x)*b/(c^3*d)

________________________________________________________________________________________

Fricas [A]
time = 0.40, size = 146, normalized size = 0.94 \begin {gather*} -\frac {3 \, {\left (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 (b c^{3} x^{3} + 6 \, b c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} + 3 \, {\left (a c^{4} x^{4} + a c^{2} x^{2} - 2 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{9 \, {\left (c^{6} d x^{2} - c^{4} d\right )}} \end {gather*}

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]

-1/9*(3*(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)) - (b*c^3*x^3 + 6*b*c*x
)*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1) + 3*(a*c^4*x^4 + a*c^2*x^2 - 2*a)*sqrt(-c^2*d*x^2 + d))/(c^6*d*x^2 -
c^4*d)

________________________________________________________________________________________

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

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

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

________________________________________________________________________________________

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

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)

________________________________________________________________________________________