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

Optimal. Leaf size=150 \[ \frac {\sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{c^4 d^2}+\frac {a+b \cosh ^{-1}(c x)}{c^4 d \sqrt {d-c^2 d x^2}}-\frac {b \sqrt {d-c^2 d x^2} \tanh ^{-1}(c x)}{c^4 d^2 \sqrt {c x-1} \sqrt {c x+1}}-\frac {b x \sqrt {d-c^2 d x^2}}{c^3 d^2 \sqrt {c x-1} \sqrt {c x+1}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.39, antiderivative size = 163, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {5798, 98, 21, 74, 5733, 388, 208} \[ \frac {x^2 \left (a+b \cosh ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {2 (1-c x) (c x+1) \left (a+b \cosh ^{-1}(c x)\right )}{c^4 d \sqrt {d-c^2 d x^2}}+\frac {b x \sqrt {c x-1} \sqrt {c x+1}}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {c x-1} \sqrt {c x+1} \tanh ^{-1}(c x)}{c^4 d \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 388

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

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 \frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{3/2}} \, dx &=-\frac {\left (\sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{(-1+c x)^{3/2} (1+c x)^{3/2}} \, dx}{d \sqrt {d-c^2 d x^2}}\\ &=\frac {x^2 \left (a+b \cosh ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {2 (1-c x) (1+c x) \left (a+b \cosh ^{-1}(c x)\right )}{c^4 d \sqrt {d-c^2 d x^2}}+\frac {\left (b c \sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {2-c^2 x^2}{c^4-c^6 x^2} \, dx}{d \sqrt {d-c^2 d x^2}}\\ &=\frac {b x \sqrt {-1+c x} \sqrt {1+c x}}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {x^2 \left (a+b \cosh ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {2 (1-c x) (1+c x) \left (a+b \cosh ^{-1}(c x)\right )}{c^4 d \sqrt {d-c^2 d x^2}}+\frac {\left (b c \sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {1}{c^4-c^6 x^2} \, dx}{d \sqrt {d-c^2 d x^2}}\\ &=\frac {b x \sqrt {-1+c x} \sqrt {1+c x}}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {x^2 \left (a+b \cosh ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {2 (1-c x) (1+c x) \left (a+b \cosh ^{-1}(c x)\right )}{c^4 d \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {-1+c x} \sqrt {1+c x} \tanh ^{-1}(c x)}{c^4 d \sqrt {d-c^2 d x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 97, normalized size = 0.65 \[ \frac {-a c^2 x^2+2 a+b \left (2-c^2 x^2\right ) \cosh ^{-1}(c x)+b c x \sqrt {c x-1} \sqrt {c x+1}+b \sqrt {c x-1} \sqrt {c x+1} \tanh ^{-1}(c x)}{c^4 d \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 429, normalized size = 2.86 \[ \left [-\frac {4 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} b c x - 4 \, {\left (b c^{2} x^{2} - 2 \, b\right )} \sqrt {-c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) + {\left (b c^{2} x^{2} - b\right )} \sqrt {-d} \log \left (-\frac {c^{6} d x^{6} + 5 \, c^{4} d x^{4} - 5 \, c^{2} d x^{2} - 4 \, {\left (c^{3} x^{3} + c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} \sqrt {-d} - d}{c^{6} x^{6} - 3 \, c^{4} x^{4} + 3 \, c^{2} x^{2} - 1}\right ) - 4 \, {\left (a c^{2} x^{2} - 2 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{4 \, {\left (c^{6} d^{2} x^{2} - c^{4} d^{2}\right )}}, -\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} b c x + {\left (b c^{2} x^{2} - b\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} c \sqrt {d} x}{c^{4} d x^{4} - d}\right ) - 2 \, {\left (b c^{2} x^{2} - 2 \, b\right )} \sqrt {-c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) - 2 \, {\left (a c^{2} x^{2} - 2 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{2 \, {\left (c^{6} d^{2} x^{2} - c^{4} d^{2}\right )}}\right ] \]

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)^(3/2),x, algorithm="fricas")

[Out]

[-1/4*(4*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*b*c*x - 4*(b*c^2*x^2 - 2*b)*sqrt(-c^2*d*x^2 + d)*log(c*x + sqr
t(c^2*x^2 - 1)) + (b*c^2*x^2 - b)*sqrt(-d)*log(-(c^6*d*x^6 + 5*c^4*d*x^4 - 5*c^2*d*x^2 - 4*(c^3*x^3 + c*x)*sqr
t(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*sqrt(-d) - d)/(c^6*x^6 - 3*c^4*x^4 + 3*c^2*x^2 - 1)) - 4*(a*c^2*x^2 - 2*a)
*sqrt(-c^2*d*x^2 + d))/(c^6*d^2*x^2 - c^4*d^2), -1/2*(2*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*b*c*x + (b*c^2*
x^2 - b)*sqrt(d)*arctan(2*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*c*sqrt(d)*x/(c^4*d*x^4 - d)) - 2*(b*c^2*x^2 -
2*b)*sqrt(-c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 - 1)) - 2*(a*c^2*x^2 - 2*a)*sqrt(-c^2*d*x^2 + d))/(c^6*d^2*x
^2 - c^4*d^2)]

________________________________________________________________________________________

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)^(3/2),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

________________________________________________________________________________________

maple [B]  time = 0.54, size = 313, normalized size = 2.09 \[ -\frac {a \,x^{2}}{c^{2} d \sqrt {-c^{2} d \,x^{2}+d}}+\frac {2 a}{d \,c^{4} \sqrt {-c^{2} d \,x^{2}+d}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \mathrm {arccosh}\left (c x \right ) x^{2}}{c^{2} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x +1}\, \sqrt {c x -1}\, x}{c^{3} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \mathrm {arccosh}\left (c x \right )}{c^{4} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \ln \left (c x +\sqrt {c x -1}\, \sqrt {c x +1}-1\right )}{c^{4} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \ln \left (1+c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{c^{4} d^{2} \left (c^{2} x^{2}-1\right )} \]

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)^(3/2),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.93, size = 157, normalized size = 1.05 \[ -\frac {1}{2} \, b c {\left (\frac {2 \, \sqrt {-d} x}{c^{4} d^{2}} + \frac {\sqrt {-d} \log \left (c x + 1\right )}{c^{5} d^{2}} - \frac {\sqrt {-d} \log \left (c x - 1\right )}{c^{5} d^{2}}\right )} - b {\left (\frac {x^{2}}{\sqrt {-c^{2} d x^{2} + d} c^{2} d} - \frac {2}{\sqrt {-c^{2} d x^{2} + d} c^{4} d}\right )} \operatorname {arcosh}\left (c x\right ) - a {\left (\frac {x^{2}}{\sqrt {-c^{2} d x^{2} + d} c^{2} d} - \frac {2}{\sqrt {-c^{2} d x^{2} + d} c^{4} d}\right )} \]

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)^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

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)^(3/2),x)

[Out]

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

________________________________________________________________________________________

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

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)**(3/2),x)

[Out]

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

________________________________________________________________________________________