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

Optimal. Leaf size=158 \[ \frac {2 c^2 x \left (a+b \cosh ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}-\frac {a+b \cosh ^{-1}(c x)}{d x \sqrt {d-c^2 d x^2}}+\frac {b c \log (x) \sqrt {d-c^2 d x^2}}{d^2 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b c \sqrt {d-c^2 d x^2} \log \left (1-c^2 x^2\right )}{2 d^2 \sqrt {c x-1} \sqrt {c x+1}} \]

[Out]

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

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Rubi [A]  time = 0.40, antiderivative size = 159, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {5798, 103, 12, 39, 5733, 446, 72} \[ \frac {2 c^2 x \left (a+b \cosh ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}-\frac {a+b \cosh ^{-1}(c x)}{d x \sqrt {d-c^2 d x^2}}-\frac {b c \sqrt {c x-1} \sqrt {c x+1} \log (x)}{d \sqrt {d-c^2 d x^2}}-\frac {b c \sqrt {c x-1} \sqrt {c x+1} \log \left (1-c^2 x^2\right )}{2 d \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcCosh[c*x])/(d*x*Sqrt[d - c^2*d*x^2])) + (2*c^2*x*(a + b*ArcCosh[c*x]))/(d*Sqrt[d - c^2*d*x^2]) - (
b*c*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*Log[x])/(d*Sqrt[d - c^2*d*x^2]) - (b*c*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*Log[1 - c
^2*x^2])/(2*d*Sqrt[d - c^2*d*x^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 39

Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> Simp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d
*x]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 103

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))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 446

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

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

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Mathematica [A]  time = 0.09, size = 114, normalized size = 0.72 \[ \frac {4 a c^2 x^2-2 a-b c x \sqrt {c x-1} \sqrt {c x+1} \log \left (1-c^2 x^2\right )+2 b \left (2 c^2 x^2-1\right ) \cosh ^{-1}(c x)-2 b c x \sqrt {c x-1} \sqrt {c x+1} \log (x)}{2 d x \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b \operatorname {arcosh}\left (c x\right ) + a\right )}}{c^{4} d^{2} x^{6} - 2 \, c^{2} d^{2} x^{4} + d^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-c^2*d*x^2 + d)*(b*arccosh(c*x) + a)/(c^4*d^2*x^6 - 2*c^2*d^2*x^4 + d^2*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.36, size = 242, normalized size = 1.53 \[ -\frac {a}{d x \sqrt {-c^{2} d \,x^{2}+d}}+\frac {2 a \,c^{2} x}{d \sqrt {-c^{2} d \,x^{2}+d}}-\frac {2 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \mathrm {arccosh}\left (c x \right ) c}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \mathrm {arccosh}\left (c x \right ) x \,c^{2}}{\left (c^{2} x^{2}-1\right ) d^{2}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \mathrm {arccosh}\left (c x \right )}{x \left (c^{2} x^{2}-1\right ) d^{2}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \ln \left (\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{4}-1\right ) c}{d^{2} \left (c^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.74, size = 144, normalized size = 0.91 \[ \frac {1}{2} \, b c {\left (\frac {\sqrt {-d} \log \left (c x + 1\right )}{d^{2}} + \frac {\sqrt {-d} \log \left (c x - 1\right )}{d^{2}} + \frac {2 \, \sqrt {-d} \log \relax (x)}{d^{2}}\right )} + {\left (\frac {2 \, c^{2} x}{\sqrt {-c^{2} d x^{2} + d} d} - \frac {1}{\sqrt {-c^{2} d x^{2} + d} d x}\right )} b \operatorname {arcosh}\left (c x\right ) + {\left (\frac {2 \, c^{2} x}{\sqrt {-c^{2} d x^{2} + d} d} - \frac {1}{\sqrt {-c^{2} d x^{2} + d} d x}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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