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

Optimal. Leaf size=71 \[ -\frac {\sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{d x}-\frac {b c \sqrt {c x-1} \sqrt {c x+1} \log (x)}{\sqrt {d-c^2 d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 79, normalized size of antiderivative = 1.11, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {5798, 5724, 29} \[ -\frac {(1-c x) (c x+1) \left (a+b \cosh ^{-1}(c x)\right )}{x \sqrt {d-c^2 d x^2}}-\frac {b c \sqrt {c x-1} \sqrt {c x+1} \log (x)}{\sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 5724

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d1_) + (e1_.)*(x_))^(p_.)*((d2_) + (e2_.)*(x
_))^(p_.), x_Symbol] :> Simp[((f*x)^(m + 1)*(d1 + e1*x)^(p + 1)*(d2 + e2*x)^(p + 1)*(a + b*ArcCosh[c*x])^n)/(d
1*d2*f*(m + 1)), x] + Dist[(b*c*n*(-(d1*d2))^IntPart[p]*(d1 + e1*x)^FracPart[p]*(d2 + e2*x)^FracPart[p])/(f*(m
+ 1)*(1 + c*x)^FracPart[p]*(-1 + c*x)^FracPart[p]), Int[(f*x)^(m + 1)*(-1 + c^2*x^2)^(p + 1/2)*(a + b*ArcCosh
[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d1, e1, d2, e2, f, m, p}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2,
0] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1] && IntegerQ[p + 1/2]

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

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Mathematica [A]  time = 0.07, size = 71, normalized size = 1.00 \[ \frac {\sqrt {c x-1} \sqrt {c x+1} \left (\frac {\sqrt {c x-1} \sqrt {c x+1} \left (a+b \cosh ^{-1}(c x)\right )}{x}-b c \log (x)\right )}{\sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.58, size = 265, normalized size = 3.73 \[ \left [-\frac {b c \sqrt {-d} x \log \left (\frac {c^{2} d x^{6} + c^{2} d x^{2} - d x^{4} + \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} {\left (x^{4} - 1\right )} \sqrt {-d} - d}{c^{2} x^{4} - x^{2}}\right ) + 2 \, \sqrt {-c^{2} d x^{2} + d} b \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) + 2 \, \sqrt {-c^{2} d x^{2} + d} a}{2 \, d x}, \frac {b c \sqrt {d} x \arctan \left (\frac {\sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} {\left (x^{2} + 1\right )} \sqrt {d}}{c^{2} d x^{4} - {\left (c^{2} + 1\right )} d x^{2} + d}\right ) - \sqrt {-c^{2} d x^{2} + d} b \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) - \sqrt {-c^{2} d x^{2} + d} a}{d x}\right ] \]

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)^(1/2),x, algorithm="fricas")

[Out]

[-1/2*(b*c*sqrt(-d)*x*log((c^2*d*x^6 + c^2*d*x^2 - d*x^4 + sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*(x^4 - 1)*sq
rt(-d) - d)/(c^2*x^4 - x^2)) + 2*sqrt(-c^2*d*x^2 + d)*b*log(c*x + sqrt(c^2*x^2 - 1)) + 2*sqrt(-c^2*d*x^2 + d)*
a)/(d*x), (b*c*sqrt(d)*x*arctan(sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*(x^2 + 1)*sqrt(d)/(c^2*d*x^4 - (c^2 + 1
)*d*x^2 + d)) - sqrt(-c^2*d*x^2 + d)*b*log(c*x + sqrt(c^2*x^2 - 1)) - sqrt(-c^2*d*x^2 + d)*a)/(d*x)]

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

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)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.38, size = 219, normalized size = 3.08 \[ -\frac {a \sqrt {-c^{2} d \,x^{2}+d}}{d x}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \mathrm {arccosh}\left (c x \right ) c}{d \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \mathrm {arccosh}\left (c x \right ) x \,c^{2}}{\left (c^{2} x^{2}-1\right ) d}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \mathrm {arccosh}\left (c x \right )}{x \left (c^{2} x^{2}-1\right ) d}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \ln \left (1+\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}\right ) c}{d \left (c^{2} x^{2}-1\right )} \]

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)^(1/2),x)

[Out]

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

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maxima [C]  time = 0.54, size = 116, normalized size = 1.63 \[ -\frac {{\left (c^{2} d \sqrt {-\frac {1}{c^{4} d}} \log \left (x^{2} - \frac {1}{c^{2}}\right ) + i \, \left (-1\right )^{-2 \, c^{2} d x^{2} + 2 \, d} \sqrt {d} \log \left (-2 \, c^{2} d + \frac {2 \, d}{x^{2}}\right )\right )} b c}{2 \, d} - \frac {\sqrt {-c^{2} d x^{2} + d} b \operatorname {arcosh}\left (c x\right )}{d x} - \frac {\sqrt {-c^{2} d x^{2} + d} a}{d x} \]

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)^(1/2),x, algorithm="maxima")

[Out]

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

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

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)^(1/2)),x)

[Out]

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

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

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)**(1/2),x)

[Out]

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

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