3.113 \(\int \frac {a+b \cosh ^{-1}(c x)}{x^4 \sqrt {d-c^2 d x^2}} \, dx\)

Optimal. Leaf size=155 \[ -\frac {2 c^2 \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{3 d x}-\frac {\sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{3 d x^3}+\frac {b c \sqrt {c x-1} \sqrt {c x+1}}{6 x^2 \sqrt {d-c^2 d x^2}}-\frac {2 b c^3 \sqrt {c x-1} \sqrt {c x+1} \log (x)}{3 \sqrt {d-c^2 d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.51, antiderivative size = 171, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {5798, 5748, 5724, 29, 30} \[ -\frac {2 c^2 (1-c x) (c x+1) \left (a+b \cosh ^{-1}(c x)\right )}{3 x \sqrt {d-c^2 d x^2}}-\frac {(1-c x) (c x+1) \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3 \sqrt {d-c^2 d x^2}}+\frac {b c \sqrt {c x-1} \sqrt {c x+1}}{6 x^2 \sqrt {d-c^2 d x^2}}-\frac {2 b c^3 \sqrt {c x-1} \sqrt {c x+1} \log (x)}{3 \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

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

Rule 30

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

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 5748

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)/(d1*
d2*f*(m + 1)), x] + (Dist[(c^2*(m + 2*p + 3))/(f^2*(m + 1)), Int[(f*x)^(m + 2)*(d1 + e1*x)^p*(d2 + e2*x)^p*(a
+ b*ArcCosh[c*x])^n, x], 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, p}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 +
c*d2, 0] && GtQ[n, 0] && LtQ[m, -1] && IntegerQ[m] && 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^4 \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^4 \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 )}{3 x^3 \sqrt {d-c^2 d x^2}}-\frac {\left (b c \sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {1}{x^3} \, dx}{3 \sqrt {d-c^2 d x^2}}+\frac {\left (2 c^2 \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}{3 \sqrt {d-c^2 d x^2}}\\ &=\frac {b c \sqrt {-1+c x} \sqrt {1+c x}}{6 x^2 \sqrt {d-c^2 d x^2}}-\frac {(1-c x) (1+c x) \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3 \sqrt {d-c^2 d x^2}}-\frac {2 c^2 (1-c x) (1+c x) \left (a+b \cosh ^{-1}(c x)\right )}{3 x \sqrt {d-c^2 d x^2}}-\frac {\left (2 b c^3 \sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {1}{x} \, dx}{3 \sqrt {d-c^2 d x^2}}\\ &=\frac {b c \sqrt {-1+c x} \sqrt {1+c x}}{6 x^2 \sqrt {d-c^2 d x^2}}-\frac {(1-c x) (1+c x) \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3 \sqrt {d-c^2 d x^2}}-\frac {2 c^2 (1-c x) (1+c x) \left (a+b \cosh ^{-1}(c x)\right )}{3 x \sqrt {d-c^2 d x^2}}-\frac {2 b c^3 \sqrt {-1+c x} \sqrt {1+c x} \log (x)}{3 \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 174, normalized size = 1.12 \[ -\frac {\sqrt {d-c^2 d x^2} \left (4 a c^2 x^2 \sqrt {c x-1} \sqrt {c x+1}+2 a \sqrt {c x-1} \sqrt {c x+1}+6 b c^3 x^3-4 b c^3 x^3 \log (c x-1)-4 b c^3 x^3 \log \left (\frac {1}{c x-1}+1\right )+2 b \sqrt {c x-1} \sqrt {c x+1} \left (2 c^2 x^2+1\right ) \cosh ^{-1}(c x)+b c x\right )}{6 d x^3 \sqrt {c x-1} \sqrt {c x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.86, size = 479, normalized size = 3.09 \[ \left [-\frac {2 \, {\left (2 \, b c^{4} x^{4} - b c^{2} x^{2} - b\right )} \sqrt {-c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) + 2 \, {\left (b c^{5} x^{5} - b c^{3} x^{3}\right )} \sqrt {-d} \log \left (\frac {c^{2} d x^{6} + c^{2} d x^{2} - d x^{4} + \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} {\left (x^{4} - 1\right )} \sqrt {-d} - d}{c^{2} x^{4} - x^{2}}\right ) - \sqrt {-c^{2} d x^{2} + d} {\left (b c x^{3} - b c x\right )} \sqrt {c^{2} x^{2} - 1} + 2 \, {\left (2 \, a c^{4} x^{4} - a c^{2} x^{2} - a\right )} \sqrt {-c^{2} d x^{2} + d}}{6 \, {\left (c^{2} d x^{5} - d x^{3}\right )}}, \frac {4 \, {\left (b c^{5} x^{5} - b c^{3} x^{3}\right )} \sqrt {d} \arctan \left (\frac {\sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} {\left (x^{2} + 1\right )} \sqrt {d}}{c^{2} d x^{4} - {\left (c^{2} + 1\right )} d x^{2} + d}\right ) - 2 \, {\left (2 \, b c^{4} x^{4} - b c^{2} x^{2} - b\right )} \sqrt {-c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) + \sqrt {-c^{2} d x^{2} + d} {\left (b c x^{3} - b c x\right )} \sqrt {c^{2} x^{2} - 1} - 2 \, {\left (2 \, a c^{4} x^{4} - a c^{2} x^{2} - a\right )} \sqrt {-c^{2} d x^{2} + d}}{6 \, {\left (c^{2} d x^{5} - d x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(c*x))/x^4/(-c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(c*x))/x^4/(-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^4), x)

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maple [B]  time = 0.66, size = 854, normalized size = 5.51 \[ -\frac {a \sqrt {-c^{2} d \,x^{2}+d}}{3 d \,x^{3}}-\frac {2 a \,c^{2} \sqrt {-c^{2} d \,x^{2}+d}}{3 d x}-\frac {4 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^{3}}{3 d \left (c^{2} x^{2}-1\right )}-\frac {2 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x^{3} \left (c x -1\right ) \left (c x +1\right ) c^{6}}{3 d \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right )}+\frac {2 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x^{5} c^{8}}{3 d \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right )}+\frac {2 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x^{2} \mathrm {arccosh}\left (c x \right ) \sqrt {c x -1}\, \sqrt {c x +1}\, c^{5}}{d \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right )}-\frac {2 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x^{3} \mathrm {arccosh}\left (c x \right ) c^{6}}{d \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x \left (c x -1\right ) \left (c x +1\right ) c^{4}}{3 d \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x^{3} c^{6}}{3 d \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right )}+\frac {2 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \mathrm {arccosh}\left (c x \right ) \sqrt {c x -1}\, \sqrt {c x +1}\, c^{3}}{3 d \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x \,\mathrm {arccosh}\left (c x \right ) c^{4}}{3 d \left (3 c^{4} x^{4}-2 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}\, c^{3}}{2 d \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x \,c^{4}}{3 d \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right )}+\frac {4 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \mathrm {arccosh}\left (c x \right ) c^{2}}{3 d \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right ) x}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x +1}\, \sqrt {c x -1}\, c}{6 d \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right ) x^{2}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \mathrm {arccosh}\left (c x \right )}{3 d \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right ) x^{3}}+\frac {2 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^{3}}{3 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^4/(-c^2*d*x^2+d)^(1/2),x)

[Out]

-1/3*a/d/x^3*(-c^2*d*x^2+d)^(1/2)-2/3*a*c^2/d/x*(-c^2*d*x^2+d)^(1/2)-4/3*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^3-2/3*b*(-d*(c^2*x^2-1))^(1/2)/d/(3*c^4*x^4-2*c^2*x^2-1)*x^3*(c*x
-1)*(c*x+1)*c^6+2/3*b*(-d*(c^2*x^2-1))^(1/2)/d/(3*c^4*x^4-2*c^2*x^2-1)*x^5*c^8+2*b*(-d*(c^2*x^2-1))^(1/2)/d/(3
*c^4*x^4-2*c^2*x^2-1)*x^2*arccosh(c*x)*(c*x-1)^(1/2)*(c*x+1)^(1/2)*c^5-2*b*(-d*(c^2*x^2-1))^(1/2)/d/(3*c^4*x^4
-2*c^2*x^2-1)*x^3*arccosh(c*x)*c^6-1/3*b*(-d*(c^2*x^2-1))^(1/2)/d/(3*c^4*x^4-2*c^2*x^2-1)*x*(c*x-1)*(c*x+1)*c^
4-1/3*b*(-d*(c^2*x^2-1))^(1/2)/d/(3*c^4*x^4-2*c^2*x^2-1)*x^3*c^6+2/3*b*(-d*(c^2*x^2-1))^(1/2)/d/(3*c^4*x^4-2*c
^2*x^2-1)*arccosh(c*x)*(c*x-1)^(1/2)*(c*x+1)^(1/2)*c^3+1/3*b*(-d*(c^2*x^2-1))^(1/2)/d/(3*c^4*x^4-2*c^2*x^2-1)*
x*arccosh(c*x)*c^4-1/2*b*(-d*(c^2*x^2-1))^(1/2)/d/(3*c^4*x^4-2*c^2*x^2-1)*(c*x+1)^(1/2)*(c*x-1)^(1/2)*c^3-1/3*
b*(-d*(c^2*x^2-1))^(1/2)/d/(3*c^4*x^4-2*c^2*x^2-1)*x*c^4+4/3*b*(-d*(c^2*x^2-1))^(1/2)/d/(3*c^4*x^4-2*c^2*x^2-1
)/x*arccosh(c*x)*c^2-1/6*b*(-d*(c^2*x^2-1))^(1/2)/d/(3*c^4*x^4-2*c^2*x^2-1)/x^2*(c*x+1)^(1/2)*(c*x-1)^(1/2)*c+
1/3*b*(-d*(c^2*x^2-1))^(1/2)/d/(3*c^4*x^4-2*c^2*x^2-1)/x^3*arccosh(c*x)+2/3*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^3

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maxima [A]  time = 0.97, size = 134, normalized size = 0.86 \[ \frac {1}{6} \, {\left (\frac {4 \, c^{2} \sqrt {-d} \log \relax (x)}{d} - \frac {\sqrt {-d}}{d x^{2}}\right )} b c - \frac {1}{3} \, b {\left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} c^{2}}{d x} + \frac {\sqrt {-c^{2} d x^{2} + d}}{d x^{3}}\right )} \operatorname {arcosh}\left (c x\right ) - \frac {1}{3} \, a {\left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} c^{2}}{d x} + \frac {\sqrt {-c^{2} d x^{2} + d}}{d x^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(c*x))/x^4/(-c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {acosh}\left (c\,x\right )}{x^4\,\sqrt {d-c^2\,d\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(c*x))/(x^4*(d - c^2*d*x^2)^(1/2)),x)

[Out]

int((a + b*acosh(c*x))/(x^4*(d - c^2*d*x^2)^(1/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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