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

Optimal. Leaf size=119 \[ -\frac{\left (d-c^2 d x^2\right )^{3/2} \left (a+b \cosh ^{-1}(c x)\right )}{3 d x^3}-\frac{b c \sqrt{d-c^2 d x^2}}{6 x^2 \sqrt{c x-1} \sqrt{c x+1}}-\frac{b c^3 \log (x) \sqrt{d-c^2 d x^2}}{3 \sqrt{c x-1} \sqrt{c x+1}} \]

[Out]

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

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Rubi [A]  time = 0.285478, antiderivative size = 127, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {5798, 5724, 14} \[ -\frac{(1-c x) (c x+1) \sqrt{d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac{b c \sqrt{d-c^2 d x^2}}{6 x^2 \sqrt{c x-1} \sqrt{c x+1}}-\frac{b c^3 \log (x) \sqrt{d-c^2 d x^2}}{3 \sqrt{c x-1} \sqrt{c x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.121119, size = 88, normalized size = 0.74 \[ \frac{\sqrt{d-c^2 d x^2} \left (\frac{(c x-1)^{3/2} (c x+1)^{3/2} \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac{1}{3} b c \left (c^2 \log (x)+\frac{1}{2 x^2}\right )\right )}{\sqrt{c x-1} \sqrt{c x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.286, size = 1017, normalized size = 8.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a/d/x^3*(-c^2*d*x^2+d)^(3/2)+2/3*b*(-d*(c^2*x^2-1))^(1/2)/(c*x-1)^(1/2)/(c*x+1)^(1/2)*arccosh(c*x)*c^3-b*
(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-3*c^2*x^2+1)*x^4/(c*x+1)^(1/2)/(c*x-1)^(1/2)*arccosh(c*x)*c^7+b*(-d*(c^2*x^2
-1))^(1/2)/(3*c^4*x^4-3*c^2*x^2+1)*x^5/(c*x+1)/(c*x-1)*arccosh(c*x)*c^8-1/6*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^
4-3*c^2*x^2+1)*x^3*c^6+1/6*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-3*c^2*x^2+1)*x^5/(c*x+1)/(c*x-1)*c^8+b*(-d*(c^2
*x^2-1))^(1/2)/(3*c^4*x^4-3*c^2*x^2+1)*x^2/(c*x+1)^(1/2)/(c*x-1)^(1/2)*arccosh(c*x)*c^5-3*b*(-d*(c^2*x^2-1))^(
1/2)/(3*c^4*x^4-3*c^2*x^2+1)*x^3/(c*x+1)/(c*x-1)*arccosh(c*x)*c^6+1/6*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-3*c^
2*x^2+1)*x*c^4-1/2*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-3*c^2*x^2+1)*x^2/(c*x+1)^(1/2)/(c*x-1)^(1/2)*c^5-1/3*b*
(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-3*c^2*x^2+1)*x^3/(c*x+1)/(c*x-1)*c^6-1/3*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4
-3*c^2*x^2+1)/(c*x+1)^(1/2)/(c*x-1)^(1/2)*arccosh(c*x)*c^3+10/3*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-3*c^2*x^2+
1)*x/(c*x+1)/(c*x-1)*arccosh(c*x)*c^4+1/2*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-3*c^2*x^2+1)/(c*x+1)^(1/2)/(c*x-
1)^(1/2)*c^3+1/6*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-3*c^2*x^2+1)*x/(c*x+1)/(c*x-1)*c^4-5/3*b*(-d*(c^2*x^2-1))
^(1/2)/(3*c^4*x^4-3*c^2*x^2+1)/x/(c*x+1)/(c*x-1)*arccosh(c*x)*c^2-1/6*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-3*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)/(3*c^4*x^4-3*c^2*x^2+1)/x^3/(c*x+1)/(c
*x-1)*arccosh(c*x)-1/3*b*(-d*(c^2*x^2-1))^(1/2)/(c*x-1)^(1/2)/(c*x+1)^(1/2)*ln((c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2
))^2+1)*c^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.26496, size = 986, normalized size = 8.29 \begin{align*} \left [\frac{2 \,{\left (b c^{4} x^{4} - 2 \, 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 ) +{\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 (a c^{4} x^{4} - 2 \, a c^{2} x^{2} + a\right )} \sqrt{-c^{2} d x^{2} + d}}{6 \,{\left (c^{2} x^{5} - x^{3}\right )}}, -\frac{2 \,{\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 (b c^{4} x^{4} - 2 \, 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 (a c^{4} x^{4} - 2 \, a c^{2} x^{2} + a\right )} \sqrt{-c^{2} d x^{2} + d}}{6 \,{\left (c^{2} x^{5} - x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-c^{2} d x^{2} + d}{\left (b \operatorname{arcosh}\left (c x\right ) + a\right )}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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