3.129 \(\int \frac{a+b \cosh ^{-1}(c x)}{(d-c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=162 \[ \frac{2 x \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 \sqrt{d-c^2 d x^2}}+\frac{x \left (a+b \cosh ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{b \sqrt{c x-1} \sqrt{c x+1} \log \left (1-c^2 x^2\right )}{3 c d^2 \sqrt{d-c^2 d x^2}}+\frac{b \sqrt{c x-1} \sqrt{c x+1}}{6 c d \left (d-c^2 d x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.268182, antiderivative size = 189, normalized size of antiderivative = 1.17, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {5713, 5691, 5688, 260, 261} \[ \frac{2 x \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 \sqrt{d-c^2 d x^2}}+\frac{x \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 (1-c x) (c x+1) \sqrt{d-c^2 d x^2}}+\frac{b \sqrt{c x-1} \sqrt{c x+1}}{6 c d^2 \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}}-\frac{b \sqrt{c x-1} \sqrt{c x+1} \log \left (1-c^2 x^2\right )}{3 c d^2 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCosh[c*x])/(d - c^2*d*x^2)^(5/2),x]

[Out]

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

Rule 5713

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((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[(1 + c*x)^p*(-1 + c*x)^p*(a + b*Ar
cCosh[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[c^2*d + e, 0] &&  !IntegerQ[p]

Rule 5691

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((d1_) + (e1_.)*(x_))^(p_)*((d2_) + (e2_.)*(x_))^(p_), x_Symbol]
:> -Simp[(x*(d1 + e1*x)^(p + 1)*(d2 + e2*x)^(p + 1)*(a + b*ArcCosh[c*x])^n)/(2*d1*d2*(p + 1)), x] + (Dist[(2*
p + 3)/(2*d1*d2*(p + 1)), Int[(d1 + e1*x)^(p + 1)*(d2 + e2*x)^(p + 1)*(a + b*ArcCosh[c*x])^n, x], x] - Dist[(b
*c*n*(-(d1*d2))^(p + 1/2)*Sqrt[1 + c*x]*Sqrt[-1 + c*x])/(2*(p + 1)*Sqrt[d1 + e1*x]*Sqrt[d2 + e2*x]), Int[x*(-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}, x] && EqQ[e1,
c*d1] && EqQ[e2, -(c*d2)] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[p + 1/2]

Rule 5688

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)/(((d1_) + (e1_.)*(x_))^(3/2)*((d2_) + (e2_.)*(x_))^(3/2)), x_Sym
bol] :> Simp[(x*(a + b*ArcCosh[c*x])^n)/(d1*d2*Sqrt[d1 + e1*x]*Sqrt[d2 + e2*x]), x] + Dist[(b*c*n*Sqrt[1 + c*x
]*Sqrt[-1 + c*x])/(d1*d2*Sqrt[d1 + e1*x]*Sqrt[d2 + e2*x]), Int[(x*(a + b*ArcCosh[c*x])^(n - 1))/(1 - c^2*x^2),
x], x] /; FreeQ[{a, b, c, d1, e1, d2, e2}, x] && EqQ[e1, c*d1] && EqQ[e2, -(c*d2)] && GtQ[n, 0]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{a+b \cosh ^{-1}(c x)}{\left (d-c^2 d x^2\right )^{5/2}} \, dx &=\frac{\left (\sqrt{-1+c x} \sqrt{1+c x}\right ) \int \frac{a+b \cosh ^{-1}(c x)}{(-1+c x)^{5/2} (1+c x)^{5/2}} \, dx}{d^2 \sqrt{d-c^2 d x^2}}\\ &=\frac{x \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 (1-c x) (1+c x) \sqrt{d-c^2 d x^2}}-\frac{\left (2 \sqrt{-1+c x} \sqrt{1+c x}\right ) \int \frac{a+b \cosh ^{-1}(c x)}{(-1+c x)^{3/2} (1+c x)^{3/2}} \, dx}{3 d^2 \sqrt{d-c^2 d x^2}}+\frac{\left (b c \sqrt{-1+c x} \sqrt{1+c x}\right ) \int \frac{x}{\left (-1+c^2 x^2\right )^2} \, dx}{3 d^2 \sqrt{d-c^2 d x^2}}\\ &=\frac{b \sqrt{-1+c x} \sqrt{1+c x}}{6 c d^2 \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}}+\frac{2 x \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 \sqrt{d-c^2 d x^2}}+\frac{x \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 (1-c x) (1+c x) \sqrt{d-c^2 d x^2}}+\frac{\left (2 b c \sqrt{-1+c x} \sqrt{1+c x}\right ) \int \frac{x}{1-c^2 x^2} \, dx}{3 d^2 \sqrt{d-c^2 d x^2}}\\ &=\frac{b \sqrt{-1+c x} \sqrt{1+c x}}{6 c d^2 \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}}+\frac{2 x \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 \sqrt{d-c^2 d x^2}}+\frac{x \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 (1-c x) (1+c x) \sqrt{d-c^2 d x^2}}-\frac{b \sqrt{-1+c x} \sqrt{1+c x} \log \left (1-c^2 x^2\right )}{3 c d^2 \sqrt{d-c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0835394, size = 132, normalized size = 0.81 \[ \frac{4 a c^3 x^3-6 a c x-2 b \sqrt{c x-1} \sqrt{c x+1} \left (c^2 x^2-1\right ) \log \left (1-c^2 x^2\right )+2 b c x \left (2 c^2 x^2-3\right ) \cosh ^{-1}(c x)-b \sqrt{c x-1} \sqrt{c x+1}}{6 c d^2 \left (c^2 x^2-1\right ) \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCosh[c*x])/(d - c^2*d*x^2)^(5/2),x]

[Out]

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

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Maple [B]  time = 0.143, size = 1073, normalized size = 6.6 \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)^(5/2),x)

[Out]

1/3*a/d*x/(-c^2*d*x^2+d)^(3/2)+2/3*a/d^2*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^3/c/(c^2*x^2-1)*arccosh(c*x)+2/3*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^4
/d^3*(c*x+1)*(c*x-1)*x^5-2/3*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^6/d^3*x^7+2*b*(-d*
(c^2*x^2-1))^(1/2)/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^3/d^3*arccosh(c*x)*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^4-2*
b*(-d*(c^2*x^2-1))^(1/2)/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^4/d^3*arccosh(c*x)*x^5-5/3*b*(-d*(c^2*x^2-1))^(
1/2)/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^2/d^3*(c*x+1)*(c*x-1)*x^3+7/3*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^6*x^6-1
0*c^4*x^4+11*c^2*x^2-4)*c^4/d^3*x^5-14/3*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c/d^3*ar
ccosh(c*x)*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^2+17/3*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c
^2/d^3*arccosh(c*x)*x^3+b*(-d*(c^2*x^2-1))^(1/2)/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)/d^3*(c*x+1)*(c*x-1)*x+1/2
*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c/d^3*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^2-8/3*b*(-d*
(c^2*x^2-1))^(1/2)/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^2/d^3*x^3+8/3*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^6*x^6-10*
c^4*x^4+11*c^2*x^2-4)/c/d^3*arccosh(c*x)*(c*x+1)^(1/2)*(c*x-1)^(1/2)-4*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^6*x^6-10*
c^4*x^4+11*c^2*x^2-4)/d^3*arccosh(c*x)*x-2/3*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)/c/d^
3*(c*x-1)^(1/2)*(c*x+1)^(1/2)+b*(-d*(c^2*x^2-1))^(1/2)/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)/d^3*x+2/3*b*(-d*(c^
2*x^2-1))^(1/2)*(c*x-1)^(1/2)*(c*x+1)^(1/2)/d^3/c/(c^2*x^2-1)*ln((c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2-1)

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Maxima [A]  time = 1.22012, size = 212, normalized size = 1.31 \begin{align*} \frac{1}{6} \, b c{\left (\frac{\sqrt{-d}}{c^{4} d^{3} x^{2} - c^{2} d^{3}} + \frac{2 \, \sqrt{-d} \log \left (c x + 1\right )}{c^{2} d^{3}} + \frac{2 \, \sqrt{-d} \log \left (c x - 1\right )}{c^{2} d^{3}}\right )} + \frac{1}{3} \, b{\left (\frac{2 \, x}{\sqrt{-c^{2} d x^{2} + d} d^{2}} + \frac{x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}} d}\right )} \operatorname{arcosh}\left (c x\right ) + \frac{1}{3} \, a{\left (\frac{2 \, x}{\sqrt{-c^{2} d x^{2} + d} d^{2}} + \frac{x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}} d}\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)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-c^{2} d x^{2} + d}{\left (b \operatorname{arcosh}\left (c x\right ) + a\right )}}{c^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}, x\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)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{acosh}{\left (c x \right )}}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac{5}{2}}}\, 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)**(5/2),x)

[Out]

Integral((a + b*acosh(c*x))/(-d*(c*x - 1)*(c*x + 1))**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arcosh}\left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{5}{2}}}\,{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)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arccosh(c*x) + a)/(-c^2*d*x^2 + d)^(5/2), x)