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

Optimal. Leaf size=180 \[ \frac{3 b \text{PolyLog}\left (2,-e^{\cosh ^{-1}(c x)}\right )}{8 c d^3}-\frac{3 b \text{PolyLog}\left (2,e^{\cosh ^{-1}(c x)}\right )}{8 c d^3}+\frac{3 x \left (a+b \cosh ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac{x \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{3 \tanh ^{-1}\left (e^{\cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{4 c d^3}-\frac{3 b}{8 c d^3 \sqrt{c x-1} \sqrt{c x+1}}+\frac{b}{12 c d^3 (c x-1)^{3/2} (c x+1)^{3/2}} \]

[Out]

b/(12*c*d^3*(-1 + c*x)^(3/2)*(1 + c*x)^(3/2)) - (3*b)/(8*c*d^3*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) + (x*(a + b*ArcCo
sh[c*x]))/(4*d^3*(1 - c^2*x^2)^2) + (3*x*(a + b*ArcCosh[c*x]))/(8*d^3*(1 - c^2*x^2)) + (3*(a + b*ArcCosh[c*x])
*ArcTanh[E^ArcCosh[c*x]])/(4*c*d^3) + (3*b*PolyLog[2, -E^ArcCosh[c*x]])/(8*c*d^3) - (3*b*PolyLog[2, E^ArcCosh[
c*x]])/(8*c*d^3)

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Rubi [A]  time = 0.134077, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {5689, 74, 5694, 4182, 2279, 2391} \[ \frac{3 b \text{PolyLog}\left (2,-e^{\cosh ^{-1}(c x)}\right )}{8 c d^3}-\frac{3 b \text{PolyLog}\left (2,e^{\cosh ^{-1}(c x)}\right )}{8 c d^3}+\frac{3 x \left (a+b \cosh ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac{x \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{3 \tanh ^{-1}\left (e^{\cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{4 c d^3}-\frac{3 b}{8 c d^3 \sqrt{c x-1} \sqrt{c x+1}}+\frac{b}{12 c d^3 (c x-1)^{3/2} (c x+1)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

b/(12*c*d^3*(-1 + c*x)^(3/2)*(1 + c*x)^(3/2)) - (3*b)/(8*c*d^3*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) + (x*(a + b*ArcCo
sh[c*x]))/(4*d^3*(1 - c^2*x^2)^2) + (3*x*(a + b*ArcCosh[c*x]))/(8*d^3*(1 - c^2*x^2)) + (3*(a + b*ArcCosh[c*x])
*ArcTanh[E^ArcCosh[c*x]])/(4*c*d^3) + (3*b*PolyLog[2, -E^ArcCosh[c*x]])/(8*c*d^3) - (3*b*PolyLog[2, E^ArcCosh[
c*x]])/(8*c*d^3)

Rule 5689

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcCosh[c*x])^n)/(2*d*(p + 1)), x] + (-Dist[(b*c*n*(-d)^p)/(2*(p + 1)), Int[x*(1 + c*x)^(p + 1/2)
*(-1 + c*x)^(p + 1/2)*(a + b*ArcCosh[c*x])^(n - 1), x], x] + Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p
+ 1)*(a + b*ArcCosh[c*x])^n, x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p,
-1] && IntegerQ[p]

Rule 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 5694

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Dist[(c*d)^(-1), Subst[Int[
(a + b*x)^n*Csch[x], x], x, ArcCosh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{a+b \cosh ^{-1}(c x)}{\left (d-c^2 d x^2\right )^3} \, dx &=\frac{x \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac{(b c) \int \frac{x}{(-1+c x)^{5/2} (1+c x)^{5/2}} \, dx}{4 d^3}+\frac{3 \int \frac{a+b \cosh ^{-1}(c x)}{\left (d-c^2 d x^2\right )^2} \, dx}{4 d}\\ &=\frac{b}{12 c d^3 (-1+c x)^{3/2} (1+c x)^{3/2}}+\frac{x \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{3 x \left (a+b \cosh ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac{(3 b c) \int \frac{x}{(-1+c x)^{3/2} (1+c x)^{3/2}} \, dx}{8 d^3}+\frac{3 \int \frac{a+b \cosh ^{-1}(c x)}{d-c^2 d x^2} \, dx}{8 d^2}\\ &=\frac{b}{12 c d^3 (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac{3 b}{8 c d^3 \sqrt{-1+c x} \sqrt{1+c x}}+\frac{x \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{3 x \left (a+b \cosh ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}-\frac{3 \operatorname{Subst}\left (\int (a+b x) \text{csch}(x) \, dx,x,\cosh ^{-1}(c x)\right )}{8 c d^3}\\ &=\frac{b}{12 c d^3 (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac{3 b}{8 c d^3 \sqrt{-1+c x} \sqrt{1+c x}}+\frac{x \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{3 x \left (a+b \cosh ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac{3 \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\cosh ^{-1}(c x)}\right )}{4 c d^3}+\frac{(3 b) \operatorname{Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\cosh ^{-1}(c x)\right )}{8 c d^3}-\frac{(3 b) \operatorname{Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\cosh ^{-1}(c x)\right )}{8 c d^3}\\ &=\frac{b}{12 c d^3 (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac{3 b}{8 c d^3 \sqrt{-1+c x} \sqrt{1+c x}}+\frac{x \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{3 x \left (a+b \cosh ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac{3 \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\cosh ^{-1}(c x)}\right )}{4 c d^3}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{\cosh ^{-1}(c x)}\right )}{8 c d^3}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{\cosh ^{-1}(c x)}\right )}{8 c d^3}\\ &=\frac{b}{12 c d^3 (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac{3 b}{8 c d^3 \sqrt{-1+c x} \sqrt{1+c x}}+\frac{x \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{3 x \left (a+b \cosh ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac{3 \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\cosh ^{-1}(c x)}\right )}{4 c d^3}+\frac{3 b \text{Li}_2\left (-e^{\cosh ^{-1}(c x)}\right )}{8 c d^3}-\frac{3 b \text{Li}_2\left (e^{\cosh ^{-1}(c x)}\right )}{8 c d^3}\\ \end{align*}

Mathematica [A]  time = 1.1607, size = 316, normalized size = 1.76 \[ \frac{-\frac{3 b \left (\cosh ^{-1}(c x) \left (\cosh ^{-1}(c x)-4 \log \left (e^{\cosh ^{-1}(c x)}+1\right )\right )-4 \text{PolyLog}\left (2,-e^{\cosh ^{-1}(c x)}\right )\right )}{2 c}+\frac{3 b \left (\cosh ^{-1}(c x) \left (\cosh ^{-1}(c x)-4 \log \left (1-e^{\cosh ^{-1}(c x)}\right )\right )-4 \text{PolyLog}\left (2,e^{\cosh ^{-1}(c x)}\right )\right )}{2 c}-\frac{6 a x}{c^2 x^2-1}+\frac{4 a x}{\left (c^2 x^2-1\right )^2}-\frac{3 a \log (1-c x)}{c}+\frac{3 a \log (c x+1)}{c}+\frac{b \left (\sqrt{c x-1} \sqrt{c x+1} (c x+2)-3 \cosh ^{-1}(c x)\right )}{3 c (c x+1)^2}+\frac{b \left (\sqrt{c x-1} \sqrt{c x+1} (2-c x)+3 \cosh ^{-1}(c x)\right )}{3 c (c x-1)^2}+\frac{3 b \left (\frac{\cosh ^{-1}(c x)}{1-c x}-\frac{1}{\sqrt{\frac{c x-1}{c x+1}}}\right )}{c}+\frac{3 b \left (\sqrt{\frac{c x-1}{c x+1}}-\frac{\cosh ^{-1}(c x)}{c x+1}\right )}{c}}{16 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((4*a*x)/(-1 + c^2*x^2)^2 - (6*a*x)/(-1 + c^2*x^2) + (b*(Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(2 + c*x) - 3*ArcCosh[c*
x]))/(3*c*(1 + c*x)^2) + (b*((2 - c*x)*Sqrt[-1 + c*x]*Sqrt[1 + c*x] + 3*ArcCosh[c*x]))/(3*c*(-1 + c*x)^2) + (3
*b*(-(1/Sqrt[(-1 + c*x)/(1 + c*x)]) + ArcCosh[c*x]/(1 - c*x)))/c + (3*b*(Sqrt[(-1 + c*x)/(1 + c*x)] - ArcCosh[
c*x]/(1 + c*x)))/c - (3*a*Log[1 - c*x])/c + (3*a*Log[1 + c*x])/c - (3*b*(ArcCosh[c*x]*(ArcCosh[c*x] - 4*Log[1
+ E^ArcCosh[c*x]]) - 4*PolyLog[2, -E^ArcCosh[c*x]]))/(2*c) + (3*b*(ArcCosh[c*x]*(ArcCosh[c*x] - 4*Log[1 - E^Ar
cCosh[c*x]]) - 4*PolyLog[2, E^ArcCosh[c*x]]))/(2*c))/(16*d^3)

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Maple [A]  time = 0.106, size = 378, normalized size = 2.1 \begin{align*}{\frac{a}{16\,c{d}^{3} \left ( cx-1 \right ) ^{2}}}-{\frac{3\,a}{16\,c{d}^{3} \left ( cx-1 \right ) }}-{\frac{3\,a\ln \left ( cx-1 \right ) }{16\,c{d}^{3}}}-{\frac{a}{16\,c{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{3\,a}{16\,c{d}^{3} \left ( cx+1 \right ) }}+{\frac{3\,a\ln \left ( cx+1 \right ) }{16\,c{d}^{3}}}-{\frac{3\,{c}^{2}b{\rm arccosh} \left (cx\right ){x}^{3}}{8\,{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) }}-{\frac{3\,bc{x}^{2}}{8\,{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) }\sqrt{cx-1}\sqrt{cx+1}}+{\frac{5\,b{\rm arccosh} \left (cx\right )x}{8\,{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) }}+{\frac{11\,b}{24\,c{d}^{3} \left ({c}^{4}{x}^{4}-2\,{c}^{2}{x}^{2}+1 \right ) }\sqrt{cx-1}\sqrt{cx+1}}+{\frac{3\,b{\rm arccosh} \left (cx\right )}{8\,c{d}^{3}}\ln \left ( 1+cx+\sqrt{cx-1}\sqrt{cx+1} \right ) }+{\frac{3\,b}{8\,c{d}^{3}}{\it polylog} \left ( 2,-cx-\sqrt{cx-1}\sqrt{cx+1} \right ) }-{\frac{3\,b{\rm arccosh} \left (cx\right )}{8\,c{d}^{3}}\ln \left ( 1-cx-\sqrt{cx-1}\sqrt{cx+1} \right ) }-{\frac{3\,b}{8\,c{d}^{3}}{\it polylog} \left ( 2,cx+\sqrt{cx-1}\sqrt{cx+1} \right ) } \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)^3,x)

[Out]

1/16/c*a/d^3/(c*x-1)^2-3/16/c*a/d^3/(c*x-1)-3/16/c*a/d^3*ln(c*x-1)-1/16/c*a/d^3/(c*x+1)^2-3/16/c*a/d^3/(c*x+1)
+3/16/c*a/d^3*ln(c*x+1)-3/8*c^2*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arccosh(c*x)*x^3-3/8*c*b/d^3/(c^4*x^4-2*c^2*x^2+1)
*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^2+5/8*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arccosh(c*x)*x+11/24/c*b/d^3/(c^4*x^4-2*c^2*x
^2+1)*(c*x+1)^(1/2)*(c*x-1)^(1/2)+3/8/c*b/d^3*arccosh(c*x)*ln(1+c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))+3/8*b*polylog
(2,-c*x-(c*x-1)^(1/2)*(c*x+1)^(1/2))/c/d^3-3/8/c*b/d^3*arccosh(c*x)*ln(1-c*x-(c*x-1)^(1/2)*(c*x+1)^(1/2))-3/8*
b*polylog(2,c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))/c/d^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \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)^3,x, algorithm="maxima")

[Out]

1/2048*(18432*c^5*integrate(1/32*x^5*log(c*x - 1)/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3), x) - 48
*c^4*(2*(5*c^2*x^3 - 3*x)/(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4*d^3) + 3*log(c*x + 1)/(c^5*d^3) - 3*log(c*x - 1)/
(c^5*d^3)) - 6144*c^4*integrate(1/32*x^4*log(c*x - 1)/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3), x)
+ 18*(c*(2*(5*c^2*x^2 + 3*c*x - 6)/(c^8*d^3*x^3 - c^7*d^3*x^2 - c^6*d^3*x + c^5*d^3) - 5*log(c*x + 1)/(c^5*d^3
) + 5*log(c*x - 1)/(c^5*d^3)) + 16*(2*c^2*x^2 - 1)*log(c*x - 1)/(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4*d^3))*c^3 +
80*c^2*(2*(c^2*x^3 + x)/(c^6*d^3*x^4 - 2*c^4*d^3*x^2 + c^2*d^3) - log(c*x + 1)/(c^3*d^3) + log(c*x - 1)/(c^3*
d^3)) + 12288*c^2*integrate(1/32*x^2*log(c*x - 1)/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3), x) + 9*
(c*(2*(3*c^2*x^2 - 3*c*x - 2)/(c^6*d^3*x^3 - c^5*d^3*x^2 - c^4*d^3*x + c^3*d^3) - 3*log(c*x + 1)/(c^3*d^3) + 3
*log(c*x - 1)/(c^3*d^3)) - 16*log(c*x - 1)/(c^6*d^3*x^4 - 2*c^4*d^3*x^2 + c^2*d^3))*c - 32*(3*(c^4*x^4 - 2*c^2
*x^2 + 1)*log(c*x + 1)^2 + 6*(c^4*x^4 - 2*c^2*x^2 + 1)*log(c*x + 1)*log(c*x - 1) + 4*(6*c^3*x^3 - 10*c*x - 3*(
c^4*x^4 - 2*c^2*x^2 + 1)*log(c*x + 1) + 3*(c^4*x^4 - 2*c^2*x^2 + 1)*log(c*x - 1))*log(c*x + sqrt(c*x + 1)*sqrt
(c*x - 1)))/(c^5*d^3*x^4 - 2*c^3*d^3*x^2 + c*d^3) + 2048*integrate(-1/16*(6*c^3*x^3 - 10*c*x - 3*(c^4*x^4 - 2*
c^2*x^2 + 1)*log(c*x + 1) + 3*(c^4*x^4 - 2*c^2*x^2 + 1)*log(c*x - 1))/(c^7*d^3*x^7 - 3*c^5*d^3*x^5 + 3*c^3*d^3
*x^3 - c*d^3*x + (c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3)*sqrt(c*x + 1)*sqrt(c*x - 1)), x) - 6144*i
ntegrate(1/32*log(c*x - 1)/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3), x))*b - 1/16*a*(2*(3*c^2*x^3 -
5*x)/(c^4*d^3*x^4 - 2*c^2*d^3*x^2 + d^3) - 3*log(c*x + 1)/(c*d^3) + 3*log(c*x - 1)/(c*d^3))

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

[Out]

integral(-(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*} - \frac{\int \frac{a}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx + \int \frac{b \operatorname{acosh}{\left (c x \right )}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx}{d^{3}} \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)**3,x)

[Out]

-(Integral(a/(c**6*x**6 - 3*c**4*x**4 + 3*c**2*x**2 - 1), x) + Integral(b*acosh(c*x)/(c**6*x**6 - 3*c**4*x**4
+ 3*c**2*x**2 - 1), x))/d**3

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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 )}^{3}}\,{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)^3,x, algorithm="giac")

[Out]

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