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

Optimal. Leaf size=250 \[ \frac {3 c^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 d^3 \left (1-c^2 x^2\right )}+\frac {3 c^2 \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \cosh ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {6 c^2 \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{d^3}-\frac {2 b c^3 x}{3 d^3 \sqrt {c x-1} \sqrt {c x+1}}-\frac {5 b c^3 x}{12 d^3 (c x-1)^{3/2} (c x+1)^{3/2}}+\frac {3 b c^2 \text {Li}_2\left (-e^{2 \cosh ^{-1}(c x)}\right )}{2 d^3}-\frac {3 b c^2 \text {Li}_2\left (e^{2 \cosh ^{-1}(c x)}\right )}{2 d^3}+\frac {b c}{2 d^3 x (c x-1)^{3/2} (c x+1)^{3/2}} \]

[Out]

1/2*b*c/d^3/x/(c*x-1)^(3/2)/(c*x+1)^(3/2)-5/12*b*c^3*x/d^3/(c*x-1)^(3/2)/(c*x+1)^(3/2)+3/4*c^2*(a+b*arccosh(c*
x))/d^3/(-c^2*x^2+1)^2+1/2*(-a-b*arccosh(c*x))/d^3/x^2/(-c^2*x^2+1)^2+3/2*c^2*(a+b*arccosh(c*x))/d^3/(-c^2*x^2
+1)+6*c^2*(a+b*arccosh(c*x))*arctanh((c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2)/d^3+3/2*b*c^2*polylog(2,-(c*x+(c*x-1
)^(1/2)*(c*x+1)^(1/2))^2)/d^3-3/2*b*c^2*polylog(2,(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2)/d^3-2/3*b*c^3*x/d^3/(c*
x-1)^(1/2)/(c*x+1)^(1/2)

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Rubi [A]  time = 0.37, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {5746, 103, 12, 40, 39, 5754, 5721, 5461, 4182, 2279, 2391} \[ \frac {3 b c^2 \text {PolyLog}\left (2,-e^{2 \cosh ^{-1}(c x)}\right )}{2 d^3}-\frac {3 b c^2 \text {PolyLog}\left (2,e^{2 \cosh ^{-1}(c x)}\right )}{2 d^3}+\frac {3 c^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 d^3 \left (1-c^2 x^2\right )}+\frac {3 c^2 \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \cosh ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {6 c^2 \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{d^3}-\frac {2 b c^3 x}{3 d^3 \sqrt {c x-1} \sqrt {c x+1}}-\frac {5 b c^3 x}{12 d^3 (c x-1)^{3/2} (c x+1)^{3/2}}+\frac {b c}{2 d^3 x (c x-1)^{3/2} (c x+1)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c)/(2*d^3*x*(-1 + c*x)^(3/2)*(1 + c*x)^(3/2)) - (5*b*c^3*x)/(12*d^3*(-1 + c*x)^(3/2)*(1 + c*x)^(3/2)) - (2*
b*c^3*x)/(3*d^3*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) + (3*c^2*(a + b*ArcCosh[c*x]))/(4*d^3*(1 - c^2*x^2)^2) - (a + b*
ArcCosh[c*x])/(2*d^3*x^2*(1 - c^2*x^2)^2) + (3*c^2*(a + b*ArcCosh[c*x]))/(2*d^3*(1 - c^2*x^2)) + (6*c^2*(a + b
*ArcCosh[c*x])*ArcTanh[E^(2*ArcCosh[c*x])])/d^3 + (3*b*c^2*PolyLog[2, -E^(2*ArcCosh[c*x])])/(2*d^3) - (3*b*c^2
*PolyLog[2, E^(2*ArcCosh[c*x])])/(2*d^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 39

Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> Simp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d
*x]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0]

Rule 40

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(x*(a + b*x)^(m + 1)*(c + d*x)^(m +
1))/(2*a*c*(m + 1)), x] + Dist[(2*m + 3)/(2*a*c*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(m + 1), x], x] /; F
reeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && ILtQ[m + 3/2, 0]

Rule 103

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

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]

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 5461

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

Rule 5721

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

Rule 5746

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

Rule 5754

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

Rubi steps

\begin {align*} \int \frac {a+b \cosh ^{-1}(c x)}{x^3 \left (d-c^2 d x^2\right )^3} \, dx &=-\frac {a+b \cosh ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\left (3 c^2\right ) \int \frac {a+b \cosh ^{-1}(c x)}{x \left (d-c^2 d x^2\right )^3} \, dx+\frac {(b c) \int \frac {1}{x^2 (-1+c x)^{5/2} (1+c x)^{5/2}} \, dx}{2 d^3}\\ &=\frac {b c}{2 d^3 x (-1+c x)^{3/2} (1+c x)^{3/2}}+\frac {3 c^2 \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \cosh ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {(b c) \int \frac {4 c^2}{(-1+c x)^{5/2} (1+c x)^{5/2}} \, dx}{2 d^3}-\frac {\left (3 b c^3\right ) \int \frac {1}{(-1+c x)^{5/2} (1+c x)^{5/2}} \, dx}{4 d^3}+\frac {\left (3 c^2\right ) \int \frac {a+b \cosh ^{-1}(c x)}{x \left (d-c^2 d x^2\right )^2} \, dx}{d}\\ &=\frac {b c}{2 d^3 x (-1+c x)^{3/2} (1+c x)^{3/2}}+\frac {b c^3 x}{4 d^3 (-1+c x)^{3/2} (1+c x)^{3/2}}+\frac {3 c^2 \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \cosh ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {3 c^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 d^3 \left (1-c^2 x^2\right )}+\frac {\left (b c^3\right ) \int \frac {1}{(-1+c x)^{3/2} (1+c x)^{3/2}} \, dx}{2 d^3}+\frac {\left (3 b c^3\right ) \int \frac {1}{(-1+c x)^{3/2} (1+c x)^{3/2}} \, dx}{2 d^3}+\frac {\left (2 b c^3\right ) \int \frac {1}{(-1+c x)^{5/2} (1+c x)^{5/2}} \, dx}{d^3}+\frac {\left (3 c^2\right ) \int \frac {a+b \cosh ^{-1}(c x)}{x \left (d-c^2 d x^2\right )} \, dx}{d^2}\\ &=\frac {b c}{2 d^3 x (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac {5 b c^3 x}{12 d^3 (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac {2 b c^3 x}{d^3 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {3 c^2 \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \cosh ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {3 c^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 d^3 \left (1-c^2 x^2\right )}-\frac {\left (3 c^2\right ) \operatorname {Subst}\left (\int (a+b x) \text {csch}(x) \text {sech}(x) \, dx,x,\cosh ^{-1}(c x)\right )}{d^3}-\frac {\left (4 b c^3\right ) \int \frac {1}{(-1+c x)^{3/2} (1+c x)^{3/2}} \, dx}{3 d^3}\\ &=\frac {b c}{2 d^3 x (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac {5 b c^3 x}{12 d^3 (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac {2 b c^3 x}{3 d^3 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {3 c^2 \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \cosh ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {3 c^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 d^3 \left (1-c^2 x^2\right )}-\frac {\left (6 c^2\right ) \operatorname {Subst}\left (\int (a+b x) \text {csch}(2 x) \, dx,x,\cosh ^{-1}(c x)\right )}{d^3}\\ &=\frac {b c}{2 d^3 x (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac {5 b c^3 x}{12 d^3 (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac {2 b c^3 x}{3 d^3 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {3 c^2 \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \cosh ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {3 c^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 d^3 \left (1-c^2 x^2\right )}+\frac {6 c^2 \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right )}{d^3}+\frac {\left (3 b c^2\right ) \operatorname {Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\cosh ^{-1}(c x)\right )}{d^3}-\frac {\left (3 b c^2\right ) \operatorname {Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\cosh ^{-1}(c x)\right )}{d^3}\\ &=\frac {b c}{2 d^3 x (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac {5 b c^3 x}{12 d^3 (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac {2 b c^3 x}{3 d^3 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {3 c^2 \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \cosh ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {3 c^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 d^3 \left (1-c^2 x^2\right )}+\frac {6 c^2 \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right )}{d^3}+\frac {\left (3 b c^2\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \cosh ^{-1}(c x)}\right )}{2 d^3}-\frac {\left (3 b c^2\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \cosh ^{-1}(c x)}\right )}{2 d^3}\\ &=\frac {b c}{2 d^3 x (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac {5 b c^3 x}{12 d^3 (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac {2 b c^3 x}{3 d^3 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {3 c^2 \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \cosh ^{-1}(c x)}{2 d^3 x^2 \left (1-c^2 x^2\right )^2}+\frac {3 c^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 d^3 \left (1-c^2 x^2\right )}+\frac {6 c^2 \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right )}{d^3}+\frac {3 b c^2 \text {Li}_2\left (-e^{2 \cosh ^{-1}(c x)}\right )}{2 d^3}-\frac {3 b c^2 \text {Li}_2\left (e^{2 \cosh ^{-1}(c x)}\right )}{2 d^3}\\ \end {align*}

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Mathematica [A]  time = 3.03, size = 273, normalized size = 1.09 \[ -\frac {\frac {12 a c^2}{c^2 x^2-1}-\frac {3 a c^2}{\left (c^2 x^2-1\right )^2}+18 a c^2 \log \left (1-c^2 x^2\right )-36 a c^2 \log (x)+\frac {6 a}{x^2}+b c^2 \left (\frac {12 \cosh ^{-1}(c x)}{c^2 x^2-1}-\frac {3 \cosh ^{-1}(c x)}{\left (c^2 x^2-1\right )^2}+\frac {6 \cosh ^{-1}(c x)}{c^2 x^2}+18 \text {Li}_2\left (-e^{-2 \cosh ^{-1}(c x)}\right )-18 \text {Li}_2\left (e^{-2 \cosh ^{-1}(c x)}\right )+\frac {14 c x}{\sqrt {\frac {c x-1}{c x+1}} (c x+1)}-\frac {c x}{\left (\frac {c x-1}{c x+1}\right )^{3/2} (c x+1)^3}-\frac {6 \sqrt {\frac {c x-1}{c x+1}} (c x+1)}{c x}+36 \cosh ^{-1}(c x) \log \left (1-e^{-2 \cosh ^{-1}(c x)}\right )-36 \cosh ^{-1}(c x) \log \left (e^{-2 \cosh ^{-1}(c x)}+1\right )\right )}{12 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/12*((6*a)/x^2 - (3*a*c^2)/(-1 + c^2*x^2)^2 + (12*a*c^2)/(-1 + c^2*x^2) - 36*a*c^2*Log[x] + 18*a*c^2*Log[1 -
c^2*x^2] + b*c^2*(-((c*x)/(((-1 + c*x)/(1 + c*x))^(3/2)*(1 + c*x)^3)) + (14*c*x)/(Sqrt[(-1 + c*x)/(1 + c*x)]*
(1 + c*x)) - (6*Sqrt[(-1 + c*x)/(1 + c*x)]*(1 + c*x))/(c*x) + (6*ArcCosh[c*x])/(c^2*x^2) - (3*ArcCosh[c*x])/(-
1 + c^2*x^2)^2 + (12*ArcCosh[c*x])/(-1 + c^2*x^2) + 36*ArcCosh[c*x]*Log[1 - E^(-2*ArcCosh[c*x])] - 36*ArcCosh[
c*x]*Log[1 + E^(-2*ArcCosh[c*x])] + 18*PolyLog[2, -E^(-2*ArcCosh[c*x])] - 18*PolyLog[2, E^(-2*ArcCosh[c*x])]))
/d^3

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b \operatorname {arcosh}\left (c x\right ) + a}{c^{6} d^{3} x^{9} - 3 \, c^{4} d^{3} x^{7} + 3 \, c^{2} d^{3} x^{5} - d^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b*arccosh(c*x) + a)/(c^6*d^3*x^9 - 3*c^4*d^3*x^7 + 3*c^2*d^3*x^5 - d^3*x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {b \operatorname {arcosh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} - d\right )}^{3} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 0.65, size = 641, normalized size = 2.56 \[ -\frac {a}{2 d^{3} x^{2}}+\frac {3 c^{2} a \ln \left (c x \right )}{d^{3}}+\frac {c^{2} a}{16 d^{3} \left (c x -1\right )^{2}}-\frac {9 c^{2} a}{16 d^{3} \left (c x -1\right )}-\frac {3 c^{2} a \ln \left (c x -1\right )}{2 d^{3}}+\frac {c^{2} a}{16 d^{3} \left (c x +1\right )^{2}}+\frac {9 c^{2} a}{16 d^{3} \left (c x +1\right )}-\frac {3 c^{2} a \ln \left (c x +1\right )}{2 d^{3}}-\frac {2 c^{5} b \sqrt {c x +1}\, \sqrt {c x -1}\, x^{3}}{3 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {2 c^{6} b \,x^{4}}{3 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {3 c^{4} b \,\mathrm {arccosh}\left (c x \right ) x^{2}}{2 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {c^{3} b \sqrt {c x +1}\, \sqrt {c x -1}\, x}{4 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {4 c^{4} b \,x^{2}}{3 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {9 c^{2} b \,\mathrm {arccosh}\left (c x \right )}{4 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {c b \sqrt {c x +1}\, \sqrt {c x -1}}{2 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right ) x}+\frac {2 c^{2} b}{3 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {b \,\mathrm {arccosh}\left (c x \right )}{2 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right ) x^{2}}+\frac {3 c^{2} b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1+\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}\right )}{d^{3}}+\frac {3 b \,c^{2} \polylog \left (2, -\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}\right )}{2 d^{3}}-\frac {3 c^{2} b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1+c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d^{3}}-\frac {3 c^{2} b \polylog \left (2, -c x -\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d^{3}}-\frac {3 c^{2} b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1-c x -\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d^{3}}-\frac {3 c^{2} b \polylog \left (2, c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a/d^3/x^2+3*c^2*a/d^3*ln(c*x)+1/16*c^2*a/d^3/(c*x-1)^2-9/16*c^2*a/d^3/(c*x-1)-3/2*c^2*a/d^3*ln(c*x-1)+1/1
6*c^2*a/d^3/(c*x+1)^2+9/16*c^2*a/d^3/(c*x+1)-3/2*c^2*a/d^3*ln(c*x+1)-2/3*c^5*b/d^3/(c^4*x^4-2*c^2*x^2+1)*(c*x+
1)^(1/2)*(c*x-1)^(1/2)*x^3+2/3*c^6*b/d^3/(c^4*x^4-2*c^2*x^2+1)*x^4-3/2*c^4*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arccosh
(c*x)*x^2+1/4*c^3*b/d^3/(c^4*x^4-2*c^2*x^2+1)*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x-4/3*c^4*b/d^3/(c^4*x^4-2*c^2*x^2+1
)*x^2+9/4*c^2*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arccosh(c*x)+1/2*c*b/d^3/(c^4*x^4-2*c^2*x^2+1)/x*(c*x+1)^(1/2)*(c*x-
1)^(1/2)+2/3*c^2*b/d^3/(c^4*x^4-2*c^2*x^2+1)-1/2*b/d^3/(c^4*x^4-2*c^2*x^2+1)/x^2*arccosh(c*x)+3*c^2*b/d^3*arcc
osh(c*x)*ln(1+(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2)+3/2*b*c^2*polylog(2,-(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2)/d
^3-3*c^2*b/d^3*arccosh(c*x)*ln(1+c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))-3*c^2*b/d^3*polylog(2,-c*x-(c*x-1)^(1/2)*(c*
x+1)^(1/2))-3*c^2*b/d^3*arccosh(c*x)*ln(1-c*x-(c*x-1)^(1/2)*(c*x+1)^(1/2))-3*c^2*b/d^3*polylog(2,c*x+(c*x-1)^(
1/2)*(c*x+1)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{4} \, a {\left (\frac {6 \, c^{4} x^{4} - 9 \, c^{2} x^{2} + 2}{c^{4} d^{3} x^{6} - 2 \, c^{2} d^{3} x^{4} + d^{3} x^{2}} + \frac {6 \, c^{2} \log \left (c x + 1\right )}{d^{3}} + \frac {6 \, c^{2} \log \left (c x - 1\right )}{d^{3}} - \frac {12 \, c^{2} \log \relax (x)}{d^{3}}\right )} - b \int \frac {\log \left (c x + \sqrt {c x + 1} \sqrt {c x - 1}\right )}{c^{6} d^{3} x^{9} - 3 \, c^{4} d^{3} x^{7} + 3 \, c^{2} d^{3} x^{5} - d^{3} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a*((6*c^4*x^4 - 9*c^2*x^2 + 2)/(c^4*d^3*x^6 - 2*c^2*d^3*x^4 + d^3*x^2) + 6*c^2*log(c*x + 1)/d^3 + 6*c^2*l
og(c*x - 1)/d^3 - 12*c^2*log(x)/d^3) - b*integrate(log(c*x + sqrt(c*x + 1)*sqrt(c*x - 1))/(c^6*d^3*x^9 - 3*c^4
*d^3*x^7 + 3*c^2*d^3*x^5 - d^3*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {acosh}\left (c\,x\right )}{x^3\,{\left (d-c^2\,d\,x^2\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(c*x))/(x^3*(d - c^2*d*x^2)^3),x)

[Out]

int((a + b*acosh(c*x))/(x^3*(d - c^2*d*x^2)^3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {a}{c^{6} x^{9} - 3 c^{4} x^{7} + 3 c^{2} x^{5} - x^{3}}\, dx + \int \frac {b \operatorname {acosh}{\left (c x \right )}}{c^{6} x^{9} - 3 c^{4} x^{7} + 3 c^{2} x^{5} - x^{3}}\, dx}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a/(c**6*x**9 - 3*c**4*x**7 + 3*c**2*x**5 - x**3), x) + Integral(b*acosh(c*x)/(c**6*x**9 - 3*c**4*x*
*7 + 3*c**2*x**5 - x**3), x))/d**3

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