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

Optimal. Leaf size=230 \[ \frac {15 c^2 x \left (a+b \cosh ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac {5 c^2 x \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \cosh ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {15 c \tanh ^{-1}\left (e^{\cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3}+\frac {15 b c \text {Li}_2\left (-e^{\cosh ^{-1}(c x)}\right )}{8 d^3}-\frac {15 b c \text {Li}_2\left (e^{\cosh ^{-1}(c x)}\right )}{8 d^3}-\frac {7 b c}{8 d^3 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b c}{12 d^3 (c x-1)^{3/2} (c x+1)^{3/2}}+\frac {b c \tan ^{-1}\left (\sqrt {c x-1} \sqrt {c x+1}\right )}{d^3} \]

[Out]

1/12*b*c/d^3/(c*x-1)^(3/2)/(c*x+1)^(3/2)+(-a-b*arccosh(c*x))/d^3/x/(-c^2*x^2+1)^2+5/4*c^2*x*(a+b*arccosh(c*x))
/d^3/(-c^2*x^2+1)^2+15/8*c^2*x*(a+b*arccosh(c*x))/d^3/(-c^2*x^2+1)+b*c*arctan((c*x-1)^(1/2)*(c*x+1)^(1/2))/d^3
+15/4*c*(a+b*arccosh(c*x))*arctanh(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))/d^3+15/8*b*c*polylog(2,-c*x-(c*x-1)^(1/2)*
(c*x+1)^(1/2))/d^3-15/8*b*c*polylog(2,c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))/d^3-7/8*b*c/d^3/(c*x-1)^(1/2)/(c*x+1)^(
1/2)

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Rubi [A]  time = 0.24, antiderivative size = 230, 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, 104, 21, 92, 205, 5689, 74, 5694, 4182, 2279, 2391} \[ \frac {15 b c \text {PolyLog}\left (2,-e^{\cosh ^{-1}(c x)}\right )}{8 d^3}-\frac {15 b c \text {PolyLog}\left (2,e^{\cosh ^{-1}(c x)}\right )}{8 d^3}+\frac {15 c^2 x \left (a+b \cosh ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac {5 c^2 x \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \cosh ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {15 c \tanh ^{-1}\left (e^{\cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3}-\frac {7 b c}{8 d^3 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b c}{12 d^3 (c x-1)^{3/2} (c x+1)^{3/2}}+\frac {b c \tan ^{-1}\left (\sqrt {c x-1} \sqrt {c x+1}\right )}{d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c)/(12*d^3*(-1 + c*x)^(3/2)*(1 + c*x)^(3/2)) - (7*b*c)/(8*d^3*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) - (a + b*ArcCos
h[c*x])/(d^3*x*(1 - c^2*x^2)^2) + (5*c^2*x*(a + b*ArcCosh[c*x]))/(4*d^3*(1 - c^2*x^2)^2) + (15*c^2*x*(a + b*Ar
cCosh[c*x]))/(8*d^3*(1 - c^2*x^2)) + (b*c*ArcTan[Sqrt[-1 + c*x]*Sqrt[1 + c*x]])/d^3 + (15*c*(a + b*ArcCosh[c*x
])*ArcTanh[E^ArcCosh[c*x]])/(4*d^3) + (15*b*c*PolyLog[2, -E^ArcCosh[c*x]])/(8*d^3) - (15*b*c*PolyLog[2, E^ArcC
osh[c*x]])/(8*d^3)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 92

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

Rule 104

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] &&
IntegersQ[2*m, 2*n, 2*p]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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 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 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 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]

Rubi steps

\begin {align*} \int \frac {a+b \cosh ^{-1}(c x)}{x^2 \left (d-c^2 d x^2\right )^3} \, dx &=-\frac {a+b \cosh ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\left (5 c^2\right ) \int \frac {a+b \cosh ^{-1}(c x)}{\left (d-c^2 d x^2\right )^3} \, dx+\frac {(b c) \int \frac {1}{x (-1+c x)^{5/2} (1+c x)^{5/2}} \, dx}{d^3}\\ &=-\frac {b c}{3 d^3 (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac {a+b \cosh ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {b \int \frac {3 c+3 c^2 x}{x (-1+c x)^{3/2} (1+c x)^{5/2}} \, dx}{3 d^3}-\frac {\left (5 b c^3\right ) \int \frac {x}{(-1+c x)^{5/2} (1+c x)^{5/2}} \, dx}{4 d^3}+\frac {\left (15 c^2\right ) \int \frac {a+b \cosh ^{-1}(c x)}{\left (d-c^2 d x^2\right )^2} \, dx}{4 d}\\ &=\frac {b c}{12 d^3 (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac {a+b \cosh ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {15 c^2 x \left (a+b \cosh ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}-\frac {(b c) \int \frac {1}{x (-1+c x)^{3/2} (1+c x)^{3/2}} \, dx}{d^3}+\frac {\left (15 b c^3\right ) \int \frac {x}{(-1+c x)^{3/2} (1+c x)^{3/2}} \, dx}{8 d^3}+\frac {\left (15 c^2\right ) \int \frac {a+b \cosh ^{-1}(c x)}{d-c^2 d x^2} \, dx}{8 d^2}\\ &=\frac {b c}{12 d^3 (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac {7 b c}{8 d^3 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {a+b \cosh ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {15 c^2 x \left (a+b \cosh ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac {b \int \frac {c+c^2 x}{x \sqrt {-1+c x} (1+c x)^{3/2}} \, dx}{d^3}-\frac {(15 c) \operatorname {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\cosh ^{-1}(c x)\right )}{8 d^3}\\ &=\frac {b c}{12 d^3 (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac {7 b c}{8 d^3 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {a+b \cosh ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {15 c^2 x \left (a+b \cosh ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac {15 c \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\cosh ^{-1}(c x)}\right )}{4 d^3}+\frac {(b c) \int \frac {1}{x \sqrt {-1+c x} \sqrt {1+c x}} \, dx}{d^3}+\frac {(15 b c) \operatorname {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\cosh ^{-1}(c x)\right )}{8 d^3}-\frac {(15 b c) \operatorname {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\cosh ^{-1}(c x)\right )}{8 d^3}\\ &=\frac {b c}{12 d^3 (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac {7 b c}{8 d^3 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {a+b \cosh ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {15 c^2 x \left (a+b \cosh ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac {15 c \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\cosh ^{-1}(c x)}\right )}{4 d^3}+\frac {(15 b c) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\cosh ^{-1}(c x)}\right )}{8 d^3}-\frac {(15 b c) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\cosh ^{-1}(c x)}\right )}{8 d^3}+\frac {\left (b c^2\right ) \operatorname {Subst}\left (\int \frac {1}{c+c x^2} \, dx,x,\sqrt {-1+c x} \sqrt {1+c x}\right )}{d^3}\\ &=\frac {b c}{12 d^3 (-1+c x)^{3/2} (1+c x)^{3/2}}-\frac {7 b c}{8 d^3 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {a+b \cosh ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x \left (a+b \cosh ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {15 c^2 x \left (a+b \cosh ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac {b c \tan ^{-1}\left (\sqrt {-1+c x} \sqrt {1+c x}\right )}{d^3}+\frac {15 c \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\cosh ^{-1}(c x)}\right )}{4 d^3}+\frac {15 b c \text {Li}_2\left (-e^{\cosh ^{-1}(c x)}\right )}{8 d^3}-\frac {15 b c \text {Li}_2\left (e^{\cosh ^{-1}(c x)}\right )}{8 d^3}\\ \end {align*}

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Mathematica [A]  time = 1.78, size = 362, normalized size = 1.57 \[ \frac {-\frac {84 a c^2 x}{c^2 x^2-1}+\frac {24 a c^2 x}{\left (c^2 x^2-1\right )^2}-90 a c \log (1-c x)+90 a c \log (c x+1)-\frac {96 a}{x}+\frac {96 b c \sqrt {c^2 x^2-1} \tan ^{-1}\left (\sqrt {c^2 x^2-1}\right )}{\sqrt {c x-1} \sqrt {c x+1}}-45 b c \left (\cosh ^{-1}(c x) \left (\cosh ^{-1}(c x)-4 \log \left (e^{\cosh ^{-1}(c x)}+1\right )\right )-4 \text {Li}_2\left (-e^{\cosh ^{-1}(c x)}\right )\right )+45 b c \left (\cosh ^{-1}(c x) \left (\cosh ^{-1}(c x)-4 \log \left (1-e^{\cosh ^{-1}(c x)}\right )\right )-4 \text {Li}_2\left (e^{\cosh ^{-1}(c x)}\right )\right )-\frac {2 b c \left ((c x-2) \sqrt {c x-1} \sqrt {c x+1}-3 \cosh ^{-1}(c x)\right )}{(c x-1)^2}+\frac {2 b c \left (\sqrt {c x-1} \sqrt {c x+1} (c x+2)-3 \cosh ^{-1}(c x)\right )}{(c x+1)^2}+42 b c \left (\frac {\cosh ^{-1}(c x)}{1-c x}-\frac {1}{\sqrt {\frac {c x-1}{c x+1}}}\right )+42 b c \left (\sqrt {\frac {c x-1}{c x+1}}-\frac {\cosh ^{-1}(c x)}{c x+1}\right )-\frac {96 b \cosh ^{-1}(c x)}{x}}{96 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-96*a)/x + (24*a*c^2*x)/(-1 + c^2*x^2)^2 - (84*a*c^2*x)/(-1 + c^2*x^2) - (2*b*c*((-2 + c*x)*Sqrt[-1 + c*x]*S
qrt[1 + c*x] - 3*ArcCosh[c*x]))/(-1 + c*x)^2 + (2*b*c*(Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(2 + c*x) - 3*ArcCosh[c*x]
))/(1 + c*x)^2 - (96*b*ArcCosh[c*x])/x + 42*b*c*(-(1/Sqrt[(-1 + c*x)/(1 + c*x)]) + ArcCosh[c*x]/(1 - c*x)) + 4
2*b*c*(Sqrt[(-1 + c*x)/(1 + c*x)] - ArcCosh[c*x]/(1 + c*x)) + (96*b*c*Sqrt[-1 + c^2*x^2]*ArcTan[Sqrt[-1 + c^2*
x^2]])/(Sqrt[-1 + c*x]*Sqrt[1 + c*x]) - 90*a*c*Log[1 - c*x] + 90*a*c*Log[1 + c*x] - 45*b*c*(ArcCosh[c*x]*(ArcC
osh[c*x] - 4*Log[1 + E^ArcCosh[c*x]]) - 4*PolyLog[2, -E^ArcCosh[c*x]]) + 45*b*c*(ArcCosh[c*x]*(ArcCosh[c*x] -
4*Log[1 - E^ArcCosh[c*x]]) - 4*PolyLog[2, E^ArcCosh[c*x]]))/(96*d^3)

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b \operatorname {arcosh}\left (c x\right ) + a}{c^{6} d^{3} x^{8} - 3 \, c^{4} d^{3} x^{6} + 3 \, c^{2} d^{3} x^{4} - d^{3} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.45, size = 392, normalized size = 1.70 \[ -\frac {a}{d^{3} x}+\frac {c a}{16 d^{3} \left (c x -1\right )^{2}}-\frac {7 c a}{16 d^{3} \left (c x -1\right )}-\frac {15 c a \ln \left (c x -1\right )}{16 d^{3}}-\frac {c a}{16 d^{3} \left (c x +1\right )^{2}}-\frac {7 c a}{16 d^{3} \left (c x +1\right )}+\frac {15 c a \ln \left (c x +1\right )}{16 d^{3}}-\frac {15 b \,\mathrm {arccosh}\left (c x \right ) c^{4} x^{3}}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {7 b \sqrt {c x +1}\, \sqrt {c x -1}\, c^{3} x^{2}}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {25 b \,\mathrm {arccosh}\left (c x \right ) c^{2} x}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {23 c b \sqrt {c x +1}\, \sqrt {c x -1}}{24 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {b \,\mathrm {arccosh}\left (c x \right )}{d^{3} x \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {2 c b \arctan \left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d^{3}}+\frac {15 c b \dilog \left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{8 d^{3}}+\frac {15 c b \dilog \left (1+c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{8 d^{3}}+\frac {15 c b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1+c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{8 d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/d^3/x+1/16*c*a/d^3/(c*x-1)^2-7/16*c*a/d^3/(c*x-1)-15/16*c*a/d^3*ln(c*x-1)-1/16*c*a/d^3/(c*x+1)^2-7/16*c*a/d
^3/(c*x+1)+15/16*c*a/d^3*ln(c*x+1)-15/8*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arccosh(c*x)*c^4*x^3-7/8*b/d^3/(c^4*x^4-2*
c^2*x^2+1)*(c*x+1)^(1/2)*(c*x-1)^(1/2)*c^3*x^2+25/8*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arccosh(c*x)*c^2*x+23/24*c*b/d
^3/(c^4*x^4-2*c^2*x^2+1)*(c*x+1)^(1/2)*(c*x-1)^(1/2)-b/d^3/x/(c^4*x^4-2*c^2*x^2+1)*arccosh(c*x)+2*c*b/d^3*arct
an(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))+15/8*c*b/d^3*dilog(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))+15/8*c*b/d^3*dilog(1+c
*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))+15/8*c*b/d^3*arccosh(c*x)*ln(1+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 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2048*(92160*c^7*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) - 24
0*c^6*(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)) - 30720*c^6*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
) + 90*(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^5
+ 400*c^4*(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)) + 61440*c^4*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) +
45*(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^3 + 128*c^2*(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)) - 30720*c^
2*integrate(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) - 32*(15*(c^5*x^5 - 2*c^
3*x^3 + c*x)*log(c*x + 1)^2 + 30*(c^5*x^5 - 2*c^3*x^3 + c*x)*log(c*x + 1)*log(c*x - 1) + 4*(30*c^4*x^4 - 50*c^
2*x^2 - 15*(c^5*x^5 - 2*c^3*x^3 + c*x)*log(c*x + 1) + 15*(c^5*x^5 - 2*c^3*x^3 + c*x)*log(c*x - 1) + 16)*log(c*
x + sqrt(c*x + 1)*sqrt(c*x - 1)))/(c^4*d^3*x^5 - 2*c^2*d^3*x^3 + d^3*x) + 2048*integrate(-1/16*(30*c^5*x^4 - 5
0*c^3*x^2 - 15*(c^6*x^5 - 2*c^4*x^3 + c^2*x)*log(c*x + 1) + 15*(c^6*x^5 - 2*c^4*x^3 + c^2*x)*log(c*x - 1) + 16
*c)/(c^7*d^3*x^8 - 3*c^5*d^3*x^6 + 3*c^3*d^3*x^4 - c*d^3*x^2 + (c^6*d^3*x^7 - 3*c^4*d^3*x^5 + 3*c^2*d^3*x^3 -
d^3*x)*sqrt(c*x + 1)*sqrt(c*x - 1)), x))*b - 1/16*a*(2*(15*c^4*x^4 - 25*c^2*x^2 + 8)/(c^4*d^3*x^5 - 2*c^2*d^3*
x^3 + d^3*x) - 15*c*log(c*x + 1)/d^3 + 15*c*log(c*x - 1)/d^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {a}{c^{6} x^{8} - 3 c^{4} x^{6} + 3 c^{2} x^{4} - x^{2}}\, dx + \int \frac {b \operatorname {acosh}{\left (c x \right )}}{c^{6} x^{8} - 3 c^{4} x^{6} + 3 c^{2} x^{4} - x^{2}}\, dx}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a/(c**6*x**8 - 3*c**4*x**6 + 3*c**2*x**4 - x**2), x) + Integral(b*acosh(c*x)/(c**6*x**8 - 3*c**4*x*
*6 + 3*c**2*x**4 - x**2), x))/d**3

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