3.5.75 \(\int \frac {(d+e x^2)^2 (a+b \cosh ^{-1}(c x))}{x} \, dx\) [475]

Optimal. Leaf size=342 \[ -\frac {b e \left (16 c^2 d+3 e\right ) x \sqrt {-1+c x} \sqrt {1+c x}}{32 c^3}-\frac {b e^2 x^3 \sqrt {-1+c x} \sqrt {1+c x}}{16 c}-\frac {b e \left (16 c^2 d+3 e\right ) \cosh ^{-1}(c x)}{32 c^4}+d e x^2 \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \cosh ^{-1}(c x)\right )-\frac {i b d^2 \sqrt {1-c^2 x^2} \text {ArcSin}(c x)^2}{2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b d^2 \sqrt {1-c^2 x^2} \text {ArcSin}(c x) \log \left (1-e^{2 i \text {ArcSin}(c x)}\right )}{\sqrt {-1+c x} \sqrt {1+c x}}+d^2 \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-\frac {b d^2 \sqrt {1-c^2 x^2} \text {ArcSin}(c x) \log (x)}{\sqrt {-1+c x} \sqrt {1+c x}}-\frac {i b d^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,e^{2 i \text {ArcSin}(c x)}\right )}{2 \sqrt {-1+c x} \sqrt {1+c x}} \]

[Out]

-1/32*b*e*(16*c^2*d+3*e)*arccosh(c*x)/c^4+d*e*x^2*(a+b*arccosh(c*x))+1/4*e^2*x^4*(a+b*arccosh(c*x))+d^2*(a+b*a
rccosh(c*x))*ln(x)-1/32*b*e*(16*c^2*d+3*e)*x*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c^3-1/16*b*e^2*x^3*(c*x-1)^(1/2)*(c*x
+1)^(1/2)/c-1/2*I*b*d^2*arcsin(c*x)^2*(-c^2*x^2+1)^(1/2)/(c*x-1)^(1/2)/(c*x+1)^(1/2)+b*d^2*arcsin(c*x)*ln(1-(I
*c*x+(-c^2*x^2+1)^(1/2))^2)*(-c^2*x^2+1)^(1/2)/(c*x-1)^(1/2)/(c*x+1)^(1/2)-b*d^2*arcsin(c*x)*ln(x)*(-c^2*x^2+1
)^(1/2)/(c*x-1)^(1/2)/(c*x+1)^(1/2)-1/2*I*b*d^2*polylog(2,(I*c*x+(-c^2*x^2+1)^(1/2))^2)*(-c^2*x^2+1)^(1/2)/(c*
x-1)^(1/2)/(c*x+1)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.53, antiderivative size = 369, normalized size of antiderivative = 1.08, number of steps used = 16, number of rules used = 15, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {272, 45, 5958, 6874, 92, 54, 102, 12, 2365, 2363, 4721, 3798, 2221, 2317, 2438} \begin {gather*} d^2 \log (x) \left (a+b \cosh ^{-1}(c x)\right )+d e x^2 \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \cosh ^{-1}(c x)\right )-\frac {i b d^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{2 i \text {ArcSin}(c x)}\right )}{2 \sqrt {c x-1} \sqrt {c x+1}}-\frac {i b d^2 \sqrt {1-c^2 x^2} \text {ArcSin}(c x)^2}{2 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b d^2 \sqrt {1-c^2 x^2} \text {ArcSin}(c x) \log \left (1-e^{2 i \text {ArcSin}(c x)}\right )}{\sqrt {c x-1} \sqrt {c x+1}}-\frac {b d^2 \sqrt {1-c^2 x^2} \log (x) \text {ArcSin}(c x)}{\sqrt {c x-1} \sqrt {c x+1}}-\frac {3 b e^2 \cosh ^{-1}(c x)}{32 c^4}-\frac {3 b e^2 x \sqrt {c x-1} \sqrt {c x+1}}{32 c^3}-\frac {b d e \cosh ^{-1}(c x)}{2 c^2}-\frac {b d e x \sqrt {c x-1} \sqrt {c x+1}}{2 c}-\frac {b e^2 x^3 \sqrt {c x-1} \sqrt {c x+1}}{16 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b*d*e*x*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/c - (3*b*e^2*x*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(32*c^3) - (b*e^2*x^3
*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(16*c) - (b*d*e*ArcCosh[c*x])/(2*c^2) - (3*b*e^2*ArcCosh[c*x])/(32*c^4) + d*e*x
^2*(a + b*ArcCosh[c*x]) + (e^2*x^4*(a + b*ArcCosh[c*x]))/4 - ((I/2)*b*d^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x]^2)/(Sq
rt[-1 + c*x]*Sqrt[1 + c*x]) + (b*d^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x]*Log[1 - E^((2*I)*ArcSin[c*x])])/(Sqrt[-1 +
c*x]*Sqrt[1 + c*x]) + d^2*(a + b*ArcCosh[c*x])*Log[x] - (b*d^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x]*Log[x])/(Sqrt[-1
+ c*x]*Sqrt[1 + c*x]) - ((I/2)*b*d^2*Sqrt[1 - c^2*x^2]*PolyLog[2, E^((2*I)*ArcSin[c*x])])/(Sqrt[-1 + c*x]*Sqrt
[1 + c*x])

Rule 12

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 54

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

Rule 92

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a + b*x
)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 102

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)/(d*f*(m + n + p + 1))), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 272

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

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2317

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 2363

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-e, 2]*(x/Sqr
t[d])]*((a + b*Log[c*x^n])/Rt[-e, 2]), x] - Dist[b*(n/Rt[-e, 2]), Int[ArcSin[Rt[-e, 2]*(x/Sqrt[d])]/x, x], x]
/; FreeQ[{a, b, c, d, e, n}, x] && GtQ[d, 0] && NegQ[e]

Rule 2365

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

Rule 2438

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 3798

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[I*((c + d*x)^(m + 1)/(d*(
m + 1))), x] - Dist[2*I, Int[(c + d*x)^m*E^(2*I*k*Pi)*(E^(2*I*(e + f*x))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x))))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4721

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n*Cot[x], x], x, ArcSin[c*
x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 5958

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcCosh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(Sqrt[
1 + c*x]*Sqrt[-1 + c*x]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p] &
& (GtQ[p, 0] || (IGtQ[(m - 1)/2, 0] && LeQ[m + p, 0]))

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {\left (d+e x^2\right )^2 \left (a+b \cosh ^{-1}(c x)\right )}{x} \, dx &=d e x^2 \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \cosh ^{-1}(c x)\right )+d^2 \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-(b c) \int \frac {d e x^2+\frac {e^2 x^4}{4}+d^2 \log (x)}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx\\ &=d e x^2 \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \cosh ^{-1}(c x)\right )+d^2 \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-(b c) \int \left (\frac {d e x^2}{\sqrt {-1+c x} \sqrt {1+c x}}+\frac {e^2 x^4}{4 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {d^2 \log (x)}{\sqrt {-1+c x} \sqrt {1+c x}}\right ) \, dx\\ &=d e x^2 \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \cosh ^{-1}(c x)\right )+d^2 \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-\left (b c d^2\right ) \int \frac {\log (x)}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx-(b c d e) \int \frac {x^2}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx-\frac {1}{4} \left (b c e^2\right ) \int \frac {x^4}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx\\ &=-\frac {b d e x \sqrt {-1+c x} \sqrt {1+c x}}{2 c}-\frac {b e^2 x^3 \sqrt {-1+c x} \sqrt {1+c x}}{16 c}+d e x^2 \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \cosh ^{-1}(c x)\right )+d^2 \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-\frac {(b d e) \int \frac {1}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{2 c}-\frac {\left (b e^2\right ) \int \frac {3 x^2}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{16 c}-\frac {\left (b c d^2 \sqrt {1-c^2 x^2}\right ) \int \frac {\log (x)}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {b d e x \sqrt {-1+c x} \sqrt {1+c x}}{2 c}-\frac {b e^2 x^3 \sqrt {-1+c x} \sqrt {1+c x}}{16 c}-\frac {b d e \cosh ^{-1}(c x)}{2 c^2}+d e x^2 \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \cosh ^{-1}(c x)\right )+d^2 \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-\frac {b d^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log (x)}{\sqrt {-1+c x} \sqrt {1+c x}}-\frac {\left (3 b e^2\right ) \int \frac {x^2}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{16 c}+\frac {\left (b d^2 \sqrt {1-c^2 x^2}\right ) \int \frac {\sin ^{-1}(c x)}{x} \, dx}{\sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {b d e x \sqrt {-1+c x} \sqrt {1+c x}}{2 c}-\frac {3 b e^2 x \sqrt {-1+c x} \sqrt {1+c x}}{32 c^3}-\frac {b e^2 x^3 \sqrt {-1+c x} \sqrt {1+c x}}{16 c}-\frac {b d e \cosh ^{-1}(c x)}{2 c^2}+d e x^2 \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \cosh ^{-1}(c x)\right )+d^2 \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-\frac {b d^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log (x)}{\sqrt {-1+c x} \sqrt {1+c x}}-\frac {\left (3 b e^2\right ) \int \frac {1}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{32 c^3}+\frac {\left (b d^2 \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int x \cot (x) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {b d e x \sqrt {-1+c x} \sqrt {1+c x}}{2 c}-\frac {3 b e^2 x \sqrt {-1+c x} \sqrt {1+c x}}{32 c^3}-\frac {b e^2 x^3 \sqrt {-1+c x} \sqrt {1+c x}}{16 c}-\frac {b d e \cosh ^{-1}(c x)}{2 c^2}-\frac {3 b e^2 \cosh ^{-1}(c x)}{32 c^4}+d e x^2 \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \cosh ^{-1}(c x)\right )-\frac {i b d^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)^2}{2 \sqrt {-1+c x} \sqrt {1+c x}}+d^2 \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-\frac {b d^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log (x)}{\sqrt {-1+c x} \sqrt {1+c x}}-\frac {\left (2 i b d^2 \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \frac {e^{2 i x} x}{1-e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {b d e x \sqrt {-1+c x} \sqrt {1+c x}}{2 c}-\frac {3 b e^2 x \sqrt {-1+c x} \sqrt {1+c x}}{32 c^3}-\frac {b e^2 x^3 \sqrt {-1+c x} \sqrt {1+c x}}{16 c}-\frac {b d e \cosh ^{-1}(c x)}{2 c^2}-\frac {3 b e^2 \cosh ^{-1}(c x)}{32 c^4}+d e x^2 \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \cosh ^{-1}(c x)\right )-\frac {i b d^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)^2}{2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b d^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )}{\sqrt {-1+c x} \sqrt {1+c x}}+d^2 \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-\frac {b d^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log (x)}{\sqrt {-1+c x} \sqrt {1+c x}}-\frac {\left (b d^2 \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {b d e x \sqrt {-1+c x} \sqrt {1+c x}}{2 c}-\frac {3 b e^2 x \sqrt {-1+c x} \sqrt {1+c x}}{32 c^3}-\frac {b e^2 x^3 \sqrt {-1+c x} \sqrt {1+c x}}{16 c}-\frac {b d e \cosh ^{-1}(c x)}{2 c^2}-\frac {3 b e^2 \cosh ^{-1}(c x)}{32 c^4}+d e x^2 \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \cosh ^{-1}(c x)\right )-\frac {i b d^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)^2}{2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b d^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )}{\sqrt {-1+c x} \sqrt {1+c x}}+d^2 \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-\frac {b d^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log (x)}{\sqrt {-1+c x} \sqrt {1+c x}}+\frac {\left (i b d^2 \sqrt {1-c^2 x^2}\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 \sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {b d e x \sqrt {-1+c x} \sqrt {1+c x}}{2 c}-\frac {3 b e^2 x \sqrt {-1+c x} \sqrt {1+c x}}{32 c^3}-\frac {b e^2 x^3 \sqrt {-1+c x} \sqrt {1+c x}}{16 c}-\frac {b d e \cosh ^{-1}(c x)}{2 c^2}-\frac {3 b e^2 \cosh ^{-1}(c x)}{32 c^4}+d e x^2 \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \cosh ^{-1}(c x)\right )-\frac {i b d^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)^2}{2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b d^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )}{\sqrt {-1+c x} \sqrt {1+c x}}+d^2 \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-\frac {b d^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log (x)}{\sqrt {-1+c x} \sqrt {1+c x}}-\frac {i b d^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{2 \sqrt {-1+c x} \sqrt {1+c x}}\\ \end {align*}

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Mathematica [A]
time = 0.34, size = 228, normalized size = 0.67 \begin {gather*} a d e x^2+\frac {1}{4} a e^2 x^4+b d e x^2 \cosh ^{-1}(c x)+\frac {1}{4} b e^2 x^4 \cosh ^{-1}(c x)-\frac {b d e \left (c x \sqrt {-1+c x} \sqrt {1+c x}+2 \tanh ^{-1}\left (\sqrt {\frac {-1+c x}{1+c x}}\right )\right )}{2 c^2}-\frac {b e^2 \left (c x \sqrt {\frac {-1+c x}{1+c x}} \left (3+3 c x+2 c^2 x^2+2 c^3 x^3\right )+6 \tanh ^{-1}\left (\sqrt {\frac {-1+c x}{1+c x}}\right )\right )}{32 c^4}+\frac {1}{2} b d^2 \cosh ^{-1}(c x) \left (\cosh ^{-1}(c x)+2 \log \left (1+e^{-2 \cosh ^{-1}(c x)}\right )\right )+a d^2 \log (x)-\frac {1}{2} b d^2 \text {PolyLog}\left (2,-e^{-2 \cosh ^{-1}(c x)}\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

a*d*e*x^2 + (a*e^2*x^4)/4 + b*d*e*x^2*ArcCosh[c*x] + (b*e^2*x^4*ArcCosh[c*x])/4 - (b*d*e*(c*x*Sqrt[-1 + c*x]*S
qrt[1 + c*x] + 2*ArcTanh[Sqrt[(-1 + c*x)/(1 + c*x)]]))/(2*c^2) - (b*e^2*(c*x*Sqrt[(-1 + c*x)/(1 + c*x)]*(3 + 3
*c*x + 2*c^2*x^2 + 2*c^3*x^3) + 6*ArcTanh[Sqrt[(-1 + c*x)/(1 + c*x)]]))/(32*c^4) + (b*d^2*ArcCosh[c*x]*(ArcCos
h[c*x] + 2*Log[1 + E^(-2*ArcCosh[c*x])]))/2 + a*d^2*Log[x] - (b*d^2*PolyLog[2, -E^(-2*ArcCosh[c*x])])/2

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Maple [A]
time = 2.94, size = 225, normalized size = 0.66 \[\frac {a \,e^{2} x^{4}}{4}+a d e \,x^{2}+a \,d^{2} \ln \left (c x \right )-\frac {3 b \,e^{2} \mathrm {arccosh}\left (c x \right )}{32 c^{4}}-\frac {b d e \,\mathrm {arccosh}\left (c x \right )}{2 c^{2}}-\frac {b d e x \sqrt {c x -1}\, \sqrt {c x +1}}{2 c}-\frac {b \,e^{2} x^{3} \sqrt {c x -1}\, \sqrt {c x +1}}{16 c}-\frac {3 b \,e^{2} x \sqrt {c x -1}\, \sqrt {c x +1}}{32 c^{3}}-\frac {d^{2} b \mathrm {arccosh}\left (c x \right )^{2}}{2}+\frac {b \,\mathrm {arccosh}\left (c x \right ) e^{2} x^{4}}{4}+b \,\mathrm {arccosh}\left (c x \right ) x^{2} d e +d^{2} b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1+\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}\right )+\frac {d^{2} b \polylog \left (2, -\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}\right )}{2}\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*e^2*x^4+a*d*e*x^2+a*d^2*ln(c*x)-3/32*b*e^2*arccosh(c*x)/c^4-1/2*b*d*e*arccosh(c*x)/c^2-1/2*b*d*e*x*(c*x-
1)^(1/2)*(c*x+1)^(1/2)/c-1/16*b*e^2*x^3*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c-3/32*b*e^2*x*(c*x-1)^(1/2)*(c*x+1)^(1/2)
/c^3-1/2*d^2*b*arccosh(c*x)^2+1/4*b*arccosh(c*x)*e^2*x^4+b*arccosh(c*x)*x^2*d*e+d^2*b*arccosh(c*x)*ln(1+(c*x+(
c*x-1)^(1/2)*(c*x+1)^(1/2))^2)+1/2*b*d^2*polylog(2,-(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*x^4*e^2 + a*d*x^2*e + a*d^2*log(x) + integrate(b*x^3*e^2*log(c*x + sqrt(c*x + 1)*sqrt(c*x - 1)) + 2*b*d*
x*e*log(c*x + sqrt(c*x + 1)*sqrt(c*x - 1)) + b*d^2*log(c*x + sqrt(c*x + 1)*sqrt(c*x - 1))/x, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*x^4*e^2 + 2*a*d*x^2*e + a*d^2 + (b*x^4*e^2 + 2*b*d*x^2*e + b*d^2)*arccosh(c*x))/x, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {acosh}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*acosh(c*x))*(d + e*x**2)**2/x, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^2}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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