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

Optimal. Leaf size=264 \[ d \log (x) \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{2} e x^2 \left (a+b \cosh ^{-1}(c x)\right )-\frac {i b d \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{2 \sqrt {c x-1} \sqrt {c x+1}}-\frac {i b d \sqrt {1-c^2 x^2} \sin ^{-1}(c x)^2}{2 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b d \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )}{\sqrt {c x-1} \sqrt {c x+1}}-\frac {b d \sqrt {1-c^2 x^2} \log (x) \sin ^{-1}(c x)}{\sqrt {c x-1} \sqrt {c x+1}}-\frac {b e \cosh ^{-1}(c x)}{4 c^2}-\frac {b e x \sqrt {c x-1} \sqrt {c x+1}}{4 c} \]

[Out]

-1/4*b*e*arccosh(c*x)/c^2+1/2*e*x^2*(a+b*arccosh(c*x))+d*(a+b*arccosh(c*x))*ln(x)-1/4*b*e*x*(c*x-1)^(1/2)*(c*x
+1)^(1/2)/c-1/2*I*b*d*arcsin(c*x)^2*(-c^2*x^2+1)^(1/2)/(c*x-1)^(1/2)/(c*x+1)^(1/2)+b*d*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*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*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)

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Rubi [A]  time = 0.68, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 13, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.684, Rules used = {14, 5790, 12, 6742, 90, 52, 2328, 2326, 4625, 3717, 2190, 2279, 2391} \[ -\frac {i b d \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )}{2 \sqrt {c x-1} \sqrt {c x+1}}+d \log (x) \left (a+b \cosh ^{-1}(c x)\right )+\frac {1}{2} e x^2 \left (a+b \cosh ^{-1}(c x)\right )-\frac {i b d \sqrt {1-c^2 x^2} \sin ^{-1}(c x)^2}{2 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b d \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )}{\sqrt {c x-1} \sqrt {c x+1}}-\frac {b d \sqrt {1-c^2 x^2} \log (x) \sin ^{-1}(c x)}{\sqrt {c x-1} \sqrt {c x+1}}-\frac {b e \cosh ^{-1}(c x)}{4 c^2}-\frac {b e x \sqrt {c x-1} \sqrt {c x+1}}{4 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*e*x*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(4*c) - (b*e*ArcCosh[c*x])/(4*c^2) + (e*x^2*(a + b*ArcCosh[c*x]))/2 - ((
I/2)*b*d*Sqrt[1 - c^2*x^2]*ArcSin[c*x]^2)/(Sqrt[-1 + c*x]*Sqrt[1 + c*x]) + (b*d*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*(a + b*ArcCosh[c*x])*Log[x] - (b*d*Sqrt[1 -
c^2*x^2]*ArcSin[c*x]*Log[x])/(Sqrt[-1 + c*x]*Sqrt[1 + c*x]) - ((I/2)*b*d*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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 52

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 90

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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 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 2326

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(ArcSin[(Rt[-e, 2]*x)/S
qrt[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 2328

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_.)*(x_)]), x_Symbol] :>
Dist[Sqrt[1 + (e1*e2*x^2)/(d1*d2)]/(Sqrt[d1 + e1*x]*Sqrt[d2 + e2*x]), Int[(a + b*Log[c*x^n])/Sqrt[1 + (e1*e2*x
^2)/(d1*d2)], x], x] /; FreeQ[{a, b, c, d1, e1, d2, e2, n}, x] && EqQ[d2*e1 + d1*e2, 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 3717

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 4625

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

Rule 5790

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 6742

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 ) \left (a+b \cosh ^{-1}(c x)\right )}{x} \, dx &=\frac {1}{2} e x^2 \left (a+b \cosh ^{-1}(c x)\right )+d \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-(b c) \int \frac {e x^2+2 d \log (x)}{2 \sqrt {-1+c x} \sqrt {1+c x}} \, dx\\ &=\frac {1}{2} e x^2 \left (a+b \cosh ^{-1}(c x)\right )+d \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-\frac {1}{2} (b c) \int \frac {e x^2+2 d \log (x)}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx\\ &=\frac {1}{2} e x^2 \left (a+b \cosh ^{-1}(c x)\right )+d \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-\frac {1}{2} (b c) \int \left (\frac {e x^2}{\sqrt {-1+c x} \sqrt {1+c x}}+\frac {2 d \log (x)}{\sqrt {-1+c x} \sqrt {1+c x}}\right ) \, dx\\ &=\frac {1}{2} e x^2 \left (a+b \cosh ^{-1}(c x)\right )+d \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-(b c d) \int \frac {\log (x)}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx-\frac {1}{2} (b c e) \int \frac {x^2}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx\\ &=-\frac {b e x \sqrt {-1+c x} \sqrt {1+c x}}{4 c}+\frac {1}{2} e x^2 \left (a+b \cosh ^{-1}(c x)\right )+d \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-\frac {(b e) \int \frac {1}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{4 c}-\frac {\left (b c d \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 e x \sqrt {-1+c x} \sqrt {1+c x}}{4 c}-\frac {b e \cosh ^{-1}(c x)}{4 c^2}+\frac {1}{2} e x^2 \left (a+b \cosh ^{-1}(c x)\right )+d \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-\frac {b d \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log (x)}{\sqrt {-1+c x} \sqrt {1+c x}}+\frac {\left (b d \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 e x \sqrt {-1+c x} \sqrt {1+c x}}{4 c}-\frac {b e \cosh ^{-1}(c x)}{4 c^2}+\frac {1}{2} e x^2 \left (a+b \cosh ^{-1}(c x)\right )+d \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-\frac {b d \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log (x)}{\sqrt {-1+c x} \sqrt {1+c x}}+\frac {\left (b d \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int x \cot (x) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {b e x \sqrt {-1+c x} \sqrt {1+c x}}{4 c}-\frac {b e \cosh ^{-1}(c x)}{4 c^2}+\frac {1}{2} e x^2 \left (a+b \cosh ^{-1}(c x)\right )-\frac {i b d \sqrt {1-c^2 x^2} \sin ^{-1}(c x)^2}{2 \sqrt {-1+c x} \sqrt {1+c x}}+d \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-\frac {b d \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 \sqrt {1-c^2 x^2}\right ) \operatorname {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 e x \sqrt {-1+c x} \sqrt {1+c x}}{4 c}-\frac {b e \cosh ^{-1}(c x)}{4 c^2}+\frac {1}{2} e x^2 \left (a+b \cosh ^{-1}(c x)\right )-\frac {i b d \sqrt {1-c^2 x^2} \sin ^{-1}(c x)^2}{2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b d \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 \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-\frac {b d \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log (x)}{\sqrt {-1+c x} \sqrt {1+c x}}-\frac {\left (b d \sqrt {1-c^2 x^2}\right ) \operatorname {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 e x \sqrt {-1+c x} \sqrt {1+c x}}{4 c}-\frac {b e \cosh ^{-1}(c x)}{4 c^2}+\frac {1}{2} e x^2 \left (a+b \cosh ^{-1}(c x)\right )-\frac {i b d \sqrt {1-c^2 x^2} \sin ^{-1}(c x)^2}{2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b d \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 \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-\frac {b d \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 \sqrt {1-c^2 x^2}\right ) \operatorname {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 e x \sqrt {-1+c x} \sqrt {1+c x}}{4 c}-\frac {b e \cosh ^{-1}(c x)}{4 c^2}+\frac {1}{2} e x^2 \left (a+b \cosh ^{-1}(c x)\right )-\frac {i b d \sqrt {1-c^2 x^2} \sin ^{-1}(c x)^2}{2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b d \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 \left (a+b \cosh ^{-1}(c x)\right ) \log (x)-\frac {b d \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log (x)}{\sqrt {-1+c x} \sqrt {1+c x}}-\frac {i b d \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 = 0.27, size = 119, normalized size = 0.45 \[ \frac {1}{2} \left (2 a d \log (x)+a e x^2-\frac {b e \left (c x \sqrt {c x-1} \sqrt {c x+1}+2 \tanh ^{-1}\left (\sqrt {\frac {c x-1}{c x+1}}\right )\right )}{2 c^2}-b d \text {Li}_2\left (-e^{-2 \cosh ^{-1}(c x)}\right )+b d \cosh ^{-1}(c x) \left (\cosh ^{-1}(c x)+2 \log \left (e^{-2 \cosh ^{-1}(c x)}+1\right )\right )+b e x^2 \cosh ^{-1}(c x)\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*e*x^2 + b*e*x^2*ArcCosh[c*x] - (b*e*(c*x*Sqrt[-1 + c*x]*Sqrt[1 + c*x] + 2*ArcTanh[Sqrt[(-1 + c*x)/(1 + c*x)
]]))/(2*c^2) + b*d*ArcCosh[c*x]*(ArcCosh[c*x] + 2*Log[1 + E^(-2*ArcCosh[c*x])]) + 2*a*d*Log[x] - b*d*PolyLog[2
, -E^(-2*ArcCosh[c*x])])/2

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a e x^{2} + a d + {\left (b e x^{2} + b d\right )} \operatorname {arcosh}\left (c x\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.32, size = 130, normalized size = 0.49 \[ \frac {a \,x^{2} e}{2}+d a \ln \left (c x \right )-\frac {b \mathrm {arccosh}\left (c x \right )^{2} d}{2}-\frac {b e x \sqrt {c x -1}\, \sqrt {c x +1}}{4 c}+\frac {b \,\mathrm {arccosh}\left (c x \right ) x^{2} e}{2}-\frac {b e \,\mathrm {arccosh}\left (c x \right )}{4 c^{2}}+b d \,\mathrm {arccosh}\left (c x \right ) \ln \left (1+\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}\right )+\frac {b d \polylog \left (2, -\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2*e+d*a*ln(c*x)-1/2*b*arccosh(c*x)^2*d-1/4*b*e*x*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c+1/2*b*arccosh(c*x)*x^2*
e-1/4*b*e*arccosh(c*x)/c^2+b*d*arccosh(c*x)*ln(1+(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2)+1/2*b*d*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 \[ \frac {1}{2} \, a e x^{2} + a d \log \relax (x) + \int b e x \log \left (c x + \sqrt {c x + 1} \sqrt {c x - 1}\right ) + \frac {b d \log \left (c x + \sqrt {c x + 1} \sqrt {c x - 1}\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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