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

Optimal. Leaf size=265 \[ 3 d^2 e x \left (a+b \cosh ^{-1}(c x)\right )-\frac{d^3 \left (a+b \cosh ^{-1}(c x)\right )}{x}+d e^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{5} e^3 x^5 \left (a+b \cosh ^{-1}(c x)\right )+\frac{b e \left (1-c^2 x^2\right ) \left (15 c^4 d^2+5 c^2 d e+e^2\right )}{5 c^5 \sqrt{c x-1} \sqrt{c x+1}}+\frac{b c d^3 \sqrt{c^2 x^2-1} \tan ^{-1}\left (\sqrt{c^2 x^2-1}\right )}{\sqrt{c x-1} \sqrt{c x+1}}-\frac{b e^2 \left (1-c^2 x^2\right )^2 \left (5 c^2 d+2 e\right )}{15 c^5 \sqrt{c x-1} \sqrt{c x+1}}+\frac{b e^3 \left (1-c^2 x^2\right )^3}{25 c^5 \sqrt{c x-1} \sqrt{c x+1}} \]

[Out]

(b*e*(15*c^4*d^2 + 5*c^2*d*e + e^2)*(1 - c^2*x^2))/(5*c^5*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) - (b*e^2*(5*c^2*d + 2*
e)*(1 - c^2*x^2)^2)/(15*c^5*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) + (b*e^3*(1 - c^2*x^2)^3)/(25*c^5*Sqrt[-1 + c*x]*Sqr
t[1 + c*x]) - (d^3*(a + b*ArcCosh[c*x]))/x + 3*d^2*e*x*(a + b*ArcCosh[c*x]) + d*e^2*x^3*(a + b*ArcCosh[c*x]) +
(e^3*x^5*(a + b*ArcCosh[c*x]))/5 + (b*c*d^3*Sqrt[-1 + c^2*x^2]*ArcTan[Sqrt[-1 + c^2*x^2]])/(Sqrt[-1 + c*x]*Sq
rt[1 + c*x])

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Rubi [A]  time = 0.417573, antiderivative size = 265, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {270, 5790, 1610, 1799, 1620, 63, 205} \[ 3 d^2 e x \left (a+b \cosh ^{-1}(c x)\right )-\frac{d^3 \left (a+b \cosh ^{-1}(c x)\right )}{x}+d e^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{5} e^3 x^5 \left (a+b \cosh ^{-1}(c x)\right )+\frac{b e \left (1-c^2 x^2\right ) \left (15 c^4 d^2+5 c^2 d e+e^2\right )}{5 c^5 \sqrt{c x-1} \sqrt{c x+1}}+\frac{b c d^3 \sqrt{c^2 x^2-1} \tan ^{-1}\left (\sqrt{c^2 x^2-1}\right )}{\sqrt{c x-1} \sqrt{c x+1}}-\frac{b e^2 \left (1-c^2 x^2\right )^2 \left (5 c^2 d+2 e\right )}{15 c^5 \sqrt{c x-1} \sqrt{c x+1}}+\frac{b e^3 \left (1-c^2 x^2\right )^3}{25 c^5 \sqrt{c x-1} \sqrt{c x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e*(15*c^4*d^2 + 5*c^2*d*e + e^2)*(1 - c^2*x^2))/(5*c^5*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) - (b*e^2*(5*c^2*d + 2*
e)*(1 - c^2*x^2)^2)/(15*c^5*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) + (b*e^3*(1 - c^2*x^2)^3)/(25*c^5*Sqrt[-1 + c*x]*Sqr
t[1 + c*x]) - (d^3*(a + b*ArcCosh[c*x]))/x + 3*d^2*e*x*(a + b*ArcCosh[c*x]) + d*e^2*x^3*(a + b*ArcCosh[c*x]) +
(e^3*x^5*(a + b*ArcCosh[c*x]))/5 + (b*c*d^3*Sqrt[-1 + c^2*x^2]*ArcTan[Sqrt[-1 + c^2*x^2]])/(Sqrt[-1 + c*x]*Sq
rt[1 + c*x])

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 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 1610

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Dist[((
a + b*x)^FracPart[m]*(c + d*x)^FracPart[m])/(a*c + b*d*x^2)^FracPart[m], Int[Px*(a*c + b*d*x^2)^m*(e + f*x)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[Px, x] && EqQ[b*c + a*d, 0] && EqQ[m, n] &&  !Intege
rQ[m]

Rule 1799

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*SubstFor[x^2,
Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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]

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right )^3 \left (a+b \cosh ^{-1}(c x)\right )}{x^2} \, dx &=-\frac{d^3 \left (a+b \cosh ^{-1}(c x)\right )}{x}+3 d^2 e x \left (a+b \cosh ^{-1}(c x)\right )+d e^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{5} e^3 x^5 \left (a+b \cosh ^{-1}(c x)\right )-(b c) \int \frac{-d^3+3 d^2 e x^2+d e^2 x^4+\frac{e^3 x^6}{5}}{x \sqrt{-1+c x} \sqrt{1+c x}} \, dx\\ &=-\frac{d^3 \left (a+b \cosh ^{-1}(c x)\right )}{x}+3 d^2 e x \left (a+b \cosh ^{-1}(c x)\right )+d e^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{5} e^3 x^5 \left (a+b \cosh ^{-1}(c x)\right )-\frac{\left (b c \sqrt{-1+c^2 x^2}\right ) \int \frac{-d^3+3 d^2 e x^2+d e^2 x^4+\frac{e^3 x^6}{5}}{x \sqrt{-1+c^2 x^2}} \, dx}{\sqrt{-1+c x} \sqrt{1+c x}}\\ &=-\frac{d^3 \left (a+b \cosh ^{-1}(c x)\right )}{x}+3 d^2 e x \left (a+b \cosh ^{-1}(c x)\right )+d e^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{5} e^3 x^5 \left (a+b \cosh ^{-1}(c x)\right )-\frac{\left (b c \sqrt{-1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{-d^3+3 d^2 e x+d e^2 x^2+\frac{e^3 x^3}{5}}{x \sqrt{-1+c^2 x}} \, dx,x,x^2\right )}{2 \sqrt{-1+c x} \sqrt{1+c x}}\\ &=-\frac{d^3 \left (a+b \cosh ^{-1}(c x)\right )}{x}+3 d^2 e x \left (a+b \cosh ^{-1}(c x)\right )+d e^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{5} e^3 x^5 \left (a+b \cosh ^{-1}(c x)\right )-\frac{\left (b c \sqrt{-1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \left (\frac{e \left (15 c^4 d^2+5 c^2 d e+e^2\right )}{5 c^4 \sqrt{-1+c^2 x}}-\frac{d^3}{x \sqrt{-1+c^2 x}}+\frac{e^2 \left (5 c^2 d+2 e\right ) \sqrt{-1+c^2 x}}{5 c^4}+\frac{e^3 \left (-1+c^2 x\right )^{3/2}}{5 c^4}\right ) \, dx,x,x^2\right )}{2 \sqrt{-1+c x} \sqrt{1+c x}}\\ &=\frac{b e \left (15 c^4 d^2+5 c^2 d e+e^2\right ) \left (1-c^2 x^2\right )}{5 c^5 \sqrt{-1+c x} \sqrt{1+c x}}-\frac{b e^2 \left (5 c^2 d+2 e\right ) \left (1-c^2 x^2\right )^2}{15 c^5 \sqrt{-1+c x} \sqrt{1+c x}}+\frac{b e^3 \left (1-c^2 x^2\right )^3}{25 c^5 \sqrt{-1+c x} \sqrt{1+c x}}-\frac{d^3 \left (a+b \cosh ^{-1}(c x)\right )}{x}+3 d^2 e x \left (a+b \cosh ^{-1}(c x)\right )+d e^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{5} e^3 x^5 \left (a+b \cosh ^{-1}(c x)\right )+\frac{\left (b c d^3 \sqrt{-1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{-1+c^2 x}} \, dx,x,x^2\right )}{2 \sqrt{-1+c x} \sqrt{1+c x}}\\ &=\frac{b e \left (15 c^4 d^2+5 c^2 d e+e^2\right ) \left (1-c^2 x^2\right )}{5 c^5 \sqrt{-1+c x} \sqrt{1+c x}}-\frac{b e^2 \left (5 c^2 d+2 e\right ) \left (1-c^2 x^2\right )^2}{15 c^5 \sqrt{-1+c x} \sqrt{1+c x}}+\frac{b e^3 \left (1-c^2 x^2\right )^3}{25 c^5 \sqrt{-1+c x} \sqrt{1+c x}}-\frac{d^3 \left (a+b \cosh ^{-1}(c x)\right )}{x}+3 d^2 e x \left (a+b \cosh ^{-1}(c x)\right )+d e^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{5} e^3 x^5 \left (a+b \cosh ^{-1}(c x)\right )+\frac{\left (b d^3 \sqrt{-1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{c^2}+\frac{x^2}{c^2}} \, dx,x,\sqrt{-1+c^2 x^2}\right )}{c \sqrt{-1+c x} \sqrt{1+c x}}\\ &=\frac{b e \left (15 c^4 d^2+5 c^2 d e+e^2\right ) \left (1-c^2 x^2\right )}{5 c^5 \sqrt{-1+c x} \sqrt{1+c x}}-\frac{b e^2 \left (5 c^2 d+2 e\right ) \left (1-c^2 x^2\right )^2}{15 c^5 \sqrt{-1+c x} \sqrt{1+c x}}+\frac{b e^3 \left (1-c^2 x^2\right )^3}{25 c^5 \sqrt{-1+c x} \sqrt{1+c x}}-\frac{d^3 \left (a+b \cosh ^{-1}(c x)\right )}{x}+3 d^2 e x \left (a+b \cosh ^{-1}(c x)\right )+d e^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{5} e^3 x^5 \left (a+b \cosh ^{-1}(c x)\right )+\frac{b c d^3 \sqrt{-1+c^2 x^2} \tan ^{-1}\left (\sqrt{-1+c^2 x^2}\right )}{\sqrt{-1+c x} \sqrt{1+c x}}\\ \end{align*}

Mathematica [A]  time = 0.337214, size = 182, normalized size = 0.69 \[ 3 a d^2 e x-\frac{a d^3}{x}+a d e^2 x^3+\frac{1}{5} a e^3 x^5-\frac{b e \sqrt{c x-1} \sqrt{c x+1} \left (c^4 \left (225 d^2+25 d e x^2+3 e^2 x^4\right )+2 c^2 e \left (25 d+2 e x^2\right )+8 e^2\right )}{75 c^5}+\frac{b \cosh ^{-1}(c x) \left (15 d^2 e x^2-5 d^3+5 d e^2 x^4+e^3 x^6\right )}{5 x}-b c d^3 \tan ^{-1}\left (\frac{1}{\sqrt{c x-1} \sqrt{c x+1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*d^3)/x) + 3*a*d^2*e*x + a*d*e^2*x^3 + (a*e^3*x^5)/5 - (b*e*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(8*e^2 + 2*c^2*e*
(25*d + 2*e*x^2) + c^4*(225*d^2 + 25*d*e*x^2 + 3*e^2*x^4)))/(75*c^5) + (b*(-5*d^3 + 15*d^2*e*x^2 + 5*d*e^2*x^4
+ e^3*x^6)*ArcCosh[c*x])/(5*x) - b*c*d^3*ArcTan[1/(Sqrt[-1 + c*x]*Sqrt[1 + c*x])]

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Maple [A]  time = 0.019, size = 282, normalized size = 1.1 \begin{align*}{\frac{a{e}^{3}{x}^{5}}{5}}+ad{e}^{2}{x}^{3}+3\,a{d}^{2}ex-{\frac{a{d}^{3}}{x}}+{\frac{b{\rm arccosh} \left (cx\right ){e}^{3}{x}^{5}}{5}}+b{\rm arccosh} \left (cx\right )d{e}^{2}{x}^{3}+3\,b{\rm arccosh} \left (cx\right ){d}^{2}ex-{\frac{b{d}^{3}{\rm arccosh} \left (cx\right )}{x}}-{cb{d}^{3}\sqrt{cx-1}\sqrt{cx+1}\arctan \left ({\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}}}-{\frac{b{x}^{4}{e}^{3}}{25\,c}\sqrt{cx-1}\sqrt{cx+1}}-{\frac{b{x}^{2}d{e}^{2}}{3\,c}\sqrt{cx-1}\sqrt{cx+1}}-3\,{\frac{\sqrt{cx+1}\sqrt{cx-1}b{d}^{2}e}{c}}-{\frac{4\,b{x}^{2}{e}^{3}}{75\,{c}^{3}}\sqrt{cx-1}\sqrt{cx+1}}-{\frac{2\,bd{e}^{2}}{3\,{c}^{3}}\sqrt{cx-1}\sqrt{cx+1}}-{\frac{8\,b{e}^{3}}{75\,{c}^{5}}\sqrt{cx-1}\sqrt{cx+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*e^3*x^5+a*d*e^2*x^3+3*a*d^2*e*x-a*d^3/x+1/5*b*arccosh(c*x)*e^3*x^5+b*arccosh(c*x)*d*e^2*x^3+3*b*arccosh(
c*x)*d^2*e*x-b*arccosh(c*x)*d^3/x-c*b*(c*x-1)^(1/2)*(c*x+1)^(1/2)/(c^2*x^2-1)^(1/2)*d^3*arctan(1/(c^2*x^2-1)^(
1/2))-1/25*b/c*(c*x-1)^(1/2)*(c*x+1)^(1/2)*x^4*e^3-1/3*b/c*(c*x-1)^(1/2)*(c*x+1)^(1/2)*x^2*d*e^2-3*b/c*(c*x-1)
^(1/2)*(c*x+1)^(1/2)*d^2*e-4/75*b/c^3*(c*x-1)^(1/2)*(c*x+1)^(1/2)*x^2*e^3-2/3*b/c^3*(c*x-1)^(1/2)*(c*x+1)^(1/2
)*d*e^2-8/75*b/c^5*(c*x-1)^(1/2)*(c*x+1)^(1/2)*e^3

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Maxima [A]  time = 1.68462, size = 302, normalized size = 1.14 \begin{align*} \frac{1}{5} \, a e^{3} x^{5} + a d e^{2} x^{3} -{\left (c \arcsin \left (\frac{1}{\sqrt{c^{2}}{\left | x \right |}}\right ) + \frac{\operatorname{arcosh}\left (c x\right )}{x}\right )} b d^{3} + \frac{1}{3} \,{\left (3 \, x^{3} \operatorname{arcosh}\left (c x\right ) - c{\left (\frac{\sqrt{c^{2} x^{2} - 1} x^{2}}{c^{2}} + \frac{2 \, \sqrt{c^{2} x^{2} - 1}}{c^{4}}\right )}\right )} b d e^{2} + \frac{1}{75} \,{\left (15 \, x^{5} \operatorname{arcosh}\left (c x\right ) -{\left (\frac{3 \, \sqrt{c^{2} x^{2} - 1} x^{4}}{c^{2}} + \frac{4 \, \sqrt{c^{2} x^{2} - 1} x^{2}}{c^{4}} + \frac{8 \, \sqrt{c^{2} x^{2} - 1}}{c^{6}}\right )} c\right )} b e^{3} + 3 \, a d^{2} e x + \frac{3 \,{\left (c x \operatorname{arcosh}\left (c x\right ) - \sqrt{c^{2} x^{2} - 1}\right )} b d^{2} e}{c} - \frac{a d^{3}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*e^3*x^5 + a*d*e^2*x^3 - (c*arcsin(1/(sqrt(c^2)*abs(x))) + arccosh(c*x)/x)*b*d^3 + 1/3*(3*x^3*arccosh(c*x
) - c*(sqrt(c^2*x^2 - 1)*x^2/c^2 + 2*sqrt(c^2*x^2 - 1)/c^4))*b*d*e^2 + 1/75*(15*x^5*arccosh(c*x) - (3*sqrt(c^2
*x^2 - 1)*x^4/c^2 + 4*sqrt(c^2*x^2 - 1)*x^2/c^4 + 8*sqrt(c^2*x^2 - 1)/c^6)*c)*b*e^3 + 3*a*d^2*e*x + 3*(c*x*arc
cosh(c*x) - sqrt(c^2*x^2 - 1))*b*d^2*e/c - a*d^3/x

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Fricas [A]  time = 3.12706, size = 716, normalized size = 2.7 \begin{align*} \frac{15 \, a c^{5} e^{3} x^{6} + 75 \, a c^{5} d e^{2} x^{4} + 150 \, b c^{6} d^{3} x \arctan \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) + 225 \, a c^{5} d^{2} e x^{2} - 75 \, a c^{5} d^{3} + 15 \,{\left (5 \, b c^{5} d^{3} - 15 \, b c^{5} d^{2} e - 5 \, b c^{5} d e^{2} - b c^{5} e^{3}\right )} x \log \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) + 15 \,{\left (b c^{5} e^{3} x^{6} + 5 \, b c^{5} d e^{2} x^{4} + 15 \, b c^{5} d^{2} e x^{2} - 5 \, b c^{5} d^{3} +{\left (5 \, b c^{5} d^{3} - 15 \, b c^{5} d^{2} e - 5 \, b c^{5} d e^{2} - b c^{5} e^{3}\right )} x\right )} \log \left (c x + \sqrt{c^{2} x^{2} - 1}\right ) -{\left (3 \, b c^{4} e^{3} x^{5} +{\left (25 \, b c^{4} d e^{2} + 4 \, b c^{2} e^{3}\right )} x^{3} +{\left (225 \, b c^{4} d^{2} e + 50 \, b c^{2} d e^{2} + 8 \, b e^{3}\right )} x\right )} \sqrt{c^{2} x^{2} - 1}}{75 \, c^{5} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/75*(15*a*c^5*e^3*x^6 + 75*a*c^5*d*e^2*x^4 + 150*b*c^6*d^3*x*arctan(-c*x + sqrt(c^2*x^2 - 1)) + 225*a*c^5*d^2
*e*x^2 - 75*a*c^5*d^3 + 15*(5*b*c^5*d^3 - 15*b*c^5*d^2*e - 5*b*c^5*d*e^2 - b*c^5*e^3)*x*log(-c*x + sqrt(c^2*x^
2 - 1)) + 15*(b*c^5*e^3*x^6 + 5*b*c^5*d*e^2*x^4 + 15*b*c^5*d^2*e*x^2 - 5*b*c^5*d^3 + (5*b*c^5*d^3 - 15*b*c^5*d
^2*e - 5*b*c^5*d*e^2 - b*c^5*e^3)*x)*log(c*x + sqrt(c^2*x^2 - 1)) - (3*b*c^4*e^3*x^5 + (25*b*c^4*d*e^2 + 4*b*c
^2*e^3)*x^3 + (225*b*c^4*d^2*e + 50*b*c^2*d*e^2 + 8*b*e^3)*x)*sqrt(c^2*x^2 - 1))/(c^5*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{acosh}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{3}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*acosh(c*x))*(d + e*x**2)**3/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )}^{3}{\left (b \operatorname{arcosh}\left (c x\right ) + a\right )}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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